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[TeXmacs] GNU TeXmacs 1.0.7.5 with experimental PDF support fails to export to PDF
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From : Marc Mertens <address@hidden>- To: address@hidden
- Subject: [TeXmacs] GNU TeXmacs 1.0.7.5 with experimental PDF support fails to export to PDF
- Date: Sat, 24 Jul 2010 13:19:49 +0200
Hello,
I compiled to TeXmacs 1.0.7.5 with the experimental PDF support to see
if
the PDF hyperlinks from the content where generated. Unfortunately TeXmacs
crashed when I exported to PDF with the following error:
TeXmacs] With linked TrueType support
TeXmacs] Loading corktounicode
TeXmacs] Loading unicode-cork-oneway
TeXmacs] Loading tmuniversaltounicode
TeXmacs] Loading unicode-symbol-oneway
TeXmacs] Warning: Redefined subset of a finite set as 1.179
TeXmacs] Warning: Redefined subset of a finite set as 1.186
TeXmacs] Warning: Redefined characterization of closure as 3.16
TeXmacs] Warning: Redefined characterization of closure as 3.24
TeXmacs] Warning: Redefined subset of a finite set as 1.179
TeXmacs] Warning: Redefined subset of a finite set as 1.186
TeXmacs] Warning: Redefined characterization of closure as 3.16
TeXmacs] Warning: Redefined characterization of closure as 3.24
** WARNING ** Failed to load AGL file "pdfglyphlist.txt"...
** WARNING ** Failed to load AGL file "glyphlist.txt"...
sh: -c: line 0: syntax error near unexpected token `('
sh: -c: line 0: `kpsewhich compound-(math (math (tex cmr 10 600) (tex cmmi 10
600) (tex cmsy 10 600) (tex msam 10 600) (tex msbm 10 600) (tex stmary 10
600)
(tex wasy 10 600) (tex line 10 600) (tex cmex 10 600) (tex cmsy 10 600) (tex
eufm 10 600) (tex bbm 10 600) (tex grmn 10 600) (tex cmbx 10 600) (tex cmmib
10 600) (tex cmbsy 10 600) (virtual long 10 600) (virtual negate 10 600)
(virtual misc 10 600)) (rubber (tex-rubber rubber-cmex cmex 10 600) (tex-
rubber rubber-stmary stmary 10 600) (tex-rubber rubber-wasy wasy 10 600) (tex-
dummy-rubber (tex-rubber rubber-cmex cmex 10 600))) (ec ecrm 10 600) (ec ecrm
10 600))#address@hidden >
/home/marc/.TeXmacs/system/tmp/tmp_230942725 2>&1'
** ERROR ** TFM: Invalid TFM ID: 337794328
I have included my *.tm file so that you can repeat the error.
I fully understand that the experimental support is indeed 'experimental' so
I
will recompile TeXmacs with the default settings.
BTW I'm really impressed with the reaction of the TeXmacs developers on my
question about generating the content as hyperlinks in PDF, by bringing out
the new PDF support to solve that problem, this proves that the TeXmacs
developers really act on the questions of their users. Thanks for this folks.
Greetings
Marc Mertens
<TeXmacs|1.0.7.5>
<style|book>
<\body>
<doc-title|Differentiable Manifolds>
<\table-of-contents|toc>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|1<space|2spc>Preliminaries>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-1><vspace|0.5fn>
1.1<space|2spc>Set, relations and functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-2>
<with|par-left|1.5fn|1.1.1<space|2spc>Sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-3>>
<with|par-left|1.5fn|1.1.2<space|2spc>Relations
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-11>>
<with|par-left|1.5fn|1.1.3<space|2spc>Partial functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-16>>
<with|par-left|1.5fn|1.1.4<space|2spc>Functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-26>>
<with|par-left|1.5fn|1.1.5<space|2spc>Families
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-31>>
<with|par-left|1.5fn|1.1.6<space|2spc>Axiom of choice
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-36>>
<with|par-left|1.5fn|1.1.7<space|2spc>Order Relations
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-46>>
<with|par-left|1.5fn|1.1.8<space|2spc>The real numbers
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-57>>
<with|par-left|1.5fn|1.1.9<space|2spc>Finite and infinite sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-70>>
<with|par-left|1.5fn|1.1.10<space|2spc>Principle of recursive definition
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-75>>
<with|par-left|1.5fn|1.1.11<space|2spc>Countable sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-77>>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|2<space|2spc>Algebra>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-82><vspace|0.5fn>
2.1<space|2spc>Groups <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-83>
2.2<space|2spc>Fields <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-100>
2.3<space|2spc>Vector spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-103>
2.4<space|2spc>Algebras <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-121>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|3<space|2spc>Topology>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-124><vspace|0.5fn>
3.1<space|2spc>Topological spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-125>
3.2<space|2spc>Metric and normed spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-144>
3.3<space|2spc>Inner product spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-164>
3.4<space|2spc>Continuity <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-170>
<with|par-left|1.5fn|3.4.1<space|2spc>Linear maps and continuity
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-178>>
<with|par-left|1.5fn|3.4.2<space|2spc>Separation
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-193>>
<with|par-left|1.5fn|3.4.3<space|2spc>Topological Vector Spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-203>>
<with|par-left|1.5fn|3.4.4<space|2spc>Bundles
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-206>>
3.5<space|2spc>Compact Spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-211>
<with|par-left|1.5fn|3.5.1<space|2spc>Filter bases in Spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-215>>
<with|par-left|1.5fn|3.5.2<space|2spc>Product of compacts subsets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-223>>
3.6<space|2spc>Complete Spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-226>
3.7<space|2spc>Integration in Banach space
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-237>
3.8<space|2spc>Connected Sets <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-246>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|4<space|2spc>Differentiability
in normed vector spaces> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-249><vspace|0.5fn>
4.1<space|2spc>Differentiability <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-250>
4.2<space|2spc>Higher order differentiability
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-262>
4.3<space|2spc>Differentiability on general sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-268>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|5<space|2spc>Inverse
function theorem> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-270><vspace|0.5fn>
5.1<space|2spc>Intermediate value theorem
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-271>
5.2<space|2spc>The inverse function theorem
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-284>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|6<space|2spc>Differentiable
manifolds> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-287><vspace|0.5fn>
6.1<space|2spc>Manifolds <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-288>
6.2<space|2spc>Differentiable mappings between manifolds
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-300>
6.3<space|2spc>Tangent spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-303>
6.4<space|2spc>The tangent bundle <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-325>
6.5<space|2spc>Vector fields <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-327>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|7<space|2spc>Notations>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-342><vspace|0.5fn>
<vspace*|1fn><with|font-series|bold|math-font-series|bold|Index>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-343><vspace|0.5fn>
</table-of-contents>
<chapter|Preliminaries>
<section|Set, relations and functions>
<subsection|Sets>
<\definition>
<subindex|set|subset><subindex|set|intersection><subindex|set|union><subindex|set|difference>Let
A,B be sets then we define the following operations
<\eqnarray*>
<tformat|<table|<row|<cell|A\<subseteq\>B>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>x\<in\>A\<succ\>x\<in\>B>>|<row|<cell|A=B>|<cell|\<Leftrightarrow\>>|<cell|A\<subseteq\>B\<vee\>B\<subseteq\>A>>|<row|<cell|A<big|cap>B>|<cell|=>|<cell|{x\|x\<in\>A<with|mode|text|
and >x\<in\>B}>>|<row|<cell|A<big|cup>B>|<cell|=>|<cell|{x\|x\<in\>A<with|mode|text|
or >x\<in\>B}>>|<row|<cell|A\<subset\>B>|<cell|\<Leftrightarrow\>>|<cell|A\<subseteq\>B\<wedge\>A<big|cap>B=\<emptyset\>>>|<row|<cell|A<mid|\\>B>|<cell|=>|<cell|{x\|x\<in\>A<with|mode|text|
and >x\<nin\>B}>>>>
</eqnarray*>
</definition>
<\theorem>
Given <math|A,B\<subseteq\>X\<Rightarrow\>X<mid|\\>(A<big|cap>B)=(X<mid|\\>A)<big|cup>(X<mid|\\>B)>
and <math|X<mid|\\>(A<big|cup>B)=(X<mid|\\>A)<big|cap>(X<mid|\\>B)>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|x\<in\>X<space|0.2spc><mid|\\>(A<big|cap>B)\<Leftrightarrow\>x\<in\>X\<wedge\>x\<nin\>(A<big|cap>B)\<Leftrightarrow\>x\<in\>X\<wedge\>(x\<nin\>A\<vee\>x\<nin\>B)\<Leftrightarrow\>(x\<in\>X\<wedge\>x\<nin\>A)\<vee\>(x\<in\>X\<wedge\>x\<in\>B)\<Leftrightarrow\>x\<in\>(X<mid|\\>A)<big|cup>(X<mid|\\>B)>
<item><math|x\<in\>X<mid|\\>(A<big|cup>B)\<Leftrightarrow\>x\<in\>X\<wedge\>x\<nin\>(A<big|cup>B)\<Leftrightarrow\>x\<in\>X\<wedge\>(x\<nin\>A\<wedge\>x\<nin\>B)\<Leftrightarrow\>(x\<in\>X\<wedge\>x\<nin\>A)\<wedge\>(x\<in\>X\<wedge\>x\<nin\>B)\<Leftrightarrow\>x\<in\>X<mid|\\>A\<wedge\>x\<in\>X<mid|\\>B\<Leftrightarrow\>x\<in\>(X<mid|\\>A)<big|cup>(X<mid|\\>B)>
</enumerate>
</proof>
<\theorem>
Given <math|A,B\<subseteq\>X> then <math|(X<mid|\\>A)<big|cap>B=B<mid|\\>A>
</theorem>
<\proof>
<math|x\<in\>(X<mid|\\>A)<big|cap>B\<Leftrightarrow\>x\<in\>(X<mid|\\>A)\<wedge\>x\<in\>B\<Leftrightarrow\>x\<in\>X\<wedge\>x\<nin\>A\<wedge\>x\<in\>B\<Leftrightarrowlim\><rsub|B=X<big|cap>B>x\<in\>B\<wedge\>x\<nin\>A\<Leftrightarrow\>x\<in\>B<mid|\\>A>
</proof>
<\theorem>
<label|difference of difference of sets>Let
<math|A\<subseteq\>X\<Rightarrow\>X<mid|\\>(X<mid|\\>A)=A>
</theorem>
<\proof>
If <math|x\<in\>X<mid|\\>(X<mid|\\>A)\<Rightarrow\>x\<in\>X\<wedge\>x\<nin\>(X<mid|\\>A)\<Rightarrow\>x\<in\>X\<wedge\>\<thicksim\>(x\<in\>X\<wedge\>x\<nin\>A)\<Rightarrow\>x\<in\>X\<wedge\>(x\<nin\>X\<vee\>x\<in\>A)\<Rightarrow\>(x\<in\>X\<wedge\>x\<nin\>X)\<vee\>(x\<in\>X\<wedge\>x\<in\>A)\<Rightarrow\>x\<in\>X<big|cap>A=A>
If <math|x\<in\>A\<Rightarrowlim\><rsub|A\<subseteq\>B>x\<in\>A\<wedge\>x\<in\>X\<Rightarrow\>x\<in\>X\<wedge\>(x\<nin\>X\<vee\>x\<in\>A)\<Rightarrow\>x\<in\>X\<wedge\>\<sim\>(x\<in\>X\<wedge\>x\<nin\>A)\<Rightarrow\>x\<in\>X\<wedge\>x\<nin\>X<mid|\\>A\<Rightarrow\>x\<in\>X<mid|\\>(X<mid|\\>A)>
</proof>
<\theorem>
<label|union of disjoint sets>Let A,B,C be sets such that
<math|A<big|cap>B\<eqcirc\>\<emptyset\>>, <math|C=A<big|cup>B> then
<math|A=C<mid|\\>B> and <math|B=C<mid|\\>A>
<\proof>
\;
<\enumerate>
<item>Let <math|x\<in\>A\<subseteq\>A<big|cup>B=C\<Rightarrow\>x\<in\>C>
and if <math|x\<in\>B> we would have
<math|x\<in\>A<big|cap>B\<Rightarrow\>A<big|cap>B\<neq\>\<emptyset\>>
so we have <math|x\<nin\>B> and thus <math|x\<in\>C<mid|\\>B> and
thus <math|A\<subseteq\>C<mid|\\>B>
<item>Let <math|x\<in\>C<mid|\\>B\<Rightarrow\>x\<in\>C=A<big|cup>B<with|mode|text|
and >x\<nin\>B\<Rightarrow\> [x\<in\>A<with|mode|text| or
>x\<in\>B]<with|mode|text| and >x\<in\>B\<Rightarrow\>x\<in\>A>
<item>The proof that <math|B=C<mid|\\>A> is analog\
</enumerate>
</proof>
</theorem>
<\definition>
<subindex|set|powerset>Let <math|X> be a set then
<math|2<rsup|X>={A\|A\<subseteq\>X}> the set of all subsets of <math|X>
</definition>
<\definition>
<subindex|set|pair>Let A,B be two sets and <math|x\<in\>A> and
<math|x\<in\>B> then <math|(x,y)> is defined by <math|(x,y)={{x},{x,y}}>
</definition>
<\theorem>
Let A,B be two sets then <math|\<forall\>x<rsub|1>,x<rsub|2>\<in\>A,y<rsub|1>,y<rsub|2>\<in\>B\<succ\>(x<rsub|1>,y<rsub|1>)=(x<rsub|2>,y<rsub|2>)\<Leftrightarrow\>x<rsub|1>=x<rsub|2><with|mode|text|
and >y<rsub|1>=y<rsub|2>>
<\proof>
<math|(x<rsub|1>,y<rsub|1>)=(x<rsub|2>,y<rsub|2>)\<Rightarrow\>{{x<rsub|1>},{x<rsub|1>,y<rsub|1>}}\<subseteq\>{{x<rsub|2>},{x<rsub|2>,y<rsub|2>}}<with|mode|text|
and >{{x<rsub|2>},{x<rsub|2>,y<rsub|2>}}\<subseteq\>{{x<rsub|1>},{x<rsub|1>,y<rsub|1>}}>
lets discuss the different possibilities for the first inclusion
<\enumerate>
<item>We must have either <math|{x<rsub|1>}={x<rsub|2>}> or
<math|{x<rsub|1>}={x<rsub|2>,y<rsub|2>}> and this gives either\
<\enumerate>
<item><math|x<rsub|1>=x<rsub|2>>
<item><math|x<rsub|1>=x<rsub|2>=y<rsub|2> >
meaning in either case that <math|x<rsub|1>=x<rsub|2>>
</enumerate>
<item>We must have also either <math|{x<rsub|1>,y<rsub|1>}={x<rsub|2>}>
or <math|{x<rsub|1>,y<rsub|1>}={x<rsub|2>,y<rsub|2>}> and this gives
for <math|y<rsub|1>>either
<\enumerate>
<item><math|x<rsub|1>=y<rsub|1>=x<rsub|2>> this means that
<math|{x<rsub|1>,y<rsub|1>}={y<rsub|1>}> and
<math|{x<rsub|1>}={y<rsub|1>}> and the second inclusion gives then
<math|{x<rsub|2>,y<rsub|2>}={y<rsub|1>}> or
<math|y<rsub|1>=y<rsub|2>=x<rsub|2>> or <math|y<rsub|1>=y<rsub|2>>
<item><math|y<rsub|1>=x<rsub|2>> now the second inclusion gives the
following possibilities for <math|{x<rsub|2>,y<rsub|2>}>
<\enumerate>
<item><math|{x<rsub|2>,y<rsub|2>}={x<rsub|1>}\<Rightarrow\>y<rsub|2>=x<rsub|2>\<Rightarrow\>y<rsub|2>=y<rsub|1>>
<item><math|{x<rsub|2>,y<rsub|2>}={x<rsub|1>,y<rsub|1>}\<Rightarrowlim\><rsub|x<rsub|1>=x<rsub|2>>{x<rsub|2>,y<rsub|2>}={x<rsub|2>,y<rsub|1>}\<equallim\><rsub|y<rsub|1>=x<rsub|2>>{y<rsub|1>}\<Rightarrow\>y<rsub|2>=y<rsub|1>>
</enumerate>
<item><math|y<rsub|1>=y<rsub|2>>
so again in all cases we have <math|y<rsub|1>=y<rsub|2>>
</enumerate>
</enumerate>
</proof>
</theorem>
<\definition>
<subindex|set|product>Let A,B be sets then the Cartesian product of these
sets is defined by the set <math|A\<times\>B={(x,y)\|x\<in\>A<with|mode|text|
and >x\<in\>B}>
</definition>
<\remark>
If <math|A=\<emptyset\> or B=\<emptyset\>
\<Rightarrow\>A\<times\>B=\<emptyset\>>
</remark>
<\theorem>
Let <math|A,B,C> be sets then <math|(A<big|cap>B)\<times\>C=(A\<times\>C)<big|cap>(B\<times\>C)>
</theorem>
<\proof>
If <math|(x,y)\<in\>(A<big|cap>B)\<times\>C\<Rightarrow\>x\<in\>A\<wedge\>x\<in\>B>
and <math|y\<in\>C\<Rightarrow\>(x,y)\<in\>A\<times\>C\<wedge\>(x,y)\<in\>B\<times\>C\<Rightarrow\>(x,y)\<in\>(A\<times\>C)<big|cap>(B\<times\>C)>.
If <math|(x,y)\<in\>(A\<times\>C)<big|cap>(B\<times\>C)\<Rightarrow\>(x,y)\<in\>A\<times\>C\<wedge\>(x,y)\<in\>(B\<times\>C)\<Rightarrow\>x\<in\>A<big|cap>B\<wedge\>y\<in\>C\<Rightarrow\>(A<big|cap>B)\<times\>C>
</proof>
<\theorem>
If <math|A,B,C> are sets with <math|C\<subseteq\>A\<Rightarrow\>C\<times\>B\<subseteq\>A\<times\>B>
</theorem>
<\proof>
if <math|(x,y)\<in\>C\<times\>B\<Rightarrow\>x\<in\>C,y\<in\>B\<Rightarrow\>x\<in\>A,y\<in\>B\<Rightarrow\>(x,y)\<in\>A\<times\>B>
</proof>
<subsection|Relations>
<\definition>
<subindex|set|relation>A relation is a triple <math|(A,B,R)> where
<math|A,B> are sets and <math|R\<subseteq\>A\<times\>B>
</definition>
<\notation>
If <math|\<cal-R\>=(A,B,R)> is a relation then we note
<math|(a,b)\<in\>R> as <math|a\<cal-R\>b>. Also sometimes we omit the
sets <math|A,B> if they are implicitly understood and use the subset
<math|\<cal-R\>\<subseteq\>A\<times\>B> as the notation of a relation.
<math|R> is called the graph of the relation and noted by
<math|graph(\<cal-R\>)>
</notation>
<\remark>
If <math|A=\<emptyset\>> or <math|B=\<emptyset\>> in the relation
<math|(A,B,R) >then we must have <math|R=\<emptyset\>>
</remark>
<\definition>
<index|composition of relations>Let <math|\<cal-R\><rsub|1>=(A,B,R<rsub|1>)>
and <math|\<cal-R\><rsub|2>=(C,D,R<rsub|2>)> with
<math|B<big|cap>C\<neq\>\<emptyset\>> then the relation
<math|\<cal-R\><rsub|2>\<circ\>\<cal-R\><rsub|1>> is defined to be the
relation <math|(A,D,graph(\<cal-R\><rsub|2>\<circ\>\<cal-R\><rsub|1>))>
where <math|graph(\<cal-R\><rsub|2>\<circ\>\<cal-R\><rsub|1>)={(x,z)\<in\>A\<times\>D\|\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>R<rsub|1>\<wedge\>(y,z)\<in\>R<rsub|2>}>
</definition>
<\definition>
<label|equivalence relation><index|equivalence relation>Let <math|A> be a
set then a binary relation <math|\<cal-R\>=(A,A,R)> is a equivalence
relation on <math|A> if the following is fulfilled
<\enumerate>
<item><math|\<forall\>a\<in\>A> then <math|a\<cal-R\>a> (reflexivity)
<item><math|\<forall\>a,a<rprime|'>\<in\>A> then if
<math|a\<cal-R\>a<rprime|'>\<Rightarrow\>a<rprime|'>\<cal-R\>a>
(symmetry)
<item><math|\<forall\>a,b,c\<in\>A> we have if <math|a\<cal-R\>b> and
<math|b\<cal-R\>c\<Rightarrow\>a\<cal-R\>c> (transitivity)
</enumerate>
</definition>
<\definition>
<index|equivalence class>Let <math|A> be a set and <math|\<cal-R\>> a
equivalence relation on <math|A> and <math|a\<in\>A> then
<math|\<cal-R\>[a]={b\|a\<cal-R\>b}> is called a equivalence class on
<math|A>
</definition>
<\theorem>
Let <math|A> be a set and <math|\<cal-R\>> a equivalence relation on
<math|A> then if <math|a,b\<in\>A> we have that
<math|\<thicksim\>(a\<cal-R\>b)><math|\<Leftrightarrow\>\<cal-R\>[a]<big|cap>\<cal-R\>[b]=\<emptyset\>>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>> If <math|\<sim\>(a\<cal-R\>b)> then if
<math|x\<in\>\<cal-R\>[a]<big|cap>\<cal-R\>[b]> we have
<math|a\<cal-R\>x,b\<cal-R\>x\<Rightarrowlim\><rsub|symmetry>a\<cal-R\>x,x\<cal-R\>b\<Rightarrowlim\><rsub|transitivity>a\<cal-R\>b>
a contradiction so we have <math|\<cal-R\>[a]<big|cap>\<cal-R\>[b]=\<emptyset\>>\
<math|\<Leftarrow\>>If <math|><math|\<cal-R\>[a]<big|cap>\<cal-R\>[b]=\<emptyset\>>
then if <math|a\<cal-R\>b\<Rightarrowlim\><rsub|symmetry>b\<cal-R\>a\<Rightarrow\>a\<in\>\<cal-R\>[a]<big|cap>\<cal-R\>[b]>
a contradiction
</proof>
<subsection|Partial functions>
<\definition>
<index|domain><index|partial function>A partial function <math|f> between
<math|A> and <math|B> (<math|A,B> sets) is a relation
<math|\<cal-R\>=(A,B,R)> such that if <math|x\<cal-R\>y> and
<math|x\<cal-R\>z> then <math|y=z>. The set
<math|dom(f)={x\<in\>A\|\<exists\>y\<in\>B\<vdash\>x\<cal-R\>y}> is
called the domain of f and the set <math|rng(f)={y\<in\>B\|\<exists\>x\<in\>A\<vdash\>x\<cal-R\>y}>
is called the range of f. If <math|x\<in\>dom(f)> then the unique
<math|y> such that <math|x\<cal-R\>y> is noted by <math|f(x)>. <math|A>
is called the source, <math|B> is called the target or the codomain and
<math|R> is called the graph of the function and noted by
<math|graph(f)>. From this we can note a function <math|f> as follows
<math|(A,B,graph(f))>
</definition>
<\notation>
\;
If we write <math|f(x)> then we implicitly assume that
<math|x\<in\>dom(f)> and <math|(x,f(x))\<in\>graph(f)>, so that
<math|f(x)> make sense. Also if <math|f=(A,B,graph(f))> then we interpret
<math|f(x)=y> to be equivalent with <math|(x,y)\<in\>graph(f)>
A shorthand notation for a function <math|f> between A and B where
<math|f=(A,B,graph(f))> is
<math|f:C\<rightarrow\>B:x\<rightarrow\>E(x)> or sometimes
<math|f:C\<rightarrow\>B:x\<rightarrow\>f(x)=E(x)> where <math|E(x)> is a
expression based on the <math|x> and which is valid only if
<math|x\<in\>C> and we mean by this <math|dom(f)=C> and
<math|R={(x,y)\|x\<in\>C\<wedge\>y=E(x)}>
</notation>
<\remark>
If <math|dom(f)=\<emptyset\>> then <math|rng(f)=\<emptyset\>> and the
graph is the empty set.
</remark>
<\definition>
<index|image><index|preimage>Let f be a partial function between A and B
and <math|C\<subseteq\>A>, <math|D\<subseteq\>B> then
<math|f(C)={y\<in\>B\|\<exists\>x\<in\>C\<vdash\>(x,y)\<in\>graph(f)}>
and <math|f<rsup|-1>(D)={x\<in\>A\|\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(f)}>.
Note that <math|f(C)\<subseteq\>rng(f)> and that
<math|f<rsup|-1>(D)\<subseteq\>dom(f)>
</definition>
<\theorem>
<label|image of preimage>Let <math|f> be a partial function between
<math|A> and <math|B> and <math|C\<subseteq\>B> then
<math|f(f<rsup|-1>(C))\<subseteq\>C>
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>f(f<rsup|-1>(C))>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>f<rsup|-1>(C)\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>y<rprime|'>\<in\>C\<vdash\>(x,y<rprime|'>)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is a function>>|<cell|y=y<rprime|'>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>C>>>>
</eqnarray*>
\;
</proof>
<\theorem>
<label|function extension>Let <math|f<rsub|1>=(A<rsub|1>,B<rsub|1>,graph(f))>
be a partial function between <math|A<rsub|1>> and <math|B<rsub|1>> and
<math|f<rsub|2>=(A<rsub|2>,B<rsub|2>,graph(f<rsub|2>))> be a function
between <math|A<rsub|2>> and <math|B<rsub|2>> and let
<math|A<rsub|1><big|cap>A<rsub|2>\<neq\>\<emptyset\>>. Then
<math|f=(A<rsub|1><big|cup>A<rsub|2>,B<rsub|1><big|cup>B<rsub|2>),graph(f<rsub|1>)<big|cup>graph(f<rsub|2>))>
is a partial function between <math|A<rsub|1><big|cup>A<rsub|2>> and
<math|B<rsub|1><big|cup>B<rsub|2>>. (we note this definition as\
<\eqnarray*>
<tformat|<table|<row|<cell|f(x)>|<cell|=f<rsub|1>(x)>|<cell|if
x\<in\>A<rsub|1>>>|<row|<cell|>|<cell|=f<rsub|2>(x)>|<cell|if
x\<in\>A<rsub|2>>>>>
</eqnarray*>
and <math|dom(f)=dom(f<rsub|1>)<big|cup>dom(f<rsub|2>)> and
<math|rng(f)=rng(f<rsub|1>)<big|cup>rng(f<rsub|2>)>
</theorem>
<\proof>
If <math|(x,y<rsub|1>),(x,y<rsub|1>)\<in\>graph(f<rsub|1><big|cup>f<rsub|2>)=graph(f<rsub|1>)<big|cup>graph(f<rsub|2>)>
we have the following possibilities to consider\
<\enumerate>
<item><math|(x,y<rsub|1>)\<in\>graph(f<rsub|1>)\<wedge\>(x,y<rsub|2>)\<in\>graph(f<rsub|1>)<rsub|>>
then as <math|f<rsub|1>> is a function we have
<math|y<rsub|1>=y<rsub|2>>
<item><math|(x,y<rsub|1>)\<in\>graph(f<rsub|1>)\<wedge\>(x,y<rsub|2>)\<in\>graph(f<rsub|2>)>
this would mean that <math|x\<in\>A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>>
which is impossible
<item><math|(x,y<rsub|1>)\<in\>graph(f<rsub|2>)\<wedge\>(x,y<rsub|2>)\<in\>graph(f<rsub|1>)>
this would mean that <math|x\<in\>A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>>
which is impossible
<item><math|(x,y<rsub|1>)\<in\>graph(f<rsub|2>)\<wedge\>(x,y<rsub|2>)\<in\>graph(f<rsub|2>)>
then as <math|f<rsub|2>> is a function we have
<math|y<rsub|1>=y<rsub|2>>
</enumerate>
proving that <math|f> is a partial function.
Now <math|x\<in\>dom(f)\<Leftrightarrow\>\<exists\>y\<in\>B<rsub|1><big|cup>B<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|1>)<big|cup>graph(f<rsub|2>)\<Leftrightarrow\>(\<exists\>y\<in\>B<rsub|1><big|cup>B<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|1>))\<vee\>\<exists\>y\<in\>B<rsub|1><big|cup>B<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|2>))\<Leftrightarrowlim\><rsub|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>>(\<exists\>y\<in\>B<rsub|1>>
such that <math|(x,y)\<in\>graph(f<rsub|1>))\<vee\>(\<exists\>y\<in\>B<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|2>))\<Leftrightarrow\>x\<in\>dom(f<rsub|1>)<big|cup>dom(f<rsub|2>)>
Also <math|y\<in\>rng(f)\<Leftrightarrow\>\<exists\>x\<in\>A<rsub|1><big|cup>A<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|1>)<big|cup>graph(f<rsub|2>)\<Leftrightarrow\>(\<exists\>x\<in\>A<rsub|1>>
such that <math|(x,y)\<in\>graph(f<rsub|1>))\<vee\>(\<exists\>x\<in\>A<rsub|2>>
such that <math|(x,y)\<in\>graph(f<rsub|2>))\<Leftrightarrow\>y\<in\>rng(f<rsub|1>)<big|cup>rng(f<rsub|2>)>
</proof>
<\theorem>
Let f be a partial function between A and B and let g be function between
C and D where <math|B<big|cap>C\<neq\>\<emptyset\>> then
<math|g\<circ\>f> is a partial function between <math|A> and <math|D>
with <math|dom(g\<circ\>f)\<subseteq\>dom(f)> [we have
<math|dom(g\<circ\>f)=dom(f)> if <math|rng(f)\<subseteq\>dom(g)]> and
<math|\<forall\>x\<in\>dom(g\<circ\>f)\<succ\>(g\<circ\>f)(x)=g(f(x))>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y),(x,y<rprime|'>)\<in\>graph(g\<circ\>f)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z<rsub|1>,z<rsub|2>\<in\>B\<vdash\>(x,z<rsub|1>),(x,z<rsub|2>)\<in\>graph(f)\<wedge\>(z<rsub|1>,y),(z<rsub|2>,y<rprime|'>)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is a function>>|<cell|z<rsub|1>=z<rsub|2>\<Rightarrow\>(z<rsub|1>,y),(z<rsub|21>,y<rprime|'>)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|g
is a function>>|<cell|y=y<rprime|'>>>>>
</eqnarray*>
To proof that <math|dom(g\<circ\>f)\<subseteq\>dom(f)>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>dom(g\<circ\>f)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(g\<circ\>f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f)\<wedge\>(z,y)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)>>>>
</eqnarray*>
If <math|rng(f)\<subseteq\>dom(g)> then\
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>dom(f)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>rng(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>dom(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>D\<vdash\>(y,z)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(g\<circ\>f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(g\<circ\>f)>>>>
</eqnarray*>
Now if <math|x\<in\>dom(g\<circ\>f)> then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(g\<circ\>f)>|<cell|\<Rightarrow\>>|<cell|y=(g\<circ\>f)(x)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f),(z,y)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|z=f(x),y=g(z)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y=g(f(x))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(g\<circ\>f)(x)=g(f(x))>>>>
</eqnarray*>
</proof>
</theorem>
<\theorem>
<label|composition is associative>Let <math|f> be a partial function
between <math|A> and <math|B>, <math|g> be a partial function between
<math|C> and <math|D>, <math|h> a partial function between <math|E> and
<math|F> with <math|B<big|cap>C\<neq\>\<emptyset\>,
D<big|cap>F\<neq\>\<emptyset\>> then <math|(h\<circ\>(g\<circ\>f))=(h\<circ\>g)\<circ\>f>
(associativity of composition). As it does not matter where we place the
parentheses we can use the notation <math|h\<circ\>g\<circ\>f>\
</theorem>
<\proof>
We have
<\eqnarray*>
<tformat|<table|<row|<cell|g\<circ\>f>|<cell|=>|<cell|(A,D,graph(g\<circ\>f))
>>|<row|<cell|h\<circ\>(g\<circ\>f)>|<cell|=>|<cell|(A,F,graph(h\<circ\>(g\<circ\>f)))
>>|<row|<cell|h\<circ\>g>|<cell|=>|<cell|(C,F,graph(h\<circ\>g))
>>|<row|<cell|(h\<circ\>g)\<circ\>f>|<cell|=>|<cell|(A,F,graph((h\<circ\>g)\<circ\>f))
>>>>
</eqnarray*>
So we must proof that <math|graph(h\<circ\>(g\<circ\>f))=graph(h\<circ\>g)\<circ\>f)>\
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y)\<in\>graph(h\<circ\>(g\<circ\>f))>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>D\<vdash\>(x,z)\<in\>graph(g\<circ\>f)\<wedge\>(z,y)\<in\>graph(h)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z<rprime|'>\<in\>B\<vdash\>(x,z<rprime|'>)\<in\>graph(f)\<wedge\>(z<rprime|'>,z)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z<rprime|'>,y)\<in\>graph(h\<circ\>g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph((h\<circ\>g)\<circ\>f)>>|<row|<cell|(x,y)\<in\>graph((h\<circ\>g)\<circ\>f)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>f\<wedge\>(z,y)\<in\>graph(h\<circ\>g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z<rprime|'>\<in\>D\<vdash\>(z,z<rprime|'>)\<in\>graph(g)\<wedge\>(z<rprime|'>,y)\<in\>graph(h)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z<rprime|'>)\<in\>graph(g\<circ\>f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(h\<circ\>(g\<circ\>f))>>>>
</eqnarray*>
</proof>
<\theorem>
<label|preimage of composition>Let <math|f> be a partial function between
<math|A> and <math|B>, <math|g> a partial function between<math| C> and
<math|D> with <math|B <big|cap>C\<neq\>\<emptyset\>> then if
<math|E\<subseteq\>D> \ we have that <math|(g\<circ\>f)<rsup|-1>(E)=f<rsup|-1>(g<rsup|-1>(E))>
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>(g\<circ\>f)<rsup|-1>(E)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>E\<vdash\>(x,y)\<in\>graph(g\<circ\>f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f)\<wedge\>(z,y)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|z\<in\>g<rsup|-1>(E)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(g<rsup|-1>(E))>>>>
</eqnarray*>
</proof>
<\theorem>
<label|preimage of union or intersection>Let f be a function between A
and B and <math|X,T\<subseteq\>B> then
<math|f<rsup|-1>(X<big|cap>Y)=f<rsup|-1>(X)<big|cap>f<rsup|-1>(Y)> and
<math|f<rsup|-1>(X<big|cup>Y)=f<rsup|-1>(X)<big|cup>f<rsup|-1>(Y)>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>f<rsup|-1>(X<big|cap>Y)>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>y\<in\>X<big|cap>Y\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[y\<in\>X\<wedge\>(x,y)\<in\>graph(f)]\<wedge\>[y\<in\>Y\<wedge\>(x,y)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(X)\<wedge\>x\<in\>f<rsup|-1>(Y)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(X)<big|cap>f<rsup|-1>(Y)>>|<row|<cell|x\<in\>f<rsup|-1>(X)<big|cap>f<rsup|-1>(Y)>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(X)\<wedge\>x\<in\>f<rsup|-1>(Y)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[\<exists\>y\<in\>X\<vdash\>(x,y)\<in\>graph(f)]\<wedge\>[\<exists\>y<rprime|'>\<in\>Y\<vdash\>(x,y<rprime|'>)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is a function>>|<cell|y=y<rprime|'>\<Rightarrow\>y\<in\>X<big|cap>Y>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(X<big|cap>Y)>>>>
</eqnarray*>
</proof>
</theorem>
<\theorem>
<label|image of union or intersection>Let f be a partial function between
A and B and <math|X,Y\<in\>A> then <math|f(A<big|cup>B)=f(A)<big|cup>f(B)>
and <math|f(A<big|cap>B)\<subseteq\>f(A)<big|cap>f(B)>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>f(A<big|cup>B)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>A<big|cup>B\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[\<exists\>x\<in\>A\<vdash\>(x,y)\<in\>graph(f)]\<vee\>[\<exists\>x\<in\>B\<vdash\>(x,y)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(A)\<vee\>y\<in\>f(B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(A)<big|cup>f(B)>>|<row|<cell|y\<in\>f(A)<big|cup>f(B)>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(A)\<vee\>y\<in\>f(B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[\<exists\>x\<in\>A\<vdash\>(x,y)\<in\>graph(f)]\<vee\>[\<exists\>x\<in\>B\<vdash\>(x,y)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>A<big|cup>B\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(A<big|cup>B)>>|<row|<cell|y\<in\>f(A<big|cap>B)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>A<big|cap>B\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[x\<in\>A\<vdash\>(x,y)\<in\>graph(f)]\<wedge\>[x\<in\>B\<vdash\>(x,y)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f(A)\<wedge\>x\<in\>f(B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f(A)<big|cap>f(B)>>>>
</eqnarray*>
</proof>
</theorem>
<\theorem>
<label|restricted function><index|restricted function>Let <math|f> be a
partial function between A and B so <math|f=(A,B,graph(f))> and let
<math|C\<subseteq\>A> then <math|f<rsub|\|C>=(C,B,graph(f)<big|cap>(C\<times\>B))
> is a partial function, called the partial function induced by f on C
(or the function f restricted to <math|C>) between <math|C> and <math|B>.
Also <math|dom(f<rsub|\|C>)=dom(f)<big|cap>C>. In other words
<math|f<rsub|\|C>> is the function for which we have
<math|\<forall\>x\<in\>dom(f)<big|cap>C> that <math|f<rsub|\|C>(x)=f(x)>.
Note that by definition of <math|f<rsub|\|C>> we have
<math|graph(f<rsub|\|C>)=graph(f)<big|cap>(C\<times\>B)\<subseteq\>graph(f)>
<\proof>
\;
First <math|f<rsub|\|C>> is a function as if
<math|(x,y<rsub|1>),(x,y<rsub|2>)\<in\>graph(f)<big|cap>(C\<times\>B)\<Rightarrow\>(x,y<rsub|1>),(x,y<rsub|2>)\<in\>graph(f)\<Rightarrowlim\><rsub|f
is a function>y<rsub|1>=y<rsub|2>> and thus <math|f<rsub|\|C>> is a
function.\
Second\
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>dom(f<rsub|\|C>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>x\<in\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)<big|cap>C>>|<row|<cell|x\<in\>dom(f)<big|cap>C>|<cell|\<Rightarrow\>>|<cell|[\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>graph(f)\<subseteq\>A\<times\>B]\<wedge\>x\<in\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f<rsub|\|C>)>>>>
</eqnarray*>
</proof>
</theorem>
<\theorem>
<label|image of a restricted function>Let <math|f> be a partial function
between <math|A> and <math|B> and <math|D\<subseteq\>C\<subset\>A> then
<math|f<rsub|\|C>(D)=f(D<big|cap>C)> and
<math|f<rsub|\|D>=(f<rsub|\|C>)<rsub|\|D>> .
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>f<rsub|\|C>(D)>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>x\<in\>D\<vdash\>(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>D<big|cap>C\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(D<big|cap>C)>>|<row|<cell|y\<in\>f(D<big|cap>C)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>D<big|cap>C\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f<rsub|\|C>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f<rsub|C>(D)>>|<row|<cell|>|<cell|>|<cell|>>|<row|<cell|f<rsub|\|D>>|<cell|=>|<cell|(D,B,graph(f)<big|cap>(D\<times\>B))>>|<row|<cell|f<rsub|\|C>>|<cell|=>|<cell|(C,B,graph(f)<big|cap>(C\<times\>B))>>|<row|<cell|(f<rsub|\|C>)<rsub|\|D>>|<cell|=>|<cell|(D,B,(graph(f)<big|cap>(C\<times\>B))<big|cap>(D\<times\>B))>>|<row|<cell|>|<cell|\<equallim\><rsub|(D\<times\>B)\<subseteq\>(C\<times\>B)>>|<cell|(D,B,graph(f)<big|cap>(D\<times\>B))>>|<row|<cell|>|<cell|=>|<cell|f<rsub|\|D>>>>>
</eqnarray*>
</proof>
<\theorem>
<label|restriction and preimage>Let <math|f> be a partial function
between <math|A> and <math|B> and let
<math|C\<subseteq\>A,D\<subseteq\>B> then
<math|(f<rsub|\|C>)<rsup|-1>(D)=C<big|cap>f<rsup|-1>(D)>
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>(f<rsub|\|C>)<rsup|-1>(D)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>C\<wedge\>x\<in\>f<rsup|-1>(D)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>C<big|cap>f<rsup|-1>(D)>>|<row|<cell|x\<in\>C<big|cap>f<rsup|-1>(D)>|<cell|\<Rightarrow\>>|<cell|x\<in\>C\<wedge\>[\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(f)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>(f<rsub|\|C>)<rsup|-1>(D)>>>>
</eqnarray*>
</proof>
<\theorem>
<label|restriction of composition>Let <math|f> be a partial function
between <math|A> and <math|B> and let <math|g> be a partial function
between <math|C> and <math|D> with <math|C<big|cap>D\<neq\>\<emptyset\>>,
<math|E\<subseteq\>A> such that <math|f(E)<big|cap>C\<neq\>\<emptyset\>>
then <math|(g\<circ\>f)<rsub|\|E>=g<rsub|\|f(E)>\<circ\>f<rsub|\|E>>
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|g\<circ\>f>|<cell|=>|<cell|(A,D,graph(g\<circ\>f))>>|<row|<cell|(g\<circ\>f)<rsub|\|E>>|<cell|=>|<cell|(E,D,(graph(g\<circ\>f)<big|cap>(E\<times\>D))>>|<row|<cell|f<rsub|\|E>>|<cell|=>|<cell|(E,B,graph(f)<big|cap>(E\<times\>B))>>|<row|<cell|g<rsub|\|f(E)>>|<cell|=>|<cell|(f(E),D,graph(g)<big|cap>(f(E)\<times\>D))>>|<row|<cell|g<rsub|\|f(E)>\<circ\>f<rsub|\|E>>|<cell|=>|<cell|(E,D,(graph(g<rsub|\|f(E)>\<circ\>f<rsub|\|E>))>>>>
</eqnarray*>
\ So we must proof that <with|mode|math|graph(g\<circ\>f)<big|cap>(E\<times\>D)=graph(g<rsub|\|f(E)>\<circ\>f<rsub|\|E>)>
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y)\<in\>graph(g\<circ\>f)<big|cap>(E\<times\>D)>|<cell|\<Rightarrow\>>|<cell|x\<in\>E\<wedge\>[\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f)\<wedge\>(z,y)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|z\<in\>f(E)\<wedge\>(x,z)\<in\>graph(f)<big|cap>(E\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z,y)\<in\>graph(g)<big|cap>(f(E)\<times\>D)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f<rsub|\|E>)\<wedge\>(z,y)\<in\>graph(g<rsub|\|f(E)>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(g<rsub|\|f(E)>\<circ\>f<rsub|\|E>)>>|<row|<cell|(x,y)\<in\>graph(g<rsub|\|f(E)>\<circ\>f<rsub|\|E>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f<rsub|\|E>)\<wedge\>(z,x)\<in\>graph(g<rsub|\|f(E)>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f)<big|cap>(E\<times\>B)\<wedge\>(z,x)\<in\>graph(g)<big|cap>(f(E)*\<times\>D)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(g\<circ\>f)<big|cap>(E\<times\>D)>>>>
</eqnarray*>
</proof>
<\definition>
<subindex|injection|function>A partial function f between A and B is a
injection if \ from <math|(x,y),(x<rprime|'>,y)\<in\>graph(f)> if follows
that <math|x=x<rprime|'>> or in other words
<math|f(x)=f(y)\<Rightarrow\>x=y>
</definition>
<\theorem>
<label|image by a injective function of a intersection>If <math|f> is a
injection between <math|A> and <math|B>. Then if
<math|C<rsub|1>,C<rsub|2>\<subseteq\>A> then
<math|f(C<rsub|1><big|cap>C<rsub|2>)=f(C<rsub|1>)<big|cap>f(C<rsub|2>)>
</theorem>
<\proof>
We have already proved that <math|f(C<rsub|1><big|cap>C<rsub|2>)\<subseteq\>f(C<rsub|1>)<big|cap>f(C<rsub|2>)>.
Now if <math|y\<in\>f(C<rsub|1>)<big|cap>f(C<rsub|2>)\<Rightarrow\>\<exists\>c<rsub|1>\<in\>C<rsub|1>,c<rsub|2>\<in\>C<rsub|2>>
such that <math|(c<rsub|1>,y),(c<rsub|2>,y)\<in\>graph(f)\<Rightarrowlim\><rsub|injectivity>c<rsub|1>=c<rsub|2>\<in\>C<rsub|1><big|cap>C<rsub|2>>
</proof>
<\theorem>
<label|inverse function>If f is a injection between A and B then there
exists a unique (meaning there exists only one function g fulfilling
1,2,3,4) partial function g between B and A such that\
<\enumerate-numeric>
<item><math|\<forall\>x\<in\>dom(f)\<succ\>g(f(x)=x>
<item><math|\<forall\>y\<in\>dom(g)\<succ\>f(g(y))=y>
<item><math|dom(f)=rng(g)>
<item><math|rng(f)=dom(g)>
<item><math|g> is a injection
</enumerate-numeric>
The function g is called the inverse of <math|g> and noted as
<math|f<rsup|-1>>
<\proof>
Define then <math|g=(B,A,graph(f<rsup|-1>))> where
<math|graph(f<rsup|-1>)={(x,y)\<in\>B\<times\>A\|(y,x)\<in\>graph(f)}>
then we have the following\
<\enumerate-roman>
<item>If <math|(y,x<rsub|1>),(y,x<rsub|2>)\<in\>graph(f<rsup|-1>)\<Rightarrow\>(x<rsub|1>,y),(x<rsub|2>,y)\<in\>graph(f)\<Rightarrowlim\><rsub|f
is a injection>x<rsub|1>=x<rsub|2>> and thus g is a partial function
as required
<item>If <math|(y<rsub|1>,x),(y<rsub|2>,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>(x,y<rsub|1>),(x,y<rsub|2>)\<in\>graph(f)\<Rightarrowlim\><rsub|f
is function>y<rsub|1>=y<rsub|2>> and thus g is a injection
<item>If <math|x\<in\>dom(f)\<Rightarrow\>\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>graph(f)\<Rightarrow\>(y,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>x\<in\>rng(g)>.
If <math|x\<in\>rng(g)\<Rightarrow\>\<exists\>y\<in\>B\<vdash\>(y,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>(x,y)\<in\>graph(f)\<Rightarrow\>x\<in\>dom(f)>
<item>If <math|y\<in\>rng(f)=\<exists\>x\<in\>A\<vdash\>(x,y)\<in\>graph(f)\<Rightarrow\>(y,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>y\<in\>dom(g)>.
If <math|y\<in\>dom(g)\<Rightarrow\>\<exists\>x\<in\>A\<vdash\>(y,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>(x,y)\<in\>graph(f)\<Rightarrow\>y\<in\>rng(f)>
<item><math|\<forall\>x\<in\>dom(f)> we have
<math|(x,y)\<in\>graph(f)\<Rightarrow\>(y,x)\<in\>graph(f<rsup|-1>)>
and as <math|(x,y)\<in\>graph(f)> is noted as <math|y=f(x)> and
<math|(y,x)\<in\>graph(f<rsup|-1>)> is noted as <math|x=g(y)> we have
thus <math|x=g(f(x))>
<item><math|\<forall\>y\<in\>dom(g)> we have
<math|(y,x)\<in\>graph(f<rsup|-1>)\<Rightarrow\>(x,y)\<in\>graph(f)>
and as <math|(y,x)\<in\>graph(f<rsup|-1>)> is noted as <math|x=g(y)>
and <math|(x,y)\<in\>graph(f)> is noted as <math|y=f(x)> we have
<math|y=f(g(y))>
<item>Assume that there exists <math|g=(B,A,R<rprime|'>)> and
<math|f=(B,A,R<rprime|''>)> such that
<math|dom(f)=rng(g)=rng(k),rng(f)=dom(g)=dom(k)> and
<math|\<forall\>x\<in\>dom(f)\<succ\>g(f(x))=k(f(x))=x,\<forall\>y\<in\>dom(g)=dom(k)\<succ\>f(g(y))=f(k(y))=y>.
If we proof then that <math|R<rprime|'>=R<rprime|'><rprime|'>> we
have proved uniqueness.
<\eqnarray*>
<tformat|<table|<row|<cell|(y,x)\<in\>R<rprime|'>>|<cell|\<Rightarrow\>>|<cell|y\<in\>dom(g)=dom(k)=rng(f)\<wedge\>x=g(y)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>dom(k)\<wedge\>f(x)=f(g(y))=y>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|k(y)=k(f(x))=x>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>R<rprime|''>>>|<row|<cell|(y,x)\<in\>R<rprime|''>>|<cell|\<Rightarrow\>>|<cell|y\<in\>dom(k)=dom(g)=rng(f)\<wedge\>x=k(y)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>dom(g)\<wedge\>f(x)=f(k(y))=y>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|g(y)=g(f(x))=x>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>R<rprime|'>>>>>
</eqnarray*>
</enumerate-roman>
</proof>
</theorem>
<\theorem>
<label|preimage of injective mapping>Let <math|f> be a injective partial
function between <math|A> and <math|B> and
<math|C\<subseteq\>A,D\<subseteq\>B> then
<\enumerate>
<item><math|(f<rsup|-1>)<rsup|-1>(C)=f(C)>
<item><math|f<rsup|-1>(D)=(f<rsup|-1>)(D)> (here with
<math|f<rsup|-1>(D)> we mean the preimage of <math|f> and
<math|(f<rsup|-1>)(D)> the image of <math|f<rsup|-1>>
</enumerate>
</theorem>
<\proof>
\;
(1)
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>(f<rsup|-1>)<rsup|-1>(C)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>C\<vdash\>(x,y)\<in\>graph(f<rsup|-1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f(C)>>|<row|<cell|x\<in\>f(C)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>C\<vdash\>(y,x)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f<rsup|-1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>(f<rsup|-1>)<rsup|-1>(C)>>>>
</eqnarray*>
\ Another way to prove this is using the notation
<math|f<rsup|-1>(x)\<equiv\>(x,f<rsup|-1>(x))\<in\>graph(f<rsup|-1>)> and
<math|f(x)\<equiv\>(x,f(x))\<in\>graph(f)> a way of proof that we will
use regularly in this document
\;
<\enumerate>
<item>If <math|x\<in\>(f<rsup|-1>)<rsup|-1>(C)> then
<math|f<rsup|-1>(x)\<in\>C> so <math|x=f(f<rsup|-1>(x))\<in\>f(C)> so
<math|(f<rsup|-1>)<rsup|-1>(C)\<subseteq\>f(C)>
<item>If <math|y\<in\>f(C)> then <math|y=f(x), x\<in\>C> and so
<math|f<rsup|-1>(y)=f<rsup|-1>(f(x))=x\<in\>C\<Rightarrow\>y\<in\>(f<rsup|-1>)<rsup|-1>(C)>
</enumerate>
</proof>
(2)
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>f<rsup|-1>(D)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f<rsup|-1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>(f<rsup|-1>)(D)>>|<row|<cell|x\<in\>(f<rsup|-1>)(D)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>D\<vdash\>(y,x)\<in\>graph(f<rsup|-1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(D)>>>>
</eqnarray*>
Another way to proof is\
<\enumerate>
<item><math|x\<in\>f<rsup|-1>(D)> then <math|f(x)\<in\>D> so
<math|x=f<rsup|-1>(f(x))\<in\>(f<rsup|-1>)(D)>
<item><math|x\<in\>(f<rsup|-1>)(D)\<Rightarrow\>\<exists\>y\<in\>D\<vdash\>x=f<rsup|-1>(y)\<Rightarrow\>f(x)=y\<Rightarrow\>x\<in\>f<rsup|-1>(D)>
</enumerate>
<\theorem>
<label|composition of injections is injective>Let <math|f> be a partial
function between A and B and <math|g> a partial function between
<math|B<rprime|'>> and C with <math|B\<subseteq\>B<rprime|'>> then
<math|f\<circ\>g> is a injection
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|(f\<circ\>g)(x)=(f\<circ\>g)(x<rprime|'>)>|<cell|\<Rightarrow\>>|<cell|f(g(x))=f(g(x<rprime|'>))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|g(x)=g(x<rprime|'>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x=x<rprime|'>>>>>
</eqnarray*>
</proof>
<\theorem>
<label|preimage of difference of sets>Let f be a partial function between
A and B and <math|C\<subseteq\>B> then
<math|f<rsup|-1>(B<mid|\\>C)=dom(f)<mid|\\>f<rsup|-1>(C)>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>f<rsup|-1>(B<mid|\\>C)>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>B<mid|\\>C\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>y\<nin\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<subset\>dom(f)\<wedge\>x\<nin\>f<rsup|-1>(C)
[otherwise we would have \<exists\>y<rprime|'>\<in\>C\<vdash\>(x,y<rprime|'>)\<in\>graph(f)\<Rightarrowlim\><rsub|f
is a partial function>y<rprime|'>=y\<Rightarrow\>y\<in\>C a
contradiction]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)<mid|\\>f<rsup|-1>(C)>>|<row|<cell|x\<in\>dom(f)<mid|\\>f<rsup|-1>(C)>|<cell|\<Rightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>x\<nin\>f<rsup|-1>(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>B\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>
>|<cell|y\<nin\>C [otherwise x\<in\>f<rsup|-1>(C)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>B<mid|\\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(B<mid|\\>C)>>>>
</eqnarray*>
</proof>
</theorem>
<\definition>
<subindex|surjection|function>A partial function f between A and B is a
surjection if <math|\<forall\>y\<in\>B> there exists a
<math|x\<in\>A\<vdash\>(x,y)\<in\>graph(f)> In other word
<math|\<forall\>y\<in\>B\<succ\>\<exists\>x\<in\>A\<vdash\>f(x)=y>
</definition>
<\theorem>
If a partial function between <math|A> and <math|B> is surjective and
<math|C\<subseteq\>B> then <math|f<rsup|>(f<rsup|-1>(C))=C>
</theorem>
<\proof>
We have already <math|f(f<rsup|-1>(C))\<subseteq\>C> (see
<reference|image of preimage>) so we must prove that
<math|C\<subseteq\>f(f<rsup|-1>(C))>
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>C>|<cell|\<Rightarrowlim\><rsub|f is
a surjection>>|<cell|\<exists\>x\<in\>A\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>f<rsup|-1>(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>f(f<rsup|-1>(C))>>|<row|<cell|>|<cell|>|<cell|>>>>
</eqnarray*>
\;
</proof>
<\theorem>
<label|composition of surjections is surtjective>Let <math|f> be a
partial function between A and B, g a partial function between <math|B>
and C where both f and g are surjection's then <math|f\<circ\>g> is a
surjection
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|z\<in\>C>|<cell|\<Rightarrowlim\><rsub|g is
surjective>>|<cell|\<exists\>y\<in\>B\<vdash\>(y,z)\<in\>graph(g)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is surjective>>|<cell|\<exists\>x\<in\>A\<vdash\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>A\<vdash\>(x,z)\<in\>graph(g\<circ\>f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|f\<circ\>g
is a surjection>>>>
</eqnarray*>
\;
</proof>
<\definition>
<subindex|bijection|function>A partial function f between A and B is a
bijection if f is a injection and a surjection
</definition>
<\definition>
Let <math|A> be a set then <math|i<rsub|A>=(A,A,graph(i<rsub|A>))> with
<math|graph(i<rsub|A>)={(x,y)\<in\>A\<times\>A\|x=y}> . From this it
follows that <math|dom(i<rsub|A>)=rng(i<rsub|A>)=A> and that
<math|i<rsub|A>> is a bijection
</definition>
<\theorem>
Let <math|f> a partial function between <math|A> and <math|B> then
<math|f=f\<circ\>i<rsub|A>=i<rsub|B>\<circ\>f>
</theorem>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|f>|<cell|=>|<cell|(A,B,graph(f))>>|<row|<cell|i<rsub|A>>|<cell|=>|<cell|(A,A,graph(i<rsub|A>))>>|<row|<cell|i<rsub|B>>|<cell|=>|<cell|(B,B,graph(i<rsub|B>))>>|<row|<cell|f\<circ\>i<rsub|A>>|<cell|=>|<cell|(A,B,graph(f\<circ\>i<rsub|A>))>>|<row|<cell|i<rsub|B>\<circ\>f>|<cell|=>|<cell|(A,B,graph(i<rsub|B>\<circ\>f)>>>>
</eqnarray*>
So our proof is complete if we have that
<math|graph(f)=graph(i<rsub|B>\<circ\>f)=graph(f\<circ\>i<rsub|A>)>
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y)\<in\>graph(f)>|<cell|\<Leftrightarrow\>>|<cell|(x,y)\<in\>graph(f)\<wedge\>(y,y)\<in\>graph(i<rsub|B>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|(x,y)\<in\>graph(i<rsub|b>\<circ\>f)>>|<row|<cell|(x,y)\<in\>graph(f)>|<cell|\<Leftrightarrow\>>|<cell|(x,x)\<in\>graph(i<rsub|A>)\<wedge\>(x,y)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|(x,y)\<in\>graph(f\<circ\>i<rsub|A>)>>>>
</eqnarray*>
\;
</proof>
<\theorem>
<label|composition of bijective partial functions is bijective>Let
<math|f> be a bijective partial function between A and B, g a bijective
partial function between B and C then <math|g\<circ\>f> is a bijective
partial function between A and C
</theorem>
<\proof>
This follows from <reference|composition of injections is injective> and
<reference|composition of surjections is surtjective>
</proof>
<\theorem>
<label|restriction of injection>Let f be a injection between A and B and
<math|C\<subseteq\>A> then <math|f<rsub|\|C>> is a injection and
<math|graph((f<rsub|\|C>)<rsup|-1>)=graph((f<rsup|-1>)<rsub|\|f(C)>)>
</theorem>
<\proof>
\;
First we prove injectivity, if <math|f<rsub|\|C>(x)=f<rsub|\|C>(y)\<Rightarrow\>f(x)=f(y)\<Rightarrow\>x=y>
Second we have to proof that <math|graph((f<rsub|\|C>)<rsup|-1>)=graph((f<rsup|-1>)<rsub|\|f(C)>)>\
<\eqnarray*>
<tformat|<table|<row|<cell|(y,x)\<in\>graph((f<rsub|\|C>)<rsup|-1>)>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f<rsub|\|C>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)\<wedge\>y\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f<rsup|-1>)\<wedge\>x\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>A)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph((f<rsup|-1>)<rsub|\|f(C)>)>>|<row|<cell|(y,x)\<in\>graph((f<rsup|-1>)<rsub|\|f(C)>)>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>A)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)\<wedge\>y\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f)\<wedge\>[\<exists\>x<rprime|'>\|x<rprime|'>\<in\>C\<wedge\>f(x<rprime|'>)=y]>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|injectivity\<Rightarrow\>x=x<rprime|'>>>|<cell|(x,y)\<in\>graph(f)\<wedge\>x\<in\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f)<big|cap>(C\<times\>B)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,x)\<in\>graph(f<rsub|\|C>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph((f<rsub|\|C>)<rsup|-1>)>>>>
</eqnarray*>
\;
</proof>
<\definition>
<index|<math|f<rsup|\|C>>><label|image restriction>Let <math|f> be a
partial function between <math|A> and <math|B> (or
<math|f=(A,B,graph(f)))> and <math|C\<subseteq\>B> then
<math|f<rsup|\|C>=(A,C,graph(f)<big|cap>(A\<times\>C))>. As <math|f> is a
function we have trivial that <math|f<rsup|\|C>> is a function between
<math|A> and <math|C>. Also it is trivial that
<math|graph(f<rsup|\|C>)\<subseteq\>graph(f)> and
<math|graph(f)=graph(f<rsup|\|f(C)>)>
</definition>
<\theorem>
<label|image restriction of a injection>Let <math|f> be a injection
between <math|A> and <math|B> then <math|f<rsup|\|f(A)>> is a bijection
between <math|A> and <math|f(A)>
</theorem>
<\proof>
\;
First note that if <math|f<rsup|\|C>(x)=f<rsup|\|C>(y)\<Rightarrow\>f(x)=f(y)\<Rightarrow\>x=y>
so we have that <math|f<rsup|\|C>> is a injection
Second, surjectivity is trivial as if
<math|y\<in\>f(A)\<Rightarrow\>\<exists\>x\<in\>A\<vdash\>y=f(x)=f<rsup|<rsub|\|f(A)>>(x)>
</proof>
<\theorem>
<label|restriction of a bijection>Let f be a bijection between A and B
and <math|\<emptyset\>\<neq\>C\<subseteq\>A> then
<math|(f<rsub|\|C>)<rsup|\|f(C)> is a bijection between C> and
<math|f(C)=B<mid|\\>f(A<mid|\\>C)>
</theorem>
<\proof>
Now to prove that it is a injection and surjection. By
<reference|restriction of injection> we have already that
<math|f<rsub|\|C>> is a injection, surjectivity follows from
<math|f<rsub|\|C>(C)=f(C)> and the previous theorem
(<reference|restriction of a injection>). Now to prove that
<math|f(C)=B<mid|\\>f(A<mid|\\>C)> first take
<math|y\<in\>f(C)\<subseteq\>B\<Rightarrow\>\<exists\>c\<in\>C> such that
<math|f(c)=y> and assume that <math|y\<in\>f(A<mid|\\>C)> then there
exists a <math|x\<in\>A,x\<neq\>c> such that <math|f(x)=y> then because
of injectivity we have <math|c=x\<neq\>c> a contradiction, so
<math|y\<nin\>f(A<mid|\\>C)\<Rightarrow\>y\<in\>B<mid|\\>f(A<mid|\\>C)\<Rightarrow\>f(C)\<subseteq\>B<mid|\\>f(A<mid|\\>C)>.
On the other hand if <math|y\<in\>B<mid|\\>f(A<mid|\\>C)> then there by
surjectivity of <math|f> a <math|x\<in\>A> such that <math|f(x)=y>,
assume now that <math|x\<nin\>C\<Rightarrow\>x\<in\>A<mid|\\>C\<Rightarrow\>y=f(x)\<in\>f(A<mid|\\>C)>
a contradiction so we have that <math|x\<in\>C\<Rightarrow\>y=f(x)\<in\>f(C)\<Rightarrow\>B<mid|\\>f(A<mid|\\>C)\<subseteq\>f(C)>
</proof>
<\theorem>
<label|injection extension>For two injections, <math|f<rsub|1>> between
<math|A<rsub|1>> and <math|B<rsub|1>> and <math|f<rsub|2>> between
<math|A<rsub|2>> and <math|B<rsub|2>> where
<math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> and
<math|B<rsub|1><big|cap>B<rsub|2>=\<emptyset\>> then
<math|f=f<rsub|1><big|cup>f<rsub|2>> is a injection
</theorem>
<\proof>
By <reference|function extension> we have that <math|f> is a function now
now if <math|(x<rsub|1>,y),(x<rsub|2>,y)\<in\>graph(f)> then as
<math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> then we have the
following possibilities\
<\enumerate>
<item><math|(x<rsub|1>,y)\<in\>graph(f<rsub|1>)\<vee\>(x<rsub|2>,y)\<in\>graph(f<rsub|1>)>
then as <math|f<rsub|1>> is a injection we have
<math|x<rsub|1>=x<rsub|2>>
<item><math|(x<rsub|1>,y)\<in\>graph(f<rsub|1>)\<vee\>(x<rsub|2>,y)\<in\>graph(f<rsub|2>)>
which does not count because <math|B<rsub|1><big|cap>B<rsub|2>=\<emptyset\>>
<item><math|(x<rsub|1>,y)\<in\>graph(f<rsub|2>)\<vee\>(x<rsub|2>,y)\<in\>graph(f<rsub|1>)>
which does not count because <math|B<rsub|1><big|cap>B<rsub|2>=\<emptyset\>>
<item><math|(x<rsub|1>,y)\<in\>graph(f<rsub|2>)\<vee\>(x<rsub|2>,y)\<in\>graph(f<rsub|2>)>
then as <math|f<rsub|2>> is a injection we have
<math|x<rsub|1>=x<rsub|2>>\
</enumerate>
</proof>
<\theorem>
<label|surjection extension>For two surjection's, <math|f<rsub|1>>
between <math|A<rsub|1>> and <math|B<rsub|1>> and <math|f<rsub|2>>
between <math|A<rsub|2>> and <math|B<rsub|2>> where
<math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> then
<math|f=f<rsub|1><big|cup>f<rsub|2>> is a surjection between
<math|A<rsub|1><big|cup>A<rsub|2>\<rightarrow\>B<rsub|1><big|cup>B<rsub|2>>
</theorem>
<\proof>
Let <math|y\<in\>B<rsub|1><big|cup>B<rsub|2>> then either
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>B<rsub|1>>|<cell|\<Rightarrowlim\><rsub|f<rsub|1>
is surjective>>|<cell|\<exists\>x\<in\>A<rsub|1>\<vdash\>y=f<rsub|1>(x)=f(x)>>|<row|<cell|y\<in\>B<rsub|2>>|<cell|\<Rightarrowlim\><rsub|f<rsub|2>
is surjective>>|<cell|\<exists\>x\<in\>A<rsub|2>\<vdash\>y=f<rsub|2>(x)=f(x)>>>>
</eqnarray*>
proving surjectivity.
</proof>
<\theorem>
<label|bijection extension>For two bijection's, <math|f<rsub|1>> between
<math|A<rsub|1>> and <math|B<rsub|1>> and <math|f<rsub|2>> between
<math|A<rsub|2>> and <math|B<rsub|2>> where
<math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> and
<math|B<rsub|1><big|cap>B<rsub|2>=\<emptyset\>> then
<math|f=f<rsub|1><big|cup>f<rsub|2>> is a bijection
</theorem>
<\proof>
Trivial by using <reference|injection extension> and
<reference|surjection extension>
</proof>
<subsection|Functions>
<\definition>
<index|mapping><index|function>A function is a partial function
<math|f=(A,B,graph(f))> such that <math|dom(f)=A>. Sometimes we refer to
a function also via its synonym 'mapping'. We use also say <math|f> is a
function (mapping) from <math|A> to <math|B>
</definition>
<\example>
<math|i<rsub|A>> is a function that is also a bijection
</example>
<\remark>
If f is a function from A to B (<math|f:A\<rightarrow\>B)> then
<math|rng(f)=f(A)> and <math|f<rsup|-1>(B)=A>
</remark>
Note that everything that we have already proved is valid for functions
(mappings) is still valid as functions (mappings) are partial functions.
Also if <math|f=(A,B,graph(f))> is a function, so that <math|dom(f)=A>, we
don't have to use the notation <math|(A,B,graph(f))> anymore as
<math|f:A\<rightarrow\>B> contains all the info we need (with the exception
of the graph, which is either not important or is defined in the context or
is noted by <math|graph(f)>). For the rest of the text if we are dealing
with a function we use the notation <math|f:A\<rightarrow\>B> and
<math|graph(f)> to specify the function, mapping. So for a function
<math|f:A\<rightarrow\>B> means <math|f=(A,B.graph(f))>.
<\theorem>
<label|restriction of a mapping>Let <math|f:A\<rightarrow\>B> be a
function (or mapping) and <math|C\<subseteq\>A,D\<subseteq\>B> then
<math|f<rsub|\|C>> and <math|(f<rsup|\|D>)<rsub|\|f<rsup|-1>(D)>> are
functions (mappings)
</theorem>
<\proof>
\;
First if <math|x\<in\>C\<Rightarrowlim\><rsub|f is a
function>\<exists\>y\<in\>B> such that
<math|(x,y)\<in\>graph(f)\<Rightarrow\>(x,y)\<in\>graph(f)<big|cap>(C\<times\>B)=graph(f<rsub|\|C>)\<Rightarrow\>x\<in\>dom(f<rsub|\|C>)\<subseteq\>C\<Rightarrow\>dom(f<rsub|\|C>)=C>
Second if <math|x\<in\>f<rsup|-1>(D)\<Rightarrow\>\<exists\>y\<in\>D\<succ\>(x,y)\<in\>graph(f)\<Rightarrow\>(x,y)\<in\>graph(f)<big|cap>(A,D)=graph(f<rsup|\|D>)\<Rightarrow\>x\<in\>f<rsup|\|D>>
</proof>
<\notation>
<label|shorthand notation of function>If <math|f:A\<rightarrow\>B> is a
function then if <math|C\<subseteq\>A,D\<subseteq\>B> if we define the
graph of <math|f:C\<rightarrow\>D> by
<math|graph(f:C\<rightarrow\>D)=graph(f)<big|cap>(C\<times\>D)> then we
can write.
<\enumerate>
<item><math|f> as <math|f:A\<rightarrow\>B>
<item><math|f<rsub|\|C>> as <math|f:C\<rightarrow\>B> (which is a
function by the previous theorem)
<item><math|f<rsup|\|D>> as <math|f:f<rsup|-1>(D)\<rightarrow\>D>
(again a function by the previous theorem)
<item><math|(f<rsub|\|C>)<rsup|\|D>> as <math|f:C\<rightarrow\>D> (a
function if <math|C\<subseteq\>f<rsup|-1>(D)> for if
<math|x\<in\>C\<Rightarrow\>\<exists\>y\<in\>D\<vdash\>(x,y)\<in\>graph(f)\<Rightarrow\>(x,y)\<in\>graph(f)<big|cap>(C\<times\>D)\<Rightarrow\>(x,y)\<in\>(graph(f)<big|cap>(C\<times\>B))<big|cap>(C\<times\>D)=graph((f<rsub|\|C>)<rsup|\|D>))>
</enumerate>
For the rest of the text, this is the way we will note a restriction
(image restriction) of a given function (mapping).
</notation>
<\definition>
If <math|f> is a function from <math|A> to <math|B> which is injective
then <math|f<rsup|-1>:rng(f)\<rightarrow\>A> is defined and we note the
function <math|f<rsup|-1>:rng(f)\<rightarrow\>A> the inverse mapping.\
</definition>
<\note>
If <math|f> is surjective then <math|rng(f)=B=dom(f<rsup|-1>)> and thus
the inverse of <math|f> is a mapping and equal to the inverse mapping.
Note the difference between 'inverse' and 'inverse mapping'
</note>
<\theorem>
<label|restriction of composition of a mapping>Let
<math|f:A\<rightarrow\>B> and <math|g:C\<rightarrow\>D> be functions
where <math|C<big|cap>B=\<emptyset\>> then <math|g\<circ\>f> is a
function if and only if <math|f(A)\<subseteq\>C>. Also if
<math|E\<subseteq\>A> then <math|(g\<circ\>f)<rsub|\|E>> is a function if
and only if <math|f(E)\<subseteq\>C>
</theorem>
<\proof>
First we prove that <math|g\<circ\>f> is a mapping
<math|\<Leftrightarrow\>f(A)\<subseteq\>C>
<\enumerate>
<item>Assume that <math|g\<circ\>f> is a mapping and take
<math|y\<in\>f(A)> then <math|\<exists\>x\<in\>A> such that
<math|(x,y)\<in\>graph(f)>, now because <math|g\<circ\>f> is a mapping
there exists a <math|z\<in\>D> such that
<math|(x,z)\<in\>graph(g\<circ\>f)> and thus there exists a
<math|y<rprime|'>\<in\>B> such that
<math|(x,y<rprime|'>)\<in\>graph(f)> and
<math|(y<rprime|'>,z)\<in\>graph(g)> then because <math|f> is a
function we have <math|y=y<rprime|'>> so <math|(y,z)\<in\>graph(g)> and
thus <math|y\<in\>dom(g)=C>
<item>Assume <math|f(A)\<subseteq\>C> then if <math|x\<in\>A> we have
because <math|f> is a mapping a <math|y\<in\>B> such that
<math|(x,y)\<in\>graph(f)\<Rightarrow\>y\<in\>f(A)\<Rightarrow\>y\<in\>C>
and then because <math|g> is a mapping the existence of a
<math|z\<in\>D> such that <math|(y,z)\<in\>graph(g)\<Rightarrow\>(x,z)\<in\>g\<circ\>f>
<item>As we have <math|(g\<circ\>f)<rsub|\|E>=g<rsub|\|f(E)>\<circ\>f<rsub|\|E>>
where <math|g<rsub|\|f(E)>>,<math|f<rsub|\|E>> are mappings \ and
<math|f<rsub|\|E>(E)=f(E)> we have by (2) that
<math|(g\<circ\>f)<rsub|\|E>> is a mapping
</enumerate>
</proof>
<\theorem>
<label|restriction and composition of functions>If
<math|f:A\<rightarrow\>B>, <math|g:C\<rightarrow\>D> and
<math|B<big|cap>C\<neq\>\<emptyset\>> are functions and
<math|D\<subseteq\>A> is such that <math|f(D)\<subseteq\>C> then
<math|g\<circ\>f:D\<rightarrow\>g(f(D))> is a function and is the
composition of <math|f:D\<rightarrow\>f(D)> and
<math|g:f(D)\<rightarrow\>g(f(D))>
</theorem>
<\proof>
Using the notation <reference|shorthand notation of function> we have
that\
<\eqnarray*>
<tformat|<table|<row|<cell|(f<rsub|\|D>)<rsup|\|f(D)>=f:D\<rightarrow\>f(D)>|<cell|=>|<cell|(D,f(D),graph(f)<big|cap>(D\<times\>f(D)))>>|<row|<cell|(g<rsub|\|f(D)>)<rsup|\|g(f(D))>=g:f(D)\<rightarrow\>g(f(D))>|<cell|=>|<cell|(f(D),g(f(D)),graph(g)<big|cap>(f(D)\<times\>g(f(D)))>>|<row|<cell|g\<circ\>f:D\<rightarrow\>g(f(D))>|<cell|=>|<cell|(D,g(f(D)),graph(g\<circ\>f)<big|cap>(D\<times\>g(f(D)))>>|<row|<cell|(g<rsub|\|f(D)>)<rsup|\|g(f(D))>\<circ\>(f<rsub|\|D>)<rsup|\|f(D)>
(function by <reference|restriction of composition of a
mapping>)>|<cell|=>|<cell|(D,g(f(D)),graph((g<rsub|\|f(D)>)<rsup|\|g(f(D))>\<circ\>(f<rsub|\|D>)<rsup|\|f(D)>))>>>>
</eqnarray*>
so we must prove that <math|graph(g\<circ\>f)<big|cap>(D\<times\>g(f(D)))=graph((g<rsub|\|f(D)>)<rsup|\|g(f(D))>\<circ\>(f<rsub|\|D>)<rsup|\|f(D)>)>
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y)\<in\>graph(g\<circ\>f)<big|cap>(D\<times\>g(f(D)))>|<cell|\<Rightarrow\>>|<cell|[\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f)\<wedge\>(z,y)\<in\>graph(g)]\<wedge\>x\<in\>D\<wedge\>y\<in\>g(f(D))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|z\<in\>f(D)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f)<big|cap>(D\<times\>f(D))=graph((f<rsub|\|D>)<rsup|\|f(D)>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z,y)\<in\>graph(g)<big|cap>(f(D)\<times\>g(f(D)))=graph((g<rsub|\|f(D)>)<rsup|\|g(f(D))>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph((g<rsub|\|f(D)>)<rsup|\|g(f(D))>\<circ\>(f<rsub|\|D>)<rsup|\|f(D)>)>>|<row|<cell|(x,y)\<in\>graph((g<rsub|\|f(D)>)<rsup|\|g(f(D))>\<circ\>(f<rsub|\|D>)<rsup|\|f(D)>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>f(D)\<vdash\>(x,z)\<in\>graph(f)<big|cap>(D\<times\>f(D))\<wedge\>(z,x)\<in\>graph(g)<big|cap>(f(D)\<times\>g(f(D)))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(g\<circ\>f)<big|cap>(D\<times\>g(f(D))>>>>
</eqnarray*>
</proof>
<\theorem>
<label|image of inversion of injective function>If <math|f> is a
injective function from <math|A> to <math|B> then if
<math|C\<subseteq\>A> we have <math|f<rsup|-1>(f<rsup|>(C))=C> (there is
a slight confusion of notation here, we don't use <math|f<rsup|-1>> as
the preimage, but as the inverse)
</theorem>
<\proof>
\;
<\enumerate>
<item>If <math|x\<in\>f<rsup|-1>(f(C))> then <math|x=f<rsup|-1>(y)>
where <math|y\<in\>f(C)> and then <math|y=f(z)> where <math|z\<in\>C>
and thus <math|x=f<rsup|-1>(f(z))=z\<in\>C> so
<math|f<rsup|-1>(f(C))\<subseteq\>C>
<item>If <math|x\<in\>C> then <math|f(x)\<in\>f(C)> (<math|f> is a
function) and as <math|x=f<rsup|-1>(f(x))\<in\>f<rsup|-1>(f(C))\<Rightarrow\>C\<subseteq\>f<rsup|-1>(f(C))>
</enumerate>
</proof>
<\theorem>
<label|mapping extension>Let <math|f<rsub|1>:A<rsub|1>\<rightarrow\>B<rsub|1>>
and <math|f<rsub|2>:A<rsub|2>\<rightarrow\>B<rsub|2>> be functions and
let <math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> then
<math|f<rsub|1><big|cup>f<rsub|2>:A<rsub|1><big|cup>A<rsub|2>\<rightarrow\>B<rsub|1><big|cup>B<rsub|2>>
is a function
</theorem>
<\proof>
By <reference|function extension> <math|f<rsub|1><big|cup>f<rsub|2>> is a
function and <math|dom(f)=dom(f<rsub|1>)<big|cup>dom(f<rsub|2>)\<equallim\><rsub|f<rsub|1>,f<rsub|2>
are functions>A<rsub|1><big|cup>A<rsub|2>>
</proof>
<\theorem>
<label|image and preimage>Let <math|f> be a function \ from <math|A> to
<math|B> then if <math|C\<subseteq\>A,D\<subseteq\>B> is such that
<math|f(C)\<subseteq\>D> then <math|C\<subseteq\>f<rsup|-1>(D)>
</theorem>
<\proof>
If <math|x\<in\>C\<subseteq\>A=dom(f)\<Rightarrow\>\<exists\>y> such that
<math|(x,y)\<in\>graph(f)> and as <math|f(C)\<subseteq\>D> we have
<math|y\<in\>D\<Rightarrow\>x\<in\>f<rsup|-1>(D)>
</proof>
<\theorem>
<label|preimage of difference>If <math|f:A\<rightarrow\>B> is a function
(mapping) then <math|f<rsup|-1>(B<mid|\\>C)=A<mid|\\>f<rsup|-1>(C)>
</theorem>
<\proof>
This follows from <reference|preimage of difference of sets> and the fact
that <math|dom(f)=A>
</proof>
<\definition>
<index|bijective sets>If <math|f:A\<rightarrow\>B> is a function from A
to B that is a bijection then we say that A and B are bijective
</definition>
<\theorem>
<label|characterization of a bijective mapping>For a function
<math|f:A\<rightarrow\>B> we have the following\
<\eqnarray*>
<tformat|<table|<row|<cell|f<with|mode|text| is a
bijection>>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>g:B\<rightarrow\>A
such that f\<circ\>g=i<rsub|B><with|mode|text| and
>g\<circ\>f=i<rsub|A> >>|<row|<cell|g>|<cell|=>|<cell|f<rsup|-1>>>>>
</eqnarray*>
<\proof>
\;
First if f is a bijection (thus injective) then by <reference|inverse
function> <math|\<exists\>g> injection from <math|B> to A such that
<math|dom(f)=rng(g)>, <math|rng(f)=dom(g)>,
<math|\<forall\>x\<in\>dom(f)> we have <math|g(f(x))=x> and
<math|\<forall\>y\<in\>rng(f)> we have <math|f(g(y))=y>, now because f
is a bijective function we have <math|dom(f)=A> and <math|rng(f)=B> so
that <math|\<exists\>g> injective function (mapping) from B to A such
that <math|A=rng(g)>, <math|B=dom(g)> (which makes g a bijection) and
<math|\<forall\>x\<in\>A,\<forall\>y\<in\>B> we have
<math|g(f(x))=x,f(g(y))=y> which means that
<math|g\<circ\>f=i<rsub|A>,f\<circ\>g=i<rsub|B>>
Finally if <math|f\<circ\>g=i<rsub|B>> and <math|g\<circ\>f=i<rsub|A>>
then if <math|f(x<rsub|1>)=f(x<rsub|2>)> we have that
<math|g(f(x<rsub|1>))=g(f(x<rsub|2>))\<Rightarrow\>i<rsub|A>(x<rsub|1>)=i<rsub|A>(x<rsub|2>)\<Rightarrow\>x<rsub|1>=x<rsub|2>>.
and thus f is injective. Also if <math|y\<in\>B> then
<math|y=i<rsub|B>(y)=f(g(y))\<Rightarrow\>y=f(x)> with <math|x=g(y)> so
f is a bijection
The fact that <math|g=f<rsup|-1>> follows from <reference|inverse
function> and <math|g(f(x))=i<rsub|A>(x)=y> and
<math|f(g(y))=i<rsub|B>(y)> for all
<math|x\<in\>A=dom(f),y\<in\>B=dom(g)>
</proof>
</theorem>
<\theorem>
<label|inverse of bijective mapping>If <math|f:A\<rightarrow\>B> is a
bijective function then <math|f<rsup|-1>:B\<rightarrow\>A> is also a
bijective function
</theorem>
<\proof>
As <math|f> is bijective it is a injection and we can use
<reference|inverse function> to prove that <math|f<rsup|-1>> exists and
is injective. It also a function because <math|dom(f<rsup|-1>)=rng(f)=B>
and it is surjective because <math|rng(f<rsup|-1>)=dom(g)=A>
</proof>
<\theorem>
<label|composition of bijective mappings>If <math|f:A\<rightarrow\>B> is
a bijective function and if <math|g:B\<rightarrow\>C> is a bijective
function then <math|f\<circ\>g:A\<rightarrow\>C> is a bijection and
<math|(f\<circ\>g)<rsup|-1>=g<rsup|-1>\<circ\>f<rsup|-1>>
</theorem>
<\proof>
Take <math|f<rsup|-1>\<circ\>g<rsup|-1>> then <math|\<forall\>c\<in\>C>
we have <math|(g\<circ\>f)\<circ\>(f<rsup|-1>\<circ\>g<rsup|-1>)(c)=g(f(f<rsup|-1>(g<rsup|-1>(c))))=g(g<rsup|-1>(c))=c>
and <math|\<forall\>a\<in\>A> we have
<math|(f<rsup|-1>\<circ\>g<rsup|-1>)\<circ\>(g\<circ\>f)(a)=f<rsup|-1>(g<rsup|-1>(g(f(a)))=f<rsup|-1>(f(a))=a>
so <math|(g\<circ\>f)\<circ\>(f<rsup|-1>\<circ\>g<rsup|-1>)=i<rsub|C>>
and <math|(f<rsup|-1>\<circ\>g<rsup|-1>)\<circ\>(g\<circ\>f)=i<rsub|A>>
and thus <math|g\<circ\>f> is bijective
</proof>
<\theorem>
<label|restriction of a bijective mapping>If <math|f:A\<rightarrow\>B> is
a injective function and <math|C\<subseteq\>A> then
<math|f:C\<rightarrow\>f(C)> is a bijective function with inverse
function <math|f<rsup|-1>:f(C)\<rightarrow\>C>
</theorem>
<\proof>
We use <reference|characterization of a bijective mapping> to proof our
theorem\
<\eqnarray*>
<tformat|<table|<row|<cell|f<rsub|1>=f:C\<rightarrow\>f(C)>|<cell|=>|<cell|{C,f(C),graph(f)<big|cap>(C\<times\>f(C))>>|<row|<cell|f<rsub|2>=f<rsup|-1>:f(C)\<rightarrow\>C>|<cell|=>|<cell|{f(C),C,graph(f<rsup|-1>)<big|cap>(f(C)\<times\>C)>>|<row|<cell|f<rsub|1>\<circ\>f<rsub|2>>|<cell|=>|<cell|{f(C),f(C),graph(f<rsub|1>\<circ\>f<rsub|2>)}>>|<row|<cell|f<rsub|2>\<circ\>f<rsub|1>>|<cell|=>|<cell|{C,C,graph(f<rsub|1>\<circ\>f<rsub|2>)}>>>>
</eqnarray*>
So we must proof that <math|graph(f<rsub|1>\<circ\>f<rsub|2>)=graph(i<rsub|f(C)>),graph(f<rsub|2>\<circ\>f<rsub|1>)=graph(i<rsub|\|C>)>
<\eqnarray*>
<tformat|<table|<row|<cell|(x,y)\<in\>graph(f<rsub|1>\<circ\>f<rsub|2>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>C\<vdash\>(x,z)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>C)\<wedge\>(z,y)\<in\>graph(f)<big|cap>(C\<times\>f(C))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z,x)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is a function (partial)>>|<cell|x=y\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(i<rsub|\|f(C)>)>>|<row|<cell|(x,y)\<in\>i<rsub|f(C)>>|<cell|\<Rightarrow\>>|<cell|x=y\<wedge\>x\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>C\<vdash\>(z,x)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,z)=(x,z)\<in\>graph(f<rsup|-1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z,x)\<in\>graph(f)<big|cap>(C\<times\>f(C))=graph(f<rsub|1>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f<rsub|1>\<circ\>f<rsub|2>)>>|<row|<cell|(x,y)\<in\>graph(f<rsub|2>\<circ\>f<rsub|1>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>z\<in\>C\<vdash\>(x,z)\<in\>graph(f)<big|cap>(C\<times\>f(C)\<wedge\>(z,y)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(y,z)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is injective>>|<cell|x=y\<in\>C>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(i<rsub|C>)>>|<row|<cell|(x,y)\<in\>graph(i<rsub|C>)>|<cell|\<Rightarrow\>>|<cell|x=y\<in\>C>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|f
is a function>>|<cell|\<exists\>z\<in\>B\<vdash\>(x,z)\<in\>graph(f)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|z\<in\>f(C)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,z)\<in\>graph(f)<big|cap>(C\<times\>f(C))=f<rsub|1>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(z,x)\<in\>graph(f<rsup|-1>)<big|cap>(f(C)\<times\>C)=f<rsub|2>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(x,y)\<in\>graph(f<rsub|2>\<circ\>f<rsub|1>)>>|<row|<cell|>|<cell|>|<cell|>>>>
</eqnarray*>
</proof>
<\notation>
<index|permutation>A bijective mapping <math|h:A\<rightarrow\>A> is
called a permutation
</notation>
<\theorem>
<label|intermediate bijective function>If <math|f:A\<rightarrow\>B> and
<math|g:C\<rightarrow\>D> are bijective mappings with
<math|B<big|cap>C\<neq\>\<emptyset\>>, then if
<math|D\<subseteq\>B<big|cap>C> and <math|h:D\<rightarrow\>D> is a
bijective mapping then the bijective mapping
<math|g\<circ\>f:f<rsup|-1>(D)\<rightarrow\>g(D)> is equal to
<math|g\<circ\>h\<circ\>h<rsup|-1>\<circ\>f:f<rsup|-1>(D)\<rightarrow\>g(D)>
and equal to <math|g\<circ\>h<rsup|-1>\<circ\>h\<circ\>f:f<rsup|-1>(D)\<rightarrow\>g(D)>
</theorem>
<\proof>
Using the fact that <math|f(f<rsup|-1>(D))\<equallim\><rsub|f is a
surjection>D\<subseteq\>C>, and <reference|restriction and composition of
functions> we have that <math|g\<circ\>f:f<rsup|-1>(D)\<rightarrow\>g(D)>
\ is <math|g<rprime|'>\<circ\>f<rprime|'>> where
<math|f<rprime|'>=f:f<rsup|-1>(D)\<rightarrow\>D> and
\ <math|g<rprime|'>=g:D\<rightarrow\>g(D)>. From this we have
<math|g\<circ\>f=g<rprime|'>\<circ\>i<rsub|D>\<circ\>f<rprime|'>=(g<rprime|'>\<circ\>i<rsub|D>)\<circ\>f<rprime|'>>
now by <reference|restriction and composition of functions> we have
<math|g\<circ\>i<rsub|D>:D\<rightarrow\>g(D)=(g<rprime|'>\<circ\>i<rsub|D>)>
and thus <math|(g\<circ\>i<rsub|D>)\<circ\>f=(g<rprime|'>\<circ\>i<rsub|D>)\<circ\>f<rprime|'>=g\<circ\>f>
and then from <math|i<rsub|D>=h\<circ\>h<rsup|-1>=h<rsup|-1>\<circ\>h>
and associativity of composition we have proved our theorem.\
</proof>
<\theorem>
<label|composition of bijective function and inverse bijective
function>Let <math|X,Y> be sets, <math|f:A<rsub|1>\<rightarrow\>B<rsub|1>>
and <math|g:A<rsub|2>\<rightarrow\>B<rsub|2>> bijective mappings with
<math|A<rsub|1>,A<rsub|2>\<subseteq\>X,B<rsub|1>,B<rsub|2>\<subseteq\>Y>
and <math|A<rsub|1><big|cap>A<rsub|2>> then
<math|f\<circ\>g<rsup|-1>:g(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>f(A<rsub|1><big|cap>A<rsub|2>)>
and <math|g\<circ\>f<rsup|-1>:f(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>g(A<rsub|1><big|cap>A<rsub|2>)>
are bijective functions and are each other inverses
</theorem>
<\proof>
\;
Using <reference|restriction of a bijective mapping> and
<math|A<rsub|1><big|cap>A<rsub|2>\<subseteq\>A<rsub|1>,A<rsub|2>> we have
that <math|f<rprime|'>=f:A<rsub|1><big|cap>A<rsub|2>\<rightarrow\>f(A<rsub|1><big|cap>A<rsub|2>),g<rprime|'>=g:A<rsub|1><big|cap>A<rsub|2>\<rightarrow\>g(A<rsub|1><big|cap>A<rsub|2>)>
are bijective functions with inverses
<math|f<rprime|'><rsup|-1>=f<rsup|-1>:f(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>A<rsub|1><big|cap>A<rsub|2>>
and <math|g<rprime|'><rsup|-1>:g<rsup|-1>:g(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>A<rsub|1><big|cap>A<rsub|2>>
and using <reference|restriction and composition of functions> and
<reference|composition of bijective partial functions is bijective> we
have that <math|f<rsub|g>=f\<circ\>g<rsup|-1>:g(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>f(A<rsub|1><big|cap>A<rsub|2>)>
and <math|g<rsub|\|f>=g\<circ\>f<rsup|-1>:f(A<rsub|1><big|cap>A<rsub|2>)\<rightarrow\>g(A<rsub|1><big|cap>A<rsub|2>)>.
The fact that they are each inverses follows from
<math|f<rsub|g>=f<rprime|'>\<circ\>g<rprime|'><rsup|-1>,g<rsub|f>=g<rprime|'>\<circ\>f<rprime|'><rsup|-1>>
and thus \ using associativity<math|f<rsub|g>\<circ\>g<rsub|f>=f<rprime|'>\<circ\>g<rprime|'><rsup|-1>\<circ\>g<rprime|'>\<circ\>f<rprime|'><rsup|-1>=f<rprime|'>\<circ\><rsub|>i<rsub|A<rsub|1><big|cap>A<rsub|<rsub|2>>>\<circ\>f<rprime|'>\<circ\>f<rprime|'><rsup|-1>=i<rsub|f(A<rsub|1><big|cap>A<rsub|2>)>>
and <math|g<rsub|f>\<circ\>f<rsub|g>=g<rprime|'>\<circ\>f<rprime|'><rsup|-1>\<circ\>f<rprime|'>\<circ\>g<rprime|'><rsup|-1>=g<rprime|'>\<circ\>i<rsub|A<rsub|1><big|cap>A<rsub|2>>\<circ\>g<rprime|'><rsup|-1>=g<rprime|'>\<circ\>g<rprime|'><rsup|-1>=i<rsub|g<rsub|1>(A<rsub|1><big|cap>A<rsub|2>)>>
</proof>
<subsection|Families>
<\definition>
<label|definition of a family><index|family>Let <math|I,X> be sets then a
family <math|{A<rsub|i>}<rsub|i\<in\>I>> is a function
<math|I\<rightarrow\>X>, <math|I> is called the index set and we note
<math|A<rsub|i>> for <math|{A<rsub|i>}<rsub|i\<in\>I>(i)>. In many cases
we implicitly assume the existence of <math|X>. Also sometimes we if we
have a family <math|{x<rsub|i>}<rsub|i\<in\>I>> then we use the shorthand
of <math|x> as the name of the function
<math|{x<rsub|i>}<rsub|i\<in\>I>:I\<rightarrow\>X>
</definition>
<\definition>
<label|union of families>Let <math|\<cal-A\>={A<rsub|i>}<rsub|i\<in\>I>>
and <math|\<cal-B\>={B<rsub|i>}<rsub|i\<in\>J>> be two families (defined
by the functions <math|\<cal-A\>:I\<rightarrow\>X,\<cal-B\>:J\<rightarrow\>Y>)
then the family <math|\<cal-A\><big|cup>\<cal-B\>> is defined by the
function <math|\<cal-A\><big|cup>\<cal-B\>:({0}\<times\>I)<big|cup>({1}*\<times\>J)\<rightarrow\>(X<big|cup>Y)>
defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|(\<cal-A\><big|cup>\<cal-B\>)(x)>|<cell|=>|<cell|\<cal-A\>(i)
if x=(0,i)\<in\>{0}\<times\>I>>|<row|<cell|>|<cell|=>|<cell|\<cal-B\>(i)
if x=(1,i)\<in\>{1}\<times\>J>>>>
</eqnarray*>
which is well defined because <math|({0}\<times\>I)<big|cap>({1}*\<times\>J)=\<emptyset\>>
</definition>
<\note>
<label|union of families (1)>Let <math|\<cal-A\>={A<rsub|i>}<rsub|i\<in\>I>,\<cal-B\>={B<rsub|i>}<rsub|i\<in\>J>>
then if <math|\<cal-A\><big|cup>\<cal-B\>={C<rsub|i>}<rsub|i\<in\>K>>
then we have <math|\<forall\>k\<in\>K> there exists either a
<math|i\<in\>I> such that <math|C<rsub|k>=A<rsub|i>> or a <math|j\<in\>J>
such that <math|C<rsub|k>=B<rsub|j>>. Also if <math|i\<in\>I> <math|(or
j\<in\>J)> then there exists a <math|k\<in\>K> such that
<math|A<rsub|i>=C<rsub|k>> (or <math|B<rsub|j>=C<rsub|k>>)
</note>
<\proof>
By the definition of union of families we have that
<math|K=({0}\<times\>I)<big|cup>({1}\<times\>J)> and if <math|k\<in\>K>
then either <math|k=(0,i)\<in\>{0}\<times\>I> and thus
<math|C<rsub|k>=C(k)=(\<cal-A\><big|cup>\<cal-B\>)(k)=\<cal-A\>(i)=A<rsub|i>>
or <math|k=(1,j)\<in\>{1}\<times\>J> and thus
<math|C<rsub|k>=C(k)=(\<cal-A\><big|cup>\<cal-B\>)(k)=\<cal-B\>(j)=B<rsub|j>>.
If <math|i\<in\>I> (or <math|j\<in\>J>) then <math|k=(0,i)\<in\>K> (or
<math|k=(1,j)\<in\>K>) and <math|><math|C<rsub|k>=C(k)=(\<cal-A\><big|cup>\<cal-B\>)(k)=\<cal-A\>(i)=A<rsub|i>>
(or <math|C<rsub|k>=C(k)=(\<cal-A\><big|cup>\<cal-B\>)(k)=\<cal-B\>(j)=B<rsub|j>>).
</proof>
<\remark>
If <math|\<cal-A\>={A<rsub|i>}<rsub|i\<in\>I>> is a family then
<math|{A<rsub|i>\|i\<in\>I}={\<cal-A\>(i)\|i\<in\>I}=\<cal-A\>(I)>
</remark>
<\note>
<label|set indexed by itself>Let <math|X> be a set then we can write
<math|X={x<rsub|i>\|i\<in\>I}> where we have the family
<math|i<rsub|X>={x<rsub|i>}<rsub|i\<in\>I>> defined by the index set
<math|X=I> and the identity map <math|i<rsub|X>:X\<rightarrow\>X>
</note>
<\definition>
<label|family of sets><index|family of sets>Let
<verbatim|<samp|<samp|<em|<name|<strong|<em|<verbatim|<samp|>>>>>>>>><math|\<cal-G\>>
be a set of sets and I a set then a function
<math|A:I\<rightarrow\>\<cal-G\>> is called a family of sets and is noted
as <math|{A<rsub|i>}<rsub|i\<in\>I>> where by <math|A<rsub|i>> we mean
<math|A(i)>. I is called the index set, the set of sets is in many cases
implicitly understood. Of course this definition is a more specific
version of the general definition of a family
</definition>
<\note>
If <math|\<cal-G\>> be a set of sets then the family of sets defined by
<math|i<rsub|\<cal-A\>>:\<cal-G\>\<rightarrow\>\<cal-G\>> is noted by
<math|{A}<rsub|A\<in\>\<cal-A\>>>, this is a family indexed by itself.
</note>
<\definition>
Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be a family of sets then we define
<math|<big|cap><rsub|i\<in\>I>A<rsub|i>={x\|\<forall\>i\<in\>I\<succ\>x\<in\>A<rsub|i>}>
and <math|<big|cup><rsub|i\<in\>I>A<rsub|i>={x\|\<exists\>i\<in\>I\<vdash\>x\<in\>A<rsub|i>}>
</definition>
<\definition>
Let <math|{A<rsub|i>}<rsub|i\<in\>A>> be a family of nonempty sets then
we define <math|<big|prod><rsub|i\<in\>I>A<rsub|i>={f\|I\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>>
such that <math|f(i)\<in\>A<rsub|i>}>, note that the f used in the
notation is actually the graph of the function f.
</definition>
<\note>
If <math|{A<rsub|i>}<rsub|i\<in\>A>> is a family of sets then if there
exists a <math|i\<in\>A> such that <math|A<rsub|i>=\<emptyset\>> then we
can not find a <math|f:A\<rightarrow\><big|cup><rsub|i\<in\>A>A<rsub|i>>
such that <math|f(i)\<in\>A<rsub|i>> so that
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>=\<emptyset\>>. To avoid this
situation we assume always that none of the <math|A<rsub|i><rprime|'>s>
are empty when we define the product of sets (allowing this would make
the projection function (see later) non surjective.
</note>
<\theorem>
<label|families and surjective mappings>Let
<math|{A<rsub|i>}<rsub|i\<in\>I>> be a family of sets and
<math|b:J\<rightarrow\>I> a surjective function then
<math|<big|cup><rsub|i\<in\>I>A<rsub|i>\<equallim\><rsub|note><big|cup><rsub|i\<in\>I>A(i)=<big|cup><rsub|j\<in\>J>(A\<circ\>b)(j)\<equallim\><rsub|note><big|cup>A<rsub|b<rsub|j>><rsub|><rsub|>,
<big|cap><rsub|i\<in\>I>A<rsub|i>\<equallim\><rsub|note><big|cap><rsub|i\<in\>I>A(i)=<big|cap><rsub|j\<in\>J>A(b(j))\<equallim\><rsub|note><big|cap><rsub|j\<in\>J>A<rsub|b<rsub|j>>>
</theorem>
<\proof>
\;
<\enumerate>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|cap><rsub|i\<in\>I>A(i)>|<cell|\<Rightarrow\>>|<cell|\<forall\>i\<in\>I\<succ\>x\<in\>A(i)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<forall\>k\<in\>J\<succ\>b
(k)\<in\>I>>|<cell|\<forall\>k\<in\>J\<succ\>x\<in\>A(b(k))=(A\<circ\>b)(k)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|cap><rsub|k\<in\>J>(A\<circ\>b)(k)>>|<row|<cell|x\<in\><big|cap><rsub|k\<in\>J>(A\<circ\>b)(k)>|<cell|\<Rightarrow\>>|<cell|\<forall\>k\<in\>J\<succ\>x\<in\>(A\<circ\>b)(k)=A(b(k))>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<forall\>i\<in\>I\<succ\>\<exists\>k\<in\>J\<succ\>b(k)=i>>|<cell|\<forall\>i\<in\>I\<succ\>x\<in\>A(i)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|cap><rsub|i\<in\>I>A(i)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|cup><rsub|i\<in\>I>A(i)>|<cell|\<Rightarrow\>>|<cell|\<exists\>i\<in\>I\<succ\>x\<in\>A(i)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<forall\>i\<in\>I\<succ\>k\<in\>J\<succ\>b(k)=i>>|<cell|\<exists\>k\<in\>J\<succ\>x\<in\>A(b(k))>>|<row|<cell|>|<cell|>|<cell|\<exists\>k\<in\>J\<succ\>x\<in\>(A\<circ\>b)(k)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|cup><rsub|k\<in\>J>(A\<circ\>b)(k)>>|<row|<cell|x\<in\><big|cup><rsub|k\<in\>J>(A\<circ\>b)(k)>|<cell|\<Rightarrow\>>|<cell|\<exists\>k\<in\>J\<succ\>x\<in\>A(b(k))>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<forall\>k\<in\>J\<succ\>b(k)\<in\>I>>|<cell|\<exists\>i\<in\>I\<succ\>x\<in\>A(i)
[i=b(k)]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|cup><rsub|i\<in\>I>A(i)>>>>
</eqnarray*>
</enumerate>
</proof>
<\theorem>
<label|union of disjoint sets and difference and exclusion>Let
<math|{A<rsub|i>}<rsub|i\<in\>I>> be a family of disjoint sets (meaning
that <math|\<forall\>i,j\<in\>I\<vdash\>i\<neq\>j> we have
<math|A<rsub|i><big|cap>A<rsub|j>=\<emptyset\>> then
<math|(<big|cup><rsub|i\<in\>I>A<rsub|i>)<mid|\\>A<rsub|j>=<big|cup><rsub|i\<in\>I<mid|\\>{j}>A<rsub|i>>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|a\<in\><big|cup><rsub|i\<in\>I<mid|\\>{j}>A<rsub|i>\<Rightarrow\>>
there exists a <math|i\<in\>I<mid|\\>{j}(\<Rightarrow\>i\<neq\>j)> such
that <math|a\<in\>A<rsub|i>> so if also <math|a\<in\>A<rsub|j>> then
<math|a\<in\>A<rsub|i><big|cap>A<rsub|j>> contradicting the fact that
the intersection is empty, so we reach the conclusion that
<math|a\<nin\>A<rsub|j>>, this together with the fact that
<math|<big|cup><rsub|i\<in\>I<mid|\\>{j}>A<rsub|i>\<subseteq\><big|cup><rsub|i\<in\>I>A<rsub|i>>
gives that <math|a\<in\>(<big|cup><rsub|i\<in\>I>A<rsub|i>)<mid|\\>A<rsub|j>>
<item>Let <math|a\<in\>(<big|cup><rsub|i\<in\>I>A<rsub|i>)<mid|\\>A<rsub|j>>
then <math|\<exists\>i\<in\>I> such that <math|a\<in\>A<rsub|i>> and
<math|a\<nin\>A<rsub|j>> which excludes that <math|i=j> so
<math|i\<in\>I<mid|\\>{j}><math|> and thus
<math|a\<in\><big|cup><rsub|i\<in\>I<mid|\\>{j}>A<rsub|i>>\
</enumerate>
</proof>
<\notation>
<index|product of family of sets>If <math|\<cal-G\>={A}> then if
<math|{A<rsub|i>}<rsub|i\<in\>I>> is a family of sets such that
<math|\<forall\>i\<in\>I> we have <math|A<rsub|i>=A> then we note
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>> as <math|A<rsup|I>>
</notation>
<\theorem>
<label|equivalence of product>Let <math|I={1,2}> and let
<math|A={A<rsub|i>}<rsub|i\<in\>I>> be a family of sets then
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>> is bijective with
<math|A(1)\<times\>A(2)>
</theorem>
<\proof>
Define <math|\<iota\>> from <math|<big|prod><rsub|i\<in\>{1,2}>A<rsub|i>\<rightarrow\>A(1)\<times\>A(2)>
by <math|\<forall\>i\<in\><big|prod><rsub|k\<in\>{1,2}>><math|\<iota\>(i)=(i(1),i(2))\<in\>A(1)\<times\>A(2)>
then\
<\enumerate-numeric>
<item>If <math|(i,y<rsub|1>),(i,y<rsub|2>)\<in\>\<iota\>> then
<math|y<rsub|1>=(i(1),i(2))> and also <math|y<rsub|2>=(i(1),i(2))> and
thus <math|y<rsub|1>=y<rsub|2>> and <math|\<iota\>> is thus a partial
function and also a function as it is defined for all of
<math|<big|prod><rsub|k\<in\>{1,2}>A<rsub|i>>
<item>If <math|\<iota\>(i)=\<iota\>(j)\<Rightarrow\>(i(1),i(2))=(j(1),j(2))\<Rightarrow\>i(1)=j(1)\<wedge\>i(2)=j(2)\<Rightarrow\>i=j>
and thus <math|i=j>
<item>If <math|(x,y)\<in\>A(1)\<times\>A(2)> then define
<math|i:{1,2}\<rightarrow\>A(1)<big|cup>A(2)> by <math|i(1)=x> and
<math|i(2)=y> which is obviously a function and
<math|\<iota\>(i)=(i(1),i(2))=(x,y)>
</enumerate-numeric>
\;
</proof>
<\notation>
<math|y\<in\><big|prod><rsub|i\<in\>I>A<rsub|i>> is noted by
<math|(y<rsub|i>)<rsub|i\<in\>I>>, if <math|I={1,\<ldots\>,n}> then we
note <math|y=(y<rsub|1>,\<ldots\>,y<rsub|n>)> and
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>> as
<math|A<rsub|1>\<times\>\<ldots\>.\<times\>A<rsub|n>>. Also if f is a
partial function (or function (mapping))
from<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>> to B then we
note <math|f((y<rsub|1>,\<ldots\>,y<rsub|n>))> as
<math|f(y<rsub|1>,\<ldots\>,y<rsub|n>)>, also
<math|(y<rsub|1>,y<rsub|2>)> is a shorthand for
<math|(y<rsub|1>,\<ldots\>,y<rsub|2>)>.
</notation>
<\note>
If <math|y,y<rprime|'>\<in\><big|prod><rsub|i\<in\>I>A<rsub|i>> then
<math|y=y<rprime|'>\<Leftrightarrow\>\<forall\>i\<in\>I\<vdash\>y<rsub|i>=y<rprime|'><rsub|i>>
</note>
<\proof>
This is trivial if you think of <math|y,y<rprime|'>> as the graph of a
function from <math|I> to <math|<big|cup><rsub|i>A<rsub|i>>
</proof>
<\theorem>
If <math|{B<rsub|i>}<rsub|i\<in\>I>> and
<math|{A<rsub|i>}<rsub|i\<in\>I>> are two families of sets with the same
index set where in addition we have <math|\<forall\>i\<in\>I\<succ\>B<rsub|i>\<subseteq\>A<rsub|i>>
then <math|<big|prod><rsub|i\<in\>I>B<rsub|i>\<subseteq\><big|prod><rsub|i\<in\>I>A<rsub|i>>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|prod><rsub|i\<in\>I>B<rsub|i>>|<cell|\<Rightarrow\>>|<cell|x:I\<rightarrow\><big|cup><rsub|i\<in\>I>B<rsub|i>\<wedge\>\<forall\>i\<in\>I\<succ\>f(i)\<in\>B<rsub|i>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|B<rsub|i>\<subseteq\>A<rsub|i>
<with|mode|text|and >x<with|mode|text| is the
graph>>>|<cell|x:I\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>\<wedge\>\<forall\>i\<in\>I\<succ\>f(i)\<in\>A<rsub|i>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|prod><rsub|i\<in\>I>A<rsub|i>>>>>
</eqnarray*>
\;
</proof>
</theorem>
<\definition>
<index|projective mapping>Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be a
family of sets then we define <math|\<forall\>j\<in\>I> the projection
<math|\<pi\><rsub|j>:<big|prod><rsub|i\<in\>I>\<rightarrow\>A<rsub|j>>
defined by <math|\<pi\><rsub|j>(y)=y(j)=y<rsub|i>> (where
<math|y\<in\><big|prod><rsub|j\<in\>I>A<rsub|j>>). Also
<math|\<pi\><rsub|j>> is a surjection
</definition>
<\proof>
Of course we must proof that <math|\<pi\><rsub|j>> is a function, it is
evidently defined for all <math|y\<in\><big|prod><rsub|ij\<in\>I>A<rsub|j>>
so we must only prove that it is a partial function, for this assume that
<math|(x,y<rsub|1>),(x,y<rsub|2>)\<in\>\<pi\><rsub|j>\<Rightarrow\>(j,y<rsub|1>),(j,y<rsub|2>)\<in\>x\<Rightarrowlim\><rsub|x<with|mode|text|
is a function>>y<rsub|1>=y<rsub|2>> and thus <math|\<pi\><rsub|j>> is a
partial function.
To proof surjectivity let <math|i\<in\>I> and <math|x\<in\>A<rsub|i>> and
take then <math|{A<rprime|'><rsub|i>}<rsub|i\<in\>I>> where\
<\eqnarray*>
<tformat|<table|<row|<cell|A<rprime|'><rsub|j>>|<cell|=>|<cell|A<rsub|j>,
j\<neq\>i>>|<row|<cell|A<rsub|j><rprime|'>>|<cell|=>|<cell|{x},j=i>>>>
</eqnarray*>
then from the axiom of choice and the fact that
<math|A<rsub|j><rprime|'>s> are not empty we find a function <math|f>
such that <math|f(j)\<in\>A<rsub|j>> if <math|j\<neq\>i> \ and
<math|f(i)\<in\>{x}\<in\>A<rsub|i>> so that <math|\<pi\><rsub|i>(f)=x>
proving surjectivity.
</proof>
<\theorem>
<label|inverse of projection>Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be a
family of sets and <math|U\<subseteq\>A<rsub|i>> then
<math|\<pi\><rsub|i><rsup|-1>(U)=<big|prod><rsub|i\<in\>I>B<rsub|i>>
where\
<\eqnarray*>
<tformat|<table|<row|<cell|B<rsub|j>>|<cell|=A<rsub|j>>|<cell|i\<neq\>j>>|<row|<cell|>|<cell|=U>|<cell|i=j>>>>
</eqnarray*>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|prod><rsub|j\<in\>I>B<rsub|j>>|<cell|\<Rightarrow\>>|<cell|x:I\<rightarrow\><big|cup><rsub|j\<in\>I>B<rsub|j>\<wedge\>\<forall\>j\<in\>I\<succ\>x(j)\<in\>B<rsub|j>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<pi\><rsub|i>(x)=x(i)\<in\>B<rsub|i>=U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>\<pi\><rsub|i><rsup|-1>(U)>>|<row|<cell|x\<in\>\<pi\><rsub|i><rsup|-1>(U)>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|prod><rsub|j\<in\>I>A<rsub|j>\<wedge\>\<pi\><rsub|i>(x)\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x:I\<rightarrow\><big|cup><rsub|j\<in\>I>A<rsub|j>\<wedge\>\<forall\>j\<in\>I\<succ\>x(j)\<in\>A<rsub|j>\<wedge\>\<pi\><rsub|i>(x)\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x:I\<rightarrow\><big|cup><rsub|j\<in\>I>A<rsub|j>\<wedge\>\<forall\>j\<in\>I\<succ\>x(j)\<in\>A<rsub|j>\<wedge\>x(i)\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x:I\<rightarrow\><big|cup><rsub|j\<in\>I>A<rsub|j>\<wedge\>\<forall\>j\<in\>I\<succ\>x(j)\<in\>B<rsub|j>
\ (x(i)\<in\>A<rsub|i><big|cap>U=U)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|x(i)\<in\>U=B<rsub|i>\<subseteq\>A<rsub|i>>>|<cell|x:I\<rightarrow\><big|cup><rsub|j\<in\>I>B<rsub|j>\<wedge\>\<forall\>j\<in\>I\<succ\>x(j)\<in\>B<rsub|j>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\><big|prod><rsub|j\<in\>I>B<rsub|j>>>>>
</eqnarray*>
\;
</proof>
</theorem>
<\definition>
Let <math|\<cal-G\>> be a set of sets, <math|I<rsub|1>,I<rsub|2>> two
sets and <math|A={A<rsub|(i<rsub|1>,i<rsub|2>)>}<rsub|(i<rsub|1>,i<rsub|2>)\<in\>I<rsub|1>\<times\>I<rsub|2>>>
(so <math|A:(I<rsub|1>,I<rsub|2>)\<rightarrow\>\<cal-G\>>) we define then
<math|\<forall\>i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>
<math|A<rsub|1,i>:I<rsub|2>\<rightarrow\>\<cal-G\>> and
<math|A<rsub|2,j>:I<rsub|1>\<rightarrow\>\<cal-G\>> by
<math|A<rsub|1,i>(k)=A(i,k)> and <math|><math|A<rsub|2,j>(k)=A(k,j)>. We
use also the following notation <math|A<rsub|1,i>\<equallim\><rsub|note>{A<rsub|(i,j)>}<rsub|j\<in\>I<rsub|2>>>
and <math|A<rsub|2,j>\<equallim\><rsub|note>{A<rsub|(i,j)>}<rsub|i\<in\>I<rsub|1>><rsub|>>
or an even simpler notation <math|{A<rsub|(i<rsub|1>,i<rsub|2>)>}<rsub|(i<rsub|1>,i<rsub|2>)\<in\>I<rsub|1>\<times\>I<rsub|2>\<equallim\><rsub|note>>{A<rsub|i<rsub|1>,i<rsub|2>>}<rsub|i<rsub|1>\<in\>I<rsub|1>,i<rsub|2>\<in\>I<rsub|2>>>,
<math|{A<rsub|(i,j)>}<rsub|j\<in\>I<rsub|2>>\<equallim\><rsub|note>{A<rsub|i,j>}<rsub|j\<in\>I<rsub|2>>>
and <math|{A<rsub|(i,j)>}<rsub|i\<in\>I<rsub|1>>\<equallim\><rsub|note>{A<rsub|i,j>}<rsub|j\<in\>I<rsub|2>>>.
We note also <math|<big|cap><rsub|i\<in\>I<rsub|1>\<times\>I<rsub|2>>A<rsub|i>\<equallim\><rsub|note><big|cap><rsub|(i<rsub|1>,i<rsub|2>)\<in\>I<rsub|1>\<times\>I<rsub|2>>A<rsub|(i<rsub|1>,i<rsub|2>)>\<equallim\><rsub|note><big|cap><rsub|i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>A<rsub|i,j>>,
<math|<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>={x\|\<forall\>i\<in\>I<rsub|1>\<succ\>x\<in\>A<rsub|i,j>}>,
<math|<big|cap><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>={x\|\<forall\>j\<in\>I<rsub|2>\<succ\>x\<in\>A<rsub|i,j>}>,
<math|<big|cup><rsub|(i,j)\<in\>I<rsub|1>\<times\>I<rsub|2>>A<rsub|(i,j)>\<equallim\><rsub|note><big|cup><rsub|i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>A<rsub|i,j>>
<math|<big|cup><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>={x\|\<exists\>i\<in\>I<rsub|1>\<vdash\>x\<in\>A<rsub|i,j>}>,
<math|<big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>={x\|\<exists\>j\<in\>I<rsub|2>\<vdash\>x\<in\>A<rsub|i,j>},><math|<big|prod><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>={f\|f:I<rsub|1>\<rightarrow\><big|cup><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>\<wedge\>\<forall\>i\<in\>I<rsub|1>\<succ\>f(i)\<in\>A<rsub|i,j>}>
and <with|mode|math|<big|prod><rsub|i\<in\>I<rsub|2>>A<rsub|i,j>={f\|f:I<rsub|2>\<rightarrow\><big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>\<wedge\>\<forall\>j\<in\>I<rsub|2>\<succ\>f(j)\<in\>A<rsub|i,j>}>
</definition>
<\theorem>
<label|preimage image of union intersection of a family of sets>Let
<math|f:A\<rightarrow\>B> be a partial function and
<math|{B<rsub|i>}<rsub|i\<in\>I> be a family of subsets of >B then
<math|f<rsup|-1>(<big|cap><rsub|i\<in\>I>B<rsub|i>)=<big|cap><rsub|i\<in\>I>f<rsup|-1>(B<rsub|i>)>,<math|f<rsup|-1>(<big|cup><rsub|i\<in\>I>B<rsub|i>)=<big|cup><rsub|i\<in\>I>f<rsup|-1>(B<rsub|i>)>.
And if <math|{A<rsub|i>}<rsub|i\<in\>I>> is a family of subsets then
<math|f(<big|cap><rsub|i\<in\>I>A<rsub|i>)\<subseteq\><big|cap><rsub|i\<in\>I>f(A<rsub|i>)>
and <math|f(<big|cup><rsub|i\<in\>I>A<rsub|i>)=<big|cup><rsub|i\<in\>I>f(A<rsub|i>)>
</theorem>
<\proof>
\;
<\enumerate>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>f<rsup|-1>(<big|cap><rsub|i\<in\>I>B<rsub|i>)>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>f(x)\<in\><big|cap><rsub|i\<in\>I>B<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>(\<forall\>i\<in\>I\<succ\>f(x)\<in\>B<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I\<succ\>[x\<in\>dom(f)\<wedge\>f(x)\<in\>B<rsub|i>]>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cap><rsub|i\<in\>I>f<rsup|-1>(B<rsub|i>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>f<rsup|-1>(<big|cup><rsub|i\<in\>I>B<rsub|i>)>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>f(x)\<in\><big|cup><rsub|i\<in\>I>B<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>dom(f)\<wedge\>\<exists\>i\<in\>I\<succ\>f(x)\<in\>B<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I\<succ\>[x\<in\>dom(f)\<wedge\>f(x)\<in\>B<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cup><rsub|i\<in\>I>f<rsup|-1>(B<rsub|i>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>f(<big|cap><rsub|i\<in\>I>A<rsub|i>)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\><big|cap><rsub|i\<in\>I>A<rsub|i>\<succ\>y=f(x)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<forall\>i\<in\>I\<succ\>x\<in\>A<rsub|i>\<wedge\>y=f(x)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<subset\><big|cap><rsub|i\<in\>I>f(A<rsub|i>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>f(<big|cup><rsub|i\<in\>I>A<rsub|i>)>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>x\<in\><big|cup><rsub|i\<in\>I>A<rsub|i>\<succ\>y=f(x)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I\<succ\>(\<exists\>x\<in\>A<rsub|i>\<succ\>y=f(x))>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I\<succ\>y\<in\>f(A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|y\<in\><big|cup><rsub|i\<in\>I>f(A<rsub|i>)>>>>
</eqnarray*>
</enumerate>
</proof>
<\theorem>
<label|union and intersection of a family>Let <math|\<cal-G\>> be a set
of sets, <math|I<rsub|1>,I<rsub|2>> two sets and
<math|{A<rsub|(i,j)>}<rsub|i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>> be a
family defined on the product of <math|I<rsub|1>> and <math|I<rsub|2>>
then
<\enumerate-numeric>
<item><math|<big|cap><rsub|i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>A<rsub|i,j>=<big|cap><rsub|i\<in\>I<rsub|1>>(<big|cap><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)=<big|cap><rsub|j\<in\>I<rsub|2>>(<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>
<item><math|<big|cup><rsub|i\<in\>I<rsub|1>,j\<in\>I<rsub|2>>A<rsub|i,j>=<big|cup><rsub|i\<in\>I<rsub|1>>(<big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)=<big|cup><rsub|j\<in\>I<rsub|2>>(<big|cup><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>
<item><math|A<big|cap>(<big|cup><rsub|i\<in\>I<rsub|>>A<rsub|i>)=<big|cup><rsub|i\<in\>I<rsub|>>(A<big|cap>A<rsub|i>)>
<item><math|A<big|cup>(<big|cap><rsub|i\<in\>I>A<rsub|i>)=<big|cap><rsub|i\<in\>I>(A<big|cup>A<rsub|i>)>
<item><math|<big|cap><rsub|i\<in\>I<rsub|1>>(<big|prod><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)=<big|prod><rsub|j\<in\>I<rsub|2>>(<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>
</enumerate-numeric>
<\proof>
\;
<\enumerate-numeric>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|cap><rsub|i<rsub|1>\<in\>I<rsub|1>,i<rsub|2>\<in\>I<rsub|2>>A<rsub|i<rsub|1>,i<rsub|1>>>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\><big|prod><rsub|k\<in\>{1,2}>I<rsub|k>\<succ\>x\<in\>A<rsub|i>=A<rsub|i<rsub|1>,i<rsub|2>>>>|<row|<cell|>|<cell|\<Leftrightarrowlim\><rsub|<reference|equivalence
of product>>>|<cell|\<forall\>(i,j)\<in\>I<rsub|1>\<times\>I<rsub|2>\<succ\>x\<in\>A<rsub|\<iota\><rsup|-1>((i,j))>=A<rsub|i<rsup|-1>((i,j))(1),i<rsup|-1>((i,j))(2)>=A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1><rsup|>\<succ\>(\<forall\>j\<in\>I<rsub|2>\<succ\>x\<in\>A<rsub|i,j>)
(or \<forall\>j\<in\>I<rsub|2>\<Rightarrow\>(\<forall\>i\<in\>I<rsub|1>\<succ\>x\<in\>A<rsub|i,j>))>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1>\<succ\>x\<in\><big|cap><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>
(or \<forall\>j\<in\>I<rsub|2>\<succ\>x\<in\><big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cap><rsub|i\<in\>I<rsub|1>>(<big|cap><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)
(or x\<in\><big|cap><rsub|j\<in\>I<rsub|2>>(<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|cup><rsub|i<rsub|1>\<in\>I<rsub|1>,i<rsub|2>\<in\>I<rsub|2>>A<rsub|i<rsub|1>,i<rsub|2>>>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\><big|prod><rsub|k\<in\>{1,2}>I<rsub|k>\<vdash\>x\<in\>A<rsub|i>=A<rsub|i<rsub|1>,i<rsub|2>>>>|<row|<cell|>|<cell|\<Leftrightarrowlim\><rsub|<reference|equivalence
of product>>>|<cell|\<exists\>(i,j)\<in\>I<rsub|1>\<times\>I<rsub|2>\<vdash\>x\<in\>A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I<rsub|1>\<vdash\>(\<exists\>j\<in\>I<rsub|2>\<vdash\>x\<in\>A<rsub|i,j>)
(or \<exists\>j\<in\>I<rsub|2>\<vdash\>(\<exists\>i\<in\>I<rsub|1>\<vdash\>x\<in\>A<rsub|i,j)>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I<rsub|1>\<vdash\>(x\<in\><big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)
\ (or \<exists\>j\<in\>I<rsub|2>\<vdash\>(\<exists\>\<in\>I<rsub|1>\<vdash\>x\<in\>A<rsub|i,j>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cup><rsub|i\<in\>I<rsub|1>>(<big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)
(or x\<in\><big|cup><rsub|j\<in\>I<rsub|2>>(<big|cup><rsub|j\<in\>I<rsub|1>>A<rsub|i,j>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><rsub|>A<big|cap>(<big|cup><rsub|i\<in\>I>A<rsub|i>)>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>A\<wedge\>\<in\><big|cup><rsub|i\<in\>I<rsub|>>A<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>A\<wedge\>(\<exists\>i\<in\>I<rsub|>\<vdash\>x\<in\>A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I\<vdash\>(x\<in\>A\<wedge\>x\<in\>A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<exists\>i\<in\>I\<vdash\>x\<in\>(A<big|cap>A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cup><rsub|i\<in\>I>(A<big|cap>A<rsub|i>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\>A<big|cup>(<big|cap><rsub|i\<in\>I>A<rsub|i>)>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>A\<wedge\>x\<in\><big|cap><rsub|i\<in\>I>A<rsub|i>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\>A\<wedge\>(\<forall\>i\<in\>I\<succ\>x\<in\>A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I\<succ\>(x\<in\>A\<wedge\>x\<in\>A<rsub|i>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|cap><rsub|i\<in\>I>(A<big|cap>A<rsub|i>)>>>>
</eqnarray*>
<item>
<\eqnarray*>
<tformat|<table|<row|<cell|x\<in\><big|cap><rsub|i\<in\>I<rsub|1>>(<big|prod><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>)>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1>\<succ\>x\<in\><big|prod><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1>\<succ\>x:I<rsub|2>\<rightarrow\><big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>\<wedge\>\<forall\>j\<in\>I<rsub|2>\<succ\>x(j)\<in\>A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1>\<succ\>x:I<rsub|2>\<rightarrow\><big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>\<wedge\>\<forall\>i\<in\>I<rsub|1>\<succ\>(\<forall\>j\<in\>I<rsub|2>\<succ\>x(j)\<in\>A<rsub|i,j>)>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>I<rsub|1>\<succ\>x:I<rsub|2>\<rightarrow\><big|cup><rsub|j\<in\>I<rsub|2>>A<rsub|i,j>\<wedge\>\<forall\>j\<in\>I<rsub|2>\<succ\>x(j)\<in\><big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x:I<rsub|2>\<rightarrow\><big|cup><rsub|j\<in\>I<rsub|2>>(<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)\<wedge\>\<forall\>j\<in\>I<rsub|2>\<succ\>x(j)\<in\><big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|x\<in\><big|prod><rsub|j\<in\>I<rsub|2>>(<big|cap><rsub|i\<in\>I<rsub|1>>A<rsub|i,j>)>>>>
</eqnarray*>
</enumerate-numeric>
\;
</proof>
</theorem>
<\theorem>
<label|injection and projection>Let <math|f:X\<rightarrow\>Y<rsub|1>\<times\>Y<rsub|2>>
then\
<\enumerate>
<item>If <math|f<rsub|1>=\<pi\><rsub|1>\<circ\>f> or
<math|f<rsub|2>=\<pi\><rsub|2>\<circ\>f> is injective then <math|f> is
injective
<item>If <math|f<rsub|1>=\<pi\><rsub|1>\<circ\>f> or
<math|f<rsub|2>=\<pi\><rsub|2>\<circ\>f> is injective and
<math|g:X\<rightarrow\>Y<rsub|1>\<times\>Y<rsub|2>> is then if
<math|g<rsub|1>=\<pi\><rsub|1>\<circ\>g,g<rsub|2>=\<pi\><rsub|2>\<circ\>g>
then <math|\<pi\><rsub|1>\<circ\>g\<circ\>f<rsup|-1>=g<rsub|1>\<circ\>f<rsup|-1><rsub|1>
or \<pi\><rsub|2>\<circ\>g\<circ\>f<rsub|2><rsup|-1>>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>(Bijectivity) if <math|f(x)=f(x<rprime|'>)> then if
<math|f<rsub|1>> is injective we have
<math|f<rsub|1>(x)=\<pi\><rsub|1>(f(x))=\<pi\><rsub|1>(f(x<rprime|'>))=f<rsub|2>(x<rprime|'>)\<Rightarrow\>x=x<rprime|'>>,
if <math|f<rsub|2>> is injective then
<math|f<rsub|2>(x)=\<pi\><rsub|2>(f(x))=\<pi\><rsub|2>(f(x<rprime|'>))=f<rsub|2>(x<rprime|'>)\<Rightarrow\>x=x<rprime|'>>
proving injectivity. \ and thus <math|f<rsub|1>(x)=\<pi\><rsub|1>(f(x))=y<rsub|1>,f<rsub|2>(x)=\<pi\><rsub|2>(f(x))>
then we have the following possible cases
<\enumerate>
<item><math|f<rsub|1>> is a injection so
<math|x=f<rsup|-1><rsub|1>(y<rsub|1>)> and
<math|g<rsub|1>(f<rsup|-1><rsub|1>(y<rsub|1>))=g<rsub|1>(x)=\<pi\><rsub|1>(g(x))=\<pi\><rsub|1>(g(f<rsup|-1>(y<rsub|1>,y<rsub|2>)))>
<item><math|f<rsub|2>> is a injection so
<math|x=f<rsub|2><rsup|-1>(y<rsub|2>)> and
<math|g<rsub|2>(f<rsub|2><rsup|-1>(y<rsub|2>))=g<rsub|2>(x)=\<pi\><rsub|2>(g(x))=\<pi\><rsub|2>(g(f<rsup|-1>(y<rsub|1>,y<rsub|2>)))>
</enumerate>
</enumerate>
</proof>
<subsection|Axiom of choice>
<\definition>
<index|preorder>Let X be a set then a binary relation <math|\<prec\>> (a
subset of <math|X\<times\>X>) is called a preorder if and only if. As set
together with a preorder is called a preordered set
<\enumerate-numeric>
<item><math|\<forall\>x\<in\>X\<succ\>(x,x)\<in\>\<prec\>>(reflectivity)
<item><math|(\<forall\>x,y,z\<vdash\>(x,y)\<in\>\<prec\>\<wedge\>(y,z)\<in\>\<prec\>
)\<succ\> (x,z)\<in\>\<prec\>>(transitivity)
</enumerate-numeric>
</definition>
<\notation>
If <math|\<prec\>> is a preorder then we note <math|(x,y)\<in\>\<prec\>>
as <math|x\<prec\>y> so that we have <math|\<forall\>x\<in\>X> we have
<math|x\<prec\>x> and if <math|x\<prec\>y> and <math|y\<prec\>z> then
<math|x\<prec\>z>
</notation>
<\definition>
<index|maximal element><index|upperbound><index|chain>Let
<math|X,\<prec\>> be a preordered set then\
<\enumerate-numeric>
<item><math|m\<in\>X> is a maximal element of
X<math|\<Leftrightarrow\>>if <math|m\<prec\>x\<Rightarrow\>x\<prec\>m>
<item>Given a <math|B\<subseteq\>X>, <math|u> is a upperbound of
B<math|\<Leftrightarrow\>\<forall\>b\<in\>B\<succ\>b\<prec\>u>
<item><math|B\<subseteq\>X> is called a chain in
X<math|\<Leftrightarrow\>\<forall\>x,y\<in\>B\<succ\>x\<prec\>y\<vee\>y\<prec\>x>
</enumerate-numeric>
</definition>
<\definition>
<index|partially ordered set>Let <math|><math|X,\<prec\>>be a preordered
set then <math|\<prec\>>is called a partially ordered set (and
<math|\<prec\>> a partial order) if <math|(\<forall\>x,y\<in\>X\<vdash\>(x\<prec\>y)\<wedge\>(y\<prec\>x))\<succ\>x=y>
(antisymmetry)
</definition>
<\definition>
<index|totally ordered set>A partially ordered set <math|W,\<prec\>> that
is also a chain is called totally ordered.\
</definition>
<\definition>
<index|well ordered set>A partially ordered set <math|W,\<prec\>>is
called a well ordered (or an ordinal) set if
<math|\<forall\>\<emptyset\>\<neq\>B\<subseteq\>W\<succ\>(\<exists\>b\<in\>B\<succ\>(\<forall\>x\<in\>B\<succ\>b\<prec\>x))>
(B has a first element)
</definition>
<\axiom>
<label|axiom of choice>(Axiom of choice). <index|axiom of choice>Given
any family of sets <math|{A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>>
such that <math|\<cal-A\>\<neq\>\<emptyset\>\<wedge\>(\<forall\>\<alpha\>\<in\>\<cal-A\>\<succ\>A<rsub|\<alpha\>>\<neq\>\<emptyset\>\<wedge\>((\<forall\>\<alpha\>,\<beta\>\<in\>\<cal-A\>\<succ\>\<alpha\>\<neq\>\<beta\>)\<Rightarrow\>A<rsub|a><big|cap>A<rsub|\<beta\>>=\<emptyset\>>
(essentially every non empty family of pairwise disjoint non empty sets)
then there exists a set set <math|S\<subseteq\><big|cup><rsub|\<alpha\>\<in\>\<cal-A\>>A<rsub|\<alpha\>>>
such that <math|\<forall\>\<alpha\>\<in\>\<cal-A\>\<succ\>(\<exists\>a\<in\>A<rsub|\<alpha\>>\<succ\>S<big|cap>A<rsub|\<alpha\>>={a})>
(in other words there exists a set S that consists of exactly one element
of each <math|A<rsub|a>>)
</axiom>
<\theorem>
<label|choice function>(Choice Function). Let <math|\<cal-A\>> be a
collection of nonempty sets then there exists a function
<math|c:\<cal-A\>\<rightarrow\><big|cup><rsub|A\<in\>\<cal-A\>>A\<vdash\>(\<forall\>A\<in\>\<cal-A\>\<succ\>c(A)\<in\>A)>.
c is called a choice function.
<\proof>
<math|\<forall\>A\<in\>\<cal-A\>> define
<math|X<rsub|A>={(A,x)\|x\<in\>A}> then\
<\enumerate-alpha>
<item>Since <math|A\<neq\>\<emptyset\>\<Rightarrow\>\<exists\>x\<in\>A\<Rightarrow\>\<exists\>(A,x)\<in\>A<rprime|'>\<Rightarrow\>X<rsub|A>\<neq\>\<emptyset\>>
<item><math|X<rsub|A>\<subseteq\>\<cal-A\>\<times\>(<big|cup><rsub|A\<in\>\<cal-A\>>A)>
<item>If <math|A\<neq\>B> then if
<math|X<rsub|A><big|cap>X<rsub|B>\<neq\>\<emptyset\>\<Rightarrow\>\<exists\>x<rsub|1>\<in\>A,\<exists\>x<rsub|2>\<in\>B\<Rightarrow\>(x<rsub|1>,A)=(x<rsub|2,>B)\<Rightarrow\>A=B>
a contradiction so <math|X<rsub|A><big|cap>X<rsub|B>\<neq\>\<emptyset\>>
</enumerate-alpha>
</proof>
then <math|{X<rsub|A>}<rsub|A\<in\>\<cal-A\>>> is a nonempty collection
of nonempty pairwise sets so we can apply the axiom of choice
<reference|Axiom of choice> and find a set
<math|c\<subseteq\><big|cup><rsub|A\<in\>\<cal-A\>>X<rsub|A>\<subseteq\>\<cal-A\>\<times\>(<big|cup><rsub|A\<in\>\<cal-A\>>A)>
such that <math|\<forall\>A\<in\>\<cal-A\>\<succ\>(\<exists\>x\<in\>X<rsub|A>\<succ\>c<big|cap>X<rsub|A>={x}\<Rightarrow\>\<forall\>A\<in\>\<cal-A\>\<succ\>(\<exists\>a\<in\>A\<succ\>c<big|cap>X<rsub|A>=(A,a)>.
We prove now that <math|c\<subseteq\>\<cal-A\>\<times\>(<big|cup><rsub|A\<in\>\<cal-A\>>A)>
is indeed a function.
First if <math|(A,a<rsub|1>),(A,a<rsub|2>)\<in\>c> then
<math|\<exists\>A<rsub|1>,A<rsub|2>\<in\>\<cal-A\>> such that
<math|(A,a<rsub|1>)\<in\>X<rsub|A<rsub|1><rsub|>>>,
<math|(A,a<rsub|2>)\<in\>X<rsub|A<rsub|2>>\<Rightarrow\>A=A<rsub|1>,A=A<rsub|2>>
and <math|a\<in\>A<rsub|1>=A> and <math|a\<in\>A<rsub|2>=A><math|\<Rightarrow\>(A,a<rsub|1>),(A,a<rsub|2>)\<in\>X<rsub|A>\<Rightarrow\>(A,a<rsub|1>),(A,a<rsub|2>)\<in\>X<rsub|A><big|cap>c\<Rightarrowlim\><rsub|c<big|cap>X<rsub|A>
is a singleton>(A,a<rsub|1>)=(A,a<rsub|2>)\<Rightarrow\>a<rsub|1>-a<rsub|2>>.
So <math|c> is indeed a partial function.
Second if <math|A\<in\>\<cal-A\>\<Rightarrow\>\<exists\>x\<in\>X<rsub|A>\<succ\>c<big|cap>X<rsub|A>={x}\<Rightarrow\>\<exists\>x\<in\>X<rsub|A>\<succ\>x\<in\>c\<Rightarrow\>\<exists\>a\<in\>A\<succ\>(A,a)\<in\>c\<Rightarrow\>dom(c)=\<cal-A\>>,
so <math|c> is a function.
Third if <math|(A,a)\<in\>c\<Rightarrow\>\<exists\>B\<in\>\<cal-A\>\<succ\>(A,a)\<in\>X<rsub|B>\<Rightarrow\>A=B\<wedge\>a\<in\>B\<Rightarrow\>a\<in\>A>
</theorem>
<\theorem>
<label|choice function 1>(Choice Function 1). Let
<math|\<iota\>={A<rsub|i>}<rsub|i\<in\>I>> be a family of non empty sets
(a function <math|\<iota\>:I\<rightarrow\>\<cal-A\>)> then there exists a
function <math|c:I\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>\<vdash\>(\<forall\>i\<in\>I\<succ\>c(i)\<in\>A<rsub|i>)>
</theorem>
<\proof>
Then if <math|\<cal-A\>> is a collection of non empty sets (implicitly
defined by the family) and thus by <reference|choice function> there
exists a <math|c<rprime|'>:\<cal-A\>\<rightarrow\><big|cup><rsub|A\<in\>\<cal-A\>>A=<big|cup><rsub|i\<in\>I>A<rsub|i>>
such that <math|\<forall\>A\<in\>\<cal-A\>\<succ\>c<rprime|'>(A)\<in\>A>.
Define then <math|c=c<rprime|'>\<circ\>\<iota\>>. Then
<math|c:I\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>> and
<math|c(i)=c<rprime|'>(\<iota\>(i))=c<rprime|'>(A<rsub|i>)\<in\>A<rsub|i>>
</proof>
<\theorem>
<label|image of projection>Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be a
family of non empty sets then <math|\<forall\>i\<in\>I> we have for the
projection <math|\<pi\><rsub|i>:<big|prod><rsub|k\<in\>I>A<rsub|k>\<rightarrow\>A<rsub|i>>
has <math|\<pi\><rsub|i>(<big|prod><rsub|k\<in\>I>A<rsub|k>)=A<rsub|i>>
(or <math|rng(\<pi\><rsub|i>)=A<rsub|i>>)<math|>
</theorem>
<\proof>
As we have <math|\<pi\><rsub|i>(\<Pi\><rsub|k\<in\>I>A<rsub|k>)\<subseteq\>A<rsub|i>>
we only have to proof the opposite inclusion. Now by <reference|choice
function 1> we have the existence of a function (mapping)
<math|c:I\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>> such that
<math|\<forall\>i\<in\>I\<succ\>c(i)\<in\>A<rsub|i>>. Let
<math|a\<in\>A<rsub|i>> define then <math|c<rsub|j>> by
<math|c<rsub|j>={(x,y)\<in\>c\|x\<neq\>j}<big|cup>{(j,a)}\<subset\>I\<times\>(<big|cup><rsub|i\<in\>I>A<rsub|i>)>
or in function notation
<\eqnarray*>
<tformat|<table|<row|<cell|c<rsub|j>(i)>|<cell|=c(A<rsub|i>)>|<cell|i\<neq\>j>>|<row|<cell|>|<cell|=a>|<cell|i=j>>>>
</eqnarray*>
Then if <math|(i,x<rsub|1>),(i,x<rsub|2>)\<in\>c<rsub|j>> then either
<math|i\<neq\>j> and thus <math|(A<rsub|i>,x<rsub|1>),(A<rsub|i>,x<rsub|2>)\<in\>c>
and as <math|c> is a function then <math|x<rsub|1>=x<rsub|2>>, and if
<math|i=j> then <math|x<rsub|1>=x<rsub|2>=a> and thus <math|c<rsub|j>> is
indeed a partial function. As <math|c> is a function it is defined for
<math|i\<neq\>j> and we have defined <math|c<rsub|j>> for <math|i=j> so
<math|c<rsub|j>> is indeed a function. So
<math|c<rsub|j>\<in\><big|prod><rsub|k\<in\>I>A<rsub|k>> and as
<math|\<pi\><rsub|i>(c<rsub|j>)=c<rsub|j>(i)=a> we have proved that
<math|A<rsub|i>\<subseteq\>\<pi\><rsub|i>(<big|prod><rsub|i\<in\>I>A<rsub|i>)>
\;
</proof>
<\remark>
The family of sets in the previous theorem must be non empty, but as we
have remarked earlier in the definition of a product of sets, we have
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>=\<emptyset\>>, so we have no
projection operator at all.\
</remark>
<\theorem>
<label|zorn's lemma><dueto|Zorn's Lemma or Zormelo's
theorem><index|Zorn's lemma>The following are equivalent\
<\enumerate-numeric>
<item>The axiom of choice
<item>Zorn's Lemma: Let X be a preordered set, if every chain in X has
an upper bound then X has at least one maximal element
<item>Zormelo's theorem: Every set can be well ordered
</enumerate-numeric>
<\proof>
(To be provided)
</proof>
</theorem>
<subsection|Order Relations>
<\definition>
<label|linear order><index|strict linear order relation>Let S be a set
then a relation <math|\<less\> \<subseteq\>S\<times\>S> is a strict
linear order relation if the following conditions are true
(<math|(x,y)\<in\>\<less\> \<equallim\><rsub|note>x\<less\>y>,
<math|(x,y)\<nin\>\<less\>\<equallim\><rsub|note>x\<nless\>y>,
<math|x\<less\>y\<equallim\><rsub|note>y\<gtr\>x>)
<\enumerate-numeric>
<item><math|\<forall\>x,y\<in\>S\<vdash\>x\<neq\>y> we have either
<math|x\<less\>y> or <math|y\<less\>x>
<item><math|\<forall\>x\<in\>S\<succ\>x\<nless\>x>
<item><math|\<forall\>x,y,z\<in\>S\<vdash\>x\<less\>y\<wedge\>y\<less\>z>
we have <math|x\<less\>z>
</enumerate-numeric>
The set S together with a strict linear order S,\<less\> is called a
strict linear ordered set
</definition>
<\theorem>
If <math|S,\<less\>> is a linear ordered set and <math|A\<subseteq\>S>
then <math|\<less\><rsub|A>={(x,y)\<in\>A\<times\>A\|(x,y)\<in\>\<less\>}=\<less\><big|cap>(A\<times\>A)>
is a strict linear order and if there is no confusion we note it as
\<less\> also.
</theorem>
<\proof>
<math|\<forall\>x,y\<in\>A\<vdash\>x\<neq\>y> we have
<math|x,y\<in\>S\<wedge\>x\<neq\>y\<Rightarrow\>x\<less\>y\<vee\>y\<less\>x\<Rightarrow\>x\<less\><rsub|A>y\<vee\>y\<less\><rsub|A>x>,
<math|\<forall\>x\<in\>A\<Rightarrow\>x\<in\>S\<Rightarrow\>x\<nless\>x\<Rightarrow\>x\<nless\><rsub|A>x>,<math|\<forall\>x,y,z\<in\>A\<vdash\>x\<less\><rsub|A>y\<wedge\>y\<less\><rsub|A>z\<Rightarrow\>x\<less\>y\<wedge\>y\<less\>z\<Rightarrow\>x\<less\>z\<Rightarrow\>x\<less\><rsub|A>z>
</proof>
<\definition>
Let <math|S,\<less\>> be a strict linear ordered set then the
relation<math|\<leqslant\>>is defined by
<math|\<leqslant\>=\<less\><big|cup>{(x,x)\|x\<in\>S}> or in other words
<math|(x,y)\<in\>\<leqslant\>\<Leftrightarrow\>(x,y)\<subset\>\<less\>\<wedge\>(x=y)>
</definition>
<\theorem>
Let <math|S,\<less\>>be a strict linear ordered set then
<math|\<leqslant\>> fulfills\
<\enumerate-numeric>
<item><math|\<forall\>x,y\<in\>S\<succ\>><math|x\<leqslant\>y\<wedge\>y\<leqslant\>x>
<item><math|\<forall\>x\<in\>S\<succ\>x\<leqslant\>x>
<item><math|\<forall\>x,y\<in\>S\<vdash\>x\<leqslant\>y\<wedge\>y\<leqslant\>x>
we have <math|x=y>
<item><math|\<forall\>x,y,z\<in\>S\<vdash\>x\<leqslant\>y\<wedge\>y\<leqslant\>z>
we have <math|x\<leqslant\>z>
<item><math|\<neg\>(x\<less\>y)\<Leftrightarrow\>y\<leqslant\>x> and
<math|\<neg\>(x\<leqslant\>y)\<Leftrightarrow\>y\<less\>x>
</enumerate-numeric>
So <math|S,\<leqslant\>>is also a totally ordered set and a preordered
set
</theorem>
<\proof>
\;
<\enumerate>
<item>If <math|x,y\<in\>S> then either <math|x=y> (and thus
<math|x\<leqslant\>y>) or <math|x\<neq\>y> then either
<math|x\<less\>y> or <math|y\<less\>x> and thus <math|x\<leqslant\>y>
or <math|y\<leqslant\>x>
<item>If <math|x\<in\>S> then <math|x=x\<Rightarrow\>x\<leqslant\>x>
<item>If <math|x\<leqslant\>y> and <math|y\<leqslant\>x> then if
<math|x\<neq\>y> then we have <math|x\<less\>y> and
<math|y\<less\>x\<Rightarrow\>x\<less\>x> which is a contradiction so
we conclude that <math|x=y>\
<item>If <math|x\<leqslant\>y> and <math|y\<leqslant\>z> then\
<\enumerate>
<item><math|x=y,y=z\<Rightarrow\>x=z\<Rightarrow\>x\<leqslant\>z>
<item><math|x=y,y\<less\>z\<Rightarrow\>x\<less\>z\<Rightarrow\>x\<leqslant\>z>
<item><math|x\<less\>y,y=z\<Rightarrow\>x\<less\>z\<Rightarrow\>x\<leqslant\>z>
<item><math|x\<less\>y,y\<less\>z\<Rightarrow\>x\<less\>z\<Rightarrow\>x\<leqslant\>z>
</enumerate>
<item>If <math|x\<less\>y> is false then we must have either <math|x=y>
or if <math|x\<neq\>y> that <math|y\<less\>x> and thus
<math|y\<leqslant\>x>. If <math|x\<leqslant\>y> is false then we have
<math|\<neg\>(x\<less\>y\<vee\>x=y)\<Leftrightarrow\>\<neg\>(x\<less\>y)\<wedge\>\<neg\>(x=y)\<Leftrightarrow\>(x\<leqslant\>y)\<wedge\>(x\<neq\>y)\<Leftrightarrow\>(x\<less\>y\<vee\>x=y)\<wedge\>(x\<neq\>y)\<Leftrightarrow\>x\<less\>y><math|>
</enumerate>
</proof>
<\definition>
<index|maximum><index|minimum><index|upper bound><index|lower
bound><index|supremum><index|infimum>Let <math|S,\<less\>> be a strictly
linear ordered set and <math|\<emptyset\>\<neq\>A\<subseteq\>S> then we
define (if they exists)\
<\enumerate>
<item><math|m=max(A)\<Leftrightarrow\>m\<in\>A\<wedge\>\<forall\>a\<in\>A\<succ\>a\<leqslant\>m>
<item><math|m=min(A)\<Leftrightarrow\>me\<in\>A\<wedge\>\<forall\>a\<in\>A\<succ\>m\<leqslant\>a>
<item>A is bounded above by <math|u\<in\>S> (u is called the upper
bound of A)<math|\<Leftrightarrow\>\<forall\>a\<in\>A\<succ\>a\<leqslant\>u>
<item>A is bounded below by <math|l\<in\>S> (l is called the lower
bound of A)<math|\<Leftrightarrow\>\<forall\>a\<in\>A\<succ\>l\<leqslant\>a>
<item><math|s=sup(A)\<Leftrightarrow\>>s is a upper bound of A and
<math|\<forall\>u\<in\>S\<succ\>>u is a upper bound of A then
<math|s\<leqslant\>u> (otherwise s is the least upperbound), s is
called the supremum of A
<item><math|i=inf(A)\<Leftrightarrow\>>i is a lower bound of A and
<math|\<forall\>l\<in\>S\<succ\>l> is a lower bound of A then
<math|l\<leqslant\>i> (otherwise i is the greatest lowerbound), i is
called the infimum of A
</enumerate>
</definition>
<\proof>
Of course we have to proof (in 1,2) that we have only one maximum
(minimum), supremum (infimum)
If <math|m<rsub|1>,m<rsub|2>> are maximum's of A then
<math|m<rsub|1,>m<rsub|2>\<in\>A\<Rightarrow\>m<rsub|1>\<leqslant\>m<rsub|2>\<wedge\>m<rsub|2>\<leqslant\>m<rsub|1>\<Rightarrow\>m<rsub|1>=m<rsub|2>>.\
If <math|m<rsub|1>,m<rsub|2>> are minimum's of A then
<math|m<rsub|2>\<leqslant\>m<rsub|1>\<wedge\>m<rsub|1>\<leqslant\>m<rsub|2>\<Rightarrow\>m<rsub|1>=m<rsub|2>>.
If <math|s<rsub|1>,s<rsub|2>> are supremum's of A then because of totally
ordering of <math|\<leqslant\>> either
<math|s<rsub|1>\<leqslant\>s<rsub|2>> or <math|s<rsub|2>\<leqslant\>s>.
If <math|s<rsub|1>\<leqslant\>s<rsub|2>> then because <math|s<rsub|1>> is
a upper bound of A then <math|s<rsub|2>\<leqslant\>s<rsub|1>> and thus
<math|s<rsub|1>=s<rsub|2>>, else if <math|s<rsub|2>\<leqslant\>s<rsub|1>>
then because <math|s<rsub|2>> is a upper bound of A then
<math|s<rsub|1>\<leqslant\>s<rsub|2>> and thus
<math|s<rsub|1>=s<rsub|2>>.
If <math|i<rsub|1>,i<rsub|2>> are infimum's of A then because of totally
ordering of <math|\<leqslant\>> either
<math|i<rsub|1>\<leqslant\>i<rsub|2>> or
<math|i<rsub|2>\<leqslant\>i<rsub|1>>. If
<math|i<rsub|1>\<leqslant\>i<rsub|2><rsub|>> then because
<math|i<rsub|2>> is a lower bound of A then
<math|i<rsub|2>\<leqslant\>i<rsub|1>\<Rightarrow\>i<rsub|1>=i<rsub|2>>,
else if <math|i<rsub|2>\<leqslant\>i<rsub|1>> then because
<math|i<rsub|1>> is a lower bound we have
<math|i<rsub|1>\<leqslant\>i<rsub|2>\<Rightarrow\>i<rsub|1>=i<rsub|2>>.<math|><math|>
</proof>
<\remark>
Let <math|S,\<less\>> be a strict linear ordered set, then as we have
already remarked <math|S,\<leqslant\>>is a totally ordered set with a
preorder <math|\<leqslant\>>. We proof now that the definition of maximal
element is the same as max here and that the definition of a upper bound
for a preorder is the same as the definition of a upper bound for a
linear order.
<\proof>
\;
<\enumerate>
<item>If m is a maximal element of A then if <math|x\<in\>A> we have
either <math|x\<leqslant\>m> or <math|m\<leqslant\>x>. If
<math|m\<leqslant\>x> then (by maximality) we have
<math|x\<leqslant\>m>. So in all cases we have <math|x\<leqslant\>m>
and thus m is a maximum of A<math|>. If m is the maximum of A then if
<math|x\<in\>A\<wedge\>m\<leqslant\>x\<Rightarrow\>x\<leqslant\>m\<wedge\>m\<leqslant\>x>
and thus m is a maximal element of A.
<item>This is trivial for the definitions are exactly the same.
</enumerate>
</proof>
</remark>
\;
<\theorem>
<label|inf,sup condition>Let <math|S,\<less\>> be a strict linear ordered
set and <math|\<emptyset\>\<neq\>A\<subseteq\>S> assume now that there
exist a <math|s=sup(A)> (or <math|i=inf(A)>) then we have
<math|\<forall\>x\<in\>S\<vdash\>x\<less\>s\<succ\>\<exists\>a\<in\>A>
such that <math|x\<less\>a\<leqslant\>s> (or
<math|\<forall\>x\<in\>S\<vdash\>i\<less\>x\<succ\>\<exists\>a\<in\>A>
such that <math|i\<leqslant\>a\<less\>x>)
</theorem>
<\proof>
We prove this by contradiction
<\itemize-dot>
<item>Assume that <math|\<exists\>x\<in\>S\<vdash\>x\<less\>s> such
that <math|\<forall\>a\<in\>A> we have <math|x\<nless\>a> or
<math|a\<nleqslant\>s> now as s is a upper bound we must have
<math|a\<leqslant\>s> so we have <math|x\<nless\>a> and thus
<math|a\<leqslant\>x> and thus x is also a upper bound and thus
<math|s\<leqslant\>x> and as <math|x=s> gives <math|x\<less\>x> a
contradiction we must have <math|s\<less\>x> but this gives
<math|s\<less\>s> again a contradiction.
<item>Assume that <math|\<exists\>x\<in\>S\<vdash\>i\<less\>x> such
that <math|\<forall\>a\<in\>A> we have <math|i\<nleqslant\>a> or
<math|a\<nless\>x> then as i is a lower bound we must have
<math|i\<leqslant\>a> and thus <math|a\<nless\>x> or
<math|x\<leqslant\>a> and thus <math|x\<leqslant\>i> and as <math|x=i>
gives x\<less\>x a contradiction we must have <math|x\<less\>i> but
then <math|i\<less\>i> again a contradiction.
</itemize-dot>
</proof>
<\theorem>
<label|inf\<less\>sup>Let <math|S,\<less\>> be a strict linear ordered
set and A be a non empty subset of S which has a inf and a sup then
<math|inf(A)\<leqslant\>sup(A)>
</theorem>
<\proof>
Assume that <math|s=sup(A)\<less\>inf(A)=i> then as A is non empty, s is
a upperbound and l is a lowerbound <math|\<exists\>a\<in\>A> such that
<math|s\<leqslant\>a\<wedge\>a\<leqslant\>i> if <math|a=s\<less\>i> then
i is not a lowerbound anymore and if <math|s\<less\>i=a> then <math|s> is
not a upperbound anymore, but then we are left with
<math|s\<less\>a\<wedge\>a\<less\>i> meaning that <math|s> (i) are not
upperbound (lowerbound) so we end up with a contradiction and must thus
have <math|\<neg\>(s\<less\>i)\<Leftrightarrow\>i\<leqslant\>s>
</proof>
<\definition>
<index|upper bound property><index|lower bound property>Let
<math|S,\<less\>> be a strict linear ordered set then S has the least
upper bound property (greatest lower bound property) if every non empty
subset of S that is bounded above (below) has a least upper bound
(greatest lower bound).
</definition>
<\theorem>
<label|upper bound property implies lower bound property>Let
<math|S,\<less\>> be a strict linear ordered set that has the upper bound
property (lower bound property) then it has the lower bound property
(upper bound property).
</theorem>
<\proof>
\;
<\itemize-dot>
<item>If <math|S,\<less\>> has the upper bound property, then if A is
non empty (<math|\<exists\>a\<in\>A)> and bounded below then
<math|B={b\<in\>S\|\<forall\>a\<in\>A\<succ\>b\<leqslant\>a}> is non
empty and B is bounded above by a, so <math|s=sup(B)> exists. Now
<math|\<forall\>x\<in\>A <with|mode|text|we have
>\<forall\>b\<in\>B\<succ\>b\<leqslant\>x> so <math|x> is a upper bound
of B and thus <math|s\<leqslant\>x> proving that s is a lower bound of
A. Now if <math|l> is a lower bound of A then <math|l\<in\>B> and thus
<math|l\<leqslant\>s> so <math|s=sup(B)>.
<item>If <math|S,\<less\>> has the lower bound property, then if A is
non empty (<math|\<exists\>a\<in\>A>) and is bounded above then
<math|B={u\<in\>S\|\<forall\>a\<in\>A\<succ\>a\<leqslant\>u}> is non
empty and B is bounded below by a, so <math|i=inf(B)> exists. Now
<math|\<forall\>x\<in\>A> we have <math|\<forall\>b\<in\>B\<succ\>x\<leqslant\>b>
so <math|x> is a lower bound of B and thus <math|x\<leqslant\>i>
proving that <math|i> is a upper bound of A. Now if u is a upper bound
of A then <math|u\<in\>B> and thus <math|i\<leqslant\>u> so
<math|i=inf(B)>.
</itemize-dot>
</proof>
<\theorem>
<label|maximum of two sets>Let <math|S,\<less\>> be a strict linear
ordered set and let <math|A,B> be two sets such that
<math|\<forall\>a\<in\>A> there <math|\<exists\>b\<subset\>B> such that
<math|a\<leqslant\>b> (or <math|b\<leqslant\>a>) and let <math|A,B> has a
maximum (or minimum) <math|max(A)\<leqslant\>max(B)> (or
<math|min(B)\<leqslant\>min(A)>)
</theorem>
<\proof>
\;
<\enumerate>
<item>Assume that <math|max(B)\<less\>max(A)> then
<math|\<exists\>b\<in\>B> such that
<math|max(B)\<less\>max(A)\<leqslant\>b> contradicting the fact that
<math|max(B)> is the maximal element of <math|B>
<item>Assume that <math|min(A)\<less\>min(B)> then
<math|\<exists\>b\<in\>B> such that
<math|b\<leqslant\>min(A)\<less\>min(B)> contradicting the minimality
of <math|min(B)>
</enumerate>
</proof>
<\definition>
<index|generalized interval>Let <math|S,\<less\>> be a strict linear
ordered set then <math|\<forall\>a,b\<in\>S\<vdash\>a\<less\>b>
<\enumerate>
<item><math|[a,b]={x\<in\>S\|a\<leqslant\>x\<wedge\>x\<leqslant\>b}>
<item><math|]a,b]={x\<in\>S\|a\<less\>x\<wedge\>x\<leqslant\>b}>
<item><math|[a,b[={x\<in\>S\|a\<leqslant\>x\<wedge\>x\<less\>b}>
<item><math|]a,b[={x\<in\>S\|a\<less\>x\<wedge\>x\<less\>b}>
<item><math|]-\<infty\>,b]={x\<in\>S\|x\<leqslant\>b}>
<item><math|]-\<infty\>,b[={x\<in\>S\|x\<less\>b}>
<item><math|[a,\<infty\>[={x\<in\>S\|x\<geqslant\>a}>
<item><math|]a,\<infty\>[={x\<in\>S\|x\<gtr\>a}>
<item><math|]-\<infty\>,\<infty\>[=S>
<item><math|]a,a[=\<emptyset\>>
<item><math|[a,a]={a}>
</enumerate>
A set that takes one of the forms mentioned is called a generalized
interval.
</definition>
<\theorem>
<label|generalized intervals>Let <math|S,\<less\>> be a strictly linear
ordered set that has the upper bound property then for a subset A the
following are equivalent
<\enumerate>
<item>A is a generalized interval
<item><math|\<forall\>x,y\<in\>A> with <math|x\<less\>y> we have
<math|[x,y]\<subseteq\>A>
<item><math|\<forall\>x,y\<in\>A> with <math|x\<less\>y> we have that
<math|\<forall\>z\<in\>S\<vdash\>x\<less\>z\<less\>y> it follows that
<math|z\<in\>A>
<item><math|\<forall\>x,y\<in\>A> with <math|x\<leqslant\>y> we have
that <math|\<forall\>z\<in\>S\<vdash\>x\<leqslant\>z\<leqslant\>y> (or
<math|z\<in\>[x,y]> it follows that <math|z\<in\>A>
</enumerate>
</theorem>
<\proof>
The proof is easy but rather tedious
<\enumerate>
<item><math|1\<Rightarrow\>2> Let A be a generalized interval then we
have either\
<\enumerate>
<item><math|A=\<emptyset\>> so 2 is vacuous satisfied.
<item>A=<math|{a}> then if <math|x,y\<in\>a> we have <math|x=y> so
there is no <math|x,y\<in\>A> with <math|x\<less\>y> so 2 is vacuous
satisfied
<item><math|A=S> then <math|[x,y]\<subseteq\>S> for all
<math|x,y\<in\>S> with <math|x\<less\>y>
<item><math|A=[a,b]> then <math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y>
we have if <math|z\<in\>[x,y]\<Rightarrow\>a\<leqslant\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z>
and <math|z\<leqslant\>y\<wedge\>y\<leqslant\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>[a,b]\<Rightarrow\>[x,y]\<subseteq\>[a,b]>
<item><math|A=]a,b]> then <math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y>
we have if <math|z\<in\>[x,y]\<Rightarrow\>a\<less\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z>
and <math|z\<leqslant\>y\<wedge\>y\<leqslant\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>]a,b]\<Rightarrow\>[x,y]\<in\>]a,b]>
<item><math|A=[a,b[> then <math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y>
we have if <math|z\<in\>[x,y]\<Rightarrow\>a\<leqslant\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z>
and <math|z\<leqslant\>y\<wedge\>y\<less\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>[a,b[\<Rightarrow\>[x,y]\<subseteq\>[a,b[>
<item><math|A=]a,b[> then <math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y>
we have if <math|z\<in\>[x,y]\<Rightarrow\>a\<less\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z>
and <math|z\<leqslant\>y\<wedge\>y\<less\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>]a,b[\<Rightarrow\>[x,y]\<subseteq\>]a,b[>
<item><math|A=]\<infty\>,b]> then
<math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y> we have if
<math|z\<in\>[x,y]\<Rightarrow\>z\<leqslant\>y\<wedge\>y\<leqslant\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>]\<infty\>,b]\<Rightarrow\>[x,y]\<subseteq\>]\<infty\>,b]>
<item><math|A=]\<infty\>,b[ <with|mode|text|then
<math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y> we have if
<math|z\<in\>[x,y]\<Rightarrow\>z\<leqslant\>y\<wedge\>y\<less\>b\<Rightarrow\>z\<leqslant\>b\<Rightarrow\>z\<in\>]\<infty\>,b[\<Rightarrow\>[x,y]\<subseteq\>]\<infty\>,b[>>>
<item><math|A=[a,\<infty\>[> then
<math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y> we have if
<math|z\<in\>[x,y]\<Rightarrow\>a\<leqslant\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z\<Rightarrow\>z\<in\>[a,\<infty\>[\<Rightarrow\>[x,y]\<subseteq\>[a,\<infty\>[>
<item><math|A=]a,\<infty\>[> then
<math|\<forall\>x,y\<in\>A\<vdash\>x\<less\>y> we have if
<math|z\<in\>[x,y]\<Rightarrow\>a\<less\>x\<wedge\>x\<leqslant\>z\<Rightarrow\>a\<leqslant\>z\<Rightarrow\>z\<in\>]a,\<infty\>[\<Rightarrow\>[x,y]\<subseteq\>]a,\<infty\>[>
</enumerate>
<item><math|2\<Rightarrow\>3>Assume 2 true and let
<math|x,y\<in\>A\<vdash\>x\<less\>y> and take <math|z\<in\>S> such that
<math|x\<less\>z\<less\>y> then <math|x\<leqslant\>z\<leqslant\>y\<Rightarrow\>z\<in\>[x,y]\<subseteq\>A>
<item><math|3\<Rightarrow\>1>If <math|A=\<emptyset\>> it is a
generalized interval then if <math|A={a}> it is again a generalized
interval so that we only have to prove the the case where A has at
least two elements. Now assume that A has at least two elements
<math|a<rsub|1>,a<rsub|2>> and we can assume
<math|a<rsub|1>\<less\>a<rsub|2>> (the proof is the same if
<math|a<rsub|2>\<less\>a<rsub|1>>) and assume that <math|4> is true
then we have the following four possibilities.
<\enumerate>
<item>There exist no upper or lower bound for A
We prove now that <math|A=S>, as <math|A\<subseteq\>S> we have to
prove only that <math|S\<subseteq\>A>. Now if <math|x\<in\>S> then we
have the following possibilities\
<\enumerate>
<item><math|x\<in\>[a<rsub|1>,a<rsub|2>]>
(<math|a<rsub|1>\<leqslant\>x\<wedge\>x\<leqslant\>a<rsub|2>)>, we
could then have <math|x=a<rsub|1>\<in\>A> or
<math|x=a<rsub|2>\<in\>A> or <math|a<rsub|1>\<less\>x\<less\>a<rsub|2>\<Rightarrow\><rsub|3>x\<in\>A>
<item><math|x\<less\>a<rsub|1>> then as there is no lower bound for
A <math|\<exists\>a\<in\>A\<vdash\>a\<less\>x> (otherwise <math|x>
is a lowerbound) so that <math|x\<in\>]a,a<rsub|1>[\<subseteq\>[a,a<rsub|1>]\<subseteq\><rsub|(3)>A>
<item><math|a<rsub|2>\<less\>x> then as their is no upper bound for
A <math|\<exists\>a\<in\>A\<vdash\>x\<less\>a> (otherwise <math|x>
is a upperbound) so that <math|x\<in\>]a<rsub|2>,a[\<subseteq\>[a<rsub|2>,a]\<subseteq\><rsub|3>A>
</enumerate>
<item>There exist a lower and upper bound for A
We have then the following possibilities for <math|s=sup(A)> en
<math|i=inf(A)>, taking into account (<reference|inf\<less\>sup>)
that <math|i\<leqslant\>s>
<\enumerate>
<item><math|s,i\<in\>A>, in this case we prove that <math|A=[i,s]>\
First if <math|x\<in\>[i,s]> then either <math|x=i\<in\>A> or
<math|x=s\<in\>A> or <math|i\<less\>x\<less\>s\<Rightarrow\><rsub|3>x\<in\>A>
proving that <math|[i,s]\<subseteq\>A>
Second to prove that <math|A\<subseteq\>[i,s]> lets take
<math|x\<in\>A> then we could have the following possibilities :\
<\enumerate>
<item><math|><math|x\<in\>[i,s]> then by (4) we have
<math|x\<in\>A>
<item><math|x\<less\>i> this is impossible because it would mean
that i is not a lowerbound, so this case does not happen.
<item><math|s\<less\>x> this is impossible because it would mean
that s is not a upperbound, so this case does not happen.
</enumerate>
\ proving that <math|A\<subseteq\>[i,s]>
<item><math|s\<in\>A,i\<nin\>A>, in this case we prove that
<math|A=]i,s]>
First if <math|x\<in\>]i,s]> then either <math|x=s\<in\>A> and thus
<math|s\<in\>A> or <math|i\<less\>x\<less\>s\<Rightarrow\><rsub|<reference|inf,sup
condition>>\<exists\>a\<in\>A\<vdash\>i\<leqslant\>a\<less\>x\<less\>s\<Rightarrow\><rsub|3>x\<in\>A>
and thus <math|]i,s]\<subseteq\>A> <math|>
Second to prove that <math|A\<subseteq\>]i,s]> lets take
<math|x\<in\>A> then we could have the following possibilities :\
<\enumerate>
<item><math|x\<in\>]i,s]> then as we just have proved that
<math|]i,s]\<subseteq\>A> we have <math|x\<in\>A>
<item><math|x\<leqslant\>i> then either <math|x\<less\>i> or
<math|x=i>, <math|x\<less\>i> is impossible because it would mean
that i is not a lower bound, <math|x=i> is also impossible
because it would mean <math|i\<in\>A>
<item><math|s\<less\>x> this is impossible because it would mean
that s is not a upperbound anymore
</enumerate>
Proving that <math|A\<subseteq\>]i,s]>
<item><math|s\<nin\>A,i\<in\>A>, in this case we prove that
<math|A=[i,s[>
First if <math|x\<in\>[><math|i,s[> then either <math|x=i\<in\>A>
or <math|><math|i\<less\>x\<less\>s\<Rightarrow\><rsub|<reference|inf,sup
condition>>\<exists\>a\<in\>A\<vdash\>i\<less\>x\<less\>a\<leqslant\>s\<Rightarrow\><rsub|3>x\<in\>A>
proving that <math|[i,s[\<subseteq\>A>
Second to prove that <math|A\<subseteq\>[i,s[> lets take
<math|x\<in\>A> then we could have the following possibilities :
<\enumerate>
<item><math|x\<in\>[i,s[> then as we have already proved
<math|[i,s[\<subseteq\>A\<Rightarrow\>x\<in\>A>
<item><math|x\<less\>i> this is impossible because i would cease
to be a lowerbound
<item><math|s\<leqslant\>x> then either <math|s\<less\>x> or
<math|s=x>, <math|s\<less\>x> is impossible because s is a upper
bound so we are left with <math|x=s> which is impossible because
it would mean <math|s\<in\>A>
</enumerate>
Proving that <math|A\<subseteq\>[i,s[>
<item><math|s\<nin\>A,i\<nin\>A>, in this case we prove that
<math|A=]i,s[>
First if <math|x\<in\>]i,s[\<Rightarrow\>i\<less\>x\<less\>s\<Rightarrow\><rsub|<reference|inf,sup
condition>>\<exists\>a<rsub|1>,a<rsub|2>\<in\>A\<vdash\>i\<leqslant\>a<rsub|1>\<less\>x\<less\>a<rsub|2>\<leqslant\>s\<Rightarrow\><rsub|3>x\<in\>A>
thus <math|]i,s[\<subseteq\>A>
Second to prove that <math|A\<subseteq\>]i,s[> then we could have
the following possibilities :\
<\enumerate>
<item><math|x\<in\>]i,s[ then as we have already proved that
><math|]i,s[\<subseteq\>A\<Rightarrow\>x\<in\>A>
<item><math|x\<leqslant\>i\<Rightarrow\>> either
<math|x\<less\>i>, impossible because i is a lowerbound or
<math|x=i> impossible because <math|i\<nin\>A>
<item><math|s\<leqslant\>x\<Rightarrow\>> either
<math|s\<less\>x>, impossible because s is a upperbound or
<math|x=s> impossible because <math|s\<nin\>A>
</enumerate>
Proving that <math|A\<subseteq\>]i,s[>
</enumerate>
<item>There exists a lower bound but no upper bound for A. Then we
have for <math|i=inf(A)> the following cases:\
<\enumerate>
<item><math|i\<in\>A>, in this case we prove that
<math|A=[i,\<infty\>[>.\
First if <math|x\<in\>[i,\<infty\>[> then either <math|x=i\<in\>A>
or we have <math|x\<gtr\>i> and as there is no upper bound
<math|\<exists\>a\<in\>A\<vdash\>a\<gtr\>x> so
<math|i\<less\>x\<less\>a\<Rightarrow\><rsub|3>x\<in\>A> proving
that <math|[i,\<infty\>[\<subseteq\>A>
Second to prove that <math|A\<subseteq\>[i,\<infty\>[> let
<math|x\<in\>A> we have then the following cases :
<\enumerate>
<item><math|x\<in\>[i,\<infty\>[> then as we have already proved
that <math|[i,\<infty\>[\<subseteq\>A> we have <math|x\<in\>A>
<item><math|x\<less\>i> which is impossible because <math|i> is a
lowerbound
</enumerate>
Proving that <math|A\<subseteq\>[i,\<infty\>[>
<item><math|i\<nin\>A><math|> in this case we prove that
<math|A=]i,\<infty\>[>
First if <math|x\<in\>]i,\<infty\>[> then <math|x\<gtr\>i> and as
there is no upper bound <math|\<exists\>a<rsub|2>\<in\>A\<vdash\>i\<less\>x\<less\>a<rsub|2>\<Rightarrow\><rsub|<reference|inf,sup
condition>>\<exists\>a<rsub|1>\<in\>A\<vdash\>i\<leqslant\>a<rsub|1>\<less\>x\<less\>a<rsub|2>\<Rightarrow\><rsub|3>x\<in\>A>
proving that <math|]i,\<infty\>[\<subseteq\>A>
Secondly to prove that <math|A\<subseteq\>]i,\<infty\>[> let
<math|x\<in\>A> we have then the following cases :
<\enumerate>
<item><math|x\<in\>]i,\<infty\>[> then because we have already
proved that <math|]i,\<infty\>[\<subseteq\>A> that
<math|x\<in\>A>
<item><math|x\<leqslant\>i> then either <math|x=i> impossible
because <math|i\<nin\>A> or <math|x\<less\>i> impossible because
<math|i> is a lower bound
</enumerate>
Proving that <math|A\<subseteq\>]i,\<infty\>[>
</enumerate>
<item>There exists a no lower bound but there exists a upper bound
for A. Then we have for <math|s=sup(A)> the following cases:\
<\enumerate>
<item><math|s\<in\>A> In this case we prove that
<math|]-\<infty\>,s]=A>
First if <math|x\<in\>]-\<infty\>,s]> then <math|x\<leqslant\>s> so
either <math|x=s\<in\>A> or <math|x\<less\>s> and as there is no
lower bound then we have <math|\<exists\>a\<in\>A\<vdash\>a\<less\>x\<less\>s\<Rightarrow\><rsub|3>x\<in\>A>
\
Second to prove that <math|A\<subseteq\>]-\<infty\>,s]> let
<math|x\<in\>A> then we have the following cases :\
<\enumerate>
<item><math|x\<in\>]-\<infty\>,s]> then as we have already proved
that <math|]-\<infty\>,s]\<subseteq\>A> we have <math|x\<in\>A>
<item><math|s\<less\>x> which is impossible because s is a upper
bound
</enumerate>
Proving that <math|A\<subseteq\>]-\<infty\>,s]>
<item><math|s\<nin\>A> In this case we prove that
<math|]-\<infty\>,s[=A>
First if <math|x\<in\>]-\<infty\>,s[> then <math|x\<less\>s> then
as there is no lower bound <math|\<exists\>a<rsub|1>\<in\>A\<vdash\>a<rsub|1>\<less\>x\<less\>s\<Rightarrow\><rsub|<reference|inf,sup
condition>>\<exists\>a\<in\>A\<vdash\>a<rsub|1>\<less\>x\<less\>a<rsub|2>\<leqslant\>s\<Rightarrow\><rsub|3>x\<in\>A>
Second to prove that <math|A\<subseteq\>]-\<infty\>,s[> let
<math|x\<in\>A> then we have the following possibilities :\
<\enumerate>
<item><math|x\<in\>]-\<infty\>,s[> then as we have already proved
that <math|]-\<infty\>,s]\<subseteq\>A> we have <math|x\<in\>A>
<item><math|s\<leqslant\>x> then either <math|s=x> which is
impossible because <math|s\<nin\>A> or <math|s\<less\>x> which is
impossible because s is a upperbound
</enumerate>
Proving that <math|A\<subseteq\>]-\<infty\>,s[>
</enumerate>
</enumerate>
\
</enumerate>
</proof>
<subsection|The real numbers>
<\definition>
<label|real numbers><index|<math|\<bbb-R\>>><index|real numbers>We assume
the existence of a set <math|\<bbb-R\>>, the set of real numbers with the
following properties
<\enumerate>
<item>There exists two binary operators <math|+,.> which are functions
from <math|\<bbb-R\>\<times\>\<bbb-R\>\<rightarrow\>\<bbb-R\>> (where
we write <math|+((x,y))=<rsub|note>x+y> and
<math|.((x,y))\<equallim\><rsub|note>x.y>) with the following
properties\
<\enumerate>
<item><math|\<forall\>x,y,z\<in\>\<bbb-R\>\<succ\>(x+y)+z=x+(y+z)>
(associativity)
<item><math|\<forall\>x,y,z\<in\>\<bbb-R\>\<succ\>(x.y).z=x.(y.z)>
(associativity)
<item><math|\<forall\>x,y\<in\>\<bbb-R\>\<succ\>x+y=y+x>
(commutativity)
<item><math|\<forall\>x,y\<in\>\<bbb-R\>\<succ\>x.y=y.x>
(commutativity)
<item><math|\<exists\>0\<in\>\<bbb-R\>\<vdash\>\<forall\>x\<in\>\<bbb-R\>\<succ\>x+0=x>
<math|>(neutral element zero)
<item><math|\<exists\>1\<in\>\<bbb-R\>\<vdash\>1\<neq\>0\<vdash\>\<forall\>x\<in\>\<bbb-R\>\<succ\>x.1=x>
(neutral element one)
<item><math|\<forall\>x\<in\>\<bbb-R\>\<succ\>\<exists\>y\<in\>\<bbb-R\>\<succ\>x+y=0>\
<item><math|\<forall\>x\<in\>\<bbb-R\>\<vdash\>x\<neq\>0\<succ\>\<exists\>y\<in\>\<bbb-R\>\<succ\>x.y=1>\
<item><math|\<forall\>x,y,z\<in\>\<bbb-R\>\<succ\>x.(y+z)=(x.y)+(y.z)>
(distributivity)(we give the operator . a higher precedence then
<math|+> so we can write also <math|x.(y+z)=z.y+x.z>
</enumerate>
<item>There exist a order relation <math|\<less\>> with
<\enumerate>
<item><math|\<less\>> is a strict linear order relation
<item><math|\<forall\>x,y,z\<in\>\<bbb-R\>\<succ\>x\<less\>y\<Rightarrow\>x+z\<less\>y+z>
<item><math|\<forall\>x,y,z\<in\>\<bbb-R\>\<vdash\>z\<gtr\>0\<succ\>x\<less\>y\<Rightarrow\>x.z\<less\>y.z>
<item><math|\<bbb-R\>,\<less\>> has the upper bound property
</enumerate>
</enumerate>
</definition>
<\theorem>
<math|0> and <math|1> are the only elements such that <math|x+0=x> and
<math|x.1=x>
</theorem>
<\proof>
\;
<\enumerate>
<item>Suppose that <math|\<forall\>x\<in\>\<bbb-R\>> we have
<math|x+0<rsub|1>=x> then <math|0<rsub|1>=0<rsub|1>+0=0+0<rsub|1>=0>
<item>Suppose that <math|\<forall\>x\<in\>\<bbb-R\>> we have
<math|x.1<rsub|1>=x> then <math|1<rsub|1>=1<rsub|1>.1=1.1<rsub|1>=1>
</enumerate>
</proof>
<\theorem>
<label|uniqueness of inverse, negative>
<\enumerate>
<item><math|\<forall\>x\<in\>\<bbb-R\>> there exists only one <math|y>
such that <math|x+y=0>
<item><math|><math|\<forall\>x\<in\>\<bbb-R\>> there exists only one y
such that <math|x.y=1>
</enumerate>
So we for each x note the unique y such that x+y=0 as <math|\<um\>x> and
the unique y such that <math|x.y=1> as <math|x<rsup|-1>>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|x+y<rsub|1>=0=x+y<rsub|2>> then
<math|y<rsub|1>=y<rsub|1>+0=y<rsub|1>+(x+y<rsub|2>)=(y<rsub|1>+x)+y<rsub|2>=y<rsub|2>+(x+y<rsub|1>)=y<rsub|2>+0=y<rsub|2>>
<item>Let <math|x.y<rsub|1>=1=x.y<rsub|2>> then
<math|y<rsub|1>=y<rsub|1>.1=y<rsub|1>.(x.y<rsub|2>)=(y<rsub|1>.x).y<rsub|2>=y<rsub|2>.(y<rsub|1>.x)=y<rsub|2>.(x.y<rsub|1>)=y<rsub|2>.1=1>
</enumerate>
</proof>
<\theorem>
<label|inverse of product><math|\<forall\>x,y\<in\>\<bbb-R\><rsub|0>> we
have <math|(x.y)<rsup|-1>=x<rsup|-1>.y<rsup|-1>>
</theorem>
<\proof>
<math|><math|x.y.(x<rsup|-1>.y<rsup|-1>)=(x.x<rsup|-1>).(y.y<rsup|-1>)=1.1=1>
and then using <reference|uniqueness of inverse, negative> we have the
desired result.
</proof>
<\theorem>
<label|product with 0><math|\<forall\>x\<in\>\<bbb-R\>> we have
<math|x.0=0>
</theorem>
<\proof>
<math|\<forall\>x\<in\>\<bbb-R\>\<succ\>0=x+(\<um\>x)=1.x+(\<um\>x)=(1+0).x+(\<um\>x)=1.x+0.x+(\<um\>x)=x+0.x+(\<um\>x)=0.x>
</proof>
<\theorem>
<math|\<forall\>x,y,z\<in\>\<bbb-R\>\<succ\>x\<leqslant\>y\<Rightarrow\>x+z\<leqslant\>y+z>
and <math|\<forall\>x,y,z\<in\>\<bbb-R\>\<vdash\>z\<geqslant\>0\<succ\>x\<leqslant\>y\<Rightarrow\>x.z\<leqslant\>y.z>
</theorem>
<\proof>
\;
If <math|x\<leqslant\>y\<Rightarrow\>x=y\<vee\>x\<less\>y\<Rightarrow\>x+z=y+z\<vee\>x+y\<less\>y+z\<Rightarrow\>x+z\<leqslant\>y+z>.
If <math|x\<leqslant\>y> and <math|z\<geqslant\>0> then either
<math|z=0\<Rightarrow\>x.0=0=y.0\<Rightarrow\>x.0\<leqslant\>y.0>, or
<math|z\<gtr\>0> then <math|x\<leqslant\>y\<Rightarrow\>x=y\<vee\>x\<less\>y\<Rightarrow\>x.z=y.z\<vee\>x.z\<less\>y.z\<Rightarrow\>x.z\<leqslant\>y.z>
</proof>
<\theorem>
<math|\<forall\>x\<in\>\<bbb-R\>> then <math|\<um\>x=(\<um\>1).x>
</theorem>
<\proof>
<math|x+(\<um\>1).x=1.x+(\<um\>1).x=(1+(\<um\>1)).x=0.x=0>
</proof>
<\theorem>
We have <math|(\<um\>1).(\<um\>1)=1> and
<math|\<forall\>x\<in\>\<bbb-R\>\<succ\>-(\<um\>x)=x>
<\proof>
\;
<math|x=x+0=x+((\<um\>x)+(\<um\>(\<um\>x)))=(x+(\<um\>x))+(\<um\>(\<um\>x))=0+(\<um\>(\<um\>x))=\<um\>(\<um\>x)>
<math|1=\<um\>(\<um\>1)=(\<um\>1).(\<um\>1)>
\;
</proof>
</theorem>
<\theorem>
<math|\<forall\>x,y,z\<in\>\<bbb-R\>\<vdash\>z\<less\>0> we have
<math|x\<less\>y\<Rightarrow\>\<um\>y\<less\>\<um\>x> and
<math|x\<less\>y\<Rightarrow\>y.z\<less\>x.z>\
</theorem>
<\proof>
\;
<math|x\<less\>y\<Rightarrow\>x+(\<um\>x)\<less\>y+(-x)\<Rightarrow\>0\<less\>y+(-x)\<Rightarrow\>0+(-y)\<less\>(-x)+y+(\<um\>y)\<Rightarrow\>(\<um\>y)\<less\>(\<um\>x)>
<math|z\<less\>0\<Rightarrow\>z+(\<um\>z)\<less\>0+(\<um\>z)\<Rightarrow\>0\<less\>(\<um\>z)>
now if <math|x\<less\>y\<Rightarrow\>x.(-z)\<less\>y.(\<um\>z)\<Rightarrow\>\<um\>(y.(\<um\>z))\<less\>\<um\>(x.(\<um\>z))\<Rightarrow\>(\<um\>1).y.(\<um\>1).z\<less\>(\<um\>1).x.(\<um\>1).z\<Rightarrow\>y\<less\>x>
</proof>
<\theorem>
<math|1\<gtr\>0> and <math|\<um\>1\<less\>0>
</theorem>
<\proof>
By definition <math|1\<neq\>0> so assume that
<math|1\<less\>0\<Rightarrow\>1+(\<um\>1)\<less\>(\<um\>1)\<Rightarrow\>0\<less\>(\<um\>1)\<Rightarrow\>0.(-1)\<less\>(\<um\>1).(\<um\>1)\<Rightarrow\>0\<less\>1>
a contradiction. Finally <math|1\<gtr\>0\<Rightarrow\>\<um\>1\<less\>\<um\>0=(\<um\>1).0=0>
</proof>
<\theorem>
<label|inverse of number bigger then one>If
<math|1\<less\>x\<Rightarrow\>x<rsup|-1>\<less\>1>
</theorem>
<\proof>
First if <math|x<rsup|-1>=1> then we would have
<math|1\<less\>x\<Rightarrow\>1.1\<less\>x\<Rightarrow\>1.1\<less\>x.1\<Rightarrow\>1\<less\>x.x<rsup|-1>\<Rightarrow\>1\<less\>1>
a contradiction, so <math|x<rsup|-1>\<neq\>1>. Now if
<math|1\<less\>x<rsup|-1>\<Rightarrowlim\><rsub|0\<less\>1\<less\>x>1.x\<less\>x<rsup|-1>.x\<Rightarrow\>x\<less\>1>
which contradicts the assumption that <math|1\<less\>x> so we must have
<math|x<rsup|-1>\<less\>1>
</proof>
<\note>
We note <math|1+1=2>
</note>
<\theorem>
<math|0\<less\>1\<less\>2>
</theorem>
<\proof>
<math|0\<less\>1\<Rightarrow\>0\<less\>1=0+1\<less\>1+1=2>
</proof>
<\theorem>
<math|0\<less\><frac|1|2>\<less\>1> where <math|<frac|1|2>=2<rsup|-1>>
</theorem>
<\proof>
Using <math|1\<less\>2> and <reference|inverse of number bigger then one>
we have <math|<frac|1|2>\<less\>1> and if
<math|<frac|1|2>\<leqslant\>0\<Rightarrow\>1=2.<frac|1|2>\<leqslant\>2.0=0>
a contradiction
</proof>
<\theorem>
If <math|0\<less\>x> then <math|\<exists\>y\<in\>\<bbb-R\>> such that
<math|0\<less\>y\<less\>x>
</theorem>
<\proof>
As let <math|y=<frac|1|2>.x> then <math|0\<less\><frac|1|2>\<less\>1\<Rightarrow\>0=0.<frac|1|2>\<less\><frac|1|2>.x=y>
and <math|y=x.<frac|1|2>\<less\>1.x=x>
</proof>
<\corollary>
<label|number between two different numbers>If <math|x\<less\>y> then
<math|\<exists\>z\<in\>\<bbb-R\>> we have <math|x\<less\>z\<less\>y>
</corollary>
<\proof>
As <math|x\<less\>y\<Rightarrow\>0\<less\>y-x\<equallim\><rsub|previous
theorem>\<exists\>z<rprime|'>\<in\>\<bbb-R\>\<succ\>0\<less\>z<rprime|'>\<less\>y-x\<Rightarrow\>x=0+x\<less\>z=z<rprime|'>+x\<less\>y-x+x=y>
</proof>
<\theorem>
<label|kwadrat is positive>if <math|x\<neq\>0> then
<math|x<rsup|2>=x.x\<gtr\>0>
</theorem>
<\proof>
We have either <math|x\<gtr\>0\<Rightarrow\>x.x\<gtr\>0> or
<math|x\<less\>0\<Rightarrow\>0\<less\>\<um\>x\<Rightarrow\>0\<less\>(\<um\>x)(-x)=(\<um\>1)(\<um\>1).x.x=x.x>
</proof>
<\theorem>
<label|product of non zero numbers>If
<math|x,y\<neq\>0\<Rightarrow\>x.y\<neq\>0>
</theorem>
<\proof>
If <math|x,y\<neq\>0> then if <math|x.y=0\<Rightarrow\>0=x<rsup|-1>(x.y)=y>
a contradiction
</proof>
<\theorem>
<label|zero sum>Let <math|x,y\<in\>\<bbb-R\>> with
<math|x,y\<geqslant\>0> then <math|x+y\<Rightarrow\>x=0,y=0>
</theorem>
<\proof>
We consider the following cases\
<\enumerate>
<item><math|0\<less\>x\<Rightarrow\>0\<leqslant\>y\<less\>x+y=0> a
contradiction
<item><math|0\<less\>y\<Rightarrow\>0\<leqslant\>x\<less\>x+y=0> a
contradiction
<item><math|x=0,y=0> this is exactly what we have to prove
</enumerate>
</proof>
<\definition>
<label|absolute value of real numbers><index|absolute value>Let
<math|x\<in\>\<bbb-R\>\<Rightarrow\>\|x\|=x> (if <math|x\<geqslant\>0>)
or <math|\|x\|=\<um\>x> (if <math|x\<leqslant\>0>)
</definition>
<\theorem>
<label|bigger numbers bigger square>If
<math|0\<less\>x\<less\>y\<Rightarrow\>x<rsup|2>\<less\>y<rsup|2>>
</theorem>
<\proof>
<math|x\<less\>y\<Rightarrow\>x<rsup|2>=x.x\<less\>x.y> and
<math|x.y\<less\>y.y=y<rsup|2>> and thus
<math|x<rsup|2>\<less\>y<rsup|2>>
</proof>
<\theorem>
Let <math|\<bbb-R\><rsub|+,0>={x\<in\>\<bbb-R\>\|x\<geqslant\>0}> then
the function <math|.<rsup|2>:\<bbb-R\><rsub|+,0>\<rightarrow\>\<bbb-R\><rsub|+,0>>
where <math|.<rsup|2>(x)=x<rsup|2>> is a bijective function. We call the
inverse function of <math|.<rsup|2>> the square root function
<math|<sqrt|.>:\<bbb-R\><rsub|+,0>\<rightarrow\>\<bbb-R\><rsub|+,0>>. So
<math|\<forall\>x\<in\>\<bbb-R\><rsub|+,0> >we have
<math|(<sqrt|x>)<rsup|2>=x> and <math|<sqrt|x<rsup|2>>=x> (inverse of the
inverse of a function is the function)
</theorem>
<\proof>
\;
<\enumerate>
<item>Injectivity, let <math|x,y\<geqslant\>0> with
<math|x<rsup|2>=y<rsup|2>> then either\
<\enumerate>
<item><math|x=0\<Rightarrow\>0=x<rsup|2>=y<rsup|2>\<Rightarrowlim\><rsub|<reference|kwadrat
is positive>>y=0=x>
<item><math|y=0\<Rightarrow\>0=y<rsup|2>=x<rsup|2>\<Rightarrowlim\><rsub|<reference|kwadrat
is positive>>x=0>
<item><math|x,y\<gtr\>0> then assume that <math|x\<neq\>y> then we
have the following cases
<\enumerate>
<item><math|x\<less\>y\<Rightarrowlim\><rsub|<reference|bigger
numbers bigger square>>><math|x<rsup|2>\<less\>y<rsup|2>> a
contradiction
<item><math|><math|y\<less\>x\<Rightarrowlim\><rsub|<reference|bigger
numbers bigger square>>><math|y<rsup|2>\<less\>x<rsup|2>> a
contradiction
</enumerate>
</enumerate>
<item>Surjectivity, now given <math|x\<geqslant\>0> then we have either\
<\enumerate>
<item><math|x=0> then <math|0<rsup|2>=0=x>
<item><math|0\<less\>x> consider the set
<math|\<cal-S\>={y\<in\>\<bbb-R\>\|y\<geqslant\>0,y<rsup|2>\<leqslant\>x}>
then as <math|0<rsup|2>=0\<less\>x\<Rightarrow\>0\<in\>\<cal-S\>> and
<math|\<cal-S\>> is not empty. We prove now that <math|\<cal-S\>> has
a upper bound, we have the following cases\
<\enumerate>
<item><math|x\<less\>1> suppose then that <math|y\<in\>\<cal-S\>>
and that <math|1\<less\>y\<Rightarrowlim\><rsub|x\<less\>1>x\<less\>y>
and also <math|1\<less\>y\<Rightarrow\>y\<leqslant\>y<rsup|2>\<Rightarrow\>x\<less\>y<rsup|2>>
which contradicts the fact that <math|y\<in\>\<cal-S\>> so we must
have <math|y\<leqslant\>1> and <math|1> is a upper bound of
<math|\<cal-S\>>\
<item><math|1\<less\>x> then let <math|y\<in\>\<cal-S\>> and assume
that <math|x\<less\>y> then as \ <math|0\<less\>1\<less\>x\<less\>y>
we have \ <math|y=1.y\<less\>y.y=y<rsup|2>> so
<math|x\<less\>y\<less\>y<rsup|2>> and thus
<math|y\<nin\>\<cal-S\>><math|> and thus <math|x> is a upper bound
of <math|\<cal-S\>>
</enumerate>
So as <math|\<cal-S\>> is not empty and has a upper bound, by the
upper bound property there exists <math|s=sup(\<cal-S\>)>. We prove
now <math|\<cal-S\>> contains a <math|y\<gtr\>0> such that
<math|s\<gtr\>0>, we have two cases here\
<\enumerate>
<item><math|1\<less\>x> then <math|1.1=1\<less\>x\<Rightarrow\>1\<in\>\<cal-S\>>
with <math|0\<less\>1>
<item><math|x\<less\>1> then <math|x.x\<less\>x.1=x\<Rightarrow\>x\<in\>\<cal-S\>>
with <math|0\<less\>x>
</enumerate>
Next we prove that <math|s<rsup|2>=x> as <math|0\<less\>s> then using
<reference|number between two different numbers> there exists a
<math|\<varepsilon\><rsub|0>> such that
<math|0\<less\>\<varepsilon\><rsub|0>\<less\>s>. Then for all
<math|\<varepsilon\>> with <math|0\<less\>\<varepsilon\>\<less\>s> we
have <math|0\<less\>s-\<varepsilon\>\<less\>s\<less\>s+\<varepsilon\>>
so <math|(s-\<varepsilon\>)<rsup|2>\<less\>s<rsup|2>\<less\>(s+\<varepsilon\>)<rsup|2>\<Rightarrow\>(s-\<varepsilon\>)<rsup|2>\<less\>s<rsup|2>\<less\>(s+\<varepsilon\>)<rsup|2>>.
Now by the definition of <math|s>
<math|s+\<varepsilon\>\<nin\>\<cal-S\>\<Rightarrow\>(s+\<varepsilon\>)<rsup|2>\<gtr\>x>
and also <math|\<exists\>f\<in\>\<cal-S\>> such that
<math|s-\<varepsilon\>\<less\>f\<leqslant\>s> and thus using
<math|(s-\<varepsilon\>)<rsup|2>\<less\>f<rsup|2>\<less\>x\<Rightarrow\>(s-\<varepsilon\>)<rsup|2>\<less\>x\<less\>(s+\<varepsilon\>)<rsup|2>>
and thus <math|(s-\<varepsilon\>)<rsup|2>-(s+\<varepsilon\>)<rsup|2>\<less\>s<rsup|2>-x\<less\>-[(s-\<varepsilon\>)<rsup|2>-(s+\<varepsilon\>)<rsup|2>]\<Rightarrow\>-4.\<varepsilon\>\<less\>s<rsup|2>-x\<less\>4.\<varepsilon\>>
consider now the following possibilities\
<\enumerate>
<item><math|s<rsup|2>-x\<less\>0\<Rightarrow\>0\<less\>\<delta\>=x-s<rsup|2>\<less\>4.\<varepsilon\>>
and choose then <math|\<varepsilon\>=min(<frac|\<delta\>|4>,\<varepsilon\><rsub|0>)\<gtr\>0>
then <math|0\<less\>\<varepsilon\>\<less\>s> and
<math|4.\<varepsilon\>\<less\>\<delta\>\<less\>4.\<varepsilon\>> a
contradiction.
<item><math|0\<less\>s<rsup|2>-x\<Rightarrow\>0\<less\>\<delta\>=s<rsup|2>-x\<less\>4.\<varepsilon\>>
and chose then <math|\<varepsilon\>=min(<frac|\<delta\>|4>,\<varepsilon\><rsub|0>)\<gtr\>0>
then <math|0\<less\>\<varepsilon\>\<less\>s> and
<math|4.\<varepsilon\>\<less\>\<delta\>\<less\>4.\<varepsilon\>> a
contradiction
<item><math|0=s<rsup|2>-x\<Rightarrow\>x=s<rsup|2>>
</enumerate>
<math|><math|>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|squareroot is monotone><index|square root>If
<math|x,y\<in\>\<bbb-R\><rsub|+,0>> then if
<math|x\<less\>y\<Rightarrow\><sqrt|x>\<less\><sqrt|y>>
</theorem>
<\proof>
Assume that <math|<sqrt|y>\<leqslant\><sqrt|x>> then as
<math|0\<leqslant\><sqrt|y>,<sqrt|x>\<Rightarrow\>y=<sqrt|y>.<sqrt|y>\<leqslant\><sqrt|x>.<sqrt|y>>
and <math|<sqrt|x>.<sqrt|y>\<leqslant\><sqrt|x>.<sqrt|x>=x\<Rightarrow\>y\<leqslant\>x>
contradiction <math|x\<less\>y>
</proof>
<\definition>
<index|inductive set>A subset <math|A\<subseteq\>\<bbb-R\>> of the real
numbers is said to be inductive iff
<\enumerate>
<item><math|1\<in\>A>
<item><math|\<forall\>x\<in\>A\<succ\>x+1\<in\>A>
</enumerate>
</definition>
<\definition>
<index|<math|\<bbb-N\><rsub|0>>><index|natural numbers>Let
<math|\<bbb-A\>={A\<subseteq\>\<bbb-R\>\| A is inductive}> be the set of
all inductive subsets of <math|\<bbb-R\>> then we define the set of
natural numbers as the set <math|\<bbb-N\><rsub|0>=<big|cap><rsub|A\<in\>\<bbb-A\>>A>
</definition>
<\theorem>
<math|1\<in\>\<bbb-N\><rsub|0>> (so <math|\<bbb-N\><rsub|0>> is not
empty) and <math|1> is the minimum of <math|\<bbb-N\><rsub|0>>
</theorem>
<\proof>
\;
As <math|\<forall\>A\<in\>\<bbb-A\>\<succ\>1\<in\>A\<Rightarrow\>1\<in\><big|cap><rsub|A\<in\>\<bbb-A\>>A=\<bbb-N\><rsub|0>>
so we have proved that <math|1\<in\>\<bbb-N\>+0>.
Let <math|\<bbb-R\><rsub|1>={x\<in\>\<bbb-R\>\|1\<leqslant\>x}> then of
course <math|1\<in\>\<bbb-R\><rsub|1>, now if x\<in\>\<bbb-R\><rsub|1>>
then <math|1\<leqslant\>x\<Rightarrow\>1+1\<leqslant\>x+1> also as
<math|0\<less\>1\<Rightarrow\>0+1\<less\>1+1\<Rightarrow\>1\<leqslant\>1+1\<leqslant\>x+1\<Rightarrow\>x+1\<in\>\<bbb-R\><rsub|1>>
so <math|\<bbb-R\><rsub|1>>is inductive and we have thus
<math|\<bbb-N\><rsub|0>=<big|cap><rsub|A\<in\>\<bbb-A\>>A\<subseteq\>\<bbb-R\><rsub|1>>
so if <math|x\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>x\<in\>\<bbb-R\><rsub|1>\<Rightarrow\>x\<geqslant\>1>
so 1 so as <math|1\<in\>\<bbb-N\><rsub|0>> we have that <math|1> is the
minimum of <math|\<bbb-N\><rsub|0>>.
</proof>
<\theorem>
<math|\<bbb-N\><rsub|0>> is inductive
</theorem>
<\proof>
We have already proved that <math|1\<in\>\<bbb-N\><rsub|0>> so if
<math|n\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>\<forall\>A\<in\>\<bbb-A\>\<succ\>n\<in\>\<bbb-A\>\<Rightarrow\>n+1\<in\>\<bbb-A\>\<Rightarrow\>n+1\<in\>\<bbb-N\><rsub|0>>
</proof>
<\theorem>
<label|inductive subsets of natural numbers>If <math|A> is a inductive
set and <math|A\<subseteq\>\<bbb-N\><rsub|0>\<Rightarrow\>A=\<bbb-N\><rsub|0>>
(the set of positive integers is the smallest inductive set)
</theorem>
<\proof>
Let <math|A> be a inductive subset of <math|\<bbb-N\><rsub|0>> then
<math|A\<subseteq\>\<bbb-N\><rsub|0>=<big|cap><rsub|A\<in\>\<bbb-N\><rsub|0>>A\<subseteq\>A\<Rightarrow\>\<bbb-N\><rsub|0>=A>
</proof>
<\theorem>
<label|property of natural numbers>If
<math|n\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>n+(\<um\>1)\<equallim\><rsub|note>n-1\<in\>\<bbb-N\>>
</theorem>
<\proof>
Let <math|N={n\<in\>\<bbb-N\><rsub|0>\|n-1\<in\>\<bbb-N\>}> then of
course <math|\<bbb-N\>\<subseteq\>\<bbb-N\><rsub|0>>, further as
<math|1-1=1+(\<um\>1)=0\<Rightarrow\>1\<in\>N> finally let
<math|n\<in\>N\<Rightarrow\>n-1\<in\>\<bbb-N\>> and thus either
<math|n-1=0> or <math|n-1\<in\>\<bbb-N\><rsub|0>>. If
<math|n-1=0\<Rightarrow\>(n+1)-1=(n-1)+1=0+1=1\<in\>\<bbb-N\><rsub|0>\<subseteq\>\<bbb-N\>\<Rightarrow\>n+1\<in\>\<bbb-N\>>.
If <math|n-1\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>(n+1)-1=n\<in\>\<bbb-N\><rsub|0>>
so <math|N> is inductive and thus <math|N=\<bbb-N\>\<Rightarrow\>if
n\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>n\<in\>N\<Rightarrow\>n-1\<in\>\<bbb-N\>>
</proof>
<\definition>
<index|<math|\<bbb-Z\>>><index|integer numbers>We define the set of
integers <math|\<bbb-Z\>> as <math|\<bbb-Z\>=\<bbb-N\><rsub|0><big|cup>{0}<big|cup>{\<um\>n\|n\<in\>\<bbb-N\><rsub|0>}>
</definition>
<\theorem>
<label|positive integers are natural numbers>If <math|n\<in\>\<bbb-Z\>>
such that <math|0\<less\>n> then <math|n\<in\>\<bbb-N\><rsub|0>>
</theorem>
<\proof>
Let <math|z\<in\>\<bbb-Z\>> such that <math|z\<gtr\>0> evidently
<math|z\<neq\>0> assume now that <math|z=\<um\>n,
n\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>0\<less\>\<um\>n\<Rightarrow\>(\<um\>1).(\<um\>n)\<less\>0\<Rightarrow\>n\<less\>0\<less\>1\<Rightarrow\>1>
is not the minimum, a contradiction so the only possibility left is
<math|z\<in\>\<bbb-Z\>>.
</proof>
<\theorem>
<label|natural numbers has a minimal separation of
1><math|\<forall\>n\<in\>\<bbb-N\><rsub|0> there exists
<with|math-font-series|bold|no> ><math|a\<in\>\<bbb-N\><rsub|0>> such
that <math|n-1\<less\>a\<less\>n>
</theorem>
<\proof>
Let <math|A={i\<in\>\<bbb-N\><rsub|0>\|\<neg\>(\<exists\>a\<in\>\<bbb-N\><rsub|0>\<succ\>i-1\<less\>a\<less\>i)}={i\<in\>\<bbb-N\><rsub|0>\|\<forall\>a\<in\>\<bbb-N\><rsub|0>\<succ\>a\<leqslant\>i-1\<vee\>i\<leqslant\>a}>.
Then as 1 is a minimum element of <math|\<bbb-N\><rsub|0>> we have
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<succ\>1\<leqslant\>n\<Rightarrow\>1\<in\>A>.
Now assume that <math|n\<in\>A> and assume <math|\<exists\>a\<in\>A> such
that <math|(n+1)-1\<less\>a\<less\>n+1\<Rightarrow\>(n+1)-1\<less\>a\<wedge\>a\<less\>n+1\<Rightarrow\>(n-1)+1\<less\>a\<wedge\>a\<less\>n+1\<Rightarrow\>n-1\<less\>a-1\<wedge\>a-1\<less\>n>
(a). Now if <math|a\<in\>\<bbb-N\><rsub|0>\<Rightarrowlim\><rsub|<reference|property
of natural numbers>>>either <math|a-1=0> or
<math|a-1\<in\>\<bbb-N\><rsub|0>>, if
<math|a-1=0\<Rightarrow\>n-1\<less\>0\<wedge\>0\<less\>n\<Rightarrow\>n\<less\>1\<wedge\>0\<less\>n>
which is impossible because <math|n\<less\>1> would mean that 1 is not
the minimal element of <math|\<bbb-N\><rsub|0>> so we must have
<math|a-1\<in\>\<bbb-N\><rsub|0>> but this means by (a) that
<math|\<exists\>a<rprime|'>=a-1\<in\>\<bbb-N\><rsub|0>> such that
<math|n-1\<less\>a<rprime|'>\<wedge\>a<rprime|'>\<less\>n\<Rightarrow\>n-1\<less\>a<rprime|'>\<less\>n\<Rightarrow\>n\<nin\>A>
a contradiction so we must have <math|\<neg\>(n+1)-1\<less\>a\<less\>n+1)\<Rightarrow\>n+1\<in\>A>
</proof>
<\theorem>
<math|\<forall\>n\<in\>\<bbb-Z\>> there exists no <math|a\<in\>\<bbb-Z\>>
such that <math|n\<less\>a\<less\>n+1>
</theorem>
<\proof>
We prove this by contradiction, so assume that
<math|\<exists\>n\<in\>\<bbb-Z\>> such that
<math|\<exists\>a\<in\>\<bbb-Z\>> where we have that
<math|n\<less\>a\<less\>n+1>, we have then the following cases for
<math|n >:
<\enumerate>
<item><math|0\<less\>n\<Rightarrow\>0+1\<less\>n+1\<Rightarrow\>0\<less\>1\<less\>n+1\<Rightarrowlim\><rsub|<reference|positive
integers are natural numbers>>n+1\<in\>\<bbb-N\><rsub|0>> also from
<math|0\<less\>n\<less\>a\<less\>n+1\<Rightarrowlim\><rsub|<reference|positive
integers are natural numbers>>a\<in\>\<bbb-N\><rsub|0>> and thus using
<math|n\<less\>a\<less\>n+1\<Rightarrow\>(n+1)-1\<less\>a\<less\>(n+1)>
we have the existence a <math|n<rprime|'>=n+1\<in\>\<bbb-N\><rsub|0>>
such we have a <math|a\<in\>\<bbb-N\><rsub|0>> such that
<math|n<rprime|'>-1\<less\>a\<less\>n<rprime|'>> which is impossible by
<reference|natural numbers has a minimal separation of 1> so we end up
with a contradiction.
<item><math|n=0> then we have <math|0\<less\>a\<less\>1\<Rightarrow\>1\<less\>a+1\<less\>1+1\<Rightarrow\>1+1\<less\>(a+1)+1\<less\>(1+1)+1>
now as <math|1,n\<in\>\<bbb-N\><rsub|0>\<Rightarrowlim\><rsub|\<bbb-N\><rsub|0>
is inductive>1+1,a+1\<in\>\<bbb-N\><rsub|0>\<Rightarrowlim\><rsub|\<bbb-N\><rsub|0>
is inductive>(1+1)+1,(a+1)+1\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>\<exists\>n<rprime|'>=(1+1)=1,a<rprime|'>=(a+1)+1\<in\>\<bbb-N\><rsub|0>
such that 1+1=n<rprime|'>-1\<less\>(a+1)+1\<less\>n<rprime|'>=(1+1)+1>
which contradicts <reference|natural numbers has a minimal separation
of 1><math|>
<item><math|n=\<um\>1> then we have
<math|\<um\>1\<less\>a\<less\>(\<um\>1)+1=0\<Rightarrow\>\<um\>1\<less\>a\<less\>0\<Rightarrow\>(\<um\>1)+1\<less\>a+1\<less\>0+1\<Rightarrow\>0\<less\>a+1\<less\>0+1>
which we just have proved in (2) to be impossible
<item><math|n\<less\>\<um\>1\<less\>0> then using <reference|positive
integers are natural numbers> <math|n=-n<rprime|'>=(\<um\>1).n<rprime|'>>
where <math|n<rprime|'>\<in\>\<bbb-N\><rsub|0>>, then
<math|n\<less\>a\<less\>n+1\<Rightarrow\>\<um\>n<rprime|'>\<less\>a\<wedge\>a\<less\>\<um\>n<rprime|'>+1\<Rightarrow\>\<um\>a\<less\>n<rprime|'>\<wedge\>\<um\>(\<um\>n<rprime|'>+1)\<less\>\<um\>a>
so using <math|a<rprime|'>=\<um\>a> we have
<math|n<rprime|'>-1\<less\>a<rprime|'>\<less\>n<rprime|'>> now as
<math|n\<less\>\<um\>1\<Rightarrow\>a\<less\>n+1\<less\>(\<um\>1)+1=0\<Rightarrow\>a<rprime|'>=\<um\>a\<gtr\>0\<Rightarrowlim\><rsub|<reference|positive
integers are natural numbers>>a<rprime|'>\<in\>\<bbb-N\><rsub|0>> so we
have a <math|n<rprime|'>,a<rprime|'>\<in\>\<bbb-N\><rsub|0>> such that
<math|n<rprime|'>-1\<less\>a<rprime|'>\<less\>n<rprime|'>> which is
impossible by <reference|natural numbers has a minimal separation of
1><math|> giving a contradiction<math|>
</enumerate>
</proof>
<\theorem>
<label|negative of a integer is a integer>If
<math|z\<in\>\<bbb-Z\>\<Rightarrow\>\<um\>z\<in\>\<bbb-Z\>>
</theorem>
<\proof>
We have the following cases for <math|z>:\
<\enumerate>
<item><math|z=0> then <math|\<um\>z=(-1).z=(-1).0=0\<in\>\<bbb-Z\>>
<item><math|z\<less\>0> then <math|z=\<um\>z<rprime|'>,
z<rprime|'>\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>\<um\>z=z<rprime|'>\<in\>\<bbb-N\><rsub|0>\<subseteq\>\<bbb-Z\>>
<item><math|z\<gtr\>0> this follows directly from the definition of
<math|\<bbb-Z\>>
</enumerate>
</proof>
<\theorem>
<math|\<forall\>x,y\<in\>\<bbb-Z\>\<succ\>(x+y)\<in\>\<bbb-Z\>\<wedge\>(x.y)\<in\>\<bbb-Z\>>
(the sum and product of integers is again a integer)
</theorem>
<\proof>
Let <math|n,m\<in\>\<bbb-Z\>> then we must prove that
<math|n+m\<in\>\<bbb-Z\>> and that <math|n.m\<in\>\<bbb-Z\>>\
<\enumerate>
<item><math|n+m\<in\>\<bbb-Z\>>, here we have the following cases:\
<\enumerate>
<item><math|n=0> then <math|n+m=0+m=m\<in\>\<bbb-Z\>>
<item><math|m=0> then <math|n+m=n+0=n\<in\>\<bbb-Z\>>
<item><math|0\<less\>n\<wedge\>0\<less\>m> then by
<reference|positive integers are natural numbers> we have
<math|n,m\<in\>\<bbb-N\><rsub|0>> then define
<math|A={i\<in\>\<bbb-N\><rsub|0>\|n+i\<in\>\<bbb-N\><rsub|0>}\<subseteq\>\<bbb-N\><rsub|0>>
then as <math|\<bbb-N\><rsub|0>> is inductive we have
<math|n+1\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>1\<in\>A>. Now if
<math|i\<in\>A> then <math|n+i\<in\>\<bbb-N\><rsub|0>> and because
<math|\<bbb-N\><rsub|0>> is inductive we have
<math|(n+1)+i=(n+i)+1\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>n+1\<in\>A>
so A is inductive and thus <math|A=\<bbb-N\><rsub|0>> so if
<math|m\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>m\<in\>A\<Rightarrow\>n+m\<in\>\<bbb-N\><rsub|0>\<subseteq\>\<bbb-Z\>>
<item><math|0\<less\>n\<wedge\>m\<less\>0> then by
<reference|positive integers are natural numbers> we have
<math|n\<in\>\<bbb-N\><rsub|0>>, define then
<math|A={i\<in\>\<bbb-N\><rsub|0>\|n+(\<um\>i)\<in\>\<bbb-Z\>}\<subseteq\>\<bbb-N\><rsub|0>>
then by <reference|property of natural numbers> we have
<math|n-1\<in\>\<bbb-N\><rsub|>\<subseteq\>\<bbb-Z\>> and thus
<math|1\<in\>A>, further if <math|i\<in\>A> then
<math|n+(-i)\<in\>\<bbb-Z\>> we have then the following cases for
<math|n+(\<um\>i))>\
<\enumerate>
<item><math|n+(-i)\<gtr\>0> in this case by <reference|property of
natural numbers> we have <math|n+(\<um\>(i+1))=(n+(\<um\>i))-1\<in\>\<bbb-N\>\<subseteq\>\<bbb-Z\>>
<item><math|n+(-i)=0> then <math|n+(\<um\>(i+1))=(n+(\<um\>1))-1=0-1=(\<um\>1)\<in\>\<bbb-Z\>>
<item><math|n+(-i)\<less\>0> then <math|n+(\<um\>i)=(\<um\>z),
z\<in\>\<bbb-N\><rsub|0>> and thus as <math|\<bbb-N\><rsub|0>> is
inductive <math|z+1\<in\>\<bbb-N\>\<Rightarrow\>n+(\<um\>(i+1))=n+(\<um\>i)-1=(\<um\>z)-1=\<um\>(z+1)\<in\>\<bbb-Z\>>
</enumerate>
proving that A is inductive and thus that <math|A=\<bbb-N\><rsub|0>>
so as <math|m=\<um\>m<rprime|'>,m<rprime|'>\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>m<rprime|'>\<in\>A\<Rightarrow\>n+m=n+(-m<rprime|'>)\<in\>\<bbb-Z\>>
<math|>
<item><math|n\<less\>0\<wedge\>0\<less\>m> to proof this just take
<math|n<rprime|'>=m> and <math|m<rprime|'>=n> then
<math|0\<less\>n<rprime|'>\<wedge\>m<rprime|'>\<less\>0> and using
(d) we have <math|\<bbb-Z\>\<ni\>n<rprime|'>+m<rprime|'>=m+n=n+m>
<item><math|n\<less\>0\<wedge\>m\<less\>0> here we have
<math|n=\<um\>n<rprime|'>> and <math|m=-m<rprime|'>> where
<math|n<rprime|'>,m<rprime|'>\<in\>\<bbb-N\><rsub|0>> and
<math|n<rprime|'>\<gtr\>0\<wedge\>m<rprime|'>\<gtr\>0\<Rightarrowlim\><rsub|(c)>n<rprime|'>+m<rprime|'>\<in\>\<bbb-Z\>\<Rightarrowlim\><rsub|<reference|negative
of a integer is a integer>>n+m=-(n<rprime|'>+m<rprime|'>)\<in\>\<bbb-Z\>>
</enumerate>
<item><math|n.m\<in\>\<bbb-Z\>> here we can consider three cases for m\
<\enumerate>
<item><math|m=0> then <math|n.m=0\<in\>\<bbb-Z\>>
<item><math|0\<less\>m> then by <reference|positive integers are
natural numbers> <math|m\<in\>\<bbb-N\><rsub|0>> consider then
<math|A={i\<in\>\<bbb-N\><rsub|0>\|m.i\<in\>\<bbb-Z\>}\<subseteq\>\<bbb-N\><rsub|0>>
then as <math|n.1=n\<in\>\<bbb-Z\>> we have <math|1\<in\>A> and if
<math|i\<in\>A> then <math|n.(i+1)=n.i+n\<in\>\<bbb-Z\>> (because
<math|n.i\<in\>\<bbb-Z\> (i\<in\>A)> and <math|n\<in\>\<bbb-Z\>> an
using (1). So <math|A> is inductive and thus
<math|A=\<bbb-N\><rsub|0>> and as <math|m\<in\>\<bbb-N\><rsub|0>=A>
we have <math|n.m\<in\>\<bbb-Z\>>
<item><math|m\<less\>0> then <math|m=\<um\>m<rprime|'>,m<rprime|'>\<gtr\>0>
so by (b) we have <math|n.m<rprime|'>\<in\>\<bbb-Z\>\<Rightarrowlim\>n.m=n.(-m<rprime|'>)=(\<um\>1).n.m<rprime|'>=\<um\>(n.m<rprime|'>)\<in\><rsub|<reference|negative
of a integer is a integer>>\<bbb-Z\>>\
</enumerate>
</enumerate>
</proof>
<\definition>
<index|segment>If <math|n\<in\>\<bbb-N\><rsub|0>> then we define
<math|S<rsub|n>={m\<in\>\<bbb-N\><rsub|0>\|m\<less\>n}>, <math|S<rsub|n>>
are called segments.
</definition>
<\definition>
If <math|n,m\<in\>\<bbb-N\><rsub|0>> then we define
<math|{n,\<ldots\>,m}={i\<in\>\<bbb-N\><rsub|0>\|n\<leqslant\>i\<leqslant\>m}>.
Of course <math|S<rsub|n+1>={1,\<ldots\>.,n}> and if <math|m\<less\>n>
then <math|{n,\<ldots\>,m}=\<emptyset\>>.
</definition>
<\remark>
As <math|1> is the minimum of <math|\<bbb-N\><rsub|0>> we have
<math|S<rsub|1>=\<emptyset\>={1,\<ldots\>0}>
<\remark>
We sometimes write <math|{n,\<ldots\>,m}> where
<math|n,m\<in\>\<bbb-Z\>> to mean <math|{z\<in\>\<bbb-Z\>\|n\<leqslant\>z\<leqslant\>m}>
of course these will not always be real segments.
</remark>
</remark>
\;
<\lemma>
<label|smallest element of segment><math|\<forall\>n\<in\>\<bbb-N\><rsub|0>>
every nonempty subset of <math|S<rsub|n+1>> has a minimum
</lemma>
<\proof>
Let <math|A={i\<in\>\<bbb-N\><rsub|0>\|\<forall\>A\<subseteq\>S<rsub|i+1>\<vdash\>\<emptyset\>\<neq\>A\<succ\>A>
has a minimum<math|}\<subseteq\>A>.Now because of <reference|natural
numbers has a minimal separation of 1> there exist no natural number
between 1 and 2 and as 1 is the minimal element of
<math|\<bbb-N\><rsub|0>> we have <math|S<rsub|1+1>=S<rsub|2>={1}> and the
only non empty subset is <math|{1}> which has 1 as its minimum, so
<math|1\<in\>A>. Now assume that <math|i\<in\>A> then if
<math|\<emptyset\>\<neq\>B\<subseteq\>S<rsub|(i+1)+1>> then we have the
following possibilities:\
<\enumerate>
<item><math|B={i+1}> then <math|B> has the minimum element <math|i+1>
<item><math|\<exists\>j\<in\>B\<vdash\>j\<neq\>i+1> then as
<math|B\<subseteq\>S<rsub|(i+1)+1>\<Rightarrow\>j\<less\>(i+1)+1> and
as by <reference|natural numbers has a minimal separation of 1> there
is no element <math|n\<in\>\<bbb-N\><rsub|0>> such that
<math|i+1\<less\>n\<less\>(i+1)+1> so that
<math|j\<less\>i+1\<Rightarrow\>j\<in\>B<big|cap>S<rsub|i+1>\<Rightarrow\>\<emptyset\>\<neq\>B<big|cap>S<rsub|i+1>\<subseteq\>S<rsub|i+1>>
so because <math|i\<in\>A> we have a minimum <math|b> of
<math|B<big|cap>S<rsub|i+1>> with <math|b\<less\>i+1> (otherwise it is
not in <math|S<rsub|i+1>>). Now if <math|B=B<big|cap>S<rsub|i+1>> then
we have proved that B has a minimum, on the other hand if
<math|B\<neq\>B<big|cap>S<rsub|i+1>\<subseteq\>B> we have
<math|\<forall\>k\<in\>B<mid|\\>(B<big|cap>S<rsub|i+1>)> we have
<math|i+1\<leqslant\>k\<less\>(i+1)+1> and thus by <reference|natural
numbers has a minimal separation of 1> we have <math|k=i+1> and as
<math|b\<less\>i+1\<Rightarrow\>b\<less\>k>, if
<math|k\<in\>B<big|cap>S<rsub|i+1>> then by the definition of b
<math|b\<less\>k>, so <math|b> is a minimum of <math|B>.\
</enumerate>
Thus we have proved that <math|i+1\<in\>A> proving that <math|A> is
inductive and that <math|A=\<bbb-N\><rsub|0>> which proves our theorem.
</proof>
<\theorem>
<label|naturals are well ordered>Every nonempty subset of
<math|\<bbb-N\><rsub|0>> has a minimum.
</theorem>
<\proof>
Suppose that <math|D> is a non empty subset of <math|\<bbb-N\><rsub|0>>
then <math|\<exists\>n\<in\>D\<Rightarrowlim\><rsub|0\<less\>1\<Rightarrow\>n\<less\>n+1>n\<in\>D<big|cap>S<rsub|n+1>>
and by <reference|smallest element of segment> there exists a minimum
<math|b> in <math|D<big|cap>S<rsub|n+1>>. Now assume that there exists a
<math|b<rprime|'>\<in\>D> with <math|b<rprime|'>\<less\>b\<Rightarrowlim\><rsub|b\<less\>n+1>b<rprime|'>\<less\>n+1\<Rightarrow\>b<rprime|'>\<in\>D<big|cap>S<rsub|n+1>\<Rightarrowlim\><rsub|b
is minimum in D<big|cap>S<rsub|n+1>>b\<leqslant\>b<rprime|'>> a
contradiction so <math|b> is indeed the minimum of D
</proof>
<\remark>
The principals that we have used in our proofs with inductive sets can be
rephrased as follows. Let <math|P(n)> be a proposition and let
<math|A={n\<in\>\<bbb-N\><rsub|0>\|P(n)<space|0.2spc><with|mode|text| is
true>}> then if we can proof that <math|1\<in\>A> and if
<math|i\<in\>A\<Rightarrow\>i+1\<in\>A> then <math|P(n)> is true for
every <math|n\<in\>\<bbb-N\><rsub|0>.> Or re-framed given a proposition
<math|P(n)> if we proof that <math|P(1)> is true and if <math|P(i)> is
true then that <math|P(i+1)> is true then <math|P(i)> is true for every
<math|i\<in\>\<bbb-N\><rsub|0>>. This called the principle of induction.
</remark>
<\notation>
<math|a+(\<um\>b)\<equallim\><rsub|note>a-b>
</notation>
<\theorem>
<label|principle of strong induction>The principle of strong induction.
Let <math|A\<subseteq\>\<bbb-N\><rsub|0>> , suppose
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<vdash\>S<rsub|n>\<subseteq\>A\<succ\>n\<in\>A>
then <math|A=\<bbb-N\><rsub|0>> (in other words if for every number
<math|n> the statement <math|S<rsub|n>\<subseteq\>A> implies
<math|n\<in\>A> then <math|A=\<bbb-N\><rsub|0>)>
</theorem>
<\proof>
If <math|A\<neq\>\<bbb-N\><rsub|0>> then
<math|B={n\<in\>\<bbb-N\><rsub|0>\|n\<nin\>A}> is non empty subset of
<math|\<bbb-N\><rsub|0>> so by <reference|naturals are well ordered>
<math|B> has a minimum <math|b>. If <math|n\<in\>\<bbb-N\><rsub|0>> with
<math|n\<less\>b\<Rightarrow\>b\<nin\>B\<Rightarrow\>n\<in\>A\<Rightarrow\>S<rsub|b>\<subseteq\>A\<Rightarrowlim\><rsub|by
the hypothese>b\<in\>A> which is a contradiction, so we must have
<math|A=\<bbb-N\><rsub|0>>
</proof>
<\theorem>
<label|Archimedean ordering property>Archimedean ordering property.
<math|\<forall\>a\<in\>\<bbb-R\>\<succ\>\<exists\>n\<in\>\<bbb-N\><rsub|0>>
such that <math|a\<less\>n>
</theorem>
<\proof>
We proof this by contradiction so assume the theorem is false, then
<math|\<exists\>a\<in\>\<bbb-R\>> such that
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> such that
<math|n\<leqslant\>a>, so <math|\<bbb-N\><rsub|0>> is bounded above by
<math|a> and so <math|b=sup(\<bbb-N\><rsub|0>)> exists. Now as
<math|b-1\<less\>b\<Rightarrow\>\<exists\>n\<in\>\<bbb-N\><rsub|0>> such
that <math|b-1\<less\>n\<leqslant\>b> but then
<math|b\<less\>n+1\<in\>\<bbb-N\><rsub|0>> which means that <math|b> is
not a upper bound anymore and we reach thus a contradiction.
</proof>
<\corollary>
<label|make inverse of natural lower then
e><math|\<forall\>\<varepsilon\>\<in\>\<bbb-R\>> with
<math|\<varepsilon\>\<gtr\>0> there exists a
<math|n\<in\>\<bbb-N\><rsub|0>> such that
<math|0\<less\><frac|1|N>\<less\>\<varepsilon\>>
</corollary>
<\proof>
As <math|\<varepsilon\>\<gtr\>0> we have
<math|0\<less\><frac|1|\<varepsilon\>>> and by the Archimedean property
(<reference|Archimedean ordering property>) we find a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|0\<less\><frac|1|\<varepsilon\>>\<less\>N\<Rightarrow\><frac|1|N>\<less\>\<varepsilon\>>
</proof>
<\definition>
<label|definition of even or odd><math|<index|odd number><index|even
number>n\<in\>\<bbb-N\>> is even iff <math|\<exists\>m\<in\>\<bbb-N\>>
such that <math|n=2.m>. <math|n\<in\>\<bbb-N\><rsub|0>> is odd if it is
not even
</definition>
<\theorem>
<math|0> is even
</theorem>
<\proof>
<math|0=2.0>
</proof>
<\theorem>
<math|1> is odd
</theorem>
<\proof>
As <math|0\<less\>1\<Rightarrow\>1\<less\>1+1=2> now assume that <math|1>
is odd then <math|\<exists\>m\<in\>\<bbb-N\><rsub|0>> such that
<math|1=2.m> now from <math|1\<less\>2> it follows that
<math|m=1.m\<less\>2.m=1\<Rightarrow\>m\<less\>1> which contradicts the
fact that <math|1> is the minimum of <math|\<bbb-N\><rsub|0>>
</proof>
<\lemma>
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> we have
<math|\<exists\>m\<in\>\<bbb-N\>> such that <math|n=2.m\<vee\>n=2.m+1>
</lemma>
<\proof>
<math|Let ><math|A={n\<in\>\<bbb-N\><rsub|0>\|\<exists\>m\<in\>\<bbb-N\>\<succ\>n=2.m\<vee\>n=2.m+1}>
then as <math|1=2.0+1> we have <math|1\<in\>A> now assume that
<math|n\<in\>A> then we have that <math|\<exists\>m\<in\>\<bbb-N\>> such
that either\
<\enumerate>
<item><math|n=2.m\<Rightarrow\>n+1=2.m+1\<Rightarrow\>n+1\<in\>A>
<item><math|n=2.m+1\<Rightarrow\>n+1=2.m+1+1=2.m+2=2.(m+1)\<Rightarrow\>n+1\<in\>A>
</enumerate>
\ and thus <math|A> is inductive and thus we have proved the theorem
</proof>
<\theorem>
<label|characterization of odd><math|n\<in\>\<bbb-N\><rsub|0>> is odd
<math|\<Leftrightarrow\>><math|\<exists\>m\<in\>\<bbb-N\>> such that
<math|n=2.m+1>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>> Using the previous lemma we have
<math|\<exists\>m\<in\>\<bbb-N\>> such that either <math|n=2.m> (which
is impossible because <math|n> is odd and thus not even) or
<math|n=2.m+1> which is the only possibility left.
<item><math|\<Leftarrow\>> Assume that <math|n=2.m+1> and that <math|n>
is odd, then <math|\<exists\>m<rprime|'>\<in\>\<bbb-N\>\<vdash\>n=2.m<rprime|'>\<Rightarrow\>2.m+1=2.m<rprime|'>>
we have then the following possibilities\
<\enumerate>
<item><math|m=m<rprime|'>> then <math|1=0> which is impossible
<item><math|m\<less\>m<rprime|'>> then
<math|1=2.m<rprime|'>-2.m=2.(m<rprime|'>-m)> and <math|1> is even
which is impossible
<item><math|m<rprime|'>\<less\>m> then
<math|1=2.m<rprime|'>-2.m=2.(m<rprime|'>-m)\<less\>0> which is
impossible
</enumerate>
and we reach thus a condition, so <math|n> must be odd.
</enumerate>
</proof>
<subsection|Finite and infinite sets>
<\definition>
<label|finite sets><index|finite set>A set <math|A> is called finite if
its empty or if there exists a <math|n\<in\>\<bbb-N\><rsub|0>> such that
<math|S<rsub|n+1>={1,\<ldots\>,n}> is bijective with <math|A>.
</definition>
<\definition>
<index|infinite set>A set <math|A> is infinite if it is not finite
</definition>
<\definition>
<index|countable>A set <math|A> is countable if <math|\<bbb-N\><rsub|0>>
is bijective with <math|A>
</definition>
<\lemma>
<label|mapping with {1,..,n}>Let <math|n\<in\>\<bbb-N\><rsub|0>> and let
<math|A> be a set with <math|a\<in\>A> then there exists a bijective
function of the set <math|A> to <math|S<rsub|n+2>={1,\<ldots\>,n+1}\<Leftrightarrow\>>there
exists a bijection of <math|A<mid|\\>{a}> to
<math|S<rsub|n+1>={1,\<ldots\>,n}>
</lemma>
<\proof>
\;
<\enumerate>
<item><math|\<Leftarrow\>>If there exists a bijection
<math|f:A<mid|\\>{a}\<rightarrow\>S<rsub|n+1>> define then the function
<math|g:A\<rightarrow\>S<rsub|n+2>> by extension
<\eqnarray*>
<tformat|<table|<row|<cell|g(x)>|<cell|=f(x)>|<cell|if
\ x\<in\>A<mid|\\>{a}>>|<row|<cell|>|<cell|=n+1>|<cell|if x=a>>>>
</eqnarray*>
then because <math|f> is a bijection,<math|f<rprime|'>:{a}\<rightarrow\>{n+1}>
where <math|f<rprime|'>(a)=n+1> is a bijection,
<math|(A<mid|\\>{a})<big|cap>{a}=\<emptyset\>> and
<math|S<rsub|n><big|cap>{n+1}=\<emptyset\>> we have that <math|f> is a
bijection by <reference|bijection extension> \
<item><math|\<Rightarrow\>>If there exists a bijection
<math|f:A\<rightarrow\>S<rsub|n+2>> then we have the following cases
<\enumerate>
<item><math|f(a)=n+1> then the restriction of f to
<math|A<mid|\\>{a}> is the bijection we are looking for (use
<reference|restriction of a mapping>, <reference|restriction of a
bijection> and the fact that <math|f({a})=f(A<mid|\\>(A<mid|\\>{a}))={n+1}>)\
<item><math|f(a)\<neq\>n+1> then as <math|f> is bijection and thus
surjective we have that <math|\<exists\>b\<in\>A> such that
<math|f(b)=n+1> and as <math|f> is a function so
<math|\<exists\>m\<in\>S<rsub|n+1>> such that <math|f(a)=m>. We have
then the following possibilities:\
<\enumerate>
<item><math|A={a,b}> then <math|S<rsub|n+2>={1,2}=S<rsub|3>> then
the function <math|f<rprime|'>:{b}\<rightarrow\>{2}> defined by
<math|f<rprime|'>(b)=1> is the desired bijection<math|>
<item><math|A\<neq\>{a,b}\<Rightarrow\>\<emptyset\>\<neq\>C=A<mid|\\>{a,b}>
then using <math|A<mid|\\>C=A<mid|\\>(A<mid|\\>{a,b})={a,b}> then
by <reference|restriction of a bijection> and
<reference|restriction of a mapping> <math|f<rsub|\|C>> is a
bijective function from <math|C=A<mid|\\>{a,b}> to
<math|S<rsub|n+2><mid|\\>f(A<mid|\\>C)=S<rsub|n+2><mid|\\>f({a,b})=S<rsub|n+2><mid|\\>{n+1,m}=S<rsub|n+1><mid|\\>{m}>
then as we define the bijection
<math|f<rsub|1>:{b}\<rightarrow\>{m}> by <math|f<rsub|1>(b)=n+1>
then as <math|A<mid|\\>{a,b}<big|cap>{b}=\<emptyset\> > and
<math|(S<rsub|n+1><mid|\\>{m})<big|cap>{m}=\<emptyset\>> we can use
<reference|bijection extension> we have found the required
bijection = <math|f<rsub|\|C><big|cup>f<rsub|1>> from
<math|(A<mid|\\>{a,b}<big|cup>{b}=A<mid|\\>{a}> to
<math|(S<rsub|n+1><mid|\\>{m})<big|cup>{m}=S<rsub|n+1>>
</enumerate>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|subset of a finite set>Let A be a set and suppose that there
exists a bijective function <math|f:A\<rightarrow\>{1,\<ldots\>,n}=S<rsub|n+1>>
for some <math|n\<in\>\<bbb-N\><rsub|0>>. Let B a proper subset of A
(<math|B\<subset\>A>) then there exist no bijective function
<math|g:B\<rightarrow\>{1,\<ldots\>,n}>, but provided
<math|B\<neq\>\<emptyset\>> then <math|\<exists\>m\<in\>\<bbb-N\><rsub|0>\<vdash\>m\<less\>n>
such that there exists a bijective function
<math|h:B\<rightarrow\>{1,\<ldots\>,m}>
</theorem>
<\proof>
We have the following cases\
<\enumerate>
<item><math|B=\<emptyset\>> then as there exists no surjective function
between a empty set and another non empty set we can not have a
bijective function between <math|B> and <math|{1,\<ldots\>,n}>
<item><math|B\<neq\>\<emptyset\>> we use now induction to prove our
theorem so define <math|C={n\<in\>\<bbb-N\><rsub|0>\|> if there exists
a bijection from A to <math|{1,\<ldots\>,n}> then if
<math|B\<subset\>A> there is no bijection between <math|B> and
<math|{1,\<ldots\>,n}> and if <math|B\<neq\>\<emptyset\>> then
<math|\<exists\>m\<in\>\<bbb-N\><rsub|0>\<vdash\>m\<less\>n> such that
there exists a bijective function between B and <math|{1,\<ldots\>,m}}>
(essential <math|C> is the set of positive integers for which the
theorem assertion is true). Then we prove that C is inductive and thus
equal to <math|\<bbb-N\><rsub|0>>
<\enumerate>
<item><math|1\<in\>C>, to prove assume that there exists a bijection
<math|f:A\<rightarrow\>{1}> and thus <math|A={a}> and thus <math|B> a
proper subset of <math|{a}> must be empty and we can use (1) to prove
that <math|1\<in\>C>
<item>if <math|i\<in\>C> then <math|i+1\<in\>C>. To prove this assume
there exists a bijective function
<math|f:A\<rightarrow\>{1,\<ldots\>,i+1}> then take a proper subset
<math|B> of <math|A>, if <math|B=\<emptyset\>> we can use (1) so
assume that <math|B\<neq\>\<emptyset\>> then
<math|\<exists\>b\<in\>B> and <math|\<exists\>a\<in\>A<mid|\\>B(B\<subset\>A)>
now by <reference|mapping with {1,..,n}> there exists a bijective
function <math|g:A<mid|\\>{b}\<rightarrow\>{1,\<ldots\>,i}>, also
<math|B<mid|\\>{b}> is a proper subset of <math|A<mid|\\>{b}>
(because <math|a\<in\>A<mid|\\>B\<Rightarrow\>a\<in\>A> and
<math|a\<nin\>B\<Rightarrow\>a\<neq\>b\<Rightarrow\>a\<nin\>B\<supseteq\>B<mid|\\>{b}>
and <math|a\<in\>A<mid|\\>{b}>) and thus because <math|i\<in\>C> we
have that\
<\enumerate>
<item>There exists no bijective function between
<math|B<mid|\\>{b}> and <math|{1,.\<ldots\>,i}>
<item>If <math|B<mid|\\>{b}\<neq\>\<emptyset\>
\<Rightarrow\>\<exists\>m\<in\>\<bbb-N\><rsub|0>\<vdash\>m\<less\>i>
such that there exists a bijective function
<math|p:B<mid|\\>{b}\<rightarrow\>{1,\<ldots\>,m}>,
</enumerate>
Now using (i) and <reference|mapping with {1,..,n}> we have proved
that there does not exists a function between <math|B> and
<math|{1,\<ldots\>,i+1}> this prover first part of having
<math|i+1\<in\>C>, to prove the second part assume that
<math|B\<neq\>\<emptyset\>> then if <math|B<mid|\\>{b}=\<emptyset\>>
then <math|B={b}> and thus trivial bijective with <math|{1}>, if
<math|B<mid|\\>{b}\<neq\>\<emptyset\>> then we can use <math|(ii)>
and <reference|mapping with {1,..,n}> to prove the existence of a
bijective function between <math|B> and <math|{1,\<ldots\>,m+1}> and
as <math|m\<less\>i> we have <math|m+1\<less\>i+1> so we have finally
proved that <math|i+1\<in\>C>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|characterization of finite sets>If <math|A> is finite then there
is no bijection of <math|A> with a proper subset of itself
</theorem>
<\proof>
If <math|A=\<emptyset\>> then there is no proper subset of <math|A> and
we can not find a bijection, if <math|A> there is a bijection
<math|f:A\<rightarrow\>{1,\<ldots\>n}> then if there exists a proper
subset <math|B\<subset\>A> with a bijective function
<math|g:A\<rightarrow\>B> then <math|f\<circ\>g<rsup|-1>> is a bijective
function from <math|B> to <math|{1,\<ldots\>,n}> which is impossible by
the previous theorem <math|>\
</proof>
<\corollary>
<label|characterization of infinite sets>If A is bijective to a proper
subset of itself then <math|A> is infinite
</corollary>
<\proof>
If A is not infinite then by definition it is finite and bijective to a
proper subset of itself which by <reference|characterization of finite
sets> is a contradiction.
</proof>
<\theorem>
<label|cardinality of finite sets>If A is a finite set then if it is
bijective with <math|{1,\<ldots\>,n}> and <math|{1,\<ldots\>,m}> then
<math|n=m>.
</theorem>
<\proof>
Assume that <math|n\<neq\>m>, for example <math|n\<less\>m> then there
exists bijective functions <math|f:A\<rightarrow\>{1,\<ldots\>n},g:A\<rightarrow\>{1,\<ldots\>,m}\<Rightarrow\>g\<circ\>f<rsup|-1>:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,m}>
which is a bijection. Now as <math|{1,\<ldots\>i}> is bijective with
itself and thus finite and <math|n\<less\>m\<Rightarrow\>{1,\<ldots\>,n}\<subseteq\>{1,\<ldots\>,m}
and m\<nin\>{1,\<ldots\>,n}> so <math|g\<circ\>f<rsup|-1>> is a bijection
of proper subset of a finite set to the set which by
<reference|characterization of finite sets> gives a contradiction.
</proof>
<\definition>
<label|size of a set><index|size of a set>We define the size of a finite
sets as follows. If <math|A> is finite then either <math|A=\<emptyset\>>
and then <math|size(A)=0> or \ <math|\<exists\>!n\<in\>\<bbb-N\><rsub|0>>
(n is indeed unique by the previous theorem) such that <math|A> is
bijective with <math|{1,\<ldots\>,n}><math|> and then <math|size*(A)=n>
</definition>
<\theorem>
<label|{1,..n}*{1,..,m} is finite>Given <math|n,m\<in\>\<bbb-N\><rsub|0>>
then <math|{1,\<ldots\>,n}\<times\>{1,\<ldots\>,m}> is finite and has
size <math|n*.m>\
</theorem>
<\proof>
We proof this by induction on <math|n>
<\enumerate>
<item><math|n=1>, then define <math|i:{1,\<ldots\>,m}\<rightarrow\>{1}\<times\>{1,\<ldots\>,m}>
by <math|i(j)=(1,j)> which is trivial a bijection so
<math|{1}\<times\>{1,\<ldots\>,m}> is finite and its size is
<math|1.m=m>
<item>Assume the theorem is true for <math|n> prove it for <math|n+1>
now as <math|{1,\<ldots\>,n+1}\<times\>{1,\<ldots\>,m}={1,\<ldots\>,n}*\<times\>{1,\<ldots\>,m}<big|cup>{n+1}*\<times\>{1,\<ldots\>,m}>
(a disjoint union) and there exists a bijective function
<math|i:{1,\<ldots\>,n.m}\<rightarrow\>{1,\<ldots\>,n}\<times\>{1,\<ldots\>,m}>
by the induction hypothesis. Then define the trivial bijection
<math|j:{n.m+1,\<ldots\>,n.m+m}\<rightarrow\>{n+1}\<times\>{1,\<ldots\>,m}>
by <math|j(k)=(n+1,n.m-k)> so using <reference|bijection extension> and
<reference|mapping extension> and the fact that <math|n.m+m=(n+1).m >we
have that <math|i<big|cup>j:{1,\<ldots\>,(n+1).m}\<rightarrow\>{1,\<ldots\>,n+1}\<times\>{1,\<ldots\>,m}>
is a bijective function and thus <math|{1,\<ldots\>,n+1}\<times\>{1,\<ldots\>,m}>
is finite and has size <math|(n+1).m>
</enumerate>
</proof>
<\theorem>
<label|product of finite sets is finite>Let <math|A,B> be finite sets
then <math|A\<times\>B> is finite
</theorem>
<\proof>
As <math|A,B> are finite then there exists bijective functions
<math|h:{1,\<ldots\>,n}\<rightarrow\>A,g:{1,\<ldots\>,m}> and as
<math|{1,\<ldots\>,n}\<times\>{1,\<ldots\>,m}> is by the previous theorem
finite we have a bijective function <math|k:{1,\<ldots\>,m.n}\<rightarrow\>{1,\<ldots\>,n}\<times\>{1,\<ldots\>,m}>
define now <math|l:{1,\<ldots\>,m.n}\<rightarrow\>A\<times\>B> by
<math|l(i)=(h(\<pi\><rsub|1>(k(i))),g(\<pi\><rsub|2>(k(i))))> we prove
now that <math|l> is a bijective function
<\enumerate>
<item>Injectivity. If <math|l(i)=l(j)\<Rightarrow\>h(\<pi\><rsub|1>(k(i)))=h(\<pi\><rsub|1>(k(j)))\<wedge\>g(\<pi\><rsub|2>(k(i)))=g(\<pi\><rsub|2>(k(j)))\<Rightarrowlim\><rsub|h,g
are bijections>\<pi\><rsub|1>(k(i))=\<pi\><rsub|1>(k(j)),\<pi\><rsub|2>(k(i))=\<pi\><rsub|2>(k(j))\<Rightarrow\>k(i)=k(j)\<Rightarrowlim\><rsub|k>
is a bijection<rsub|>i=j>
<item>Surjectivity. Let <math|(a,b)\<in\>A\<times\>B> then as
<math|h,g> is bijective \ <math|\<exists\>i\<in\>{1,\<ldots\>,n},j\<in\>{1,\<ldots\>,m}\<succ\>a=h(i),b=g(j)>
and as <math|k> is bijective there exists a
<math|t\<in\>{1,\<ldots\>,m.n}> such that
<math|k(t)=(i,j)\<Rightarrow\>(a,b)=(h(\<pi\><rsub|1>(i,j)),g(\<pi\><rsub|2>(i,j)))=(h(\<pi\><rsub|1>(k(t))),g(\<pi\><rsub|2>(k(t)))=l(t)>
</enumerate>
</proof>
<\theorem>
<label|subset of a finite set>If A is finite then every subset B of A is
finite
</theorem>
<\proof>
Let <math|B\<subseteq\>A>. As <math|A> is finite then either
<math|A=\<emptyset\>\<Rightarrow\>B=\<emptyset\>\<Rightarrow\>>B is
finite or <math|\<exists\>n\<in\>\<bbb-N\><rsub|0>> such that there exist
a bijective function <math|f:A\<rightarrow\>{1,\<ldots\>,n}>. Now we have
the following cases:\
<\enumerate>
<item><math|B=A\<Rightarrow\>B> is finite
<item><math|B=\<emptyset\>\<Rightarrow\>B> is finite
<item><math|\<emptyset\>\<neq\>B\<subset\>A> (B is a non empty proper
subset of A). Then using <reference|subset of a finite set> we have the
existence of a bijection between <math|B> and
<math|{1,\<ldots\>,m},m\<less\>n> and thus <math|B> is finite.
</enumerate>
</proof>
<\theorem>
<label|natural numbers are infinite><math|\<bbb-N\><rsub|0>> is infinite
(thus not finite)
</theorem>
<\proof>
The function <math|f:\<bbb-N\><rsub|0>\<rightarrow\>\<bbb-N\><rsub|0><mid|\\>{1}:f(n)=n+1>
is a bijection to a proper subset of <math|\<bbb-N\><rsub|0>>, to prove
this note that:
<\enumerate>
<item><math|f> is a function as it is defined for all
<math|\<bbb-N\><rsub|0>>
<item>if <math|f(n)=f(m)\<Rightarrow\>n+1=m+1\<Rightarrow\>(n+1)+(\<um\>1)=(m+1)+(\<um\>1)\<Rightarrow\>n=m>
so <math|f> is injective.
<item>If <math|n\<in\>\<bbb-N\><rsub|0><mid|\\>{1}\<subseteq\>\<bbb-N\><rsub|0>>
then <reference|property of natural numbers> <math|n-1\<in\>\<bbb-N\>>,
now if <math|n-1=0> then <math|n=1> which is impossible because
<math|n\<in\>\<bbb-N\><rsub|0><mid|\\>{1}> so
<math|k=n-1\<in\>\<bbb-N\><rsub|0>> and <math|f(k)=f(n-1)=(n-1)+1=n>
and thus <math|f> is surjective
</enumerate>
\ and thus using <reference|characterization of infinite sets> we have
that <math|\<bbb-N\><rsub|0>>.
</proof>
<\theorem>
<label|test for finitness>Let <math|B> be a non empty set then the
following are equivalent
<\enumerate>
<item><math|B> is finite
<item>There is a surjective function from a section
<math|{1,\<ldots\>,n}> onto <math|B>
<item>There is a injective function from <math|B> into a section
<math|{1,\<ldots\>,n}>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|1\<Rightarrow\>2>
As <math|B> is not empty by definition we have a bijection
<math|f:B\<rightarrow\>{1,\<ldots\>,n}> and thus a bijection
<math|f<rsup|-1>:{1,\<ldots\>,n}\<rightarrow\>B> which is of course
also a surjection.
<item><math|2\<Rightarrow\>3>
If <math|f:{1,\<ldots\>,n}\<rightarrow\>B> is surjective then
<math|\<forall\>x\<in\>B\<Rightarrow\>\<emptyset\>\<neq\>f<rsup|-1>({b})\<subseteq\>\<bbb-N\><rsub|0>>,
then by <reference|naturals are well ordered> we have then existence of
a minimum of <math|f<rsup|-1>({b})> now we define
<math|h={(b,min(f<rsup|-1>({b}))\|b\<in\>B}> which is partial function
(because the minimum of a set is unique) and is thus a function. Now if
<math|h(x)=h(y)=i> then because the minimum of a set is in the set we
have <math|i\<in\>f<rsup|-1>({x})<big|cap>f<rsup|-1>({y})> and thus
<math|(i,x)\<in\>f> and <math|(i,y)\<in\>f> and because <math|f> is a
partial function (being a surjection) we must have <math|x=y>.
<item><math|3\<Rightarrow\>1>
Let <math|g:B\<rightarrow\>{1,\<ldots\>,n}> be a injective function
then <math|g:B\<rightarrow\>{1,\<ldots\>,n}<big|cap>g(B)> is a
bijective function and as <math|{1,\<ldots\>,n}<big|cap>g(B)\<subseteq\>{1,\<ldots\>,n}>
we have by <math|<reference|subset of a finite set>> that
<math|{1,\<ldots\>,n}<big|cap>g(b)> is a finite set (which is non empty
because <math|B> is not empty) and thus by definition there exists a
<math|m\<in\>\<bbb-N\><rsub|0>> such that there is a bijective function
from <math|k:{1,\<ldots\>,n}<big|cap>g(B)\<rightarrow\>{1,\<ldots\>,m}>
and thus we have found a bijective function
<math|><math|k\<circ\>f:B\<rightarrow\>{1,\<ldots\>,m}> and thus
<math|B> is finite\
</enumerate>
</proof>
<\theorem>
<label|injection from infinite set>Let <math|A,B> be sets <math|A> is
infinite and there exists a injective function <math|i:A\<rightarrow\>B>
then <math|B> is infinite
</theorem>
<\proof>
Assume that <math|B> is finite then <math|i(A)\<subseteq\>B> is finite
and there is a bijection <math|h:{1,\<ldots\>,n}\<rightarrow\>i(A)> and
then <math|i:A\<rightarrow\>i(A)> is a bijection so
<math|i<rsup|-1>\<circ\>h> is a bijection from
<math|{1,\<ldots\>,n}\<rightarrow\>B> making <math|B> is finite. <math|>
</proof>
<\theorem>
<label|union of finite sets>Let <math|A,B> be two finite sets then
<math|A<big|cup>B> is finite
</theorem>
<\proof>
We have the following cases:\
<\enumerate>
<item><math|A=\<emptyset\>\<Rightarrow\>A<big|cup>B=B> is empty
<item><math|B=\<emptyset\>\<Rightarrow\>A<big|cup>B=A> is empty
<item><math|A\<neq\>\<emptyset\>,B\<neq\>\<emptyset\>>, then there
exists <math|n,m\<in\>\<bbb-N\><rsub|0>> and two bijective functions
<math|f<rsub|A>:{1,\<ldots\>,n}\<rightarrow\>A,f<rsub|B>:{1,\<ldots\>,m}\<rightarrow\>B>.
Now define the set <math|{n+1,\<ldots\>,n+m}={i\<in\>\<bbb-N\><rsub|0>\|n+1\<leqslant\>i\<leqslant\>n+m}>
then <math|k:{n+1,\<ldots\>,n+m}\<rightarrow\>{1,\<ldots\>,m}> defined
by <math|k(i)=i-n> is a bijection, it is of course defined on all of
<math|{n+1,\<ldots\>,n+m}> and if <math|n+1\<leqslant\>i\<leqslant\>n+m\<Rightarrow\>1\<leqslant\>i-n\<leqslant\>m>,
if <math|i<rsub|1>-n=i<rsub|2>-n\<Rightarrow\>i<rsub|1>-n+n=i<rsub|2>-n+n\<Rightarrow\>i<rsub|1>=i<rsub|2>>
so <math|k> is injective, finally if
<math|j\<in\>{1,\<ldots\>,m}\<Rightarrow\>1\<leqslant\>j\<leqslant\>m\<Rightarrow\>n+1\<leqslant\>j+n\<leqslant\>m+n\<Rightarrow\>k(j+n)=j+n-n=j>
and thus <math|k> is surjective. Now
<math|f<rsub|A>:{1,\<ldots\>,n}\<rightarrow\>A> is a surjection (being
a bijection), <math|f<rsub|B>\<circ\>k:{n+1,\<ldots\>,n+m}\<rightarrow\>B>
is a surjection (begin bijective) and
<math|{1,\<ldots\>,n}<big|cap>{n+1,\<ldots\>,n+m}=\<emptyset\>> so
using <reference|surjection extension> we have found a surjection
<math|f<rsub|A><big|cup>(f<rsub|B>\<circ\>k)> of
<math|{1,\<ldots\>,n+m}={1,\<ldots\>,n}<big|cup>{n+1,\<ldots\>,n+m}> to
<math|B> so using <reference|test for finitness> we have proved that
<math|A<big|cup>B> is finite.
</enumerate>
</proof>
<\theorem>
<label|finite union of finite sets>(Finite union of finite sets are
empty) Let <math|I> be a finite set and <math|{A<rsub|i>}<rsub|i\<in\>I>>
be a family of finite sets then <math|<big|cup><rsub|i\<in\>I>A<rsub|i>>
is finite.
</theorem>
<\proof>
Let <math|P(n)> be the statement that for all <math|I> bijective to
<math|{1,\<ldots\>,n}> we have for every family
<math|{A<rsub|i>}<rsub|i\<in\>I>> with <math|A<rsub|i>> is finite
<math|(i=1,\<ldots\>,n)> we have <math|<big|cup><rsub|i\<in\>I>A<rsub|i>>
is finite.
Define now <math|C={n\<in\>\<bbb-N\><rsub|0>\|P(n)> is true<math|}>, if
we prove that <math|C> is inductive then we have proved our theorem, so
we prove:\
<\enumerate>
<item><math|1\<in\>C>. If <math|I> is bijective with
<math|{1,\<ldots\>,1}=1> then <math|I={i}> and
<math|<big|cup><rsub|j\<in\>I>A<rsub|j>=A<rsub|i>> which is finite.
<item>If <math|n\<in\>C> then <math|n+1\<in\>C>. So let assume
<math|P(n)> is true, then if <math|I> is bijective with
<math|{1,\<ldots\>,n+1}> and <math|{A<rsub|i>}<rsub|i\<in\>I>> be a
family of finite sets, then there exists a bijective function
<math|f:I\<rightarrow\>{1,\<ldots\>,n+1}> and if
<math|i=f<rsup|-1>(n+1)> we find using <reference|mapping with
{1,..,n}> the existence of a bijective function between
<math|I<mid|\\>{i}> and <math|{1,\<ldots\>,n}> and as <math|P(n)> is
true we have <math|<big|cup><rsub|j\<in\>I<mid|\\>{i}>A<rsub|j>> is
finite and as <math|<big|cup><rsub|j\<in\>I>A<rsub|j>=(<big|cup><rsub|j\<in\>I<mid|\\>{i}>A<rsub|j>)<big|cup>A<rsub|i>>
which is finite using <reference|union of finite sets> so we have
proved that <math|P(n+1)> is true
</enumerate>
</proof>
<\lemma>
<label|map of finite set is finite>If A is non empty finite set and
<math|f:A\<rightarrow\>B> is a function then <math|f(A)> is finite
</lemma>
<\proof>
This is trivial as <math|f:A\<rightarrow\>f(A)> is surjective and using
<reference|test for finitness> we proved our theorem
</proof>
<\theorem>
<label|maximum minimum of finite sets>Let <math|X> be a totally ordered
set and let <math|\<emptyset\>\<neq\>A\<subseteq\>X> be finite then
<math|max(A)> (<math|min(A))> is defined
</theorem>
<\proof>
As <math|A> is finite we have a bijective function
<math|a:{1,\<ldots\>,n}\<rightarrow\>A>, we prove now the theorem by
induction
<\enumerate>
<item><math|n=1> then <math|A={a(1)}> and then
<math|max(A)=min(A)=a(1)>
<item>Assume the theorem is true for <math|n> then we proof it for
<math|n+1> then as <math|A=(A<mid|\\>{a(n+1)})<big|cup>{a(n+1)}> there
exists <math|m\<in\>A<mid|\\>{a(n+1)}\<subseteq\>A> such that
<math|\<forall\>x\<in\>A<mid|\\>{a(n+1)}> we have <math|x\<leqslant\>m>
(or for min <math|m\<leqslant\>x>). Then because of totality either
<math|m\<leqslant\>a(n+1)> or <math|a(n+1)\<leqslant\>m> so we have
that either <math|a(n+1)=max(A)> or <math|m=max(A)> (or for the minimum
<math|either m=min(A)> or <math|a(n+1)=min(A)>
</enumerate>
</proof>
<\theorem>
If A is finite and let there be a function
<math|f:\<bbb-N\><rsub|0>\<rightarrow\>A> then <math|\<exists\>a\<in\>A>
such that <math|{m\<in\>\<bbb-N\><rsub|0>\|f(x)=a}> is infinite
</theorem>
<\proof>
Assume that the theorem is not valid then <math|\<forall\>a\<in\>A> we
have <math|f<rsup|-1>({a})={m\<in\>\<bbb-N\><rsub|0>\|f(m)=a}> is finite
then <math|<big|cup><rsub|a\<in\>A>f<rsup|-1>({a})=f<rsup|-1>(<big|cup><rsub|a\<in\>A>{a})=f<rsup|-1>(A)=\<bbb-N\><rsub|0>>
and as the finite union of finite sets is empty we have that
<math|\<bbb-N\><rsub|0>> is finite a contradiction
</proof>
<\corollary>
<label|mapping of N into a finite set>If A is finite and let there be a
function <math|f:\<bbb-N\><rsub|0>\<rightarrow\>A> then
<math|\<exists\>a\<in\>A> so that <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<succ\>\<exists\>m\<in\>\<bbb-N\><rsub|0>,m\<geqslant\>n>
with <math|f(m)=a>
</corollary>
<\proof>
By the previous theorem <math|\<exists\>a\<in\>A> such that
<math|B={m\<in\>\<bbb-N\><rsub|0>\|f(m)=a}> is infinite, now assume that
<math|\<exists\>n\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>m\<geqslant\>n> we have <math|f(m)\<neq\>a> then if
<math|m\<in\>B\<Rightarrow\>f(m)=a\<Rightarrow\>m\<less\>n\<Rightarrow\>B\<subseteq\>S<rsub|n>>
and thus <math|B> is finite.
</proof>
<\theorem>
<label|ordering of a finite set>Let <math|X,\<less\>> be a strictly
ordered set (which is partially ordered by <math|\<leqslant\>>) and let
<math|\<emptyset\>\<neq\>A\<subseteq\>X> be a finite set (size
<math|n\<gtr\>1>) then there exists a bijection
<math|i:{1,\<ldots\>,n}\<rightarrow\>A> such that
<math|\<forall\>k\<in\>{1,\<ldots\>,n-1}> we have that
<math|i(k)\<less\>i(k+1)>
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item>If <math|><math|n=2> then there exist a bijection
<math|a:{1,\<ldots\>,2}\<rightarrow\>A> so that (surjectivity)
<math|A={a(1),a(2)}> if <math|a(1)\<less\>a(2)> then we have found our
bijection, if <math|a(2)\<less\>a(1)> take then
<math|b:{1,\<ldots\>,2}\<rightarrow\>A> by <math|b(1)=a(2),b(2)=a(1)>
which is obviously the bijection we are searching for.
<item>Assume the theorem is true for <math|n> then prove it for
<math|n+1>. So assume that the cardinality of <math|A> is <math|n+1>
then there is a bijection <math|i:{1,\<ldots\>,n+1}\<rightarrow\>A>. As
<math|A> is finite a maximum exists so (surjectivity), if we take then
<math|A<mid|\\>{max(A)}> we can use <reference|mapping with {1,..,n}>
to find a function <math|j:{1,\<ldots\>,n}\<rightarrow\>A<mid|\\>{max(A)}>
proving that <math|A<mid|\\>{max(A)}> has cardinality <math|n>, so
using the induction hypothesis there exists a bijection
<math|i<rprime|'>:{1,\<ldots\>n}\<rightarrow\>A<mid|\\>{max(A)}> such
that <math|\<forall\>k\<in\>{1,\<ldots\>,n-1}> we have
<math|i<rprime|'>(k)\<less\>i<rprime|'>(k+1)>. We define then the
bijection <math|l:{1,\<ldots\>,n+1}\<rightarrow\>A> by
\ <math|l(k)=i<rprime|'>(k)> if <math|k\<in\>{1,\<ldots\>,n+1},l(n+1)=max(A)>
then as <math|i<rprime|'>(n)\<in\>A<mid|\\>{max(A)}\<subseteq\>A\<Rightarrow\>l(n)=i<rprime|'>(n)\<leqslant\>max(A)=l(n+1)>
which because <math|i<rprime|'>(n)\<nin\>{max(A)}\<Rightarrow\>i<rprime|'>(n)\<neq\>max(A)\<Rightarrow\>l(n)\<less\>l(n+1)>
proving that we have found the required bijection
</enumerate>
</proof>
<subsection|Principle of recursive definition>
<\lemma>
<label|recursive definition 1>Let <math|A> be a set and
<math|a<rsub|0>\<in\>A> (so A is not empty) and let
<math|\<cal-M\>={f\|\<exists\>n\<in\>\<bbb-N\><rsub|0>\<succ\>f:{1,\<ldots\>,n}\<rightarrow\>A}>
(<math|\<cal-M\>> is the set of functions of a non empty section of
positive integers to <math|A>} and let
<math|\<rho\>:\<cal-M\>\<rightarrow\>a> (<math|\<rho\>> maps each
function from a non empty set of positive integers to A to a element of
A). Then <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> there exists a
function <math|h:{1,\<ldots\>,n}\<rightarrow\>A> such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{2,\<ldots\>,n}> we have
<math|h(i)=\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>)>
</enumerate>
</lemma>
<\proof>
We proof this by induction, so let <math|P(n)> be the statement that
there exists a function <math|h:{1,\<ldots\>,n}\<rightarrow\>A> such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{2,\<ldots\>,n}> we have
<math|h(i)=\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>)>
</enumerate>
then we prove that:\
<\enumerate>
<item><math|P(1)> is true. This is easy just define
<math|h:{1}\<rightarrow\>A> by <math|h(1)=a<rsub|0>> and note that (2)
is vacuous true because <math|{2,\<ldots\>,1}=\<emptyset\>>\
<item>If <math|P(n)> is true then <math|P(n+1)> is also true. So there
exists a function <math|h:{1,\<ldots\>,n}\<rightarrow\>A> such that (1)
and (2) are true, as <math|h> is evidently a function of a non empty
segment of positive integers and thus <math|\<rho\>(h)> exists, define
then the function <math|g:{n+1}\<rightarrow\>A> by
<math|g(n+1)=\<rho\>(h)>, then using the fact that
<math|{1,.,n}<big|cap>{n+1}> and <reference|mapping extension> we have
a function <math|k=h<big|cup>g:{1,\<ldots\>,n+1}\<rightarrow\>A> with
the following properties
<\enumerate>
<item><math|k(1)=h(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{1,\<ldots\>,n}\<Rightarrow\>k(i)=h(i)\<Rightarrow\>\<forall\>i\<in\>{1,\<ldots\>,n}<with|mode|text|
we have >\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>)=\<rho\>(k<rsub|\|{1,\<ldots\>,i-1}>)>
<item><math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|k(i)=h(i)\<Rightarrow\>k<rsub|\|{1,\<ldots\>,n}>=h> and thus
<math|k(n+1)=\<rho\>(h)=\<rho\>(k<rsub|\|{1,\<ldots\>,n}>)>
</enumerate>
\ so we have indeed proved that <math|P(n+1)> is true.
</enumerate>
</proof>
<\lemma>
<label|recursive definition 2>Let <math|A> be a set and
<math|a<rsub|0>\<in\>A> (so A is not empty) and let
<math|\<cal-M\>={f\|\<exists\>n\<in\>\<bbb-N\><rsub|0>\<succ\>f:{1,\<ldots\>,n}\<rightarrow\>A}>
(<math|\<cal-M\>> is the set of functions of a non empty section of
positive integers to <math|A>} and let
<math|\<rho\>:\<cal-M\>\<rightarrow\>a> (<math|\<rho\>> maps each
function from a non empty set of positive integers to A to a element of
A). Let <math|n,m\<in\>\<bbb-N\><rsub|0>> and assume that we have two
functions <math|h:{1,\<ldots\>,n}\<rightarrow\>A,g:{1,\<ldots\>,m}\<rightarrow\>A>
such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>,g(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{2,\<ldots\>,n}> we have
<math|h(i)=\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>),
\<forall\>i\<in\>{2,\<ldots\>,m}> we have
<math|g(i)=\<rho\>(g<rsub|\|{1,\<ldots\>,i-1}>)>
</enumerate>
then we have <math|\<forall\>i\<in\>{1,\<ldots\>,n}<big|cap>{1,\<ldots\>,m}={1,\<ldots\>,min(n,m)}>
that <math|h(i)=g(i)>
</lemma>
<\proof>
we prove this by contradiction so assume that
<math|\<exists\>i\<in\>{1,\<ldots\>,min(n,m)}\<succ\>h(i)\<neq\>g(i)\<Rightarrow\>C={i\<in\>{1,\<ldots\>,min(n,m)}\|h(i)\<neq\>g(i)}\<neq\>\<emptyset\>\<Rightarrow\>i<rsub|0>=min(C)>
exists. Now <math|i<rsub|0>\<neq\>1> because
<math|h(1)=a<rsub|0>=g(1)\<Rightarrow\>i<rsub|0>\<gtr\>1> now if
<math|i\<less\>i<rsub|0>\<Rightarrowlim\><rsub|definition of
minimum>h(i)=g(i)\<Rightarrow\>h<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>=g<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>\<Rightarrow\>\<rho\>(h<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>)=\<rho\>(h<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>)\<Rightarrow\>h(i<rsub|0>)=g(i<rsub|0>)>
contradicting the fact that <math|i<rsub|0>=min(C)>
</proof>
<\lemma>
<label|unique recursive definition>Let <math|A> be a set and
<math|a<rsub|0>\<in\>A> (so A is not empty) and let
<math|\<cal-M\>={f\|\<exists\>n\<in\>\<bbb-N\><rsub|0>\<succ\>f:{1,\<ldots\>,n}\<rightarrow\>A}>
(<math|\<cal-M\>> is the set of functions of a non empty section of
positive integers to <math|A>} and let
<math|\<rho\>:\<cal-M\>\<rightarrow\>a> (<math|\<rho\>> maps each
function from a non empty set of positive integers to A to a element of
A). Then <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> there exists a
<with|font-series|bold|unique> function
<math|h:{1,\<ldots\>,n}\<rightarrow\>A> such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{2,\<ldots\>,n}> we have
<math|h(i)=\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>)>
</enumerate>
</lemma>
<\proof>
Just use the <reference|recursive definition 1> and <reference|recursive
definition 2> .
</proof>
<\lemma>
<label|principle of recursive definition>(Principle recursive definition)
<index|recursive definition>Let <math|A> be a set and
<math|a<rsub|0>\<in\>A> (so A is not empty) and let
<math|\<cal-M\>={f\|\<exists\>n\<in\>\<bbb-N\><rsub|0>\<succ\>f:{1,\<ldots\>,n}\<rightarrow\>A}>
(<math|\<cal-M\>> is the set of functions of a non empty section of
positive integers to <math|A>} and let
<math|\<rho\>:\<cal-M\>\<rightarrow\>a> (<math|\<rho\>> maps each
function from a non empty set of positive integers to A to a element of
A). Then <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> there exists a
<with|font-series|bold|unique> function
<math|h:\<bbb-N\><rsub|0>\<rightarrow\>A> such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<gtr\>1> we
have <math|h(i)=\<rho\>(h<rsub|\|{1,\<ldots\>,i-1}>)>
</enumerate>
</lemma>
<\proof>
By using <reference|unique recursive definition> we have
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> the existence of a unique
function <math|f<rsub|n>:{1,\<ldots\>,n}\<rightarrow\>A> such that\
<\enumerate>
<item><math|f<rsub|n>(1)=a<rsub|0>>
<item><math|\<forall\>i\<in\>{2,\<ldots\>,n}> we have
<math|f<rsub|n>(i)=\<rho\>(f<rsub|n\|{1,\<ldots\>,i-1}>)>
</enumerate>
define then <math|h=<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>f<rsub|i>>
and show that <math|h> is the desired function as
<math|f<rsub|n>\<subseteq\>{1,\<ldots\>,n}\<times\>A> we have
<math|h=<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>f<rsub|i>\<subseteq\>(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>{1,\<ldots\>,i})\<times\>A\<subseteq\>\<bbb-N\><rsub|0>\<times\>A>,
now if <math|(x,y<rsub|1>),(x,y<rsub|2>)\<in\>h\<Rightarrow\>\<exists\>i<rsub|1>,i<rsub|2>>
such that <math|(x,y<rsub|1>)\<in\>f<rsub|i<rsub|1>>,(x,y<rsub|2>)\<in\>f<rsub|i<rsub|2>>>
now if <math|i<rsub|1>=i<rsub|2>> then as<math|>
<math|f<rsub|i<rsub|1>>=f<rsub|i<rsub|2>>> is a function we have
<math|y<rsub|1>=y<rsub|2>>. If <math|i<rsub|1>\<neq\>i<rsub|2>> then we
have either <math|i<rsub|1>\<less\>i<rsub|2>> or
<math|i<rsub|2>\<less\>i<rsub|1>>, we can safely assume that
<math|i<rsub|1>\<less\>i<rsub|2>> (the other case is proved by reversing
the roles of <math|i<rsub|1>> and <math|i<rsub|2>>). Now as
<math|i<rsub|1>\<less\>i<rsub|2>> then using <reference|recursive
definition 2> we have <math|\<forall\>i\<in\>{1,\<ldots\>,i<rsub|1>}={1,\<ldots\>,i<rsub|1>}<big|cap>{1,\<ldots\>,i<rsub|2>}>
that <math|f<rsub|i<rsub|1>>(i)=f<rsub|i<rsub|2>>(i)> or now as
<math|(x,y<rsub|1>)\<in\>f<rsub|i<rsub|1>>\<Rightarrow\>x\<in\>{1,\<ldots\>,i<rsub|1>}\<Rightarrow\>(x,y<rsub|1>)\<in\>f<rsub|i<rsub|2>>>
and then as <math|(x,y<rsub|2>)\<in\>f<rsub|i<rsub|2>>> we have using the
fact that <math|f<rsub|i<rsub|2>>> is a function that
<math|y<rsub|1>=y<rsub|2>> and thus that <math|h> is indeed a function.
To prof that it is a function note that if
<math|n\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>n\<in\>{1,\<ldots\>,n}\<Rightarrow\>\<exists\>y\<in\>f<rsub|n>\<succ\>(n,y)\<in\>f<rsub|n>\<subseteq\><big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>f<rsub|i>>
and <math|h> is thus a function. Finally we must prove that <math|h> is
unique so assume that there is another <math|h<rprime|'>> satisfying the
hypothesis and that <math|h\<neq\>h<rprime|'>\<Rightarrow\>C={i\<in\>\<bbb-N\><rsub|0>\|h(i)\<neq\>h<rprime|'>(i)}\<neq\>\<emptyset\>>
and thus there exists a <math|i<rsub|0>=min(C)> then if
<math|i\<less\>i<rsub|0>> we have <math|h(i)=h<rprime|'>(i)> and thus
<math|h<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>=h<rprime|'><rsub|\|{1,\<ldots\>,i<rsub|0>-1}>\<Rightarrow\>h(i<rsub|0>)=\<rho\>(h<rsub|\|{1,\<ldots\>,i<rsub|0>-1}>)=\<rho\>(h<rprime|'><rsub|\|{1,\<ldots\>,i<rsub|0>-1}>)=h<rprime|'>(i<rsub|0>)>
contradicting the fact that <math|h(i<rsub|0>)=h<rprime|'>(i<rsub|0>)>
<math|i<rsub|0>> being <math|min(C)>.
</proof>
<\example>
<label|infinite subsets of natural numbers>Let <math|C> be a infinite
subset of <math|\<bbb-N\><rsub|0>> then if
<math|f:{1,\<ldots\>,n}\<rightarrow\>C> we have as
<math|f:{1,\<ldots\>,n}\<rightarrow\>f({1,\<ldots\>,n})> that
<math|f({1,\<ldots\>,n})> is finite and thus
<math|C<mid|\\>f({1,\<ldots\>,n})\<neq\>\<emptyset\>> so it has a
smallest element and we can define <math|\<rho\>:f\<rightarrow\>min(C<mid|\\>f({1,\<ldots\>,n}))>
for every <math|f:{1,\<ldots\>,n}\<rightarrow\>C> so by using the theorem
we have the existence of a function <math|h:\<bbb-N\><rsub|0>\<rightarrow\>C>
such that\
<\enumerate>
<item><math|h(1)=a<rsub|0>>
<item><math|i\<in\>{2,\<ldots\>,n}> such that
<math|h(i)=\<rho\>(f<rsub|\|{1,\<ldots\>,i-1}>)=min(C<mid|\\>f<rsub|\|{1,\<ldots\>,i-1}>({1,\<ldots\>,i-1})=min(C<mid|\\>f({1,\<ldots\>,i-1})>
</enumerate>
</example>
<\example>
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>,\<forall\>f:{1,\<ldots\>,n}\<rightarrow\>\<bbb-R\>>
given <math|a\<in\>\<bbb-R\>> define <math|\<rho\>(f)=a.f(n)> we have the
existence of a function <math|f:\<bbb-N\><rsub|0>\<rightarrow\>\<bbb-R\>>
such that\
<\enumerate>
<item><math|f(1)=a>
<item><math|n\<gtr\>1\<Rightarrow\>f(n)=\<rho\>(f<rsub|\|{1,\<ldots\>,n-1}>)=a.f(n-1)>
</enumerate>
We note <math|f(n)> as <math|a<rsup|n>>
</example>
<\remark>
<label|definition of power>If we note <math|f(n)> as <math|a<rsup|n>>
then the function found is defined by\
<\enumerate>
<item><math|a<rsup|1>=a>
<item><math|a<rsup|n>=a.a<rsup|n-1>>\
</enumerate>
to define <math|a<rsup|n>> on <math|\<bbb-N\>> we add that
<math|a<rsup|0>=1>
</remark>
<\lemma>
<label|power of number bigger then one><math|\<forall\>n\<in\>\<bbb-N\><rsub|0>>
we have if <math|a\<gtr\>1> then <math|a<rsup|n>\<gtr\>1>, so if
<math|\<exists\>n\<in\>\<bbb-N\>\<succ\>a<rsup|n>=1\<Rightarrow\>n=0>
</lemma>
<\proof>
This is easily proved by induction, first
<math|1\<less\>a=a<rsup|1>\<Rightarrow\>1\<neq\>a<rsup|1>>, assume not
that the theorem is true for <math|n\<gtr\>1> then if
<math|a<rsup|n+1>=a.a<rsup|n>\<gtr\>a<rsup|n>\<gtr\>1>
</proof>
<\lemma>
If <math|a\<in\>\<bbb-R\><rsub|0>={x\<in\>\<bbb-R\>\|x\<neq\>0}> and
<math|n\<in\>\<bbb-N\>> then <math|a<rsup|n>\<neq\>0>\
</lemma>
<\proof>
If <math|a\<neq\>0> then either <math|a\<gtr\>0> or <math|0\<less\>a>
then we have the following cases:
<\enumerate>
<item><math|n=0\<Rightarrow\>a<rsup|0>=1\<neq\>0>
<item><math|n\<in\>\<bbb-N\><rsub|0>> we proceed now by induction
<\enumerate>
<item>If <math|n=1> the lemma is true because
<math|a<rsup|1>=a\<neq\>0>
<item>If <math|n\<gtr\>1> then assume the theory to be true for
<math|n> we have <math|a<rsup|n+1>=a.a<rsup|n>> then using the fact
that <math|a\<neq\>0>, <math|a<rsup|n>\<neq\>0> and
<reference|product with 0> we have <math|a<rsup|n+1>\<neq\>0>
</enumerate>
</enumerate>
</proof>
<\definition>
If <math|a\<in\>\<bbb-R\><rsub|0>> and <math|n\<in\>\<bbb-N\>> (and thus
<math|a<rsup|n>\<neq\>0>) we define <math|a<rsup|-n>=(a<rsup|n>)<rsup|-1>>
</definition>
<\lemma>
<label|product of powers>If <math|a\<in\>\<bbb-R\><rsub|0>> and
<math|n,m\<in\>\<bbb-Z\>\<Rightarrow\>a<rsup|n>.a<rsup|m>=a<rsup|n+m>>
</lemma>
<\proof>
Note that in the following proof <math|(a)<rsup|-1>> means the inverse of
<math|a>, which is by definition equal to <math|a<rsup|-1>> (the power
<math|-1> of <math|a>). We have the following cases:
<\enumerate>
<item><math|n=0>, in this case we have
<math|a<rsup|n>.a<rsup|m>=1.a<rsup|m>=a<rsup|m>=a<rsup|0+m>>
<item><math|m=0>, in this case we have
<math|a<rsup|n>.a<rsup|m>=a<rsup|n>.1=a<rsup|n>=a<rsup|n+0>>
<item><math|0\<less\>n,0\<less\>m>, we prove this by induction on n
<\enumerate>
<item><math|>If <math|n=1> then <math|a<rsup|n>.a<rsup|m>=a.a<rsup|m>=a<rsup|m+1>=a<rsup|1+m>>
<item>If <math|n\<gtr\>1> then <math|a<rsup|n+1>.a<rsup|m>=a.a<rsup|n>.a<rsup|m>=a.a<rsup|n+m>=a<rsup|n+m+1>=a<rsup|(n+1)+m>>
</enumerate>
<item><math|0\<less\>n,m\<less\>0> then <math|m=\<um\>m<rprime|'>> with
<math|m<rprime|'>\<in\>\<bbb-N\><rsub|0>> we prove the theorem then by
induction on <math|n>, then <math|a<rsup|n>.a<rsup|m>=a<rsup|n>.a<rsup|-m<rprime|'>>=a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>>
if we prove now that <math|a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>=a<rsup|n-m<rprime|'>>(=a<rsup|n+m>)>
we have proved our theorem. To prove that
<math|a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>=a<rsup|n-m<rprime|'>>>
we proceed by induction of <math|m<rprime|'>>, so\
<\enumerate>
<item>if <math|m<rprime|'>=1> then
<math|a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>=a<rsup|n>.(a<rsup|<rsup|1>)<rsup|-1>>>
now if <math|n=1\<Rightarrow\>><math|a<rsup|n>.(a<rsup|1>)<rsup|-1>=a.a<rsup|-1>=1=a<rsup|0>=a<rsup|1-1>>,
if <math|n\<gtr\>1=a<rsup|n>.(a<rsup|1>)<rsup|1>=a.a<rsup|(n-1)>.(a)<rsup|-1>=a<rsup|(n-1>>
so in both cases we have <math|a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>=a<rsup|n-m<rprime|'>>>
<item>If m'\<gtr\>1 assume that <math|a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>>
is true then <math|a<rsup|n>.(a<rsup|m<rprime|'>+1>)<rsup|-1>=a<rsup|n>.(a.a<rsup|m<rprime|'>>)<rsup|-1>=a<rsup|n>.(a<rsup|m<rprime|'>>)<rsup|-1>.(a)<rsup|-1>=a<rsup|n-m<rprime|'>>.(a)<rsup|-1>>.
We can now consider the following cases:
<\enumerate>
<item><math|n-m<rprime|'>=0\<Rightarrow\>a<rsup|n-m<rprime|'>>.(a)<rsup|-1>=a<rsup|-1>=a<rsup|0-1>=a<rsup|n-(m<rprime|'>+1)>>
<item><math|n-m<rprime|'>=1\<Rightarrow\>a<rsup|n-m<rprime|'>>.(a)<rsup|-1>=a.(a)<rsup|-1>=1=a<rsup|0>=a<rsup|(n-(m<rprime|'>+1))>>
<item><math|n-m<rprime|'>\<gtr\>1\<Rightarrow\>a<rsup|n-m<rprime|'>>.(a)<rsup|-1>=a.a<rsup|n-m<rprime|'>-1>.(a)<rsup|-1>=a<rsup|n-m<rprime|'>-1>=a<rsup|n-(m<rprime|'>+1)>>
<item><math|n-m<rprime|'>\<less\>0\<Rightarrow\>a<rsup|n-m<rprime|'>>.(a)<rsup|-1>=(a<rsup|m<rprime|'>-n>)<rsup|-1>(a)<rsup|-1>\<equallim\><rsub|<reference|inverse
of product>>(a<rsup|m<rprime|'>-n>.a)<rsup|-1>\<equallim\><rsub|m<rprime|'>-n\<gtr\>0,3>(a<rsup|m<rprime|'>-n+1>)<rsup|-1>=a<rsup|n-m<rprime|'>-1>=a<rsup|n-(m<rprime|'>+1)><rsup|>>
</enumerate>
</enumerate>
<item><math|n\<less\>0,0\<less\>m> then if we that <math|n<rprime|'>=m>
and <math|m<rprime|'>=n> we have <math|0\<less\>n<rprime|'>,m<rprime|'>\<less\>0>
and thus \ <math|a<rsup|n>.a<rsup|m>=a<rsup|m<rprime|'>>.a<rsup|n<rprime|'>>=a<rsup|n<rprime|'>>.a<rsup|m<rprime|'>>\<equallim\><rsub|4>a<rsup|n<rprime|'>+m<rprime|'>>=a<rsup|m+n>=a<rsup|n+m>>
<item><math|n\<less\>0,m\<less\>0> then we have for
<math|n<rprime|'>=\<um\>n,m<rprime|'>=\<um\>m>
<math|a<rsup|n>.a<rsup|m>=(a<rsup|n<rprime|'>>)<rsup|-1>(a<rsup|m<rprime|'>>)<rsup|-1>\<equallim\><rsub|<reference|inverse
of product>><rsup|>(a<rsup|n<rprime|'>>.a<rsup|m<rprime|'>>)<rsup|-1>\<equallim\><rsub|4>(a<rsup|n<rprime|'>+m<rprime|'>>)<rsup|-1>=a<rsup|-n<rprime|'>-m<rprime|'>>=a<rsup|n+m>>
</enumerate>
</proof>
<\lemma>
<math|\<forall\>a\<in\>\<bbb-R\>,b\<in\>\<bbb-R\><rsub|0>> then
<math|\<forall\>n\<in\>\<bbb-Z\>> we have
<math|(a.b)<rsup|n>=a<rsup|n>.b<rsup|n>>
</lemma>
<\proof>
We have the following cases\
<\enumerate>
<item><math|n=0> then <math|1=(a.b)<rsup|0>=1.1=a<rsup|0>.b<rsup|0>>
<item><math|0\<less\>n> we prove this case by induction, if
<math|n=1\<Rightarrow\>(a.b)<rsup|1>=a.b=a<rsup|1>.b<rsup|1>> so the
case is true for <math|n=1>, assume the case is true for <math|n> then
<math|(a.b)<rsup|n+1>=(a.b).(a.b)<rsup|n>=(a.b).(a<rsup|n>.b<rsup|n>)=(a.a<rsup|n>).(b.b<rsup|n>)=a<rsup|n+1>.b<rsup|n+1>>
<item><math|><math|n\<less\>0> then for <math|n<rprime|'>=-n> we have
<math|(a.b)<rsup|n>=((a.b)<rsup|n<rprime|'>>)<rsup|-1>=(a<rsup|n<rprime|'>>.b<rsup|n<rprime|'>>)<rsup|-1>=(a<rsup|n<rprime|'>>)<rsup|-1>(b<rsup|n<rprime|'>>)<rsup|-1>=a<rsup|n>.b<rsup|n>>
</enumerate>
</proof>
<\lemma>
<label|odd power>If <math|n\<in\>\<bbb-N\><rsub|0>> then if
<math|a\<in\>\<bbb-N\>> we have that <math|a<rsup|n>> is odd implies that
<math|a> is odd
</lemma>
<\proof>
Let <math|a<rsup|n>> be odd and assume that <math|a> is even then
<math|\<exists\>m\<in\>\<bbb-N\>\<succ\>a=2.m><math|\<Rightarrow\>a<rsup|n>=2<rsup|n>.m<rsup|n>>
now as <math|n\<in\>\<bbb-N\><rsub|0>> and thus <math|n\<gtr\>0> and if
<math|n=1\<Rightarrow\>a=a<rsup|1>=2<rsup|1>.m<rsup|1>=2.m> and
<math|a<rsup|n>> is even a contradiction so we must have <math|n\<gtr\>1>
and then <math|2<rsup|n>=2.2<rsup|n-1>\<Rightarrow\>a<rsup|n>=2.2<rsup|n-1>.m<rsup|n>\<Rightarrow\>2.(2<rsup|n-1>.m<rsup|n>)\<Rightarrow\>a<rsup|n>>
again a contradiction so we must have that <math|a> is even.
</proof>
<\remark>
In general if we note for <math|f:{1,\<ldots\>,n}\<rightarrow\>A>
<math|f(i)> as <math|f<rsub|i>> and have a <math|\<rho\>> that is a
expression depending on <math|f<rsub|1>,\<ldots\>,f<rsub|n>> and
<math|a<rsub|1>,a<rsub|2>\<ldots\>a<rsub|m>> as
<math|\<rho\>(f<rsub|1>,\<ldots\>,f<rsub|n>,a<rsub|1>,\<ldots\>,a<rsub|m>)>
then we have the existence of a function <math|h> such that
(<math|h(i)\<equallim\><rsub|noted as>h<rsub|i>>
<\enumerate>
<item><math|h<rsub|1>=a<rsub|0>>
<item><math|n\<gtr\>1\<Rightarrow\>h<rsub|n>=\<rho\>(h<rsub|1>,\<ldots\>,h<rsub|n-1>,a<rsub|1>,\<ldots\>,a<rsub|m>)>
</enumerate>
In the rest of this document we will use the loose notation provided here
and do not refer to the function <math|\<rho\>> anymore. Be aware that it
is essential that in the definition of <math|h<rsub|n>> there is no
dependency on <math|h<rsub|n>> itself
</remark>
<\example>
Fibonacci numbers these are defined in the loose notation as\
<\eqnarray*>
<tformat|<table|<row|<cell|F<rsub|1>>|<cell|=>|<cell|0>>|<row|<cell|F<rsub|2>>|<cell|=
>|<cell|1>>|<row|<cell|F<rsub|3>>|<cell|=>|<cell|F<rsub|n-1>+F<rsub|n-2>>>>>
</eqnarray*>
the <math|\<rho\>> used in this case is the following if
<math|f:{1}\<rightarrow\>\<bbb-N\><rsub|0>> then <math|\<rho\>(f)=1> and
if <math|f:{1,\<ldots\>,n} (n\<gtr\>1<rsub|>)> as
<math|\<rho\>(f)=f(n)+f(n-1)> and then using the theorem we have the
existence of a <math|F> such that\
<\enumerate>
<item><math|F(1)=0>
<item><math|n\<gtr\>1><math| F(n)=\<rho\>(F<rsub|\|{1,\<ldots\>,n-1)>)=F(n-1)+F(n-2)>
</enumerate>
exactly the definition of Fibonacci numbers.\
</example>
<\theorem>
<math|\<forall\>n\<in\>\<bbb-N\>> we have that <math|n\<less\>2<rsup|n>>
so using the Archimedean property (<reference|Archimedean ordering
property>) we find that for every <math|a\<in\>\<bbb-R\>\<Rightarrow\>\<exists\>n\<in\>\<bbb-N\><rsub|0>>
with <math|a\<less\>n\<less\>2<rsup|n>>
</theorem>
<\proof>
The theorem is evidently true if <math|n=0> because
<math|0\<less\>1=2<rsup|0>>, for <math|n\<in\>\<bbb-N\><rsub|0>> we prove
this by induction, First <math|1\<less\>2=2<rsup|1>> so it is true for
<math|n=1> then if it is true for <math|n\<in\>\<bbb-N\><rsub|0>> now if
<math|n=1> then because of <math|0\<less\>1\<Rightarrow\>0\<less\>1\<less\>1+1\<Rightarrow\>1\<less\>1+1\<less\>1+1+1\<Rightarrow\>1+1\<less\>1+1+1+1=2+2=2.1+2.1=2.(1+1)=2.2=2<rsup|1>.2=2<rsup|2>>
so the theorem is true, so we only have to prove the theorem for
<math|n\<gtr\>1> \ then we have <math|n\<less\>2<rsup|n>> (induction
hypothesis) and <math|2<rsup|n+1>=2.2<rsup|n>\<Rightarrowlim\><rsub|2\<gtr\>1\<gtr\>0
and n\<less\>2<rsup|n>>2.n\<less\>2.2<rsup|n>=2<rsup|n+1>>, also from
<math|1\<less\>n\<Rightarrow\>n+1\<less\>n+n=1.n+1.n=(1+1).n=2.n\<less\>2<rsup|n+1>>
</proof>
<\lemma>
Let <math|x\<in\>\<bbb-R\>> with <math|x\<gtr\>1> then
<math|x<rsup|n>-1\<geqslant\>n.(x-1)>
</lemma>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1> In this case the proof is trivial as
<math|x<rsup|n>-1=x<rsup|1>-1=x-1>
<item>Assume the theorem is true for <math|n> then prove it for
<math|n+1.> <math|x<rsup|n+1>-1=x(x<rsup|n>)-1=x(x<rsup|n>-1)+(x-1)\<equallim\><rsub|induction
hypothese,x\<gtr\>0>\<geqslant\>x.n.(x-1)+(x-1)=(x-1)(x.n+1)\<gtr\>(x-1)(n+1)
> (as <math|x\<gtr\>1>)\
</enumerate>
</proof>
<\lemma>
Let <math|N\<in\>\<bbb-N\><rsub|0>> and <math|x\<in\>\<bbb-R\>> with
<math|><math|x\<gtr\>1> then there exists a
<math|n\<in\>\<bbb-N\><rsub|0>> such that <math|x<rsup|n>\<gtr\>N>\
</lemma>
<\proof>
If <math|N=1> then <math|N\<less\>x=x<rsup|1>> proving the theorem, so
let consider the case <math|N\<gtr\>1> then let
<math|\<delta\>=<frac|N-1|(x-1)>> then by the Archimedean property
(<reference|Archimedean ordering property>) there exist a
<math|N<rsub|0>\<in\>\<bbb-N\><rsub|0>> such that
<math|\<delta\>\<less\>N<rsub|0>> and then
<math|N-1=\<delta\>(x-1)\<less\>N<rsub|0>(x-1)> and using the previous
lemma we have <math|N<rsub|0>(x-1)\<leqslant\>x<rsup|N<rsub|0>>-1\<Rightarrow\>N-1\<less\>x<rsup|N<rsub|0>>-1\<Rightarrow\>N\<less\>x<rsup|N<rsub|0>>>
as we set out to prove.
</proof>
<\theorem>
If <math|0\<less\>x\<less\>1> and <math|n,m\<in\>\<bbb-N\><rsub|0>> with
<math|n\<less\>m> then <math|x<rsup|m>\<less\>x<rsup|n>>
</theorem>
<\proof>
We prove this by induction on <math|k=m-n>
<\enumerate>
<item><math|k=1> (our <math|m=n+1)> then
<math|x<rsup|m>=x<rsup|n+1>=x(x<rsup|n>)\<equallim\><rsub|x\<less\>1>1.x<rsup|n>=x<rsup|n>>
<item>Assume that it is proved for <math|k> then
<math|x<rsup|n+(k+1)>=x<rsup|(n+k)+1>=x(x<rsup|n+k>)\<less\>x<rsup|n+k>\<less\><rsub|induction
hypothese>x<rsup|n>>
</enumerate>
</proof>
<\theorem>
<label|power of x\<less\>1>Let <math|\<varepsilon\>\<in\>\<bbb-R\>> with
<math|\<varepsilon\>\<gtr\>0> and <math|x\<in\>\<bbb-R\>> such that
<math|0\<less\>x\<less\>1> then there exist a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|0\<less\>x<rsup|n>\<less\>\<varepsilon\>> if <math|n\<geqslant\>N>
</theorem>
<\proof>
Consider <math|<frac|1|\<varepsilon\>>\<gtr\>0> by the Archimedean
property (<reference|Archimedean ordering property>) there exists a
<math|n\<in\>\<bbb-N\><rsub|0>> such that
<math|0\<less\><frac|1|\<varepsilon\>>\<less\>n> and as
<math|<frac|1|x>\<gtr\>1> by the previous lemma there exist a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|n\<less\>(<frac|1|x>)<rsup|N>\<Rightarrow\><frac|1|\<varepsilon\>>\<less\>(<frac|1|x>)<rsup|N>\<Rightarrow\>x<rsup|N>\<less\>\<varepsilon\>>
, the rest follows from the previous theorem
</proof>
<subsection|Countable sets>
<\definition>
<index|countable infinite set>A set <math|X> is countable infinite if
there is a bijective function between <math|\<bbb-N\><rsub|0>> and X
</definition>
<\definition>
<index|countable set><index|uncountable set>A set <math|X> is countable
if it is finite or countable infinite. A uncountable set is a set that is
not countable (and thus by definition it can't be finite)
</definition>
<\theorem>
<label|every subset of a countable set is countable>Every subset of a
countable set is countable
</theorem>
<\proof>
Let <math|X> be a countable set and let <math|Y\<subseteq\>X> we have
then the following cases\
<\enumerate>
<item>X is finite, then we have by <reference|subset of a finite set>
that <math|Y> is finite and thus countable.
<item><math|X> is infinite, Y is finite then of course <math|Y> is
countable
<item><math|X> is infinite, <math|Y> is infinite, then because <math|X>
is finite we have a bijective function
<math|f:X\<rightarrow\>\<bbb-N\><rsub|0>\<Rightarrow\>f<rsub|\|Y>:Y\<rightarrow\>f(Y)>
is a bijective function between <math|Y> and
<math|f(Y)\<subseteq\>\<bbb-N\><rsub|0>> as <math|Y> is infinite,
<math|f*(Y)> is infinite and then using <reference|infinite subsets of
natural numbers> there exists a function <math|h:\<bbb-N\><rsub|0>>
such that <math|h(1)=min(f(Y))> and if <math|n\<gtr\>1> then
<math|h(n)=min(f(Y)<mid|\\>h({1,\<ldots\>,n-1}))> now we prove that
<math|h> is a bijective function. First for injectivity, if
<math|n\<neq\>m>, for example <math|n\<less\>m> (<math|m\<less\>n> is
similar proved) as <math|n\<in\>{1,\<ldots\>,m-1}\<Rightarrow\>h(n)\<in\>h({1,\<ldots\>,m-1})>
we have <math|h(n)\<nin\>f(Y)<mid|\\>h({1,\<ldots\>,m-1})\<Rightarrow\>h(n)\<neq\>min(f(Y)<mid|\\>h({1,\<ldots\>,m-1}))=h(m)>
a contradiction so we must have <math|n=m> if <math|h(n)=h(m)>. To
prove surjectivity note that if <math|y\<in\>f(Y)> then because
<math|h> is injective we have and <math|\<bbb-N\><rsub|0>> infinite we
cab not have that <math|h(\<bbb-N\><rsub|0>)\<subseteq\>{1,\<ldots\>,n}>
for a <math|n\<in\>\<bbb-N\><rsub|0>> because this would mean by
<reference|test for finitness> that <math|\<bbb-N\><rsub|0>> would be
finite. So there is a <math|n<rsub|y>\<in\>\<bbb-N\><rsub|0>> such that
<math|h(n<rsub|y>)\<gtr\>y> which means that
<math|A={n\<in\>\<bbb-N\><rsub|0>\|h(n)\<geqslant\>y}\<neq\>\<emptyset\>\<Rightarrow\>min(A)>
exists then if <math|i\<less\>min(A)> we must have
<math|h(i)\<less\>y\<Rightarrow\>y\<nin\>h({1,\<ldots\>,min(A)-1})\<Rightarrow\>y\<in\>f(Y)<mid|\\>h({1,\<ldots\>,min(A)-1})>
and since <math|h(min(A))=min(f(Y)<mid|\\>h({1,\<ldots\>,min(A)-1}))>
we have <math|h(min(A))\<leqslant\>y> and because of the definition of
a minimum we have <math|min(A)\<in\>A\<Rightarrow\>h(min(A))\<geqslant\>y\<Rightarrow\>h(min(A))=y>
proving that <math|h> is surjective. So we have found a bijection
<math|h:f(Y)\<rightarrow\>\<bbb-N\><rsub|0><rsub|>> and thus a
bijection <math|h\<circ\>f<rsub|Y>> between <math|Y> and
<math|\<bbb-N\><rsub|0>> proving that <math|Y> is countable.
</enumerate>
</proof>
<\theorem>
<label|test for countability>Let <math|X> be a nonempty set, the the
following are equivalent\
<\enumerate>
<item><math|X> is countable
<item>There is a surjective function
<math|f:\<bbb-N\><rsub|0>\<rightarrow\>X>
<item>There is a injective function
<math|f:X\<rightarrow\>\<bbb-N\><rsub|0>><math|>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|1\<Rightarrow\>2> Here we have two cases\
<\enumerate>
<item><math|X> is infinite then by definition there exists a
bijective function (and thus surjective function) from
<math|\<bbb-N\><rsub|0>> to <math|X>
<item>X is finite, then by definition
<math|\<exists\>n\<in\>\<bbb-N\><rsub|0>> such that there exists a
bijective function <math|f:{1,\<ldots\>,n}\<rightarrow\>X> , let then
<math|A={i\<in\>\<bbb-N\><rsub|0>\|i\<gtr\>n}> then we define the
surjective function <math|g:A\<rightarrow\>{f(1)}> by
<math|g(i)=f(1)> if <math|i\<gtr\>n> then by using
<reference|surjection extension> and <reference|mapping extension> we
have that <math|f<big|cup>g> a surjective function from
<math|\<bbb-N\><rsub|0>\<rightarrow\>X>
</enumerate>
<item><math|2\<Rightarrow\>3> Let <math|f:\<bbb-N\><rsub|0>\<rightarrow\>X>
be a surjection then <math|\<forall\>x\<in\>X> <math|f<rsup|-1>({x})>
is not empty and <math|\<subseteq\>\<bbb-N\><rsub|0>> so the minimum
<math|m=min(f<rsup|-1>({x}))> exists, define then <math|g> by
<math|{(x,min(f<rsup|-1>({x})))\|x\<in\>X}> which must be a partial
function because of the uniqueness of the minimum and which is a
function because the minimum is defined for every x. Assume now that
<math|g(x<rsub|1>)=g(x<rsub|2>)> now assume that
<math|x<rsub|1>\<neq\>x<rsub|2>> and
<math|y\<in\>f<rsup|-1>({x<rsub|1>})<big|cap>f<rsup|-1>({x<rsub|2>})\<Rightarrow\>f(y)=x<rsub|1>\<neq\>x<rsub|2>=f(y)>
which is impossible because <math|f> is a function thus a partial
function, so <math|f<rsup|-1>({x<rsub|1>})<big|cap>f<rsup|-1>({x<rsub|2>})=\<emptyset\>>
so they can't have the same minimum because that would be in the
intersection. So we reach a contradiction and <math|x<rsub|1>> must be
equal to <math|x<rsub|2>>.
<item><math|3\<Rightarrow\>1> Let <math|g:X\<rightarrow\>\<bbb-N\><rsub|0>>
be a injective map if <math|X> is finite then it is of course
countable, so we assume that <math|X> is infinite then we have
<math|g:X\<rightarrow\>g(X)> is a bijective function (g is injective)
and so <math|g(X)> is infinite and by <reference|every subset of a
countable set is countable> <math|g(X)> is countable so there exists a
bijective function <math|f:g(X)\<rightarrow\>\<bbb-N\><rsub|0>> and
thus <math|f\<circ\>g> is a bijective function from <math|X> to
<math|\<bbb-N\><rsub|0>> proving that <math|X> is countable.
</enumerate>
\
</proof>
<\corollary>
<label|product of naturals is countable>The set
<math|\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>> is countable infinite
</corollary>
<\proof>
Using <reference|test for countability> it is enough to construct a
injective function from <math|\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>\<rightarrow\>\<bbb-N\><rsub|0>>.
Define then <math|f:\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>\<rightarrow\>\<bbb-N\><rsub|0>>
by <math|f(n,m)=2<rsup|n>.3<rsup|m>> then if
<math|f(n<rsub|1>,m<rsub|1>)=f(n<rsub|2>,m<rsub|2>)\<Rightarrow\>2<rsup|n<rsub|1>>.3<rsup|m<rsub|1>>=2<rsup|n<rsub|2>>.3<rsup|m<rsub|2>>>
we can then consider two cases:
<\enumerate>
<item><math|m<rsub|1>=m<rsub|2>\<Rightarrow\>2<rsup|n<rsub|1>>=2<rsup|n<rsub|2>>>
now assume that <math|n<rsub|1>\<neq\>n<rsub|2>> (for example
<math|n<rsub|1>\<gtr\>n<rsub|2>)\<Rightarrow\>n<rsub|1>-n<rsub|2>\<in\>\<bbb-N\><rsub|0>>
and by multiplying both sides of the equation by
<math|2<rsup|-n<rsub|2>>> we get by <reference|product of powers>
<math|2<rsup|n<rsub|1>-n<rsub|2>>=2<rsup|0>=1> and then using
<reference|power of number bigger then one> we get
<math|n<rsub|1>-n<rsub|2>=0\<Rightarrow\>n<rsub|1>=n<rsub|2>> and we
get a contradiction, so if <math|m<rsub|1>=m<rsub|2>\<Rightarrow\>n<rsub|1>=n<rsub|2>>
<item><math|m<rsub|1>\<neq\>m<rsub|2>> (for example
<math|m<rsub|1>\<gtr\>m<rsub|2>> (the case
<math|m<rsub|2>\<gtr\>m<rsub|1>> is proved by using
<math|m<rsub|1><rprime|'>=m<rsub|2> and
m<rsub|2><rprime|'>=m<rsub|1>)>) then
<math|2<rsup|n<rsub|1>>.3<rsup|m<rsub|1>>.3<rsup|\<um\>m<rsub|2>>=2<rsup|n<rsub|2>>.3<rsup|m<rsub|2>>.3<rsup|\<um\>m<rsub|2>>=2<rsup|n<rsub|2>>.3<rsup|m<rsub|2>-m<rsub|2>>=2<rsup|n<rsub|2>>.3<rsup|0>=2<rsup|n<rsub|2>>>
and thus <math|2<rsup|n<rsub|2>>=2<rsup|n<rsub|1>>.3<rsup|m<rsub|1>-m<rsub|2>>>
now if <math|n<rsub|1>=n<rsub|2>\<Rightarrow\>1=3<rsup|m<rsub|1>-m<rsub|2>>\<Rightarrow\>m<rsub|1>-m<rsub|2>=0>
which contradicts <math|m<rsub|1>\<neq\>m<rsub|2>> so we are left with
the following two cases:\
<\enumerate>
<item><math|n<rsub|1>\<less\>n<rsub|2>> then
<math|2<rsup|n<rsub|2>-n<rsub|1>>=3<rsup|m<rsub|1>-m<rsub|2>>\<Rightarrow\>3<rsup|m<rsub|1>-m<rsub|2>>>
is even and thus <math|3=2.1+1> is even which is a contradiction
<item><math|n<rsub|2>\<less\>n<rsub|1>> then
<math|1=2<rsup|n<rsub|1>-n<rsub|2>>.3<rsup|m<rsub|1>-m<rsub|2>>>
which means that <math|1> is even which is again a contradiction
</enumerate>
</enumerate>
So case 2 always let to a contradiction, so we end with case 1 and thus
<math|m<rsub|1>=m<rsub|2>,n<rsub|1>=n<rsub|2>\<Rightarrow\>(n<rsub|1>,m<rsub|1>)=(n<rsub|2>,m<rsub|2>)>
proving that <math|f> is injective
</proof>
<\theorem>
<label|countable union of countable sets is countable>A countable union
of countable sets is countable.
</theorem>
<\proof>
Consider the countable family of countable sets
<math|{A<rsub|i>}<rsub|i\<in\>I>> (where <math|I> is countable and
<math|\<forall\>i\<in\>I\<succ\>A<rsub|i>> is countable). First define
<math|J={i\<in\>I\|A<rsub|i>\<neq\>\<emptyset\>}> then
<math|<big|cup><rsub|i\<in\>I>A<rsub|i>=<big|cup><rsub|j\<in\>J>A<rsub|j>>
<math|[x\<in\><big|cup><rsub|i\<in\>I>A<rsub|i>\<Rightarrow\>\<exists\>i\<in\>I\<succ\>x\<in\>A<rsub|i>\<Rightarrowlim\><rsub|thus
A<rsub|i>\<neq\>\<emptyset\>>x\<in\>A<rsub|i>,j\<in\>J\<Rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>\<subseteq\><big|cup><rsub|j\<in\>J>A<rsub|j>>,
<math|x\<in\><big|cup><rsub|i\<in\>I>A<rsub|i>\<Rightarrow\>\<exists\>i\<in\>I\<subseteq\>J\<succ\>x\<in\>A<rsub|i>\<Rightarrow\>x\<in\><big|cup><rsub|j\<in\>J>A<rsub|j>\<Rightarrow\><big|cup><rsub|j\<in\>J>A<rsub|j>\<subseteq\><big|cup><rsub|i\<in\>I>A<rsub|i>]>.
Now <math|\<forall\>j\<in\>J> we have <math|A<rsub|j>> is non empty and
countable so using \ <reference|test for countability> we have the
existence of a surjective function <math|f<rsub|j>:\<bbb-N\><rsub|0>\<rightarrow\>A<rsub|j>>.
Also as <math|I> is countable we have using <reference|every subset of a
countable set is countable> \ that <math|J> is countable and thus the
existence of a surjective function <math|g:\<bbb-N\><rsub|0>\<rightarrow\>J>.
Define now <math|f:\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>\<rightarrow\><big|cup><rsub|j\<in\>J>A<rsub|j>>
by <math|f(n,m)=f<rsub|g(n)>(m)>. This function is surjective because if
<math|y\<in\><big|cup><rsub|j\<in\>J>A<rsub|j>\<Rightarrow\>\<exists\>j\<in\>J\<vdash\>y\<in\>A<rsub|j>\<Rightarrowlim\><rsub|g
is surjective>\<exists\>n\<in\>\<bbb-N\><rsub|0>\<vdash\>y\<in\>A<rsub|g(n)>\<Rightarrowlim\><rsub|f<rsub|g(n)>
is surjective>\<exists\>m\<in\>\<bbb-N\><rsub|0>,n\<in\>\<bbb-N\><rsub|0>\<vdash\>y=f<rsub|g(n)>(m)=f(n,m)>
so there exists a surjective function from
<math|f:\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>\<rightarrow\><big|cup><rsub|j\<in\>J>A<rsub|j>=<big|cup><rsub|i\<in\>I>A<rsub|i>>,
and using <reference|test for countability> and <reference|product of
naturals is countable> we have the existence of a surjective function
<math|h:\<bbb-N\><rsub|0>\<rightarrow\>\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>>
so we have the existence of a surjective function
<math|f\<circ\>h:\<bbb-N\><rsub|0>\<rightarrow\><big|cup><rsub|i\<in\>I>A<rsub|i>>
and thus using <reference|test for countability> we have proved our
theorem.
</proof>
<\lemma>
<label|family of one set>Let <math|{A<rsub|i>}<rsub|i\<in\>{a}>> be a
family of sets, consisting of a single set <math|A<rsub|a>>. Then
<math|<big|prod><rsub|i\<in\>{a}>A<rsub|i>> is bijective with
<math|A<rsub|a>>
</lemma>
<\proof>
Define <math|h:<big|prod><rsub|i\<in\>{a}>A<rsub|i>\<rightarrow\>A<rsub|a>>
by <math|h(f)=f(a)> then if <math|h(f<rsub|1>)=h(f<rsub|2>)\<Rightarrow\>f<rsub|1>(a)=f<rsub|2>(a)\<Rightarrow\>f<rsub|1>=f<rsub|2>>
(because <math|f<rsub|i>:{a}\<rightarrow\><big|cup><rsub|i\<in\>{a}>A<rsub|i>=A<rsub|a>)>
and h is thus injective. If <math|y\<in\>A<rsub|a>> define
<math|f:{1}\<rightarrow\>A<rsub|a>> by <math|f(1)=y\<in\>A<rsub|a>> and
thus <math|f\<in\><big|prod><rsub|i\<in\>{a}>A<rsub|i>> and
<math|h(f)=f(a)=y> and h is thus surjective.
</proof>
<\lemma>
<label|finite product of sets>Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be a
non empty finite family of sets then <math|<big|prod><rsub|i\<in\>I>A<rsub|i>>
is bijective with <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|f(i)>>
for some <math|n\<in\>\<bbb-N\><rsub|0>> where
<math|f:{1,\<ldots\>,n}\<rightarrow\>I> is a bijective function (which
exists because <math|I> is non empty and finite).
</lemma>
<\proof>
As I is finite and non empty there is a bijection
<math|h:{1,\<ldots\>,n}\<rightarrow\>I> now define the function
<math|f:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>\<rightarrow\><big|prod><rsub|i\<in\>I>A<rsub|h(i)>>
by <math|f(g)=g\<circ\>h> and <math|f(g)(j)=g(h(j))\<in\>A<rsub|h(j)>> To
prove that f is injective assume that <math|f(g<rsub|1>)=f(g<rsub|2>)>
then <math|\<forall\>j\<in\>{1,\<ldots\>,n}> we have
<math|g<rsub|1>(h(j))=g<rsub|2>(h(j))> now if <math|i\<in\>I> there
exists a <math|j\<in\>{1,\<ldots\>,n}> such that <math|h(j)=i> and then
<math|g<rsub|1>(i)=g<rsub|2>(i)\<Rightarrow\>g<rsub|1>=g<rsub|2>>. To
prove surjectivity let <math|k\<in\><big|prod><rsub|i\<in\>I>A<rsub|i>>
define then <math|g=k\<circ\>h<rsup|-1>> and then
<math|f(g)=k\<circ\>h<rsup|-1>\<circ\>h=k>
</proof>
<\theorem>
<label|recursive finite product>Given <math|n\<in\>\<bbb-N\><rsub|0>>
there exists a bijective function between
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>A<rsub|i>> and
<math|(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>)\<times\>A<rsub|n+1>>
</theorem>
<\proof>
Define <math|h:<big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>A<rsub|i>\<rightarrow\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n>A<rsub|i>)\<times\>A<rsub|n+1>>
by <math|h(f)=(f<rsub|\|{1,\<ldots\>,n}>,f(n+1))> then <math|h> is
injective for if <math|h(f<rsub|1>)=h(f<rsub|2>)\<Rightarrow\>f<rsub|1\|{1,\<ldots\>,n}>=f<rsub|2\|{1,\<ldots\>,n}>\<wedge\>f<rsub|1>(n+1)=f<rsub|2>(n<rsub|>+1)\<Rightarrow\>f<rsub|1>(n+1)=f<rsub|2>(n+1)\<wedge\>\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>f<rsub|1>(i)=f<rsub|2>(i)\<Rightarrow\>f<rsub|1>=f<rsub|2>>
and thus <math|h> is a injective function. Now assume that
<math|(g,a)\<in\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>)\<times\>A<rsub|n+1>\<Rightarrow\>g\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>>
and <math|a\<in\>A<rsub|n+1>> define then
<math|f:{1,\<ldots\>,n+1}\<rightarrow\>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>)\<times\>A<rsub|n+1>>
by <math|f(i)=g(i)> if <math|i\<in\>{1,\<ldots\>,n}> and
<math|f(n+1)=a\<Rightarrow\>f\<in\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>)\<times\>A<rsub|n+1>><math|>
and <math|h(f)=(g,a)>
</proof>
<\lemma>
<label|product of countable sets>Let <math|A<rsub|1>,A<rsub|2>> be
countable sets then <math|A<rsub|1>\<times\>A<rsub|2>> is countable
</lemma>
<\proof>
We have the following possibilities\
<\enumerate>
<item><math|A<rsub|1>=\<emptyset\>\<Rightarrow\>A<rsub|1>\<times\>A<rsub|2>=\<emptyset\>>
and thus finite and countable
<item><math|A<rsub|2>=\<emptyset\>\<Rightarrow\>A<rsub|1>\<times\>A<rsub|2>=\<emptyset\>>
and thus finite and countable
<item><math|A<rsub|1>,A<rsub|2>\<neq\>\<emptyset\>> then using
<reference|test for countability> there exists surjective functions
<math|f<rsub|1>:\<bbb-N\><rsub|0>\<rightarrow\>A<rsub|1>,f<rsub|2>:\<bbb-N\><rsub|0>\<rightarrow\>A<rsub|2>>
define then <math|f:\<bbb-N\><rsub|0>\<times\>\<bbb-N\><rsub|0>\<rightarrow\>A<rsub|1>\<times\>A<rsub|2>>
as <math|f(n,m)=(f<rsub|1>(n),f<rsub|2>(m))>, and let
<math|(a<rsub|1>,a<rsub|2>)\<in\>A<rsub|1>\<times\>A<rsub|2>\<Rightarrow\>a<rsub|1>\<in\>A<rsub|1>,a<rsub|2>\<in\>A<rsub|2>\<Rightarrow\>\<exists\>n<rsub|1>,n<rsub|2>\<in\>\<bbb-N\><rsub|0>\<succ\>f<rsub|1>(n<rsub|1>)=a<rsub|1>,f<rsub|2>(n<rsub|2>)=a<rsub|2>\<Rightarrow\>f(n<rsub|1>,n<rsub|2>)=(a<rsub|1>,a<rsub|2>)>
</enumerate>
</proof>
<\theorem>
<label|finite product of countable sets is countable>A finite product of
countable sets is countable
</theorem>
<\proof>
Let <math|{A<rsub|i>}<rsub|i\<in\>I>> be finite family of countable sets
(<math|I> is finite and <math|\<forall\>i\<in\>I\<vdash\>A<rsub|i>> is
countable). \ We have the following cases for <math|I>
<\enumerate>
<item><math|I=\<emptyset\>\<Rightarrow\><big|prod><rsub|i\<in\>I>A<rsub|i>=\<emptyset\>>
<item><math|I\<neq\>\<emptyset\>> then as <math|I> is finite there
exists a bijective function <math|h> from <math|{1,\<ldots\>,n}> to
<math|I> and using <reference|finite product of sets> we have a
bijection <math|g:<big|prod><rsub|i\<in\>I>A<rsub|i>\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)><rsub|>>
If we prove that <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)>>
is countable then because of the bijection <math|g> we have
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>> is countable. We prove that
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)>> is countable
by induction.
<\enumerate>
<item>If <math|n=1> Using <reference|family of one set> we have
<math|<big|prod><rsub|i\<in\>{1}>A<rsub|h(i)>> is bijective with
<math|A<rsub|h(1)>> and thus as there is a surjective function of
<math|\<bbb-N\><rsub|0>> to <math|A<rsub|a>>, we have that
<math|<big|prod><rsub|i\<in\>{1}>A<rsub|h(i)>>
<item>Now assume that the theorem is true for <math|n> and prove it
for n+1 using <reference|recursive finite product> we have a
bijective function <math|k:<big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>A<rsub|h(i)>\<rightarrow\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)><rsub|>)\<times\>A<rsub|h(n+1)>>
and as both <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)>>
and <math|A<rsub|h(n+1)>> are countable we use <reference|product of
countable sets> to prove that <with|mode|math|(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|h(i)><rsub|>)\<times\>A<rsub|h(n+1)>><math|>
is countable and because of the bijection <math|k> that
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>A<rsub|h(i)>>.
</enumerate>
</enumerate>
</proof>
<\theorem>
Let <math|X={0,1}> and <math|I> be a countable then
<math|X<rsup|I>=<big|prod><rsub|i\<in\>I>X<rsub|i> (where X<rsub|i>=X)>
is not countable
</theorem>
<\proof>
Let <math|g:\<bbb-N\><rsub|0>\<rightarrow\>X<rsup|I>> then
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> we have
<math|g(n):I\<rightarrow\><big|cup><rsub|i\<in\>I>X<rsub|i>=X> with
<math|g(n)(i)\<in\>X<rsub|i>=X> define now <math|y\<in\>X<rsup|I>> by\
<\eqnarray*>
<tformat|<table|<row|<cell|y(i)>|<cell|=0>|<cell|g(i)(i)=1>>|<row|<cell|>|<cell|=1>|<cell|g(i)(i)=0>>>>
</eqnarray*>
then if <math|\<exists\>j\<in\>\<bbb-N\><rsub|0>\<succ\>g(j)=y\<Rightarrow\>>if
<math|g(j)(j)=1\<Rightarrow\>y(j)=0\<neq\>g(j)(j)=1> or if
<math|g(j)(j)=0\<Rightarrow\>y(j)=1\<neq\>g(j)(j)=0> which in both cases
is a contradiction so we can not find a <math|j\<in\>\<bbb-N\><rsub|0>>
such that <math|g(j)=y> and thus <math|g> is not surjective.
</proof>
<\definition>
<index|sequence>Given a set <math|X> then a sequence of elements of
<math|X> is a function <math|s:\<bbb-N\><rsub|0>\<rightarrow\>X> we note
the sequence <math|s> as <math|{s<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
where <math|\<forall\>i\<in\>\<bbb-N\><rsub|0><with|mode|text| then
>s<rsub|i>\<equallim\><rsub|note>s(i)>
</definition>
<chapter|Algebra>
<section|Groups>
<\definition>
<index|semigroup>A semigroup <math|S.+> is a set <math|S> together with a
map <math|+:S\<times\>S> (where w note <math|+(x,y)=x+y> such that
<math|\<forall\>x,y,z\<in\>S> we have <math|(x+y)+z=x+(y+z)>
</definition>
<\definition>
<index|group>A group <math|S,+> is a set <math|S> together with a map
<math|+:S\<times\>S\<rightarrow\>S> where we note <math|+(x,y)> as
<math|x+y> such that the following is true\
<\enumerate>
<item><math|\<forall\>x,y,z\<in\>S> we have <math|(x+y)+z=x+(y+z)>
(associativity)
<item><math|\<exists\>0\<in\>S> such that <math|\<forall\>x\<in\>S> we
have <math|x+0=0+x=x> (<math|o> is called the neutral element)
<item><math|\<forall\>x\<in\>S> we have
<math|\<exists\>(\<um\>x)\<in\>S> such that
<math|x+(\<um\>x)=(\<um\>x)+x=0> (negative element)
</enumerate>
</definition>
<\note>
A group is thus a semigroup together with the existence of a neutral
element and negative elements
</note>
<\definition>
<index|abelian group>A group <math|S,+> is abelian if also
<math|\<forall\>x,y\<in\>S> we have <math|x+y=y+x>
</definition>
<\theorem>
<label|group of products>Let <math|{S<rsub|i>,+<rsub|i>}<rsub|i\<in\>I>>
be a family of sets <math|{S<rsub|i>}<rsub|i\<in\>I>> with a map
<math|+<rsub|i>:S<rsub|i>\<times\>S<rsub|i>\<rightarrow\>S<rsub|i>> such
that <math|S<rsub|i>,+<rsub|i>> is a group then we can define a map
<math|+> from <math|S\<times\>S\<rightarrow\>S> where
\ <math|S=<big|prod><rsub|i\<in\>I>S<rsub|i> > as follows
<math|+(x,y):I\<rightarrow\><big|cup><rsub|i\<in\>I>S<rsub|i>> where
<math|(x+y)(i)=+(x,y)(i)=+<rsub|i>(x(i),y(i))=x(i)+<rsub|i>y(i)>.
<math|S,+> is then a (abelian) group.
</theorem>
<\proof>
\;
<\enumerate>
<item><math|((x+y)+z)(i)=((x+y)(i)+<rsub|i>z(i))=((x(i)+<rsub|i>y(i))+<rsub|i>z(i))\<equallim\><rsub|+<rsub|i>
is associative>(x(i)+<rsub|i>(y(i)+<rsub|i>z(i)))=(x(i)+<rsub|i>(y+z)(i))=(x+(y+z))(i)\<Rightarrow\>((x+y)+z)=(x+(y+z))>
<item>Define <math|0:I\<rightarrow\>S<rsub|i>> by <math|0(i)=0<rsub|i>>
(the neutral element in <math|S<rsub|i>)> then
<math|(x+0)(i)=x(i)+<rsub|i>0(i)=x(i)\<Rightarrow\>x+0=x>, likewise
<math|(0+x)(i)=0(i)+<rsub|i>x(i)=x(i)\<Rightarrow\>0+x=x>
<item>Let <math|x:I\<rightarrow\>S<rsub|i>> define then
<math|\<um\>x:I\<rightarrow\>S<rsub|i>> with
<math|(\<um\>x)(i)=\<um\>x(i)> (the negative of <math|x(i)> in
<math|S<rsub|i>>) then <math|(x+(\<um\>x))(i)=x(i)+<rsub|i>(\<um\>x)(i)=x(i)+<rsub|i>(\<um\>x(i))=0<rsub|i>=0(i)\<Rightarrow\>x+(\<um\>x)=0>
and likewise we prove that <math|(\<um\>x)+x=0>
<item>If <math|S<rsub|i>> are abelian then
<math|(x+y)(i)=x(i)+<rsub|i>y(i)=y(i)+<rsub|i>x(i)=(y+x)(i)\<Rightarrow\>x+y=y+x>
</enumerate>
</proof>
<\definition>
<index|<math|<big|sum><rsub|i=1><rsup|n>s<rsub|i>>>Let <math|S,+> be a
semigroup and <math|s:{1,\<ldots\>,n}\<rightarrow\>S> be a function where
we note <math|s(i)=s<rsub|i>> then <math|<big|sum><rsub|i=1><rsup|n>s(i)\<equallim\><rsub|note><big|sum><rsub|i=1><rsup|n>s<rsub|i>>
is defined by the following recursive definition
<\enumerate>
<item><math|<big|sum><rsub|i=1><rsup|n>s(i)=s(1)> if <math|n=1>
<item><math|<big|sum><rsub|i=1><rsup|n>s(i)=(<big|sum><rsub|i=1><rsup|n-1>s(i))+s(n)>
if <math|n\<gtr\>1>
</enumerate>
</definition>
<\notation>
If <math|S,+> is a group, <math|s:{1,\<ldots\>,n}\<rightarrow\>S> a
function and <math|j:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> a
bijective function (and thus a permutation) then
<math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>j)(i)=<big|sum><rsub|i=1><rsup|n>s<rsub|i<rsub|j>>>
where we note <math|(s\<circ\>j)(i)> as <math|s<rsub|j<rsub|i>>>
</notation>
<\definition>
<index|<math|<big|sum><rsub|i=n><rsup|m>s<rsub|i>>>Let <math|S,+0> be a
semigroup and <math|s:{n,\<ldots\>m}\<rightarrow\>S,n\<leqslant\>m> a
function then we define <math|<big|sum><rsub|i=n><rsup|m>s(i)> to be
equal to <math|<big|sum><rsub|i=1><rsup|m-n+1>s(i+n-1)>
</definition>
<\theorem>
<label|sum of majorants>Let <math|\<bbb-R\>,+,.> be the real field and
<math|s,f:{1,\<ldots\>,n}\<rightarrow\>\<bbb-R\>> such that
<math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>s(i)\<less\>(\<leqslant\>)f(i)\<Rightarrow\>(<big|sum><rsub|i=1><rsup|n>s(i))\<less\>(\<leqslant\>)(<big|sum><rsub|i=1><rsup|n>f(i))>
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|<big|sum><rsub|i=1><rsup|1>s(i)=s(1)\<less\>(\<leqslant\>)f(1)=<big|sum><rsub|i=1><rsup|n>f(i)>
<item>If the theorem is true for <math|n> then
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)\<leqslant\>(<big|sum><rsub|i=1><rsup|n>f(i))+s(n+1)\<leqslant\>(<big|sum><rsub|i=1><rsup|n>f(i))+f(n+1)=<big|sum><rsub|i=1><rsup|n+1>f(i)>
</enumerate>
</proof>
<\theorem>
<label|inequality and sum>Let <math|A={1,\<ldots\>,n}> and
<math|s:A\<rightarrow\>\<bbb-R\>(\<bbb-C\><rsub|\<bbb-R\>>)> be a
function then if <math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|m\<leqslant\>s(i)> (or <math|s(i)\<leqslant\>m)> then
<math|n.m\<leqslant\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}><rsup|n>s(i)>
(or <math|<big|sum><rsub|i=1><rsup|n>s(i)\<leqslant\>n.m>)
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|<big|sum><rsub|i=1><rsup|1>s(i)=s(1)\<geqslant\>m=m.1>
(or <math|\<leqslant\>m=m.1)>
<item>Assume the theorem is true for <math|n> then
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)\<geqslant\>n.m+m>
(or <math|\<leqslant\>n.m+m)><math|=(n+1)(m)>
</enumerate>
</proof>
<\lemma>
Given a section <math|S<rsub|n+1>={1,\<ldots\>,n}>
<math|n\<in\>\<bbb-N\><rsub|0>> then given a <math|i\<in\>S<rsub|n+1>>
there exists a bijective function <math|j:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}>
such that <math|j(n)=i>
</lemma>
<\proof>
We proof this by induction
<\enumerate>
<item>If <math|n=1> then <math|S<rsub|n+1>=S<rsub|2>={1}> and if
<math|i\<in\>S<rsub|n+1>\<Rightarrow\>i=1> and thus <math|i<rsub|{1}>>
the identity function fulfills the lemma
<item>Assume that the theorem is true for <math|n> and prove it for
<math|n+1> now we have the following cases for
<math|i\<in\>S<rsub|(n+1)+1>=S<rsub|n+1><big|cup>{n+1}={1,\<ldots\>,n}<big|cup>{n+1}>
<\enumerate>
<item><math|i=n+1\<Rightarrow\>i<rsub|{1,\<ldots\>,n+1}>>(the
identity function) is the desired bijective function
<item><math|i\<neq\>n+1\<Rightarrow\>i\<in\>S<rsub|n+1>\<Rightarrowlim\><rsub|using
the induction hypothesis>\<exists\>j:S<rsub|n+1>\<rightarrow\>S<rsub|n+1>>
such that <math|j> is a bijective function and <math|j(n)=i> let then
we define the function <math|h> by\
<\eqnarray*>
<tformat|<table|<row|<cell|k(m)=>|<cell|j(m)>|<cell|m\<in\>S<rsub|n+1><mid|\\>{n}>>|<row|<cell|>|<cell|n+1>|<cell|m=n>>|<row|<cell|>|<cell|i>|<cell|m=n+1>>>>
</eqnarray*>
then if <math|k(m<rsub|1>)=k(m<rsub|2>)> we have either\
<\enumerate>
<item><math|m<rsub|1>\<in\>S<rsub|n+1><mid|\\>{n}> then we can not
have <math|k(m<rsub|1>)=j(m<rsub|1>)=i> (otherwise
<math|i=j(n)=j(m<rsub|1>)> and then
<math|m<rsub|1>=n\<nin\>S<rsub|n+1><mid|\\>{n}> a contradiction),
also as <math|k(m<rsub|1>)=j(i)\<neq\>n+1> (<math|j> bijection of
<math|S<rsub|n+1>={1,\<ldots\>,n}\<rightarrow\>S<rsub|n+1>={1,\<ldots\>,n}>)
and thus <math|m<rsub|2>\<neq\>n,n+1> so
<math|m<rsub|2>\<in\>S<rsub|n+1>\<Rightarrow\>k(m<rsub|2>)=j(m<rsub|2>)\<Rightarrow\>j(m<rsub|1>)=j(m<rsub|2>)\<Rightarrow\>m<rsub|1>=m<rsub|2>>
<item><math|m<rsub|1>=n\<Rightarrow\>k(n)=n+1> then
<math|m<rsub|2>\<neq\>n+1> (otherwise
<math|k(m<rsub|2>)=i\<neq\>n+1>) also
<math|m<rsub|2>\<in\>S<rsub|n+1><mid|\\>{n}> (otherwise
<math|k(m<rsub|2>)=j(m)\<in\>S<rsub|n>\<nni\>n+1> contradicting
<math|n+1=k(n)=k(m<rsub|1>)=k(m<rsub|2>)>), so the only possibility
left is <math|m<rsub|2>=n\<Rightarrow\>m<rsub|1>=m<rsub|2>>
<item><math|m<rsub|1>=n+1\<Rightarrow\>k(m<rsub|1>)=i> then we
cannot have <math|m<rsub|2>=n> (otherwise
<math|k(m<rsub|2>)=n+1\<neq\>i=k(m<rsub|1>)>), also we can not have
<math|m<rsub|2>\<in\>S<rsub|n+1><mid|\\>{n}> (in this case as
\ <math|j(m<rsub|2>)=k(m<rsub|2>)=i=j(n)\<Rightarrow\>m<rsub|2>=n\<Rightarrow\>m<rsub|2>\<nin\>S<rsub|n+1><mid|\\>{n}>
a contradiction), so the only choice left is
<math|m<rsub|2>=n+1\<Rightarrow\>m<rsub|1>=m<rsub|2>>\
</enumerate>
so we must conclude that <math|k> is a injective function as
required. To prove surjectivity let
<math|y\<in\>S<rsub|n+2>={1,\<ldots\>,n+1}> then we have the
following cases
<\enumerate>
<item><math|y=n+1\<Rightarrow\>k(n)=n+1>
<item><math|y=n\<Rightarrow\>k(n+1)=n>
<item><math|y\<in\>S<rsub|n+1><mid|\\>{n}=S<rsub|n>> and
<math|y\<in\>S<rsub|n>\<Rightarrow\>S<rsub|n>\<neq\>\<emptyset\>\<Rightarrowlim\><rsub|j
is surjective>\<exists\>x\<in\>S<rsub|n>\<succ\>k(x)=j(x)=y>
</enumerate>
\ So <math|k> is surjective and injective and thus a bijective
function and by construction we have <math|k(n+1)=i> as required
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|sum and switch>If <math|S,+> is a abelian semigroup,
<math|s:{1,\<ldots\>,n}\<rightarrow\>S> then given
<math|m\<in\>{1,\<ldots\>,n}> there exists a bijective function
<math|h:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> such that
<math|h(n)=m> and <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)>
</theorem>
<\proof>
First of all if <math|m=n\<Rightarrow\>i<rsub|{1,\<ldots\>,n}>> has
<math|i<rsub|{1,\<ldots\>,n}>(n)=n> and
<math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>i<rsub|{1,\<ldots\>,n}>)(i)>
so <math|i<rsub|{1,\<ldots\>,n}>> fulfills the condition in this case.
Now assume that <math|m\<neq\>n> We proof this \ by using induction so
<\enumerate>
<item>if <math|n=1> then <math|m\<in\>{1}\<Rightarrow\>m=1> and
<math|h:{1}\<rightarrow\>{1}> defined by <math|h(1)=1=m> and
<math|<big|sum><rsub|i=1><rsup|1>s(i)=s(1)=(s\<circ\>h)(1)=<big|sum><rsub|i=1><rsup|1>(s\<circ\>h)(i)>
and the theorem is true for <math|n=1>\
<item>If <math|n=2> then as <math|m\<neq\>n=2> we must have <math|m=1>
then define <math|h:{1,2}\<rightarrow\>{1,2}> by <math|h(1)=2,
h(2)=1=m> and <math|<big|sum><rsub|i=1><rsup|2>s(i)=(<big|sum><rsub|i=1><rsup|1>s(i))+s(2)=s(1)+s(2)\<equallim\><rsub|commutativity>s(2)+s(1)=(s\<circ\>h)(1)+(s\<circ\>h)(2)=(<big|sum><rsub|i=1><rsup|1>(s\<circ\>h)(i))+(s\<circ\>h)(2)=<big|sum><rsub|i=1><rsup|2>(s\<circ\>h)(i)>
and the theorem is true for <math|n=2>
<item>Now assume the theorem is true for <math|n> and proof it for
<math|n+1> if <math|n+1=1,2> then using <math|(1) or (2)> we have
proved our theorem so assume that <math|n+1\<gtr\>2> and thus
<math|n\<gtr\>1> then as we assume <math|m\<neq\>n+1> we must have
<math|m\<in\>{1,\<ldots\>,n}> now <math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)=(1)>
by the induction hypothesis we have a bijective function
<math|g:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> such that
<math|g(n)=m> and <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>g)(i)>
then <math|(1)=(<big|sum><rsub|i=1><rsup|n>(s\<circ\>g)(i))+s(n+1)\<equallim\><rsub|n\<gtr\>1>((<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>g)(i))+s(g(n)))+s(n+1)\<equallim\><rsub|associativity>(<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>g)(i))+(s(m)+s(n+1)\<equallim\><rsub|commutativity>(<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>g)(i))+(s(n+1)+s(m)=(2)>
now define <math|h:{1,\<ldots\>,n+1}\<rightarrow\>{1,\<ldots\>,n+1}> by\
<\enumerate>
<item><math|h(i)=g(i)> if <math|i\<in\>{1,\<ldots\>,n-1}>
<item><math|h(n)=n+1>
<item><math|h(n+1)=m>
</enumerate>
then using the fact that <math|g(i)\<neq\>m> if
<math|i\<in\>{1,\<ldots\>,n-1)> (because <math|g(n)=m> and g is
injective) and <reference|mapping extension>,<reference|bijection
extension> we have that <math|h> is a bijective function. And
<math|(2)=(<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>h)(i))+((s\<circ\>h)(n)+(s\<circ\>h)(n+1))=(<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>h)(i)+(s\<circ\>h)(n))+(s\<circ\>h)(n+1)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)+(s\<circ\>h)(n+1)=<big|sum><rsub|i=1><rsup|n+1>(s\<circ\>h)(i)>
which proves the theorem for <math|n+1>
</enumerate>
\;
</proof>
<\theorem>
<label|commutativity sum>If <math|S,+> is a abelian semigroup,
<math|s:{1,\<ldots\>,n}\<rightarrow\>S> then given a bijective function
<math|h:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> (a permutation)
then <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)>
</theorem>
<\proof>
We do this by induction
<\enumerate>
<item>If <math|n=1> then the only permutation is the identity map and
the result is trivial
<item>Assume now that the theorem is true for <math|n> and prove it for
<math|n+1> as <math|h> is a bijection
<math|\<exists\>k\<in\>{1,\<ldots\>,n+1}> such that <math|h(k)=n+1>
then applying the previous theorem (<reference|sum and switch>) on
<math|s<rprime|'>=s\<circ\>h> we find a bijective function
<math|g:{1,\<ldots\>,n+1}\<rightarrow\>{1,\<ldots\>,n+1}> such that
<math|g(n+1)=k> and <math|<big|sum><rsub|i=1><rsup|n+1>(s\<circ\>h)(i)=<big|sum><rsub|i=1><rsup|n+1>(s\<circ\>h\<circ\>g)(i)>
\ define now <math|l=(h\<circ\>g)<rsub|\|{1,\<ldots\>,n}>> then
<math|l> is a bijective function [It is injective because <math|h,g>
are bijective and thus <math|h\<circ\>g> is bijective and thus
injective, also if <math|j\<in\>{1,\<ldots\>,n}\<Rightarrowlim\><rsub|h\<circ\>g
is bijective>\<exists\>i\<in\>{1,\<ldots\>,n+1}\<succ\>h\<circ\>g(i)=j>
and if <math|i=n+1\<Rightarrow\>j=h\<circ\>g(i)=h(g(n+1))=h(k)=n+1<rsub|>>
contradicting the fact that <math|j\<in\>{1,\<ldots\>,n}> so
<math|i\<in\>{1,\<ldots\>,n}\<Rightarrow\>l(i)=h\<circ\>g<rsub|\|{1,\<ldots\>,n}>(i)=j>
and <math|l> is surjective]. Now <math|<big|sum><rsub|i=1><rsup|n+1>(s\<circ\>h)(i)=<big|sum><rsub|i=1><rsup|n+1>(s\<circ\>h\<circ\>g)(i)=(<big|sum><rsub|i=1><rsup|n>(s\<circ\>h\<circ\>g)(i))+s(h(g(n+1)))=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h\<circ\>g)(i)+s(n+1)\<equallim\><big|sum><rsub|i=1><rsup|n>(s\<circ\>l)(i)+s(n+1)\<equallim\><rsub|apply
the induction hypothese on the bijective mapping
l><big|sum><rsub|i=1><rsup|n>s(i)+s(n+1)=<big|sum><rsub|i=1><rsup|n+1>s(i)>
proving that the theorem is true for <math|n+1>
</enumerate>
</proof>
<\lemma>
<label|alternative definition of fnite sum>Let <math|S,+> be a abelian
semigroup <math|n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1> and
<math|s{1,\<ldots\>,n}\<rightarrow\>S> be a given bijection then
<math|<big|sum><rsub|i=1><rsup|n>s(i)=s(1)+<big|sum><rsub|i=2><rsup|n>s(i)>
(where <math|<big|sum><rsub|i=2><rsup|n>s(i)\<equallim\><rsub|definition><big|sum><rsub|i=1><rsup|n-1>(s\<circ\>k)(i)>
where <math|k:{1,\<ldots\>,n-1}\<rightarrow\>{2,\<ldots\>,n}> is defined
by <math|k(i)=i+1>)<math|>
</lemma>
<\proof>
Define the bijective function <math|h:{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}>
defined by\
<\enumerate>
<item><math|h(i)=i+1> if <math|1\<leqslant\>i\<less\>n-1>
<item><math|h(n)=1>
</enumerate>
\ which is bijective because of <reference|mapping extension> and
<reference|bijection extension> then <math|h<rsub|\|{1,\<ldots\>,n-1}>=k>
now <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)=<big|sum><rsub|i=1><rsup|n-1>(s\<circ\>k)(i)+s(h(n))=<big|sum><rsub|i=2><rsup|n>s(i)+s(1)=s(1)+<big|sum><rsub|i=2><rsup|n>s(i)><math|>
</proof>
<\theorem>
<label|sum of differences>Let <math|S,+> be a abelian group and
<math|x:{1,\<ldots\>n}\<rightarrow\>S,n\<geqslant\>2> a function then
<math|x<rsub|n>-x<rsub|1>=<big|sum><rsub|i=1><rsup|n-1>(x<rsub|i+1>-x<rsub|i>)>
</theorem>
<\proof>
We prove this by induction
<\enumerate>
<item><math|n=2> then <math|x<rsub|2>-x<rsub|1>=<big|sum><rsub|i=1><rsup|1(=2-1)>(x<rsub|i+1>-x<rsub|i>)>
<item>Assume the theorem true for <math|n> and prove it for
<math|n<rsub|+1>>. Then <math|x<rsub|n>-x<rsub|1>=<big|sum><rsub|i=1><rsup|n-1>(x<rsub|i+1>-x<rsub|i>)\<Rightarrow\>x<rsub|n+1>-x<rsub|1>=(x<rsub|n+1>-x<rsub|n>)+(x<rsub|n>-x<rsub|1>)=(x<rsub|n+1>-x<rsub|n>)+<big|sum><rsub|i=1><rsup|n-1>(x<rsub|i+1>-x<rsub|i>)=<big|sum><rsub|i=1><rsup|n>(x<rsub|i+1>-x<rsub|i>)>
</enumerate>
</proof>
<\definition>
<index|<math|<big|sum><rsub|a\<in\>A>a>>Let <math|S,+> be a abelian group
and <math|A\<subseteq\>S> be a finite set then
<math|<big|sum><rsub|a\<in\>A>a> is defined by
<math|<big|sum><rsub|i=1><rsup|n>s(i)> where
<math|s:{1,\<ldots\>,n}\<rightarrow\>A> is a bijection
</definition>
<\proof>
This definition only makes sense if such a bijection exists, <math|n> is
unique and the sum does not depends on the chosen definition. The first
follows from the definition of a finite set (see <reference|finite sets>)
and the uniqueness of cardinality of finite sets (see
<reference|cardinality of finite sets>). Finally if
<math|t:{1,\<ldots\>,n}\<rightarrow\>A> is another bijection then
<math|r=s<rsup|-1>\<circ\>t> is a bijection
<math|{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> and
<math|s\<circ\>r=s\<circ\>s<rsup|-1>\<circ\>t=t> and using
<math|><reference|commutativity sum> we have
<math|<big|sum><rsub|i=1><rsup|n>t(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>r)(i)\<equallim\><rsub|<reference|commutativity
sum>><big|sum><rsub|i=1><rsup|n>s(i)>
</proof>
<\theorem>
<label|sum of partitions>Let <math|S,+> be a abelian group,
<math|><math|A\<subseteq\>S> a finite set and
<math|{A<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,m}>> a collection of non empty
sets such that <math|A=<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>A<rsub|i>>
and <math|A<rsub|i><big|cap>A<rsub|j>=\<emptyset\>> (if <math|i\<neq\>j>)
then <math|<big|sum><rsub|a\<in\>A>a=<big|sum><rsub|i=1><rsup|m>(<big|sum><rsub|b\<in\>A<rsub|i>>b)>
</theorem>
<\proof>
We prove this by induction on <math|m>\
<\enumerate>
<item><math|m=1> then <math|A=<big|cup><rsub|i\<in\>{1,,1}>A<rsub|i>=A<rsub|1>>
and <math|<big|sum><rsub|a\<in\>A>a=<big|sum><rsub|i=1><rsup|1>(<big|sum><rsub|\<alpha\>\<in\>A>a)=<big|sum><rsub|i=1><rsup|1>(<big|sum><rsub|\<alpha\>\<in\>A<rsub|1>>a)>
<item><math|m=2> then <math|A=A<rsub|1><big|cup>A<rsub|2>>, \ now as
<math|A> is finite we have that <math|A<rsub|2>> is also finite and
thus a isomorphism <math|s:{1,\<ldots\>,n}\<rightarrow\>A<rsub|2>>
exists , we prove now (2) by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|A<rsub|2>={s<rsub|1>}> then as <math|A>
is finite we have a bijection <math|t:{1,\<ldots\>,k}\<rightarrow\>A>,
so that <math|\<exists\>m\<in\>{1,\<ldots\>,k}> such that
<math|t(m)=s<rsub|1>> and using <reference|sum and switch> there
exists a bijection <math|p:{1,\<ldots\>,k}\<rightarrow\>{1,\<ldots\>,k}>
such that <math|p(k)=m> and thus for the bijection
<math|h=t\<circ\>p:{1,\<ldots\>,k}\<rightarrow\>A> we have
<math|h(k)=t(p(k))=t(m)=s<rsub|1>> and thus
<math|<big|sum><rsub|a\<in\>A>a=<big|sum><rsub|i=1><rsup|k>h(i)=<big|sum><rsub|i=1><rsup|k-1>h(i)+h(k)\<equallim\><rsub|h<rsub|\|{1,\<ldots\>,k-1}>
:{1,\<ldots\>,k-1}\<rightarrow\>A<rsub|1> is a
bijection>(<big|sum><rsub|\<alpha\>\<in\>A<rsub|1>>a)+h(k)=(<big|sum><rsub|\<alpha\>\<in\>A<rsub|1>>a)+<big|sum><rsub|i=1><rsup|1>s<rsub|1>=(<big|sum><rsub|\<alpha\>\<in\>A<rsub|1>>a)+(<big|sum><rsub|a\<in\>A<rsub|2>>a)=<big|sum><rsub|i=1><rsup|2>(<big|sum><rsub|b\<in\>A<rsub|i>>b)>
<item>Assume the hypothesis is valid for <math|n> then prove it for
<math|n+1>. So <math|s:{1,\<ldots\>,n+1}\<rightarrow\>A<rsub|2>> is a
bijection then <math|<big|sum><rsub|a\<in\>A<rsub|2>>a=<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)=(<big|sum><rsub|a\<in\>A<rsub|2><mid|\\>{s(n+1)}>a)+s(n+1)>
and using (1) on <math|A=(A<mid|\\>{s(n+1)})<big|cup>{s<rsub|n+1>}\<equallim\><rsub|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>\<Rightarrow\>s(n+1)\<nin\>A<rsub|1>>((A<rsub|1><big|cup>(A<rsub|2><mid|\\>{s(n+1)}))<big|cup>{s(n+1)}\<Rightarrowlim\><rsub|(2
a)><big|sum><rsub|a\<in\>A>a=(<big|sum><rsub|a\<in\>A<mid|\\>{s(n+1)}>a)+<big|sum><rsub|a\<in\>{s(n+1)}>a=(<big|sum><rsub|a\<in\>A<mid|\\>{s(n+1)}>a)+s(n+1)\<equallim\><rsub|induction
hypothese on A<rsub|2><mid|\\>{s(n+1)} with cardinality
n>((<big|sum><rsub|a\<in\>A<rsub|1>>a)+(<big|sum><rsub|a\<in\>A<rsub|2><mid|\\>{s(n+1)}>a))+s(n+1)=(<big|sum><rsub|a\<in\>A<rsub|1>>a)+(<big|sum><rsub|a\<in\>A<rsub|2><mid|\\>{s(n+1)}>+s(n+1)=<big|sum><rsub|a\<in\>A<rsub|1>>a)+(<big|sum><rsub|a\<in\>A<rsub|2>>a)=<big|sum><rsub|i=1><rsup|2>(<big|sum><rsub|a\<in\>A<rsub|i>>a))
>
</enumerate>
<item>Assume the trueness for <math|m> then prove it for <math|m+1>.
<math|<big|cup><rsub|i\<in\>{1,\<ldots\>,m+1}>A<rsub|i>=(<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>A<rsub|i>)<big|cup>A<rsub|m+1>>
and <math|A<rsub|m+1><big|cap>(<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>A<rsub|i>)=<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>(A<rsub|m+1><rsup|><big|cap>A<rsub|i>)=\<emptyset\>>
and we can then use <math|(2)> to get
<math|<big|sum><rsub|a\<in\>A>a=(<big|sum><rsub|a\<in\><big|cup><rsub|i\<in\>{1,\<ldots\>,m}>A<rsub|i>)>a)+(<big|sum><rsub|a\<in\>A<rsub|m+1>>a)\<equallim\><rsub|induction
hypothese>(<big|sum><rsub|i=1><rsup|m>(<big|sum><rsub|a\<in\>A<rsub|i>>a))+(<big|sum><rsub|a\<in\>A<rsub|m+1>>a)=<big|sum><rsub|i=1><rsup|m+1>(<big|sum><rsub|a\<in\>A<rsub|i>>a)>
\
</enumerate>
</proof>
<\lemma>
<label|split of a sum>Let <math|S,+> be a abelian group
<math|s:{1,\<ldots\>,n}\<rightarrow\>A> a function and
<math|m\<in\>{1,\<ldots\>,n-1}> then <math|<big|sum><rsub|i=1><rsup|n>s(i)=(<big|sum><rsub|i=1><rsup|m>s(i))+<big|sum><rsub|i=m+1><rsup|n>s(i)>
</lemma>
<\proof>
We proof this by induction on <math|m>
<\enumerate>
<item>If <math|m=1\<in\>{1,\<ldots\>,n-1}> then
<math|(<big|sum><rsub|i=1><rsup|1>s(i))+<big|sum><rsub|i=2><rsup|n>s(i)=s(1)+<big|sum><rsub|i=2><rsup|n>s(i)\<equallim\><rsub|<reference|alternative
definition of fnite sum>><big|sum><rsub|i=1><rsup|n>s(i)>
<item>Assume the theorem is true for all
<math|m\<in\>{1,\<ldots\>,n-1}> and take
<math|m+1\<in\>{1,\<ldots\>,n-1}> then <math|m\<in\>{1,\<ldots\>,n-1)>
and applying the induction hypothesis we find
<math|(<big|sum><rsub|i=1><rsup|n>s(i)=(<big|sum><rsub|i=1><rsup|m>s(i))+(<big|sum><rsub|i=m+1><rsup|n>s(i))=(<big|sum><rsub|i=1><rsup|m>s(i))+(<big|sum><rsub|j=1><rsup|n-(m+1)+1>s(j+(m+1)-1))=(<big|sum><rsub|i=1><rsup|m>s(i))+(<big|sum><rsub|j=1><rsup|n-m>s(j+m))\<equallim\><rsub|<reference|alternative
definition of fnite sum>>(<big|sum><rsub|i=1><rsup|m>s(i))+s(m+1)+<big|sum><rsub|j=2><rsup|n-m>s(j+m)=(<big|sum><rsub|i=1><rsup|m+1>s(i))+<big|sum><rsub|k=m+2><rsup|n>s(k)>
proving the theorem for <math|m+1>
</enumerate>
</proof>
<\theorem>
<label|partioned sum>Let <math|S,+> be a abelian group
<math|s{1,\<ldots\>,n}\<rightarrow\>A> a function and
<math|{1,\<ldots\>,n}=<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>A<rsub|i>>
where <math|A<rsub|i>={a<rsub|i>,\<ldots\>,b<rsub|i>}> and
<math|\<forall\>i,j\<in\>{1,\<ldots\>,m}\<vdash\>i\<neq\>j> we have
<math|A<rsub|i><big|cap>A<rsub|j>=\<emptyset\>> then
<math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|j=1><rsup|m>(<big|sum><rsub|k=a<rsub|j>><rsup|b<rsub|j>>s(k))>
</theorem>
<\proof>
We proof this by induction on <math|m>\
<\enumerate>
<item><math|m=1>. In this case <math|A<rsub|1>={1,\<ldots\>,n}\<Rightarrow\>a<rsub|1>=1,b<rsub|1>=n>
and <math|<big|sum><rsub|j=1><rsup|1>(<big|sum><rsub|k=a<rsub|j>><rsup|b<rsub|j>>s(k))=<big|sum><rsub|k=a<rsub|1>><rsup|b<rsub|1>>s(k)=<big|sum><rsub|k=1><rsup|n>s(k)<rsub|>>
<item>Assume that it is true for <math|m> prove it for <math|m+1 (so
m\<geqslant\>2)>. First define <math|u:{1,\<ldots\>,m+1}\<rightarrow\>S>
by <math|u(i)=<big|sum><rsub|k=a<rsub|i>><rsup|b<rsub|i>>s(k)> then
<math|<big|sum><rsub|j=1><rsup|m+1>(<big|sum><rsub|k=a<rsub|j>><rsup|b<rsub|j>>s(k))=<big|sum><rsub|j=1><rsup|m+1>u(j)>.
Now as <math|n\<in\>{1,\<ldots\>,n}=<big|cup><rsub|i\<in\>{1,\<ldots\>,m+1}>A<rsub|i>>
there exists a <math|l\<in\>{1,\<ldots\>,m+1}> such that
<math|n\<in\>{a<rsub|l>,\<ldots\>,b<rsub|l>}> now as
<math|{a<rsub|l>,\<ldots\>,b<rsub|l>}=A<rsub|l>\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,m+1}>A<rsub|i>={1,\<ldots\>,n}>
we must have that <math|a<rsub|l>\<leqslant\>n\<leqslant\>b<rsub|l>\<leqslant\>n\<Rightarrow\>b<rsub|l>=n>.
Now define the bijective function <math|h:{1,\<ldots\>,m+1}\<rightarrow\>{1,\<ldots\>,m+1}>
by\
<\eqnarray*>
<tformat|<table|<row|<cell|h(i)=>|<cell|i>|<cell|i\<neq\>l,m+1>>|<row|<cell|>|<cell|m+1>|<cell|i=l>>|<row|<cell|>|<cell|l>|<cell|i=m+1>>>>
</eqnarray*>
then using <reference|commutativity sum> we have that
<math|<big|sum><rsub|j=1><rsup|m+1>u(i)=<big|sum><rsub|j=1><rsup|m+1>(u\<circ\>h)(i)>
or if we define <math|v=u\<circ\>h>
<math|<big|sum><rsub|j=1><rsup|m+1>u(i)=<big|sum><rsub|j=1><rsup|m+1>v(j)>
where <math|v(m+1)=u(h(m+1))=u(l)=<big|sum><rsub|k=a<rsub|l><rsup|>><rsup|b<rsub|l>>s(k)=<big|sum><rsub|k=a<rsub|l>><rsup|n>s(k)>.
Define now also <math|B<rsub|i>=A<rsub|h(i)>={a<rsub|h(i)>,\<ldots\>,b<rsub|h(i)>}>
then <math|B<rsub|m+1>=A<rsub|l>={a<rsub|l>,\<ldots\>,n}> (where
<math|a<rsub|l>\<gtr\>1> because <math|m\<geqslant\>2)> and
<math|<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>B<rsub|i>=<big|cup><rsub|i\<in\>{1,\<ldots\>,m+1}<mid|\\>{m+1}>B<rsub|i>\<equallim\><rsub|<reference|union
of disjoint sets and difference and
exclusion>>(<big|cup><rsub|i\<in\>{1,\<ldots\>,m+1}>B<rsub|i>)<mid|\\>B<rsub|m+1>={1,\<ldots\>,n+1}<mid|\\>B<rsub|m+1>={1,\<ldots\>,a<rsub|l>-1}>
applying the the induction hypothesis on
<math|{1,\<ldots\>,a<rsub|l>-1}=<big|cup><rsub|i\<in\>{1,\<ldots\>,m}>B<rsub|i>>
we find that <math|<big|sum><rsub|i=1><rsup|a<rsub|l>-1>s(i)=<big|sum><rsub|j=1><rsup|m>(<big|sum><rsub|k=a<rsub|h(i)>><rsup|b<rsub|h(i)>>s(k))=<big|sum><rsub|j=1><rsup|m>u(h(i))=<big|sum><rsub|j=1><rsup|m>v(i)<rsub|>>
and then using the previous lemma on
<math|a<rsub|l>-1\<in\>{1,\<ldots\>,n-1}> we have
<math|<big|sum><rsub|i=1><rsup|n>s(i)=(<big|sum><rsub|i=1><rsup|a<rsub|l>-1>s(i))+<big|sum><rsub|i=a<rsub|l>><rsup|n>s(l)=(<big|sum><rsub|j=1><rsup|m>v(i))+v(m+1)=<big|sum><rsub|j=1><rsup|m+1>v(i)=<big|sum><rsub|j=1><rsup|m+1>u(i)=<big|sum><rsub|j=1><rsup|m+1>(<big|sum><rsub|k=a<rsub|j>><rsup|b<rsub|j>>s(k))>
which proves our theorem
\
</enumerate>
\;
</proof>
<\definition>
<index|support>Let <math|S,+> be a group and <math|s:A\<rightarrow\>S> be
a function from <math|A> to <math|S> then we define the support of
<math|s> to be <math|support(s)={a\<in\>A\|s(a)\<neq\>0}\<subseteq\>A>
(so if <math|A> is finite <math|support(s)> is finite)
</definition>
<\lemma>
<label|all zeroes in sum>Let <math|S,+> be a group and
<math|s:{1,\<ldots\>,n}\<rightarrow\>S> a function. Then if
<math|support(s)=\<emptyset\>> we have
<math|<big|sum><rsub|i=1><rsup|n>s(i)=0> and if
<math|support(s)\<neq\>\<emptyset\>> we have
<math|\<exists\>m\<in\>\<bbb-N\><rsub|0>> and a bijective function
<math|h:{1,\<ldots\>,m}\<rightarrow\>support(s)> such that
<math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)>
(Note that because of <reference|cardinality of finite sets> and
<reference|commutativity sum> we have that the <math|m> is unique and
that if we had another bijection <math|k:{1,\<ldots\>,n}\<rightarrow\>S>
then <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|m>(s\<circ\>k)(i)>)
</lemma>
<\proof>
First if <math|support(s)=\<emptyset\>\<Rightarrow\>\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>s(i)=0>,
then we prove that <math|<big|sum><rsub|i=1><rsup|n>s(i)=0>, we do this
by induction:
<\enumerate>
<item>If <math|n=1> then <math|<big|sum><rsub|i=1><rsup|1>s(i)=s(o)=0>
<item>Assume that it is true for <math|n> then
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+(n+1)=0+0=0>
</enumerate>
Now for the case <math|support(s)\<neq\>\<emptyset\>> we use induction on
<math|n>
<\enumerate>
<item>If <math|n=1> then <math|support(s)={1}> and
<math|h:{1}\<rightarrow\>support(s)={1}> is such that <math|h(1)=1>
(the identity function on <math|{1}> which is bijective) and we have
<math|<big|sum><rsub|i=1><rsup|1>s(i)=s(1)=(s\<circ\>h)(1)=<big|sum><rsub|i=1><rsup|1>(s\<circ\>h)(i)>
<item>Now assume the theorem is true for <math|n> and prove it for
<math|n+1>. Now <math|(1)=<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)=(<big|sum><rsub|i=1><rsup|n>s<rsub|\|{1,\<ldots\>,n}>(i))+s(n+1)>
now by using the induction hypothesis on
<math|s<rsub|\|{1,\<ldots\>,}>> let
<math|support(s<rsub|\|{1,\<ldots\>,n}>)={i\<in\>{1,\<ldots\>,n}\|s<rsub|\|{1,\<ldots\>,n}>\<neq\>0}\<Rightarrowlim\><rsub|induction
hypothese>>we have the following possibilities
<\enumerate>
<item><math|support(s<rsub|\|{1,\<ldots\>,n}>)=\<emptyset\>> and
<math|<big|sum><rsub|i=1><rsup|n>s<rsub|\|{1,\<ldots\>,n}>(i)=<big|sum><rsub|i=1><rsup|n>s(i)=0>
, we can have then the following two possibilities for <math|s(n+1)>
<\enumerate>
<item><math|s(n+1)=0> then <math|support(s)=support(s<rsub|\|{1,\<ldots\>,n}>)=\<emptyset\>>
and we have as we have already proved
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)=0>
<item><math|s(n+1)\<neq\>0> then <math|support(s)={n+1}> take then
<math|m=1> and <math|h={1}\<rightarrow\>{n+1}> then
<math|><with|mode|math|<big|sum><rsub|i=1><rsup|n+1>s(i)\<equallim\><rsub|(1)>0+s(n+1)=(s\<circ\>h)(1)=<big|sum><rsub|i=1><rsup|1>(s\<circ\>h)(i)>
</enumerate>
<item><math|support(s<rsub|\|{1,\<ldots\>,n}>)\<neq\>\<emptyset\>>
and <math|\<exists\>m<rprime|'>\<in\>\<bbb-N\><rsub|0>> and a
bijective function <math|h<rprime|'>:{1,\<ldots\>,m<rprime|'>}\<rightarrow\>support(s<rsub|\|{1,\<ldots\>,n}>)>
such that <math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i=1><rsup|n>s<rsub|\|{1,\<ldots\>,n}>(i)=<big|sum><rsub|i=1><rsup|m<rprime|'>>(s<rsub|\|{1,\<ldots\>,n}>\<circ\>h<rprime|'>)(i)=<big|sum><rsub|i=1><rsup|m<rprime|'>>(s\<circ\>h<rprime|'>)(i)><math|><math|>,
we can have the following two possibility's for <math|s(n+1)>
<\enumerate>
<item><math|s(n+1)=0> then <math|support(s)=support(s<rsub|\|{1,\<ldots\>,n}>)>
and <math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)=<big|sum><rsub|i=1><rsup|n>s(i)\<equallim\><big|sum><rsub|i=1><rsup|m<rprime|'>>(s\<circ\>h<rprime|'>)>
<item><math|s(n+1)\<neq\>0> then
<math|support(s)=support(s<rsub|\|{1,\<ldots\>,n}>)<big|cup>{n+1}>
define then <math|h:{1,\<ldots\>,m<rprime|'>+1}\<rightarrow\>support(s)>
by\
<\enumerate>
<item><math|h(j)=h<rprime|'>(j)> if
<math|j\<in\>{1,\<ldots\>,m<rprime|'>}>
<item><math|h(m<rprime|'>+1)=n+1>
</enumerate>
which by <reference|mapping extension> and <reference|bijection
extension> is a bijective function and
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)\<equallim\><rsub|(1)>(<big|sum><rsub|i=1><rsup|m<rprime|'>>(s\<circ\>h<rprime|'>)(i))+s(n+1)=(<big|sum><rsub|i=1><rsup|m<rprime|'>>(s\<circ\>h)(i))+(s\<circ\>h)(m<rprime|'>+1)=<big|sum><rsub|i=1><rsup|m<rprime|'>+1>(s\<circ\>h)(i)>
</enumerate>
</enumerate>
</enumerate>
</proof>
<\definition>
<label|general sum><index|<math|<big|sum><rsub|a\<in\>A>a>>Let <math|S,+>
be a abelian group, <math|A> a set and let <math|s:A\<rightarrow\>S> be
such that <math|support(A)={a\<in\>A\|s(a)\<neq\>0}> is finite. If
<math|support(s)> is non empty then <math|\<exists\>n\<in\>\<exists\>h>,
h is a bijective function <math|h:{1,\<ldots\>,n}\<rightarrow\>support(s)>
we define then <math|<big|sum><rsub|a\<in\>A>s(a)=<rsub|definition><big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)>
and if <math|support(s)=\<emptyset\>> then
<math|<big|sum><rsub|a\<in\>A><rsup|>s(a)=0><math|>
</definition>
<\proof>
To ensure that the definition is unique, first note that by
<reference|cardinality of finite sets> there is only one
<math|n\<in\>\<bbb-N\><rsub|0>> such that <math|{1,\<ldots\>,n}> is
bijective with <math|support(s)>. Also assume that there is another
bijection <math|k:{1,\<ldots\>,n}\<rightarrow\>support(s)> then
<math|k<rsup|-1>\<circ\>h> is a bijective function from
<math|{1,\<ldots\>,n}\<rightarrow\>{1,\<ldots\>,n}> and using
<reference|commutativity sum> we get <math|<big|sum><rsub|i=1><rsup|n>(s\<circ\>k)(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>k\<circ\>k<rsup|-1>\<circ\>h)(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i)>
so our definition is independent of our choice of bijection <math|h>
</proof>
<\lemma>
<label|zeroes in general sums>Let <math|S,+> be a abelian group, <math|A>
a set and let <math|s:A\<rightarrow\>S> be such that <math|support(s)> is
finite then if <math|\<cal-W\>\<supseteq\>support(s)> is finite and non
empty we have for every bijective function
<math|h:{1,\<ldots\>,m}\<rightarrow\>\<cal-W\>> (which exists because
<math|\<cal-W\>> is finite and non empty) that
<math|<big|sum><rsub|a\<in\>A>s(a)=<big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)>
</lemma>
<\proof>
We have the following cases for <math|support(s)>
<\enumerate>
<item><math|support(s)=\<emptyset\>> then
<math|support(s\<circ\>h)=\<emptyset\>\<Rightarrowlim\><rsub|<reference|all
zeroes in sum>><big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)=0\<equallim\><rsub|<reference|general
sum> ><big|sum><rsub|a\<in\>A>s(a)>
<item><math|support(s)\<neq\>\<emptyset\>> and
<math|\<cal-W\>=support(s)> then <math|<big|sum><rsub|a\<in\>A>s(a)\<equallim\><rsub|<reference|general
sum>><big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)>\
<item><math|support(s)\<neq\>\<emptyset\>> and
<math|support(s)\<subset\>\<cal-W\>> (strict inclusion) then
<math|\<cal-W\><mid|\\>support(s)> is finite and not empty so
<math|h<rsup|-1>(support(s)),h<rsup|-1>(\<cal-W\><mid|\\>support(s))\<subseteq\>{1,\<ldots\>,m}>
are finite and non empty and thus there exists
<math|n<rsub|1>,n<rsub|2>\<in\>\<bbb-N\><rsub|0>> and bijective
functions <math|k<rsub|1>,k<rsub|2>> such that
<math|k<rsub|1>:{1,\<ldots\>,n<rsub|1>}\<rightarrow\>h<rsup|-1>(support(s)),k<rsub|2>:{1,\<ldots\>,n<rsub|2>}\<rightarrow\>h<rsup|-1>(\<cal-W\><mid|\\>support(s))>
then using the fact that <math|{1,\<ldots\>,m}\<equallim\><rsub|h is a
bijective mapping>h<rsup|-1>(\<cal-W\>)=h<rsup|-1>(support(s))<big|cup>h<rsup|-1>(\<cal-W\><mid|\\>support(s))>
and <reference|mapping extension>,<reference|bijection extension> we
have the existence of the bijective function
<math|k:{1,\<ldots\>,n<rsub|1>+n<rsub|2>}\<rightarrow\>{1,\<ldots\>,m}>
defined by\
<\enumerate>
<item><math|k(i)=k<rsub|1>(i)\<in\>h<rsup|-1>(support(s))> if
<math|i\<in\>{1,\<ldots\>,n<rsub|1>}>
<item><math|k(i)=k<rsub|2>(i-n<rsub|1>)\<in\>h<rsup|-1>(\<cal-W\><mid|\\>support(s))>
if <math|i\<in\>{n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>}>
</enumerate>
\ then as <math|k<rsub|>> is a bijective function we must have
<math|n<rsub|1>+n<rsub|2>=m> (otherwise we would have a finite set that
is bijective with a proper subset of it). And if
<math|i\<in\>{1,\<ldots\>,n<rsub|1>}> then <math|s(h(k(i)))\<neq\>0>
and if <math|i\<in\>{n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>}> then
<math|s(h(k(i)))=0> so <math|support(s\<circ\>h\<circ\>k)={1,\<ldots\>,n<rsub|1>}>
and using <reference|all zeroes in sum> and <reference|commutativity
sum> we have <math|<big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)=<big|sum><rsub|i=1><rsup|m>((s\<circ\>h)\<circ\>k)(i)=<big|sum><rsub|i=1><rsup|m>(s\<circ\>h\<circ\>k)(i)=<big|sum><rsub|i=1><rsup|n<rsub|1>>((s\<circ\>h\<circ\>k)\<circ\>i<rsub|{1,\<ldots\>,n<rsub|1>}>)(i)=<big|sum><rsub|i=1><rsup|n<rsub|1>>(s\<circ\>(h\<circ\>k<rsub|1>))(i)=<big|sum><rsub|a\<in\>A>s(a)>
using the fact that <math|h\<circ\>k<rsub|1>> is a bijective functions
of <math|{1,\<ldots\>,n<rsub|1>}> to <math|support(s)> and the
definition <reference|general sum>.
</enumerate>
</proof>
<\notation>
Just like we note <math|<big|sum><rsub|i=1><rsup|n>s(i)> as
<math|<big|sum><rsub|i=1><rsup|n>s<rsub|i>> (<math|s<rsub|i>=s(i)) >we
note that <math|<big|sum><rsub|a\<in\>A>s(a)> as
<math|<big|sum><rsub|a\<in\>A>s<rsub|a>> where <math|s<rsub|a>=s(a)>
</notation>
<\theorem>
<label|general sum is finite sum in the finite case>Let <math|S,+> be a
abelian group, <math|A={1,\<ldots\>,n}> and let
<math|s:{1,\<ldots\>,n}\<rightarrow\>S> then
<math|<big|sum><rsub|i=1><rsup|n>s(i)=<big|sum><rsub|i\<in\>A>s(i)>
</theorem>
<\proof>
Then using <reference|zeroes in general sums> on
<math|\<cal-W\>=A={1,\<ldots\>,n}> and <math|h=i<rsub|{1,\<ldots\>,n}>>
we get <math|<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>s(i)=<big|sum><rsub|i=1><rsup|n>(s\<circ\>i<rsub|{1,\<ldots\>,n}>)(i)=<big|sum><rsub|i=1><rsup|n>s(i)>
</proof>
<\theorem>
<label|sum of sums>Let <math|S,+> be a abelian group, <math|A> a set and
let <math|s<rsub|1>:A\<rightarrow\>S>,<math|s<rsub|2>:A\<rightarrow\>S>
such that <math|support(s<rsub|1>),support(s<rsub|2>)> are finite then
<math|support(s<rsub|1>+s<rsub|2>)> is finite and
<math|<big|sum><rsub|a\<in\>A>(s<rsub|1>+s<rsub|2>)(a)=(<big|sum><rsub|a\<in\>A>s<rsub|1>(a))+(<big|sum><rsub|a\<in\>A>s<rsub|2>(a))>
</theorem>
<\proof>
First if <math|a\<in\>support(s<rsub|1>+s<rsub|2>)> then if
<math|s<rsub|1>(a)=0> and <math|s<rsub|2>(a)=0\<Rightarrow\>s<rsub|1>(a)+s<rsub|2>(a)=0>
or either <math|s<rsub|1>(a)\<neq\>0>,<math|s<rsub|2>(a)\<neq\>0> and
thus <math|a\<in\>support(s<rsub|1>)<big|cup>support(s<rsub|2>)> so we
have <math|support(s<rsub|1>+s<rsub|2>)\<subseteq\>support(s<rsub|1>)<big|cup>support(s<rsub|2>)>
which is finite and so <math|support(s<rsub|1>+s<rsub|2>)> is finite and
the sum is defined. Now as <math|support(s<rsub|1>)<big|cup>support(s<rsub|2>)>
is finite we have the following cases
<\enumerate>
<item><math|support(s<rsub|1>)=\<emptyset\>> then
<math|s<rsub|1>(a)+s<rsub|2>(a)=s<rsub|2>(a)\<Rightarrow\><big|sum><rsub|a\<in\>A>(s<rsub|1>+s<rsub|2>)(a)=<big|sum><rsub|a\<in\>A>s<rsub|2>(a)=0+<big|sum><rsub|a\<in\>A>s<rsub|2>(a)=(<big|sum><rsub|a\<in\>A>s<rsub|1>(a))+(<big|sum><rsub|a\<in\>A>s<rsub|2>(a))>
<item><math|support(s<rsub|2>)=\<emptyset\>> then
<math|s<rsub|1>(a)+s<rsub|2>(a)=s<rsub|1>(a)\<Rightarrow\><big|sum><rsub|a\<in\>A>(s<rsub|1>+s<rsub|2>)(a)=<big|sum><rsub|a\<in\>A>s<rsub|1>(a)=<big|sum><rsub|a\<in\>A>s<rsub|1>(a)+0=(<big|sum><rsub|a\<in\>A>s<rsub|1>(a))+(<big|sum><rsub|a\<in\>A>s<rsub|2>(a))>
<item><math|support(s<rsub|1>)<big|cup>support(s<rsub|2>)\<neq\>\<emptyset\>>
then as the union is finite we have the existence of a bijective
function <math|h:{1,\<ldots\>,n}\<rightarrow\>support(s<rsub|1>)<big|cup>support(s<rsub|2>)\<supseteq\>support(s<rsub|1>+s<rsub|2>)>.
Then using <reference|zeroes in general sums> we have
<math|<big|sum><rsub|a\<in\>A>(s<rsub|1>+s<rsub|2>)(a)=<big|sum><rsub|i=1><rsup|n>((s<rsub|1>+s<rsub|2>)\<circ\>h)(i)>.
<math|We> prove now by induction on <math|n> that
<math|<big|sum><rsub|i=1><rsup|n>((s<rsub|1>+s<rsub|2>)\<circ\>h)(i)=(<big|sum><rsub|i=1><rsup|n>(s<rsub|1>\<circ\>h)(a))+(<big|sum><rsub|i=1><rsup|n>(s<rsub|2>\<circ\>h)(a))>.
<\enumerate>
<item>If <math|n=1> then <math|<big|sum><rsub|i=1><rsup|1>((s<rsub|1>+s<rsub|2>)\<circ\>h)(i)=((s<rsub|1>+s<rsub|2>)\<circ\>h)(1)=(s<rsub|1>\<circ\>h<rsub|>)(1)+(s<rsub|2>\<circ\>h)(1)=(<big|sum><rsub|i=1><rsup|1>(s<rsub|1>\<circ\>h)(i))+(<big|sum><rsub|i=1><rsup|1>(s<rsub|2>\<circ\>h)(i))>
<item>Assume the assertion is true for <math|n> and prove it for
<math|n+1> then <math|<big|sum><rsub|i=1><rsup|n+1>((s<rsub|1>+s<rsub|2>)\<circ\>h)(i)=(<big|sum><rsub|i=1><rsup|n>((s<rsub|1>+s<rsub|2>)\<circ\>h)(i))+((s<rsub|1>+s<rsub|2>)\<circ\>h)(n+1)\<equallim\><rsub|induction
hypothese>(<big|sum><rsub|i=1><rsup|n>(s<rsub|1>\<circ\>h)(i))+(<big|sum><rsub|i=1><rsup|n>(s<rsub|2>\<circ\>h)(i))+(s<rsub|1>\<circ\>h)(n+1)+(s<rsub|2>\<circ\>h)(n+1)=(<big|sum><rsub|i=1><rsup|n+1>(s<rsub|1>\<circ\>h<rsub|1>)(i))+(<big|sum><rsub|i=1><rsup|n+1>(s<rsub|2>\<circ\>h)(i))>
</enumerate>
So we have proved that <math|<big|sum><rsub|a\<in\>A>(s<rsub|1>+s<rsub|2>)(a)=(<big|sum><rsub|i=1><rsup|n>(s<rsub|1>\<circ\>h)(i))+(<big|sum><rsub|i=1><rsup|n>(s<rsub|2>\<circ\>h)(i))\<equallim\><rsub|<reference|zeroes
in general sums>>(<big|sum><rsub|a\<in\>A>s<rsub|1>(a))+(<big|sum><rsub|a\<in\>A>s<rsub|2>(a))>
</enumerate>
</proof>
<\definition>
<label|left(right) group action><index|left group action><index|right
group action>Let <math|G,+> be a group and let <math|X> be a set then we
have the following definitions
<\enumerate>
<item>A left group action is a function <math|.
:G\<times\>X\<rightarrow\>X> with <math|.(g,x)\<equallim\><rsub|note>g.x>
such that\
<\enumerate>
<item><math|\<forall\>x\<in\>X\<succ\>1<rsub|G>.x=x> where
<math|1<rsub|G>> is the unit of <math|G>
<item><math|\<forall\>g<rsub|1>,g<rsub|2>\<in\>G,\<forall\>x\<in\>X\<succ\>(g<rsub|1>+g<rsub|2>).x=g<rsub|1>.(g<rsub|2>.x)>
</enumerate>
<item>A right group action is a function <math|.
:X\<times\>G\<rightarrow\>X> with <math|.
(x,g)\<equallim\><rsub|note>x.g> such that
<\enumerate>
<item><math|\<forall\>x\<in\>X\<succ\>x.1<rsub|G>=x>
<item><math|\<forall\>g<rsub|1>,g<rsub|2>\<in\>G,\<forall\>x\<in\>X\<succ\>x.(g<rsub|1>+g<rsub|2>)=(x.g<rsub|1>).g<rsub|2>>
</enumerate>
</enumerate>
</definition>
<\definition>
<label|faithfull left action><index|faithful left
action><index|transitive left action><index|free left
action><index|faithful right action><index|transitive right
action><index|free right action>Let <math|G,+> be a group, <math|X> be a
set and <math|. :G\<times\>X\<rightarrow\>X> a left (right) action then
we have the following definitions
<\enumerate>
<item><math|.> is called faithful iff <math|.<rsub|g>:X\<rightarrow\>X>
defined by <math|.<rsub|g>(x)=g.x> (or <math|.<rsub|g>(x)=x.g>) is
equaly to <math|i<rsub|X>> if and only if <math|g=1<rsub|G>>
<item><math|.> is called transitive iff
<math|\<forall\>x<rsub|1>,x<rsub|2>\<in\>X> there exists a
<math|g\<in\>G> such that <math|g.x<rsub|1>=x<rsub|2>> (or
<math|x<rsub|1>.g=x<rsub|2>>)
<item><math|.> is free iff <math|\<forall\>x\<in\>X \ we have
{g\<in\>G\|g.x=x}={1<rsub|G>}> (or <math|{g\<in\>G\|x.g=x}={1<rsub|G>}>
</enumerate>
</definition>
<section|Fields>
<\definition>
<index|field>A field <math|F,+,.> is a set together with maps
<math|+:F\<times\>F\<rightarrow\>F> and
<math|.:F\<times\>F\<rightarrow\>F> (where we denote <math|+(x,y)> as
<math|x+y> and <math|.(x,y)> as <math|x.y>) such that the following holds
<\enumerate>
<item><math|\<forall\>x,y,z\<in\>F\<succ\>(x+y)+z=x+(y+z)>
<item><math|\<exists\>0\<in\>F\<succ\>\<forall\>x\<in\>F\<succ\>x+0=0>
(<math|0> is called the neutral element)
<item><math|\<forall\>x\<in\>F\<succ\>\<exists\>(-x)\<in\>F\<succ\>x+(-x)=0>
<item><math|\<forall\>x,y\<in\>F\<succ\>x+y=y+x>
<item><math|\<forall\>x,y,z\<in\>F\<succ\>(x.y).z=x.(y.z)>
<item><math|\<exists\>1\<in\>F> with <math|1\<neq\>0> such that
<math|\<forall\>x\<in\>S> we have <math|1.x=x> (<math|1> is called the
unit element)
<item><math|\<forall\>x\<in\>F<mid|\\>{0}\<succ\>\<exists\>x<rsup|-1>\<in\>F\<succ\>x.x<rsup|-1>=1>
<item><math|\<forall\>x,y\<in\>F\<succ\>x.y=y.x>
<item><math|\<forall\>x,y,z\<in\>F\<succ\>x.(y+z)=(x.y)+(x.z)>
</enumerate>
</definition>
<\theorem>
<label|zero is absorbing element in a field>Let <math|F,+,.> be a field
then <math|\<forall\>f\<in\>F> we have <math|f.0=0>
</theorem>
<\proof>
<math|0=0.f+(\<um\>(0.f))=(0+0).f+(\<um\>(0.f))=0.f+(0.f+(\<um\>(0.f))=0.f+0=0.f>
</proof>
<\theorem>
<label|inverse of product in a field>Let <math|F,+,.> be a field and
<math|a,b\<in\>F<mid|\\>{0}> then <math|(a.b)<rsup|-1>=a<rsup|-1>.b<rsup|-1>>
</theorem>
<\proof>
<math|(a<rsup|-1>.b<rsup|-1>)(a.b)=(a<rsup|-1>.a).(b<rsup|-1>.b)=1.1=1>
</proof>
<\note>
Looking at <math|1,2,3,4> we note that a field <math|F,+,.> is a abelian
group <math|F,+> and <math|F,.> is a abelian semigroup (here we note the
<math|<big|sum><rsub|i=1><rsup|n>s(i)> by
<math|<big|prod><rsub|i=1><rsup|n>s(i)> to make the distinction between
<math|+> and <math|.>) So <math|<big|prod><rsub|i=1><rsup|n>s(i)> is
defined by <math|<big|prod><rsub|i=1><rsup|1>s(i)=s(1)> and
<math|<big|prod><rsub|i=1><rsup|n+1>s(i)=(<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1)>
</note>
<\theorem>
<label|general product with a zero>Let <math|F,+,.> be a field and
<math|s:{1,\<ldots\>,n}\<rightarrow\>F> a function so that
<math|\<exists\>i\<in\>{1,\<ldots\>,n}\<vdash\>s(i)=0> then
<math|<big|prod><rsub|i=1><rsup|n>s(i)=0>
</theorem>
<\proof>
We prove this by induction on <math|n> so
<\enumerate>
<item><math|n=1\<Rightarrow\><big|prod><rsub|i=1><rsup|i>s(i)=s(1)=0>
<item>Assume that the theorem is true for <math|n+1> then
<math|><math|<big|prod><rsub|i=1><rsup|n+1>s(i)=(<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1)>
and then we have the following cases\
<\enumerate>
<item><math|i=n+1\<Rightarrow\>s(n+1)=0\<Rightarrow\><big|prod><rsub|i=1><rsup|n+1>s(i)=(<big|prod><rsub|i=1><rsup|n>s(i)).0\<equallim\><rsub|<reference|zero
is absorbing element in a field>>0>
<item><math|i\<in\>{1,\<ldots\>,n}\<Rightarrowlim\><rsub|induction
hypothese><big|prod><rsub|i=1><rsup|n+1>s(i)=0.s(n+1)\<equallim\><rsub|<reference|zero
is absorbing element in a field>>0>
</enumerate>
</enumerate>
</proof>
<\definition>
<index|<math|f<rsup|n>>>In a field <math|><math|F,+,.> we define
<math|\<forall\>n\<in\>\<bbb-N\>(=\<bbb-N\><rsub|0><big|cup>{0})> and
<math|\<forall\>f\<in\>F> <math|f<rsup|n>> by\
<\enumerate>
<item><math|f<rsup|0>=1> (this is the unit in <math|F> not in
<math|\<bbb-N\>>)
<item><math|f<rsup|1>=f>
<item><math|f<rsup|n+1>=f.f<rsup|n>>
</enumerate>
</definition>
<\theorem>
Let <math|F,+,.> be a field, <math|f\<in\>F> and <math|a mapping
s:{1,\<ldots\>,n}\<rightarrow\>F,n\<in\>\<bbb-N\><rsub|0>> be defined by
<math|\<forall\>i\<in\>{1,\<ldots\>,n} s(i)=f> then
<math|><math|<big|prod><rsub|i=1><rsup|n>s(i)=f<rsup|n>>
</theorem>
<\proof>
<math|>We prove this by induction on <math|n>
<\enumerate>
<item><math|<big|prod><rsub|i=1><rsup|1>s(i)=s(1)=f=f<rsup|1>>
<item>Assume the theorem is true for <math|n> then
<math|<big|prod><rsub|i=1><rsup|n+1>s(i)=(<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1)=f<rsup|n>.f=f.f<rsup|n>=f<rsup|n+1>>
</enumerate>
</proof>
<\theorem>
<label|inverse of product of many elements in a field>Let <math|F,+,.> be
a field and <math|s:{1,\<ldots\>,n}\<rightarrow\>F> such that
<math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>s(i)\<neq\>0> then
<math|(<big|prod><rsub|i=1><rsup|n>s(i))<rsup|-1>=<big|prod><rsub|i=1><rsup|n>s(i)<rsup|-1>>
(where <math|s<rsup|-1>:{1,\<ldots\>,n}\<rightarrow\>F> is defined by
<math|s<rsup|-1>(i)=(s(i))<rsup|-1>>
</theorem>
<\proof>
This is easy by induction on n
<\enumerate>
<item><math|(<big|prod><rsub|i=1><rsup|1>s(i))<rsup|-1>=s(1)<rsup|-1>=<big|prod><rsub|i=1><rsup|1>s(i)<rsup|-1>>
<item>Assume the theorem is true for <math|n> then
<math|(<big|prod><rsub|i=1><rsup|n+1>s(i))<rsup|-1>=((<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1))<rsup|-1>\<equallim\><rsub|<reference|inverse
of product in a field>>(<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1)<rsup|-1>\<equallim\><rsub|induction>(<big|prod><rsub|i=1><rsup|n>s(i)<rsup|-1>).s(n+1)<rsup|-1>=<big|prod><rsub|i=1><rsup|n+1>s(i)>
</enumerate>
</proof>
<\theorem>
<math|\<bbb-R\>,+,.> is a field
</theorem>
<\proof>
This is trivial by the definition <reference|real numbers>.
</proof>
<section|Vector spaces>
<\definition>
<index|vector space>A vector space <math|V,\<oplus\>,\<odot\>>over a
field <math|F,+,.> is a abelian group <math|V,\<oplus\>> together with a
map <math|\<odot\>:F\<times\>V\<rightarrow\>F> (where we note
<math|\<odot\>(\<alpha\>,x)> as <math|\<alpha\>\<odot\>x> (called scalar
multiplication) such that\
<\enumerate>
<item><math|\<forall\>\<alpha\>\<in\>F,\<forall\>x,y\<in\>V\<succ\>><math|\<alpha\>\<odot\>(x\<oplus\>y)=(\<alpha\>\<odot\>x)+(\<alpha\>\<odot\>y)>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x\<in\>V\<succ\>(\<alpha\>+\<beta\>)\<odot\>x=(\<alpha\>\<odot\>x)\<oplus\>(\<beta\>\<odot\>x)>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x\<in\>V\<succ\>(\<alpha\>.\<beta\>)\<odot\>x=\<alpha\>\<odot\>(\<beta\>\<odot\>x)>
<item><math|\<forall\>x\<in\>V\<succ\>1\<odot\>x=x> (1 the neutral
element in <math|V>)\
</enumerate>
</definition>
<\theorem>
<label|product of zero>Let <math|V,\<oplus\>,\<odot\>> be a vector space
over a field <math|F,+,.> then we have
<\enumerate>
<item><math|\<forall\>x\<in\>V> we have <math|O=0\<odot\>x> (<math|O>
is the neutral element in <math|V>, <math|0> is the neutral element in
<math|F>)
<item><math|\<forall\>x\<in\>V> we have
<math|(\<um\>1)\<odot\>x=\<um\>x> where <math|\<um\>1> is the inverse
element of 1 in <math|F> (considered as a group) and <math|\<um\>x> is
the inverse element of <math|x\<in\>V>.
<item><math|\<forall\>\<alpha\>\<in\>F> we have
<math|\<alpha\>\<odot\>O=0>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|O=0\<odot\>x+(\<um\>(0.x))=(0+0)\<odot\>x+(\<um\>(0.x))=0\<odot\>x+(0\<odot\>x+(\<um\>(0\<odot\>x)))=0\<odot\>x+O=0\<odot\>x>
<item><math|x\<oplus\>((\<um\>1)\<odot\>x)=(1\<odot\>x)\<oplus\>((\<um\>1)\<odot\>x)=(1+(\<um\>1))\<odot\>x=0\<odot\>x=O>
<item><math|\<alpha\>\<odot\>O=\<alpha\>\<odot\>(0\<odot\>O)=(\<alpha\>.0)\<odot\>O=0\<odot\>O=O>
</enumerate>
</proof>
<\theorem>
<label|subspace><index|subspace>Consider a vector space
<math|V,\<oplus\>,\<odot\>> over a field <math|F,+,.> then if
<math|\<emptyset\>\<neq\>W\<subseteq\>V> is a nonempty subset of
<math|V>) we have that <math|W,\<oplus\><rprime|'>,\<odot\><rprime|'>>
(where <math|><math|\<oplus\><rprime|'>=\<oplus\><rsub|\|W>> and
<math|\<odot\><rprime|'>=\<odot\><rsub|\|F\<times\>W>>) is a vector space
(called a subspace of <math|V)\<Leftrightarrow\>\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x,y\<in\>W>
we have <math|\<alpha\>\<odot\>x+\<beta\>\<circ\>y\<in\>W>. Note that we
in general use the same symbol for <math|\<oplus\>,\<oplus\><rprime|'>>
and <math|\<odot\>.\<odot\><rprime|'>>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>> This is trivial by the definition of
<math|\<oplus\><rprime|'>,\<odot\><rprime|'>>
<math|\<Leftarrow\>><math|\<forall\>x,y\<in\>W> and
<math|\<forall\>\<alpha\>,\<beta\>\<in\>F> we have
<math|x\<oplus\><rprime|'>y=x+y=(x\<odot\>1)+(y\<odot\>1)\<in\>W> so
<math|\<oplus\><rprime|'>> maps <math|W*\<times\>W> to <math|W> and
<math|\<alpha\>\<odot\><rprime|'>x=\<alpha\>\<odot\>x=(\<alpha\>+0)\<odot\>x=\<alpha\>\<odot\>x+\<alpha\>\<odot\>x\<in\>W>
so <math|\<odot\><rprime|'>> maps <math|W\<times\>W> to <math|W>. Finally
we have:\
<\enumerate>
<item><math|(x\<oplus\><rprime|'>(y\<oplus\><rprime|'>z))=(x\<oplus\>(y\<oplus\>z))=(x\<oplus\>y)\<oplus\>z=(x\<oplus\><rprime|'>y)\<oplus\><rprime|'>z>
<item>As <math|W\<neq\>\<emptyset\>\<Rightarrow\>\<exists\>w\<in\>W>
and then <math|O=w+(\<um\>w)=1\<odot\>w+(\<um\>1)\<odot\>w\<in\>W> and
<math|x\<oplus\><rprime|'>O=x\<oplus\>O=x=O\<oplus\>x=O\<oplus\><rprime|'>x>
<item>if <math|x\<in\>W\<Rightarrow\>\<um\>x=\<um\>x\<oplus\>O=(\<um\>1)\<odot\>x\<oplus\>0\<odot\>x\<in\>W>
and <math|x\<oplus\><rprime|'>(\<um\>x)=x\<oplus\>(\<um\>x)=O=(\<um\>x)\<oplus\>x=(\<um\>x)\<oplus\><rprime|'>x>
<item><math|x\<oplus\><rprime|'>y=x\<oplus\>y=y\<oplus\>x=y\<oplus\><rprime|'>x>
<item><math|\<alpha\>\<odot\><rprime|'>(x\<oplus\><rprime|'>y)=\<alpha\>\<odot\>(x\<oplus\>y)=\<alpha\>\<odot\>x\<oplus\>\<beta\>\<odot\>y=\<alpha\>\<odot\><rprime|'>x\<oplus\>\<beta\>\<odot\><rprime|'>y=(\<alpha\>\<odot\><rprime|'>x)\<oplus\><rprime|'>(\<beta\>\<odot\><rprime|'>y)>
<item><math|(\<alpha\>+\<beta\>)\<odot\><rprime|'>x=(\<alpha\>+\<beta\>)\<odot\>x=(\<alpha\>\<odot\>x)\<oplus\>(\<beta\>\<odot\>x)=(\<alpha\>\<odot\><rprime|'>x)\<oplus\>(\<beta\>\<odot\><rprime|'>x)=(\<alpha\>\<odot\><rprime|'>x)\<oplus\><rprime|'>(\<beta\>\<odot\><rprime|'>x)>
<item><math|(\<alpha\>.\<beta\>)\<odot\><rprime|'>x=(\<alpha\>.\<beta\>)\<odot\>x=\<alpha\>\<odot\>(\<beta\>\<odot\>x)=\<alpha\>\<odot\>(\<beta\>\<odot\><rprime|'>x)=\<alpha\>\<odot\><rprime|'>(\<beta\>\<odot\><rprime|'>x)>
<item><math|1\<odot\><rprime|'>x=1\<odot\>x=x>
</enumerate>
</proof>
<\theorem>
<label|product of real numbers and \<less\>=>Let
<math|s,f\<in\>\<bbb-R\><rsup|n>> (so
<math|s:{1,\<ldots\>,n}\<rightarrow\>\<bbb-R\>,
f:{1,\<ldots\>,n}\<rightarrow\>\<bbb-R\>>) and
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|s(i)\<leqslant\>f(i)> then <math|<big|prod><rsub|i=1><rsup|n>s(i)\<leqslant\><big|prod><rsub|i=1><rsup|n>f(i)>
</theorem>
<\proof>
We proof this by induction on <math|n>\
<\enumerate>
<item><math|n=1><math|<big|prod><rsub|i=1><rsup|1>s(i)=s(1)\<leqslant\>f(1)=<big|prod><rsub|i=1><rsup|1>f(i)>
<item><math|>Assume the theorem is true for <math|n> then
<math|<big|prod><rsub|i=1><rsup|n+1>s(i)=(<big|prod><rsub|i=1><rsup|n>s(i)).s(n+1)\<leqslant\>(<big|prod><rsub|i=1><rsup|n>f(i)).s(n+1)\<leqslant\>(<big|prod><rsub|i=1><rsup|n>f(i)).f(n+1)=<big|prod><rsub|i=1><rsup|n+1>f(i)>
</enumerate>
</proof>
<\theorem>
<label|product of equal real numbers>Let <math|s\<in\>\<bbb-R\><rsup|n>>
such that <math|\<exists\>i\<in\>{1,\<ldots\>,n}> and
<math|\<exists\>a\<in\>\<bbb-R\>> such that
<math|\<forall\>j\<in\>{1,\<ldots\>,n}<mid|\\>{i}> we have <math|s(j)=a>
then <math|<big|prod><rsub|j=1><rsup|n>s(j)=a<rsup|n-1>.s(i)>
</theorem>
<\proof>
We prove this by induction
<\enumerate>
<item>If <math|n=1> then <math|<big|prod><rsub|j=1><rsup|1>s(j)=s(1)=1.s(1)=a<rsup|0>.s(1)>
<item>Assume that the theorem is true for <math|n> and prove it for
<math|n+1> then if we have the following cases
<\enumerate>
<item><math|i=n+1> then <math|\<forall\>j\<in\>{1,\<ldots\>,n}> we
have <math|s(j)=a> and thus <math|<big|prod><rsub|j=1><rsup|n+1>s(j)=(<big|prod><rsub|j=1><rsup|n>s(j)).s(n+1)=a<rsup|n>.s(n+1)>
<item><math|i\<neq\>n+1\<Rightarrow\>i\<in\>{1,\<ldots\>,n}> and
<math|\<forall\>j\<in\>{1,\<ldots\>,n}<mid|\\>{i}\<subseteq\>{1,\<ldots\>,n+1}<mid|\\>{i}>
we have <math|s(j)=a> and also <math|s(n+1)=a > (as
<math|n+1\<neq\>i>) \ then <math|<big|prod><rsub|j=1><rsup|n+1>s(j)=(<big|prod><rsub|j=1><rsup|n>s(j)).s(n+1)=a<rsup|n-1>.s(i).s(n+1)=a<rsup|n-1>.s(i).a=a<rsup|n>.s(i)>
</enumerate>
</enumerate>
</proof>
\;
<\theorem>
<label|vector space of mappings><index|<math|M(X,V)>>Let
<math|V,\<oplus\>,\<odot\>> be a vector space over a a field <math|F,+,.>
and <math|X> a set then let <math|M(X,V)={f\|f:X\<rightarrow\>V}> (the
set of all functions from <math|X> to <math|V>) together with the
functions <math|\<boxplus\>:M(X,V)\<times\>M(X,V)\<rightarrow\>M(X,V)>
and <math|\<boxdot\>:F\<times\>M(X,V)\<rightarrow\>M(X,V)> defined by
<math|(f\<boxplus\>g)(x)=f(x)\<oplus\>g(x)> and
<math|(\<alpha\>\<boxdot\>f)(x)=\<alpha\>\<odot\>f(x)> forms a
vectorspace<math|>\
</theorem>
<\proof>
<math|\<forall\>\<alpha\>,\<beta\>\<in\>F>,
<math|\<forall\>f,g,h\<in\>M(X,V)> we have using the fact that <math|V>
is a vectorspace
<\enumerate>
<item><math|\<forall\>x\<in\>X> we have
<math|((f\<boxplus\>g)\<boxplus\>h)(x)=(f\<boxplus\>g)(x)\<oplus\>h(x)=((f(x)\<oplus\>g(x))\<oplus\>h(x)=f(x)\<oplus\>(g(x)\<oplus\>h(x)=(f(x)\<oplus\>(g\<boxplus\>h)(x))=(f\<boxplus\>(g\<boxplus\>h))(x)\<Rightarrow\>((f\<boxplus\>g)\<boxplus\>h)=f\<boxplus\>(g\<boxplus\>h)>
<item>If we define <math|<wide|0|\<bar\>>:X\<rightarrow\>Y> is defined
by <math|<wide|0|\<bar\>>(x)=0> then <math|\<forall\>x\<in\>X> we have
<math|(<wide|0|\<bar\>>\<boxplus\>f)(x)=<wide|0|\<bar\>>(x)+f(x)=0+f(x)=f(x)=f(x)+0=f(x)+<wide|0|\<bar\>>(x)=(f\<boxplus\><wide|0|\<bar\>>)(x)\<Rightarrow\>(<wide|0|\<bar\>>\<boxplus\>f)=f=(f\<boxplus\><wide|0|\<bar\>>)>
<item>Given <math|f\<in\>M(X,V)> then define <math|\<um\>f\<in\>M(X,V)>
by <math|(\<um\>f)(x)=\<um\>(f(x))> then <math|\<forall\>x\<in\>X> we
have <math|((\<um\>f)\<boxplus\>f)(x)=(\<um\>f)(x)\<oplus\>f(x)=(\<um\>(f(x)))\<oplus\>f(x)=0=<wide|0|\<bar\>>(x)=f(x)\<oplus\>(\<um\>(f(x)))=f(x)\<oplus\>(\<um\>f)(x)=(f\<boxplus\>(\<um\>f))(x)\<Rightarrow\>(\<um\>f)\<boxplus\>f=<wide|0|\<bar\>>=f\<boxplus\>(\<um\>f)>
<item><math|\<forall\>x\<in\>X> we have
<math|(f\<boxplus\>g)(x)=f(x)\<oplus\>g(x)=g(x)\<oplus\>f(x)=(g\<boxplus\>f)(x)\<Rightarrow\>f\<boxplus\>g=g\<boxplus\>f><math|>
</enumerate>
\ So <math|M(X,V)> forms a abelian group and further we have\
<\enumerate>
<item><math|(\<alpha\>\<boxdot\>(f\<boxplus\>g))(x)=\<alpha\>\<odot\>(f\<boxplus\>g)(x)=\<alpha\>\<odot\>(f(x)\<oplus\>g(x))=(\<alpha\>\<odot\>f(x))\<oplus\>(\<alpha\>\<odot\>g(x))=((\<alpha\>\<boxdot\>f)(x))\<oplus\>((\<alpha\>\<boxdot\>g)(x))=((\<alpha\>\<boxdot\>f)\<boxplus\>(\<alpha\>\<boxdot\>g))(x)\<Rightarrow\>\<alpha\>\<boxdot\>(f\<boxplus\>g))=(\<alpha\>\<boxdot\>f)\<boxplus\>(\<alpha\>\<boxdot\>g)>
<item><math|((\<alpha\>+\<beta\>)\<boxdot\>f)(x)=(\<alpha\>+\<beta\>)\<odot\>f(x)=(\<alpha\>\<odot\>f(x))\<oplus\>(\<beta\>\<odot\>f(x))=((\<alpha\>\<boxdot\>f)(x))\<oplus\>((\<beta\>\<boxdot\>f)(x))=((\<alpha\>\<boxdot\>f)\<boxplus\>(\<beta\>\<boxdot\>f))(x)\<Rightarrow\>(\<alpha\>+\<beta\>)\<boxdot\>f=(\<alpha\>\<boxdot\>f)\<boxplus\>(\<beta\>\<boxdot\>f)>
<item><math|((\<alpha\>.\<beta\>)\<boxdot\>f)(x)=(\<alpha\>.\<beta\>)\<odot\>f(x)=\<alpha\>\<odot\>(\<beta\>\<odot\>f(x))=\<alpha\>\<odot\>((\<beta\>\<boxdot\>f)(x))=(\<alpha\>\<boxdot\>(\<beta\>\<boxdot\>f))(x)\<Rightarrow\>(\<alpha\>,\<beta\>)\<boxdot\>f=\<alpha\>\<boxdot\>(\<beta\>\<boxdot\>f)>
<item><math|(1\<boxdot\>f)(x)=1\<odot\>f(x)=f(x)\<Rightarrow\>1\<boxdot\>f=f>
</enumerate>
</proof>
\;
<\theorem>
<label|scalair product and sum>Let <math|V,\<oplus\>,\<odot\>> be a
vector space over a field <math|F,+,.> and let <math|s:A\<rightarrow\>V>
be a function with <math|support(s)> is finite then
<math|\<forall\>\<alpha\>\<in\>F> the map
<math|\<alpha\>\<odot\>s:A\<rightarrow\>V> defined by
<math|(\<alpha\>\<odot\>s)(a)=\<alpha\>\<odot\>s(a)> has finite support
and <math|<big|sum><rsub|a\<in\>A>(\<alpha\>\<odot\>s)(a)=\<alpha\>\<odot\>(<big|sum><rsub|a\<in\>A>s(a))>
</theorem>
<\proof>
If <math|\<alpha\>=0> then <math|support(\<alpha\>\<odot\>s)=\<emptyset\>>
and we have by definition <math|<big|sum><rsub|a\<in\>A>(\<alpha\>\<odot\>s)(a)=O\<equallim\><rsub|<reference|product
of zero>>0\<odot\>(<big|sum><rsub|a\<in\>A>s(a))> so assume that
<math|\<alpha\>\<neq\>\<emptyset\>>. Then if
<math|a\<in\>support(\<alpha\>\<odot\>s)\<Rightarrow\>\<alpha\>\<odot\>s(a)\<neq\>O\<Rightarrowlim\><rsub|\<alpha\>\<neq\>0>s(a)\<neq\>O>
[for if <math|s(a)=O\<Rightarrow\>\<alpha\>\<odot\>s(a)=O> a
contradiction] so <math|support(\<alpha\>\<circ\>s)\<subseteq\>support(s)>
and thus <math|support(\<alpha\>\<odot\>s)> is finite and as
<math|support(s)> is finite and non empty we have
<math|\<exists\>m\<in\>\<bbb-N\><rsub|0>> and a bijective function
<math|h:{1,\<ldots\>,m}\<rightarrow\>support(s)>. Then using
<reference|zeroes in general sums> we have
<math|<big|sum><rsub|a\<in\>A>(\<alpha\>\<odot\>s)(a)=<big|sum><rsub|i=1><rsup|m>((\<alpha\>\<odot\>s)\<circ\>h)(i)>,
we prove now by induction on <math|n> that
<math|<big|sum><rsub|i=1><rsup|m>((\<alpha\>\<odot\>s)\<circ\>h)(i)=\<alpha\>\<odot\>(<big|sum><rsub|i=1><rsup|m>(s\<circ\>h)(i)>:
<\enumerate>
<item><math|n=1> then <math|<big|sum><rsub|i=1><rsup|1>((\<alpha\>\<odot\>s)\<circ\>h)(i)=(\<alpha\>\<odot\>s)(h(1))=\<alpha\>\<odot\>(s(h(1)))=\<alpha\>\<odot\><big|sum><rsub|i=1><rsup|1>(s\<circ\>h)(i)>
<item>Assume that the assertion is true for <math|n> and proof it then
for <math|n+1>. <math|<big|sum><rsub|i=1><rsup|n+1>((\<alpha\>\<odot\>s)\<circ\>h)(i)=(<big|sum><rsub|i=1><rsup|n>((\<alpha\>\<odot\>s)\<circ\>h)(i))+\<alpha\>\<odot\>s(h(n+1))\<equallim\><rsub|induction
hypothese>\<alpha\>\<odot\>(<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i))=\<alpha\>\<odot\>(s\<circ\>h)(n+1)=\<alpha\>\<odot\>((<big|sum><rsub|i=1><rsup|n>(s\<circ\>h)(i))+(s\<circ\>h)(n+1))=\<alpha\>\<odot\>(<big|sum><rsub|i=1><rsup|h>(s\<circ\>h)(i))>
</enumerate>
\ We finish the proof by using <reference|general sum> to get
<math|\<alpha\>\<odot\>(<big|sum><rsub|i=1><rsup|h>(s\<circ\>h)(i))=\<alpha\>\<odot\>(<big|sum><rsub|a\<in\>A>s(a))>
</proof>
<\corollary>
<label|sum of a linear combination>Let <math|V,\<oplus\>,\<oast\>> be a
vector space over a field <math|F,+,.> and let <math|s<rsub|1>,s<rsub|2>
be mappings from ><math|{1,\<ldots\>,n}\<rightarrow\>V,\<alpha\>,\<beta\>\<in\>F>
then <math|<big|sum><rsub|i=1><rsup|n>(\<alpha\>\<oast\>s<rsub|1>(i)\<oplus\>\<beta\>\<oast\>s<rsub|2>(i))=\<alpha\>\<oast\>(<big|sum><rsub|i=1><rsup|n>s<rsub|1>(i))+\<beta\>\<oast\>(<big|sum><rsub|i=1><rsup|n>s<rsub|2>(i))>
</corollary>
<\proof>
<math|<big|sum><rsub|i=1><rsup|n>(\<alpha\>\<oast\>s<rsub|1>(i)\<oplus\>\<beta\>\<oast\>s<rsub|2>(i))\<equallim\><rsub|<reference|general
sum is finite sum in the finite case>><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(\<alpha\>\<oast\>s<rsub|1>(i)\<oplus\>\<beta\>\<oast\>s<rsub|2>(i))\<equallim\><rsub|<reference|sum
of sums>>(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(\<alpha\>\<oast\>s<rsub|1>(i)))\<oplus\>(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(\<beta\>\<oast\>s<rsub|2>(i)))\<equallim\><rsub|<reference|scalair
product and sum>>\<alpha\>\<oast\>(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>s<rsub|1>(i))+\<beta\>\<oast\>(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>s<rsub|2>(i))\<equallim\><rsub|<reference|general
sum is finite sum in the finite case>>\<alpha\>\<oast\>(<big|sum><rsub|i=1><rsup|n>s<rsub|1>(i))+\<beta\>\<oast\>(<big|sum><rsub|i=1><rsup|n>s<rsub|2>(i))>
</proof>
<\theorem>
Let <math|F,+,.> be a field then <math|F,+,.> is a vectorspace over
<math|F,+,.>
</theorem>
<\proof>
By the definition of a field we have <math|\<oplus\>=+,\<odot\>=.>
<\enumerate>
<item><math|\<forall\>\<alpha\>\<in\>F,\<forall\>x,y\<in\>F> we have
<math|\<alpha\>.(x+y)=\<alpha\>.x+\<alpha\>.y>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x\<in\>F> we
have <math|(\<alpha\>+\<beta\>).x=x.(\<alpha\>+\<beta\>)=x.\<alpha\>+x.\<beta\>=\<alpha\>.x+\<beta\>.y>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x\<in\>F> we
have <math|(\<alpha\>.\<beta\>).x=\<alpha\>.(\<beta\>.x)>
<item><math|\<forall\>x\<in\>F> we have <math|1.x=x>
</enumerate>
</proof>
<\corollary>
<math|\<bbb-R\>,+,.> is a vector space
</corollary>
<\definition>
<index|<math|\<bbb-C\>>><index|complex numbers>The space
<math|\<bbb-C\>,+,.> of complex numbers is defined as the set
<math|\<bbb-R\>\<times\>\<bbb-R\>> with <math|+,.> defined by
<\enumerate>
<item><math|\<forall\>(x,y),(x<rprime|'>,y<rprime|'>)\<in\>\<bbb-C\>\<succ\>(x,y)+(x<rprime|'>,y<rprime|'>)=(x+x<rprime|'>,y+y<rprime|'>)>
<item><math|\<forall\>(x,y),(x<rprime|'>,y<rprime|'>)\<in\>\<bbb-C\>\<succ\>(x,y).(x<rprime|'>,y<rprime|'>)=(x.x<rprime|'>-y.y<rprime|'>,x.y<rprime|'>+x<rprime|'>.y)>
</enumerate>
</definition>
<\theorem>
<math|\<bbb-C\>,+,.> is a field and thus a vectorspace where the unit
element is <math|(1,0)> and the neutral element is <math|(0,0)>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|(x<rsub|,>y)+((x<rprime|'>,y<rprime|'>)+(x<rprime|''>,y<rprime|''>))=(x,y)+(x<rprime|'>+x<rprime|''>,y<rprime|'>+y<rprime|''>)=(x+x<rprime|'>+x<rprime|''>,y+y<rprime|'>+y<rprime|''>)=((x+x<rprime|'>)+x<rprime|''>,(y+y<rprime|'>)+y<rprime|''>)=((x,y)+(x<rprime|'>,y<rprime|'>))+(x<rprime|''>,y<rprime|''>)>
<item><math|(0,0)+(x,y)=(0+x,0+y)=(x,y)>
<item><math|(x,y)+(\<um\>x,-y)=(x+(\<um\>x),y+(\<um\>y))=(0,0)>
<item><math|(x,y)+(x<rprime|'>,y<rprime|'>)=(x+x<rprime|'>,y+y<rprime|'>)=(x<rprime|'>+x,y<rprime|'>+y)=(x<rprime|'>,y<rprime|'>)+(x,y)>
<item><math|(x,y)((x<rprime|'>,y<rprime|'>).(x<rprime|''>,y<rprime|''>))=(x,y).(x<rprime|'>.x<rprime|''>-y<rprime|'>.y<rprime|''>,x<rprime|'>.y<rprime|''>+x<rprime|''>.y<rprime|'>)=(x.(x<rprime|'>.x<rprime|''>-y<rprime|'>.y<rprime|''>)-y.(x<rprime|'>.y<rprime|''>+x<rprime|''>.y<rprime|'>),(x<rprime|'>.x<rprime|''>-y<rprime|'>.y<rprime|''>).y+(x<rprime|'>.y<rprime|''>+x<rprime|''>.y<rprime|'>).x)=(x.x<rprime|'>.x<rprime|''>-x.y<rprime|'>.y<rprime|''>-y.x<rprime|'>.y<rprime|''>-y.x<rprime|''>.y<rprime|'>,x<rprime|'>.x<rprime|''>.y-y<rprime|'>.y<rprime|''>.y+x<rprime|'>.y<rprime|''>.x+x<rprime|''>.y<rprime|'>.x)=(x<rprime|''>(x.x<rprime|'>-y.y<rprime|'>)-y<rprime|''>(x.y<rprime|'>+y.x<rprime|'>),x<rprime|''>.(x<rprime|'>y+y<rprime|'>.x)+y<rprime|''>.(x.x<rprime|'>-y<rprime|'>.y))=(x.x<rprime|'>-y.y<rprime|'>,x.y<rprime|'>+y.x<rprime|'>).(x<rprime|''>,y<rprime|''>)=((x,y).(x<rprime|'>,y<rprime|'>)).(x<rprime|''>,y<rprime|''>)>
<item><math|(1,0).(x,y)=(1.x-0.y,1.y+0.x)=x,y)>
<item>Let <math|(x,y)\<neq\>(0,0)\<Rightarrow\>x<rsup|3>.y+x.y<rsup|3>=x(x<rsup|2>.y+y<rsup|3>)=x.y.(x<rsup|2>+y<rsup|2>)\<neq\>0>
(see <reference|kwadrat is positive>, <reference|product of non zero
numbers>) then <math|(x,y).(<frac|x<rsup|2>.y|x<rsup|3>.y+x.y<rsup|3>>,<frac|-y<rsup|2>.x|x<rsup|3>.y+x.y<rsup|3>>)=(<frac|x<rsup|3>.y+y<rsup|3>.x|x<rsup|3>.y+x.y<rsup|3>>,<frac|x<rsup|2>.y<rsup|2>-y<rsup|2>.x<rsup|2>|x<rsup|3>.y+x.y<rsup|3>>)=(1,0)>
<item><math|(x,y).(x<rprime|'>,y<rprime|'>)=(x.x<rprime|'>-y.y<rprime|'>,x.y<rprime|'>+x<rprime|'>.y)=(x<rprime|'>.x-y<rprime|'>.y,x<rprime|'>.y+y<rprime|'>.x)=(x<rprime|'>,y<rprime|'>).(x,y)>
<item><math|(x,y).((x<rprime|'>,y<rprime|'>)+(x<rprime|''>,y<rprime|''>))=(x,y).(x<rprime|'>+x<rprime|''>,y<rprime|'>+y<rprime|''>)=(x.x<rprime|'>+x.x<rprime|''>-y.y<rprime|'>-y.y<rprime|''>,x.y<rprime|'>+x.y<rprime|''>+y.x<rprime|'>+y.x<rprime|''>)=((x.x<rprime|'>-y.y<rprime|'>)+(x.x<rprime|''>-y.y<rprime|''>),(x.y<rprime|'>+y.x<rprime|'>)+(x.y<rprime|''>+y.x<rprime|''>))=(x.x<rprime|'>-y,y<rprime|'>,x.x<rprime|'>+y.x<rprime|'>)+(x.x<rprime|''>-y.y<rprime|''>,x.y<rprime|''>+y.x<rprime|''>)=((x,y).(x<rprime|'>,y<rprime|'>))+((x,y).(x<rprime|''>,y<rprime|''>))>
</enumerate>
</proof>
<\definition>
<index|complex conjugate>Given <math|(x,y)\<in\>\<bbb-C\>> we define
<math|<wide|(x,y)|\<bar\>>=(x,\<um\>y)>
</definition>
<\lemma>
Given <math|z,z<rprime|'>\<in\>\<bbb-C\>> then
<math|<wide|z+z<rprime|'>|\<bar\>>=<wide|z|\<bar\>>+<wide|z<rprime|'>|\<bar\>>>
</lemma>
<\proof>
Let <math|z=(x,y),z<rprime|'>=(x<rprime|'>,y<rprime|'>)> then
<math|<wide|(x,y)+(x<rprime|'>,y<rprime|'>)|\<bar\>>=<wide|(x+x<rprime|'>,y+y<rprime|'>)|\<bar\>>=(x+x<rprime|'>,\<um\>(y+y<rprime|'>))=(x,\<um\>y)+(x<rprime|'>,\<um\>y<rprime|'>)=<wide|(x,y)|\<bar\>>+<wide|(x<rprime|'>,y<rprime|'>)|\<bar\>>>
</proof>
<\theorem>
<label|conjugate of sum>Given a function
<math|s:{1,\<ldots\>,n}\<rightarrow\>\<bbb-C\>> then
<math|<wide|<big|sum><rsub|i=1><rsup|n>s(i)|\<bar\>>=<big|sum><rsub|i=1><rsup|n>s(i)>
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|<big|sum><rsub|i=1><rsup|1><wide|s(i)|\<bar\>>=<wide|s(1)|\<bar\>>=<wide|<big|sum><rsub|i=1><rsup|1>s(i)|\<bar\>>>
<item>Assume the theorem is true for <math|n> then
<math|<wide|<big|sum><rsub|i=1><rsup|n+1>s(i)|\<bar\>>=<wide|(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)|\<bar\>>\<equallim\><rsub|previous
lemma><wide|<big|sum><rsub|i=1><rsup|n>s(i)|\<bar\>>+<wide|s(n+1)|\<bar\>>=(<big|sum><rsub|i=1><rsup|n><wide|s(i)|\<bar\>>)+<wide|s(n+1)|\<bar\>>=<big|sum><rsub|i=1><rsup|n+1><wide|s(i)|\<bar\>>>
</enumerate>
</proof>
<\definition>
<label|absolute value of complex numbers><math|<index|absolute value
(complex)>\<forall\>x\<in\>\<bbb-C\>> we define
<math|\|x\|=<sqrt|x<rsup|2>+y<rsup|2>>> (which is defined as
<math|x<rsup|2>+y<rsup|2>\<geqslant\>0>
</definition>
<\theorem>
<math|\<um\>(1,0)=(\<um\>1,0),(0,1)<rsup|2>=\<um\>(1,0)> and
<math|\<forall\>(x,y)\<in\>\<bbb-C\>> we have
<math|(x,y)=(x,0)+(0,1).(y,0)>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|(1,0)+(\<um\>1,0)=(0,0)\<Rightarrow\>\<um\>(1,0)>
<item><math|(0,1)<rsup|2>=(0,1).(0,1)=(0.0-1.1,0.1+1.0)=(\<um\>1,0)=\<um\>(1,0)>
<item><math|(x,0)+(0,1).(y,0)=(x,0)+(0.y-1.0,0.0+1.y)=(x,0)+(0,y)=(x,y)>
</enumerate>
</proof>
<\theorem>
Let <math|\<bbb-C\><rsub|\<bbb-R\>>={(x,y)\<in\>\<bbb-C\>\|y=0}> The
function <math|e:\<bbb-R\>\<rightarrow\>\<bbb-C\><rsub|\<bbb-R\>>>
defined by <math|e(x)=(x,0)> is a bijective function that preserves
<math|+,.\| \|>(it embeds <math|\<bbb-R\>> in <math|\<bbb-C\>>). Further
note that <math|\<forall\>z<rsub|1>,z<rsub|2>\<in\>\<bbb-C\><rsub|\<bbb-R\>>
we have ><math|z<rsub|1>.z<rsub|2>,z<rsub|1>+z<rsub|2>\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
so <math|\<bbb-C\><rsub|\<bbb-R\>>> forms a subfield of <math|\<bbb-C\>>
</theorem>
<\proof>
First we prove that <math|e> is a bijection\
<\enumerate>
<item>Injectivity, if <math|e(x)=e(x<rprime|'>)\<Rightarrow\>(x,0)=(x<rprime|'>,0)=x=x<rprime|'>>
<item>Surjectivity, then if <math|y=(x,0)\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
we have <math|e(x)=(x,0)=y>
</enumerate>
Further\
<\enumerate>
<item><math|e(x+x<rprime|'>)=(x+x<rprime|'>,0)=(x,0)+(x<rprime|'>,0)=e(x)+e(x<rprime|'>)>
<item><math|e(x.x<rprime|'>)=(x.x<rprime|'>,0)=(x.x<rprime|'>-0.0,x.0+x<rprime|'>.0)=(x,0).(x<rprime|'>,0)=e(x).e(x<rprime|'>)>
<item><math|\|e(x)\|=\|(x,0)\|=<sqrt|x<rsup|2>+0<rsup|2>>=<sqrt|x<rsup|2>>=\|x\|>
</enumerate>
</proof>
Finally we have\
<\enumerate>
<item><math|\<forall\>z<rsub|1>=(x<rsub|1>,0),z<rsub|2>=(x<rsub|2>,0)\<in\>\<bbb-C\>\<Rightarrow\>(x<rsub|1>,0).(x<rsub|2>,0)\<in\>(x<rsub|1>.x<rsub|2>+0.0,x<rsub|1>.0+0.x<rsub|2>)=(x<rsub|1>.x<rsub|2>,0)\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
<item><with|mode|math|\<forall\>z<rsub|1>=(x<rsub|1>,0),z<rsub|2>=(x<rsub|2>,0)\<in\>\<bbb-C\>\<Rightarrow\>(x<rsub|1>,0)+(x<rsub|2>,0)=(x<rsub|1>+x<rsub|2>,0)\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
</enumerate>
<\notation>
<index|real part><index|imaginary part>Motivated by the previous theorems
we can write <math|(x,y)\<in\>\<bbb-C\>> as <math|z=x+i.y> where we note
<math|x=<rsub|note>(x,0)=<rsup|><rsub|def>Re(z),>
<math|i.y=<rsub|note>(0,y)=<rsub|def>Img(z)> (called the real and
imaginary part of <math|z>) and <math|i=(0,1)> we have then
<math|i<rsup|2>=\<um\>1>. For every <math|(x,0),(y,0)\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
we note <math|(x,0)\<leqslant\>(y,0)\<Leftrightarrow\>x\<leqslant\>y>.
Using this notation we can endow <math|\<bbb-C\><rsub|\<bbb-R\>>> with
all the properties of <math|\<bbb-R\>> we have seen so far, more
specifically\
<\enumerate>
<item><math|z<rsub|1>=(x<rsub|1>,0)=<rsub|note>x<rsub|1>,z<rsub|2>=(x<rsub|2>,0)=<rsub|note>x<rsub|2>\<in\>\<bbb-C\><rsub|\<bbb-R\>>\<Rightarrow\>z<rsub|1>\<leqslant\>z<rsub|2>\<Leftrightarrow\>x<rsub|1>\<leqslant\>x<rsub|2>>
<item><math|z=(x,0)=<rsub|note>x\<in\>\<bbb-C\><rsub|\<bbb-R\>>\<Rightarrow\><sqrt|z>=<sqrt|(x,0)>=(<sqrt|x>,0)=<rsub|note><sqrt|x>>
<item><math|z=(x,0)=<rsub|note>x\<in\>\<bbb-C\><rsub|\<bbb-R\>>\<Rightarrow\>\|(x,0)\|=<sqrt|x<rsup|2>+0<rsup|2>>=<sqrt|x<rsup|2>>=\|x\|=<rsub|note>(\|x\|,0)><math|>
</enumerate>
</notation>
<\theorem>
Let <math|z,z<rprime|'>\<in\>\<bbb-C\>> then
<math|<wide|z.z<rprime|'>|\<bar\>>=<wide|z|\<bar\>>.<wide|z<rprime|'>|\<bar\>>>.
</theorem>
<\proof>
Let <math|z=(x,y),z<rprime|'>=(x<rprime|'>,y<rprime|'>)\<Rightarrow\><wide|(x,y).(x<rprime|'>,y<rprime|'>)|\<bar\>>=<wide|(x.x<rprime|'>-y.y<rprime|'>,y.x<rprime|'>+y.x<rprime|'>)|\<bar\>>=(x.x<rprime|'>-y.y<rprime|'>,-y.x<rprime|'>-y.x<rprime|'>)=(x,\<um\>y).(x<rprime|'>,\<um\>y<rprime|'>)=<wide|(x,y)|\<bar\>>.<wide|(x<rprime|'>,y<rprime|'>)|\<bar\>>>
</proof>
<\corollary>
Let <math|z\<in\>\<bbb-C\>,z\<neq\>0> then
<math|<wide|z<rsup|-1>|\<bar\>>=<wide|z|\<bar\>><rsup|-1>>
</corollary>
<\proof>
<math|(1,0)=<wide|(1,0)|\<bar\>>=<wide|z.z<rsup|-1>|\<bar\>>=<wide|z|\<bar\>>.<wide|z<rsub|<rsup|<rsup|>>><rsup|-1>|\<bar\>>\<Rightarrow\><wide|z<rsup|-1>|\<bar\>>=<wide|z|\<bar\>><rsup|-1>>
</proof>
<\theorem>
<label|Re and absolute value><math|\<forall\>z\<in\>\<bbb-C\>> then
<math|Re(z)\<leqslant\>\|y\|>
</theorem>
<\proof>
Let <math|z=(x,y)\<Rightarrow\>Re(z)=x\<leqslant\>\|x\|> and as
<math|\|x\|<rsup|2>\<leqslant\>\|x\|<rsup|2>+\|y\|<rsup|2>\<Rightarrowlim\><rsub|<reference|squareroot
is monotone>>\|x\|=<sqrt|\|x\|<rsup|2>>\<leqslant\><sqrt|\|x\|<rsup|2>+\|y\|<rsup|2>>=\|z\|\<Rightarrow\>Re(z)\<leqslant\>\|z\|>
</proof>
<\theorem>
If <math|z\<in\>\<bbb-C\>> with <math|z=<wide|z|\<bar\>>> then
<math|><math|z=(x,0),x\<in\>\<bbb-R\>> or
<math|z\<in\>\<bbb-C\><rsub|\<bbb-R\>>> and thus\
</theorem>
<\proof>
Let <math|z=(x,y)\<Rightarrow\>(x,y)=(x,-y)\<Rightarrow\>x=x> and
<math|y=-y\<Rightarrow\>2.y=0\<Rightarrow\>y=0\<Rightarrow\>z=(x,0)\<in\>\<bbb-C\><rsub|\<bbb-R\>>>
</proof>
<\theorem>
Let <math|{V<rsub|i>,\<oplus\><rsub|i>,\<odot\><rsub|i>}<rsub|i\<in\>I>>
be a family of vector spaces over a field <math|F,+,.> then
<math|<big|prod><rsub|i\<in\>I>V<rsub|i>,\<oplus\>,\<odot\>> forms a
vector space with <math|\<oplus\>:(<big|prod><rsub|i\<in\>I>V<rsub|i>)\<times\>(<big|prod><rsub|i\<in\>I>V<rsub|i>)\<rightarrow\><big|prod><rsub|i\<in\>I>V<rsub|1>>
defined by <math|\<oplus\>(x,y)(i)=x(i)\<oplus\><rsub|i>y(i)> and
<math|\<odot\>:F\<times\>(<big|prod><rsub|i\<in\>I>V<rsub|i>)\<rightarrow\><big|prod><rsub|i\<in\>I>V<rsub|i>>
is defined by <math|\<odot\>(\<alpha\>,x)(i)=\<odot\><rsub|i>(\<alpha\>,x(i))>
</theorem>
<\proof>
First using <reference|group of products> we have that
<math|<big|prod><rsub|i\<in\>I>V<rsub|i>,\<oplus\>> forms a abelian group
with neutral element <math|0> defined by <math|0(i)=0<rsub|i> (the
neutral element in V<rsub|i>)> then we prove the rest of the vector
axioms so <math|\<forall\>x,y\<in\><big|prod><rsub|i\<in\>I>V<rsub|i>,><math|\<alpha\>,\<beta\>\<in\>F>
<\enumerate>
<item><math|\<forall\>i\<in\>I\<succ\>(\<alpha\>\<odot\>(x\<oplus\>y))(i)=\<alpha\>\<odot\><rsub|i>((x\<oplus\>y)(i))=\<alpha\>\<odot\><rsub|i>(x(i)\<oplus\><rsub|i>y(i))=(\<alpha\>\<odot\><rsub|i>x(i))+(\<alpha\>\<odot\><rsub|i>y(i))=((\<alpha\>\<odot\>x)(i))+(\<alpha\>\<odot\>y)(i))=((\<alpha\>\<odot\>x)+(\<alpha\>\<odot\>y))(i)\<Rightarrow\>\<alpha\>\<odot\>(x\<oplus\>y)=(\<alpha\>\<odot\>x)+(\<alpha\>\<odot\>y)>
<item><math|\<forall\>i\<in\>I\<succ\>((\<alpha\>+\<beta\>)\<odot\>x)(i)=(\<alpha\>+\<beta\>)\<odot\><rsub|i>(x<rsub|i>)=(\<alpha\>\<odot\><rsub|i>x(i))\<oplus\><rsub|i>(\<beta\>\<odot\><rsub|i>x(i))=(\<alpha\>\<odot\>x)(i)\<oplus\><rsub|i>(\<beta\>\<odot\>y)(i)=((\<alpha\>\<odot\>x)+(\<beta\>\<odot\>y))(i)\<Rightarrow\>(\<alpha\>+\<beta\>)\<odot\>x=(\<alpha\>\<odot\>x)\<oplus\>(\<beta\>\<odot\>y)>
<item><math|\<forall\>i\<in\>I\<succ\>((\<alpha\>.\<beta\>)\<odot\>x)(i)=(\<alpha\>.\<beta\>)\<odot\><rsub|i>x(i)=\<alpha\>\<odot\><rsub|i>(\<beta\>\<odot\><rsub|i>x(i))=\<alpha\>\<odot\><rsub|i>(\<beta\>\<odot\>x)(i)=(\<alpha\>\<odot\>(\<beta\>\<odot\>x))(i)\<Rightarrow\>(\<alpha\>.\<beta\>)\<odot\>x=(\<alpha\>\<odot\>(\<beta\>\<odot\>x))>
<item><math|\<forall\>i\<in\>I\<succ\>(1\<odot\>x)(i)=1\<circ\><rsub|i>x(i)=x(i)\<Rightarrow\>1\<odot\>x=x>
</enumerate>
</proof>
<\remark>
If there is no confusion we use the same symbol <math|.> for
<math|\<odot\>,\<odot\><rsub|i>> and <math|+> for
<math|\<oplus\>,\<oplus\><rsub|i>>
</remark>
<\remark>
If <math|I={1,\<ldots\>,n}> and <math|V<rsub|i>=V> are vector spaces over
<math|F> then <math|V<rsup|n>,+,.> forms a vector space by the previous
theorem. Good examples are <math|\<bbb-R\><rsup|n>,+,.> and
<math|\<bbb-C\><rsup|n>,+,.>
</remark>
<\definition>
<index|linear combination>Let <math|V> be a vector space over a field
<math|F> and <math|A\<subseteq\>V> then <math|v\<in\>V> is a linear
combination of <math|A> if and only if there exists a function
<math|f:A\<rightarrow\>F> with <math|support(f)> is finite (and thus
<math|support(f<rsub|A>)> is finite (where <math|f<rsub|A>(a)=f(a).a> so
that <math|<big|sum><rsub|v\<in\>A>f(v).v> is defined) and
<math|v=<big|sum><rsub|v\<in\>A>f(v).v\<equallim\><rsub|note><big|sum><rsub|v\<in\>A>f<rsub|v>.v>
</definition>
<\definition>
<index|linear independent set>Let <math|V> be a vector space over a field
<math|F> and <math|A\<subseteq\>V> then <math|A> is linear independent if
for <math|\<forall\>f:A\<rightarrow\>F> with finite <math|support(f)> and
with <math|0=<big|sum><rsub|v\<in\>A>f(v).v> we have
<math|\<forall\>v\<in\>A\<succ\>f(v)=0 <with|mode|text|or in other words
>support(f)=\<emptyset\>>. In other words every linear combination of
<math|A> that is equal to zero is the trivial combination<math|>
consisting of <math|f=0 > (the constant function <math|0>)
</definition>
<\definition>
<index|linear dependent set>Let <math|V> be a vector space over a field
<math|F> and <math|A\<subseteq\>V> then <math|A> is linear dependent if
it is not independent. Say otherwise <math|\<exists\>f:A\<rightarrow\>F>
with finite <math|support(f)> and with
<math|0=<big|sum><rsub|v\<in\>A>f(v).v> so that
<math|support(f)\<neq\>\<emptyset\>> (<math|\<exists\>v\<in\>A\<vdash\>f(v)\<neq\>0>
</definition>
<\definition>
<index|spanning set>Let <math|V> be a vector space over a field <math|F>
and let <math|A\<subseteq\>V> then <math|\<cal-S\>(V)={v\|\<exists\>f:A\<rightarrow\>F
with support(f) is finite such that v=<big|sum><rsub|v\<in\>A>f(v).v}>.
<math|\<cal-S\>(V)> is called the spanning set of <math|A> and is the set
of all linear combinations of <math|A>.
</definition>
<\definition>
<index|basis>Let <math|V> be a vector space over a field <math|F> and let
<math|B\<subseteq\>V> with <math|B> is linear independent and
<math|\<cal-S\>(B)=V> then <math|B> is called a basis for <math|V>
(remark as <math|\<cal-S\>(B)\<subseteq\>V> we only have to proof that
<math|V\<subseteq\>\<cal-S\>(B)> to proof that <math|S(B)=V>)
</definition>
<\theorem>
Let <math|V> be a vector space over a field <math|F> and let
<math|B\<subseteq\>V> be a basis of <math|V> then each vector can be
written as a unique linear combination of <math|B.>\
</theorem>
<\proof>
By definition of a base every vector in <math|V> is a linear combination
of <math|B> which proves the existence, assume now that <math|v\<in\>V>
that there exists a <math|f:B\<rightarrow\>F>,
<math|f<rprime|'>:B\<rightarrow\>F> with
<math|support(f),support(f<rprime|'>)> finite and
<math|<big|sum><rsub|u\<in\>B>f(u).u=v=<big|sum><rsub|u\<in\>B>f<rprime|'>(u).u>.
Then <math|0=(<big|sum><rsub|u\<in\>B>f(u).u)-(<big|sum><rsub|u\<in\>B>f<rprime|'>(u).u)=(<big|sum><rsub|u\<in\>B>f(u).u)+(\<um\>1).<big|sum><rsub|u\<in\>B>f<rprime|'>(u).u\<equallim\><rsub|<reference|scalair
product and sum>>(<big|sum><rsub|u\<in\>B>f(u).u)+(<big|sum><rsub|u\<in\>B>(\<um\>f<rprime|'>(u)).u\<equallim\><rsub|<reference|sum
of sums>><big|sum><rsub|u\<in\>B>(f(u)-f<rprime|'>(u)).u\<Rightarrowlim\><rsub|B
is linear independent>\<forall\>u\<in\>B\<succ\>f(u)-f<rprime|'>(u)=0\<Rightarrow\>f(u)=f<rprime|'>(u)\<Rightarrow\>f=f<rprime|'>>
</proof>
<\notation>
Let <math|V> be a vector space over a field <math|F> with a finite basis
<math|B> then there exists a bijective function
<math|e:{1,\<ldots\>,n}\<rightarrow\>B> so we can note
<math|><math|B={e<rsub|1>,\<ldots\>,e<rsub|n>}> <math|(e<rsub|i>=e(i))>
and for <math|v\<in\>V> there exists a function <math|f:B\<rightarrow\>F>
such that <math|v=<big|sum><rsub|e\<in\>B>f(e).e\<equallim\><rsub|<reference|general
sum is finite sum in the finite case>><big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>=<big|sum><rsub|i=1><rsup|n>v<rsub|i>.e<rsub|1>>
here we note <math|v<rsub|i>=f(e(i))>, <math|v<rsub|i>> is called the
<math|i>-the coordinate of <math|v>.<math|>
</notation>
<\definition>
<index|basis of <math|\<bbb-R\><rsup|n>>>In <math|\<bbb-R\><rsup|n>> we
define the elements <math|e<rsub|i>\<in\>\<bbb-R\><rsup|n>,i\<in\>{1,\<ldots\>,n}>
as follows <math|\<forall\>j\<in\>{1,\<ldots\>,n}>
<\enumerate>
<item><math|e<rsub|i>(j)=1> if <math|i=j>
<item><math|e<rsub|i>(j)=0> if <math|i\<neq\>j>
</enumerate>
we note this also as <math|e<rsub|i>(j)=\<delta\><rsub|ij>> (where
<math|\<delta\><rsub|i,j>=1> if <math|i=j> and
<math|\<delta\><rsub|ij>=0> if <math|i\<neq\>j>).
</definition>
<\lemma>
<label|sum of basevectors 1>Let <math|m\<in\>\<bbb-N\><rsub|0>> then
<math|\<forall\>j\<gtr\>m> we have if
<math|f:{e<rsub|i>\|i\<in\>{1,\<ldots\>,j}}\<rightarrow\>\<bbb-R\>> then
<math|(<big|sum><rsub|i=1><rsup|m>f(e<rsub|i>).e<rsub|i>)(j)=0>
</lemma>
<\proof>
We proof this by induction on m:
<\enumerate>
<item>If <math|m=1> then if <math|m\<less\>j> then
<math|(<big|sum><rsub|i=1><rsup|1>f(e<rsub|i>).e<rsub|i>)(j)=(f(e<rsub|1>).e<rsub|1>)(j)=f(e<rsub|1>).e<rsub|1>(j)=f(e<rsub|1>).0=0>
<item>Assume that the lemma is true for <math|m> if
<math|j\<gtr\>m+1\<gtr\>m> we have <math|(<big|sum><rsub|i=1><rsup|m+1>f(e<rsub|i>).e<rsub|i>)(j)=((<big|sum><rsub|i=1><rsup|m>f(e<rsub|i>).e<rsub|i>)+f(e<rsub|m+1>).e<rsub|m+1>)(j)=(<big|sum><rsub|i=1><rsup|m>f(e<rsub|i>).e<rsub|i>)(j)+f(e<rsub|m+1>).e<rsub|m+1>(j)\<equallim\><rsub|inudction
hypothese >0+f(e<rsub|m+1>).0=0+0=0>
</enumerate>
</proof>
<\theorem>
<label|sum of basevectors 2>Let <math|f:{e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}\<rightarrow\>\<bbb-R\>>
be a map then <math|\<forall\>j\<in\>{1,\<ldots\>,n}> we have
<math|(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(j)=f(e<rsub|j>)>
</theorem>
<\proof>
We prove this theorem by induction on <math|n>
<\enumerate>
<item>If <math|n=1> then <math|j\<in\>{1,\<ldots\>,1}\<Rightarrow\>j=1>
and <math|(<big|sum><rsub|i=1><rsup|1>f(e<rsub|i>).e<rsub|i>)(1)=(f(e<rsub|1>).e<rsub|1>)(1)=f(e<rsub|1>).e<rsub|1>(1)=f(e<rsub|1>)<rsub|>>
<item>Assume now that the theorem is true for <math|n> and prove it for
<math|n+1>, so let <math|j\<in\>{1,\<ldots\>,n+1}> then
<math|(<big|sum><rsub|i=1><rsup|n+1>f(e<rsub|i>).e<rsub|i>)(j)=(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(j)+f(e<rsub|n+1>).e<rsub|n+1>(j)>,
we have now the following possibilities for <math|j>
<\enumerate>
<item><math|j=n+1> then using <math|n+1\<gtr\>n> and <reference|sum
of basevectors 1> (previous lemma) we find that
<with|mode|math|(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(n+1)+f(e<rsub|n+1>).e<rsub|n+1>(n+1)=0+f(e<rsub|n+1>).1=f(e<rsub|n+1>)\<Rightarrow\><big|sum><rsub|i=1><rsup|n+1>f(e<rsub|i>).e<rsub|i>)(n+1)=f(e<rsub|n+1>)>
<item><math|j\<less\>n+1> then <math|j\<in\>{1,\<ldots\>,n}> and we
can use the induction hypothesis to prove that
<math|(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(j)=f(e<rsub|j>)>
and <math|f(e<rsub|n+1>).e<rsub|n+1>(j)=f(e<rsub|n+1>).0=0\<Rightarrow\>(<big|sum><rsub|i=1><rsup|n+1>f(e<rsub|i>).e<rsub|i>)(j)=f(e<rsub|j>)>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|the canonical basis of product of reals>Given the vector space
<math|\<bbb-R\><rsup|n>> over the field <math|\<bbb-R\>> then
<math|\<cal-B\>={e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}> is a basis for
<math|\<bbb-R\><rsup|n>>
</theorem>
<\proof>
First note that for every <math|f:{e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}\<rightarrow\>\<bbb-R\>>
we have <math|support(f)\<subseteq\>{e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}>
which is finite so <math|f> has always finite support. Second if
<math|v\<in\>\<bbb-R\><rsup|n> then >v is a function
<math|{1,\<ldots\>,n}\<rightarrow\>\<bbb-R\>> define then
<math|f:{e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}\<rightarrow\>\<bbb-R\>> by
<math|f(e<rsub|i>)=v(i)> then we have
<math|(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(j)\<equallim\><rsub|<reference|sum
of basevectors 2>>f(e<rsub|j>)=v(j)\<Rightarrow\><big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>=v>
so <math|\<cal-S\>(\<cal-B\>)=\<bbb-R\>>. The only thing left to proof is
that if <math|f:{e<rsub|i>\|i\<in\>{1,\<ldots\>,n}}\<rightarrow\>\<bbb-R\>>
and <math|0=<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>> then
<math|\<forall\>j\<in\>{1,\<ldots\>,n}> we have
<math|0=0(j)=(<big|sum><rsub|i=1><rsup|n>f(e<rsub|i>).e<rsub|i>)(j)=f(e<rsub|j>)\<Rightarrow\>f=0>
(the constant <math|0> function).\
</proof>
<\definition>
<index|isomorphism>Let <math|X> and <math|Y> be vector spaces over a
field <math|F> then a map <math|f:X\<rightarrow\>Y> is a isomorphism if
and only if
<\enumerate>
<item><math|f> is bijective
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>x,y\<in\>X>
we have <math|f(\<alpha\>.x+\<beta\>.y)=\<alpha\>.f(x)+\<beta\>.f*(y)>
</enumerate>
</definition>
<\theorem>
<label|inverse of isomorphisme>Let <math|X,Y> be vector spaces over a
field and <math|f:X\<rightarrow\>Y> a isomorphism then <math|f<rsup|-1>>
is a isomorphism
</theorem>
<\proof>
Using <reference|inverse of bijective mapping> we have already that
<math|f<rsup|-1>> is a bijection and as
<math|f(\<alpha\>.f<rsup|-1>(x)+\<beta\>.f<rsub|><rsup|-1>(y))=\<alpha\>.f(f<rsup|-1>(x))+\<beta\>.f(f<rsup|-1>(y))=\<alpha\>.x+\<beta\>.y\<Rightarrow\>f<rsup|-1>(\<alpha\>.x+\<beta\>.y)=f<rsup|-1>(f(\<alpha\>.f<rsup|-1>(x)+\<beta\>.f<rsup|-1>(y))=\<alpha\>.f<rsup|-1>(x)+\<beta\>.f<rsup|-1>(y)>
</proof>
<\theorem>
<label|composition of isomorphismes>Let <math|X,Y,Z> be vector spaces
over a field <math|F> and <math|f<rsub|1>:X\<rightarrow\>Y,f<rsub|2>:Y\<rightarrow\>Z>
be isomorphism's then <math|f<rsub|2>\<circ\>f<rsub|1>:X\<rightarrow\>Z>
is also a isomorphism
</theorem>
<\proof>
<math|(f<rsub|2>\<circ\>f<rsub|1>)(\<alpha\>.x+\<beta\>.y)=f<rsub|2>(f<rsub|1>(\<alpha\>.x+\<beta\>.y))=f<rsub|2>(\<alpha\><rsub|1>.f(x)+\<beta\><rsub|1>.f(y))=\<alpha\><rsub|1>f<rsub|2>(f<rsub|1>(x))+\<beta\>.f<rsub|2>(f<rsub|1>(y))=\<alpha\>.(f<rsub|2>\<circ\>f<rsub|1>)(x)+\<beta\>.(f<rsub|2>\<circ\>f<rsub|1>)(y)>
</proof>
<\theorem>
<label|induced vector space structure><index|induced vector space
structure>Let <math|X,+,.> be a vector space over a field <math|F>,
<math|Y> a set and <math|f:X\<rightarrow\>Y> a bijection. If we then
define <math|+<rsub|Y>,.<rsub|Y>> by <math|+<rsub|Y>:Y\<times\>Y\<rightarrow\>Y>
where <math|y<rsub|1>+<rsub|Y>y<rsub|2>=f<rsup|>(f<rsup|-1>(y<rsub|1>)+f(f<rsup|-1>(y<rsub|2>))>
and <math|.<rsub|Y>:F\<times\>Y\<rightarrow\>Y> where
<math|\<alpha\>.<rsub|Y>y=f(\<alpha\>.f<rsup|-1>(y))>. Then
<math|Y,+<rsub|Y>,.<rsub|Y>> is a vector space (the vector space induced
by <math|f>) with the neutral element <math|f(0)>. Further <math|f> is a
isomorphism between <math|X> and <math|Y>.
</theorem>
<\proof>
\;
First we proof that <math|Y,+<rsub|1>> forms a abelian group
<\enumerate>
<item>Associativity. <math|x+<rsub|y>(y+<rsub|Y>z)=x+<rsub|Y>(f(f<rsup|-1>(y)+f<rsup|-1>(z))=f(f<rsup|-1>(x)+f<rsup|-1>(f(f<rsup|-1>(y)+f<rsup|-1>(z)))=<rsup|>f(f<rsup|-1>(x)+(f<rsup|-1>(y)+f<rsup|-1>(z)))=f((f<rsup|-1>(x)+f<rsup|-1>(y))+f<rsup|-1>(z))=f(f<rsup|-1>(f(f<rsup|-1>(x)+f<rsup|-1>(y)))+f<rsup|-1>(z))=f(f<rsup|-1>(x+<rsub|Y>y)+f<rsup|-1>(z))=(x+<rsub|Y>y)+<rsub|Y>z>
<item>Neutral element. <math|x+<rsub|Y>f(0)=f(f<rsup|-1>(x)+f<rsup|-1>(f(0)))=f(f<rsup|-1>(x)+0)=f(f<rsup|-1>(x))=x=f(f<rsup|-1>(x))+f(0+f<rsup|-1>(x))+f(f<rsup|-1>(f(0))+f<rsup|-1>(x))=f(0)+<rsub|Y><rsup|>x>
so the neutral element is <math|f(0)>
<item>Commutativity. <math|x+<rsub|Y>y=f(f<rsup|-1>(x)+f<rsup|-1>(y))=f(f<rsup|-1>(y)+f<rsup|-1>(x))=y+<rsub|Y>x>
</enumerate>
\ Next we prove the remaining requirements of a vector spaces\
<\enumerate>
<item><math|><math|\<alpha\>.<rsub|Y>(x+y)=f(\<alpha\>.f<rsup|-1>(x+y))=f(\<alpha\>.f<rsup|-1>(f(f<rsup|-1>(x)+f<rsup|-1>(y))))=f(\<alpha\>.(f<rsup|-1>(x)+f<rsup|-1>(y)))=f(\<alpha\>.f<rsup|-1>(x)+\<alpha\>.f<rsup|-1>(y))=f(f<rsup|-1>(f(\<alpha\>.f<rsup|-1>(x)))+f<rsup|-1>(f(\<alpha\>.f<rsup|-1>(y))))=f(f<rsup|-1>(\<alpha\>.<rsub|Y>x)+f<rsup|-1>(\<alpha\>.<rsub|Y>y))=\<alpha\>.<rsub|Y>x+<rsub|Y>a.<rsub|Y>y>
<item><math|(\<alpha\>+\<beta\>).<rsub|Y>x=f((\<alpha\>+\<beta\>).f<rsup|-1>(x))=f(\<alpha\>.f<rsup|-1>(x)+\<beta\>.f<rsup|-1>(x))=f(f<rsup|-1>(f(\<alpha\>.f<rsup|-1>(x)))+f<rsup|-1>(f(\<beta\>.f<rsup|-1>(x))))=f(f<rsup|-1>(\<alpha\>.<rsub|Y>x)+f<rsup|-1>(\<beta\>.<rsub|Y>x))=\<alpha\>.<rsub|Y>x+\<beta\>.<rsub|Y>x>
<item><math|(\<alpha\>.\<beta\>).<rsub|Y>x=f((\<alpha\>.\<beta\>).f<rsup|-1>(x))=f(\<alpha\>.(\<beta\>.f<rsup|-1>(x)))=f(\<alpha\>.(f<rsup|-1>(f(\<beta\>.f<rsup|-1>(x))))=\<alpha\>.<rsub|Y>(f(\<beta\>.f<rsup|-1>(x)))=\<alpha\>.<rsub|Y>(\<beta\>.<rsub|Y>x)>
<item><math|1.<rsub|Y>x=f(1.f<rsup|-1>(x))=f(f<rsup|-1>(x))=x>
</enumerate>
Finally we have <math|f(\<alpha\>.x+\<beta\>.y)=f(f<rsup|-1>(f(\<alpha\>.x)+f<rsup|-1>(f(\<beta\>.y)))=f(\<alpha\>.x)+<rsub|Y>f(\<beta\>.y)=f(\<alpha\>.f<rsup|-1>(f(x)))+<rsub|Y>f(\<beta\>.f<rsup|-1>(f(y)))=\<alpha\>.<rsub|Y>f(x)+\<beta\>.<rsub|Y>f(y)>
</proof>
<\theorem>
<label|induced basis on induced vector space>Let <math|X,+,.> be a
n-dimensional vector space with basis
<math|{e<rsub|1>,\<ldots\>,e<rsub|n>},Y> and <math|f:X\<rightarrow\>Y> a
bijection then <math|{f(e<rsub|1>),\<ldots\>,f<rsup|>(e<rsub|n>)}> is a
basis the vector space induced by f (which is then obviously a
n-dimensional vector space)
</theorem>
<\proof>
Assume that <math|<big|sum><rsub|i=1><rsup|n>\<alpha\>(i)f(e<rsub|i>)=0\<Rightarrowlim\><rsub|f
is isomorphism\<Rightarrow\>f<rsup|-1> is a
isomorphism>0=f<rsup|-1>(0)=<big|sum><rsub|i=1><rsup|n>\<alpha\>(i)f<rsup|-1>(f(e<rsub|i>))=<big|sum><rsub|i=1><rsup|n>\<alpha\>(i)e<rsub|i>\<Rightarrow\>\<forall\>i\<in\>{1,\<ldots\>,n}>
we have <math|\<alpha\>(i)=0> proving independence. Also if
<math|y\<in\>Y\<Rightarrow\>f<rsup|-1>(y)\<in\>X\<Rightarrow\>f<rsup|-1>(y)=<big|sum><rsub|i=1><rsup|n>\<alpha\>(i)e<rsub|i>\<Rightarrow\>y=f(f<rsup|-1>(y))=<big|sum><rsub|i=1><rsup|n>\<alpha\>(i)f(e<rsub|i>)>
proving that <math|{f(e<rsub|1>),\<ldots\>,f(e<rsub|n>)}> is a basis.
</proof>
<section|Algebras>
<\definition>
<label|algebra><index|algebra>A algebra over a field <math|F> is a vector
space <math|X,+,.> together with a binary operation
<math|\<bullet\>:X\<times\>X\<rightarrow\>X> such that the following
holds <math|\<forall\>x,y,z\<in\>X,\<alpha\>,\<beta\>\<in\>F>
<\enumerate>
<item><math|(x+y)\<bullet\>z=x\<bullet\>z+y\<bullet\>z> (left
distributivity)
<item><math|x\<bullet\>(y+z)=x\<bullet\>y+x\<bullet\>z> (right
distributivity)
<item><math|(\<alpha\>.x)\<bullet\>(\<beta\>.y)=(\<alpha\>.\<beta\>).(x\<bullet\>y)><math|>
(compatibility with scalars)
</enumerate>
We note the algebra as <math|X,+,.,\<bullet\>>.
The algebra is called commutative if also
4. <math|x\<bullet\>y=y\<bullet\>x>
\
</definition>
<\theorem>
<index|subalgebra><label|subalgebra>Let <math|X,+,.,\<bullet\>> be a
(commutative) algebra over a field <math|F> and let
<math|0\<neq\>Y\<subseteq\>X> define then
<math|Y,+<rsub|<rsub|Y>>,.<rsub|<rsub|Y>>,\<bullet\><rsub|<rsub|Y>>>
where <math|+<rsub|<rsub|Y>>=+<rsub|\|Y\<times\>Y>,
\ \ .<rsub|<rsub|Y>>=.<rsub|\|F\<times\>Y>> and
<math|\<bullet\><rsub|<rsub|Y>>=\<bullet\><rsub|\|Y\<times\>Y>> are the
operators restricted to Y then the following are equivalent
<\enumerate>
<item><math|Y,+<rsub|<rsub|Y>>,.<rsub|<rsub|Y>>,\<bullet\><rsub|<rsub|Y>>>
forms a (commutative) algebra over the field F (Called the subalgebra)
<item>We have the following
<\enumerate>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>F,x,y\<in\>Y> we have
<math|\<alpha\>.x+\<beta\>.y\<in\>Y>
<item><math|\<forall\>x,y\<in\>Y> we have <math|x\<bullet\>y\<in\>Y>
</enumerate>
</enumerate>
<\proof>
\;
<math|1\<Rightarrow\>2> This is trivial
<math|2\<Rightarrow\>1> Using <reference|subspace> we have already that
<math|Y,+<rsub|<rsub|Y>>,.<rsub|<rsub|Y>>> is a vector space. Further
if <math|x\<bullet\><rsub|<rsub|Y>>y=x\<bullet\>y\<in\>Y> making the
inner product a internal operation. Finally we have
<math|\<forall\>x,y,z\<in\>Y,\<alpha\>,\<beta\>\<in\>F> that
<\enumerate>
<item><math|(x+<rsub|<rsub|Y>>y)\<bullet\><rsub|<rsub|Y>>z=(x+y)\<bullet\>z=x\<bullet\>y+x\<bullet\>z=x\<bullet\><rsub|<rsub|Y>>y+<rsub|<rsub|Y>>y\<bullet\><rsub|<rsub|Y>>z>
<item><math|x\<bullet\><rsub|<rsub|Y>>(y+<rsub|<rsub|Y>>z)=x\<bullet\>(y+z)=x\<bullet\>y+x\<bullet\>z=x\<bullet\><rsub|<rsub|Y>>y+<rsub|<rsub|Y>>y\<bullet\><rsub|<rsub|Y>>z>
<item><math|(\<alpha\>.<rsub|<rsub|Y>>x)\<bullet\><rsub|<rsub|Y>>(\<beta\>.<rsub|<rsub|Y>>y)=(\<alpha\>.x)\<bullet\>(\<beta\>.y)=(\<alpha\>.\<beta\>).(x\<bullet\>y)=(\<alpha\>.\<beta\>).<rsub|<rsub|Y>>(x\<bullet\><rsub|<rsub|Y>>y)>
</enumerate>
</proof>
</theorem>
<chapter|Topology>
<section|Topological spaces>
<\definition>
<index|topology><index|topological space><index|open set>A topological
space is a set <math|X> together with a set
<math|\<cal-T\>\<subseteq\>2<rsup|X>> (a set of subsets of <math|X)>,
(elements of <math|\<cal-T\>> are called open sets) such that:
<\enumerate>
<item><math|\<emptyset\>\<in\>\<cal-T\>>
<item><math|X\<in\>\<cal-T\>>
<item><math|\<forall\>U,V\<in\>\<cal-T\>\<succ\>U<big|cap>V\<in\>\<cal-T\>>
<item>If <math|{U<rsub|i>}<rsub|i\<in\>I>> is a family of open sets
(<math|U<rsub|i>\<in\>\<cal-T\>,i\<in\>I)> then
<math|<big|cup><rsub|i\<in\>I>U<rsub|i>\<in\>\<cal-T\>>
</enumerate>
we call <math|\<cal-T\>> the topology of <math|X>
</definition>
<\theorem>
<label|intersection of finite open sets is open>Let <math|X,\<cal-T\>> be
a topological space and let <math|{U<rsub|i>}<rsub|i\<in\>I>> be a finite
family of open sets then <math|<big|cap><rsub|i\<in\>I>U<rsub|i>> is
open.\
</theorem>
<\proof>
Here <math|U:I\<rightarrow\>\<cal-T\>> is a function representing the
family (<math|U<rsub|i>=U(i))> and as <math|I> is finite, there exists a
bijective function <math|i:{1,\<ldots\>,n}\<rightarrow\>I>. so
<math|U\<circ\>i:{1,\<ldots\>,n}\<rightarrow\>\<cal-T\>> where we note
<math|(U\<circ\>i)(j)=U<rsub|i<rsub|j>>>. then
<math|<big|cap><rsub|k\<in\>I>U<rsub|k>=<big|cap><rsub|j\<in\>{1,\<ldots\>,n}>U<rsub|i<rsub|j>>>
(see <reference|families and surjective mappings>). We proof the theorem
now by induction of <math|n>
<\enumerate>
<item><math|n=1\<Rightarrow\><big|cap><rsub|j\<in\>{1}>U<rsub|i<rsub|j>>=U<rsub|>(i(1))>
which is open
<item>Assume the theorem is true for <math|n> then prove it for
<math|n+1> <math|<big|cap><rsub|j\<in\>{1,\<ldots\>,n+1}>U(i(j))=(<big|cap><rsub|j\<in\>{1,\<ldots\>,n}>U(i(j)))<big|cap>U(j(n+1))>
which is open because <math|<big|cap><rsub|j\<in\>{1,\<ldots\>,n}>U(i(j))>
is open by the induction hypothesis and <math|U(j(n+1))> and the
definition of a topology
</enumerate>
</proof>
<\theorem>
<label|subspace topology><index|subspace topology>Let <math|X,\<cal-T\>>
be a topological space and <math|A\<subseteq\>X> then if we define
<math|\<cal-T\><rsub|A>={U<big|cap>A\|U\<in\>\<cal-T\>}> then
<math|A,\<cal-T\><rsub|A>> is a topological space,
<math|\<cal-T\><rsub|A>> is called the subspace topology induced by
<math|\<cal-T\>> on <math|A> \
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<emptyset\>=\<emptyset\><big|cap>A\<in\>\<cal-T\><rsub|A>>
<item><math|A=X<big|cap>A\<in\>\<cal-T\><rsub|A>>
<item>If <math|V<rsub|1>,V<rsub|2>\<in\>\<cal-T\><rsub|A>> then
<math|\<exists\>U<rsub|1>,U<rsub|2>\<in\>\<cal-T\>> such that
<math|V<rsub|1>=U<rsub|1><big|cap>A,V<rsub|2>=U<rsub|2><big|cap>A> then
<math|V<rsub|1><big|cap>V<rsub|2>=(U<rsub|1><big|cap>U<rsub|2>)<big|cap>A\<in\>\<cal-T\><rsub|A>>
<item>If <math|{V<rsub|i>}<rsub|i\<in\>I>> is a family of open sets in
<math|\<cal-T\><rsub|A>> then there exists a family
<math|{U<rsub|i>}<rsub|i\<in\>I>> of open sets in <math|\<cal-T\>> with
<math|V<rsub|i>=U<rsub|i><big|cap>A> then
<math|<big|cup><rsub|i\<in\>I>V<rsub|i>=<big|cup><rsub|i\<in\>I>(U<rsub|i><big|cap>A)\<equallim\><rsub|<reference|union
and intersection of a family>>(<big|cup><rsub|i\<in\>I>U<rsub|i>)<big|cap>A\<in\>\<cal-T\><rsub|A>>
</enumerate>
</proof>
<\theorem>
<label|subspace topology of subspace topology>Let <math|X,\<cal-T\>> be a
topological space and <math|B\<subseteq\>A\<subseteq\>X> then given the
subspace topology induced by <math|\<cal-T\>> on <math|A> (named
<math|\<cal-T\><rsub|A>>). Then <math|\<cal-T\><rprime|'><rsub|B>> the
topology induced by <math|\<cal-T\><rsub|A>> on <math|B> is the same as
the topology <math|\<cal-T\><rsub|B>> induced by <math|\<cal-T\>> on
<math|B>
</theorem>
<\proof>
First we have
<\eqnarray*>
<tformat|<table|<row|<cell|V\<in\>\<cal-T\><rprime|'><rsub|B>>|<cell|\<Rightarrow\>>|<cell|\<exists\>W\<in\>\<cal-T\><rsub|A>\<vdash\>V=W<big|cap>B>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>U\<in\>\<cal-T\>\<vdash\>W=U<big|cap>A\<Rightarrow\>V=U<big|cap>A<big|cap>B\<equallim\><rsub|B\<subseteq\>A>U<big|cap>B>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|V\<in\>\<cal-T\><rsub|B>>>>>
</eqnarray*>
also we have\
<\eqnarray*>
<tformat|<table|<row|<cell|V\<in\>\<cal-T\><rsub|B>>|<cell|\<Rightarrow\>>|<cell|\<exists\>U\<in\>\<cal-T\>\<vdash\>V=U<big|cap>B\<equallim\><rsub|B\<subseteq\>A>U<big|cap>(B<big|cap>A)=(U<big|cap>A)<big|cap>B>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|V\<in\>\<cal-T\><rprime|'><rsub|B>
as U<big|cap>A\<in\>\<cal-T\><rsub|A>>>>>
</eqnarray*>
</proof>
<\remark>
Let <math|X,\<cal-T\>> be a topological space and <math|U\<in\>\<cal-T\>>
then if <math|V\<in\>\<cal-T\><rsub|U>> we have
<math|\<exists\>U<rprime|'>\<in\>\<cal-T\>> such that
<math|V=U<rprime|'><big|cap>U\<in\>\<cal-T\>> (all open sets in the
subspace topology induced by <math|\<cal-T\>> on a open set are open in
<math|\<cal-T\>>).
</remark>
<\definition>
<index|inner set>Let <math|X,\<cal-T\>> be a topological space then the
open set <math|A<rsup|\<circ\>>=<big|cup><rsub|U\<in\>{U\<in\>\<cal-T\>\|U\<subseteq\>A}>U>
is called the inner set of <math|A> (note that as <math|\<emptyset\>> is
open and <math|\<emptyset\>\<subseteq\>A> we have that
<math|{U\<in\>\<cal-T\>\|U\<subseteq\>A}> is not empty and that
<math|A<rsup|\<circ\>>> is thus defined)
</definition>
<\theorem>
Let <math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X>
then \ <math|A<rsup|\<circ\>>> is the biggest open set contained in A
</theorem>
<\proof>
Define <math|\<cal-U\>={U\<in\>\<cal-T\>\|U\<subseteq\>A}> then
<math|A<rsup|\<circ\>>=<big|cup><rsub|U\<in\>\<cal-U\>>U> and is open and
<math|\<subseteq\>A>, now if <math|U> is open and <math|U\<subseteq\>A>
then <math|U\<subseteq\><big|cup><rsub|U\<in\>\<cal-U\>>U=A<rsup|\<circ\>>>
</proof>
<\definition>
<index|closed set>Let <math|X,\<cal-T\>> be a topological space then the
family of closed sets <math|\<cal-C\>> is defined by
<math|\<cal-C\>={A\<in\>2<rsup|X>\|(X<mid|\\>A)\<in\>\<cal-T\>}>
</definition>
<\theorem>
Let <math|X,\<cal-T\>> be a topological space then the family of closed
sets <math|\<cal-C\>> fulfills the following conditions
<\enumerate>
<item><math|\<emptyset\>\<in\>\<cal-C\>>
<item><math|X\<in\>\<cal-C\>>
<item>If <math|A,B\<in\>\<cal-C\>> then
<math|A<big|cup>B\<in\>\<cal-C\>>
<item>If <math|{A<rsub|i>}<rsub|i\<in\>I>> is a family of closed sets
(<math|A<rsub|i>\<subseteq\>\<cal-C\>>) then
<math|<big|cap><rsub|i\<in\>I>A<rsub|i>\<in\>\<cal-C\>>
</enumerate>
furthermore when <math|\<cal-C\>\<in\>2<rsup|X>> is such that it fulfills
1,2,3 and 4 then <math|\<cal-T\>={U\|U=X<mid|\\>A> where
<math|A\<in\>\<cal-C\>}> is a topology and the closed sets in this
topology are the sets in <math|\<cal-C\>>
</theorem>
<\proof>
\;
First is <math|\<cal-T\>> be a topology on <math|X> and <math|\<cal-C\>>
its set of closed sets, then\
<\enumerate>
<item><math|X<mid|\\>\<emptyset\>=X\<in\>\<cal-T\>\<Rightarrow\>\<emptyset\>\<in\>\<cal-C\>>
<item><math|X<mid|\\>X=\<emptyset\>\<in\>\<cal-T\>\<Rightarrow\>X\<in\>\<cal-C\>>
<item><math|X<mid|\\>(A<big|cup>B)=(X<mid|\\>A)<big|cap>(X<mid|\\>B)\<in\>\<cal-T\>\<Rightarrow\>A<big|cup>B\<in\>\<cal-C\>>
<item><math|X<mid|\\>(<big|cap><rsub|i\<in\>I>A<rsub|i>)=<big|cup><rsub|i\<in\>I>(X<mid|\\>A<rsub|i>)\<in\>\<cal-T\>\<Rightarrow\><big|cap><rsub|i\<in\>I>A<rsub|i>\<in\>\<cal-C\>>
</enumerate>
Now assume that the set <math|\<cal-C\>> of subsets of <math|X> fulfills
1,2,3 and 4 and define <math|<with|mode|text|<math|\<cal-T\>={U\|U=X<mid|\\>A>
where <math|A\<in\>\<cal-C\>}>>> then\
<\enumerate>
<item><math|\<emptyset\>=X<mid|\\>X\<in\>\<cal-C\>>
<item><math|X=X<mid|\\>\<emptyset\>\<in\>\<cal-C\>>
<item><math|(X<mid|\\>A)<big|cap>(X<mid|\\>B)=X<mid|\\>(A<big|cup>B)\<in\>\<cal-T\>>
<item><math|<big|cup><rsub|i\<in\>I>(X<mid|\\>A<rsub|i>)=X<mid|\\>(<big|cap><rsub|i\<in\>I>A<rsub|i>)>
</enumerate>
\ so <math|\<cal-T\>> is a topology, now if <math|A\<in\>\<cal-C\>> then
<math|X<mid|\\>A\<in\>\<cal-T\>> and thus <math|A> is closed in
<math|\<cal-T\>>, further if <math|A> is closed in <math|\<cal-T\>> then
<math|X<mid|\\>A\<in\>\<cal-T\>> then
<math|\<exists\>A<rprime|'>\<in\>\<cal-C\>> such that
<math|X<mid|\\>A=X<mid|\\>A<rprime|'>> then using \ <reference|difference
of difference of sets> <math|A<rprime|'>=X<mid|\\>(X<mid|\\>A<rprime|'>)=X<mid|\\>(X<mid|\\>A)\<equallim\><rsub|<reference|difference
of difference of sets>>A\<Rightarrow\>A\<in\>\<cal-C\>>
</proof>
<\theorem>
Let <math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X>
then the set of closed sets in the subspace topology of <math|A> is
<math|{C<big|cap>A\|C> is closed in <math|X,\<cal-T\>}=\<cal-A\>>
</theorem>
<\proof>
Let <math|B> be closed in <math|A> then <math|A<mid|\\>B> is open in in
<math|A> so there exists <math|U\<in\>\<cal-T\>>(<math|U> is open) with
<math|A<mid|\\>B=U<big|cap>A> then <math|B=A<mid|\\>(A<mid|\\>B)=A<mid|\\>(U<big|cap>A)=(A<mid|\\>U)<big|cup>(A<mid|\\>A)=A<mid|\\>U\<Rightarrow\>(X<mid|\\>U)<big|cap>A\<in\>\<cal-A\>>
(because <math|X<mid|\\>U> is closed in <math|X,\<cal-T\>>
If <math|B\<in\>\<cal-A\>> then there exists <math|C> such that
<math|X<mid|\\>C> is open in <math|X,\<cal-T\>> and
<math|B=C<big|cap>A\<Rightarrow\>A<mid|\\>B=A<mid|\\>(C<big|cap>A)=(A<mid|\\>C)<big|cup>(A<mid|\\>A)=A<mid|\\>C=(X<mid|\\>C)<big|cap>A>
which is open in the subspace topology of A
</proof>
<\corollary>
Let <math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X>
be a closed subset then all the closed sets in the subspace topology on
<math|A> are closed in <math|X,\<cal-T\>>
</corollary>
<\definition>
<index|closure>Let <math|X,\<cal-T\>> be a topological space and
<math|A\<subseteq\>X> then <math|<wide|A|\<bar\>>=<big|cap><rsub|C\<in\>{C\|C
is closed and A\<subseteq\>C>C> is closed and is called the closure of
<math|A>. (the intersection is defined because <math|X\<supseteq\>A> and
<math|X> is closed)
</definition>
<\theorem>
Let <math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X>
then <math|A> the smallest closed set containing A is
<math|<wide|A|\<bar\>>>
</theorem>
<\proof>
Define <math|\<cal-A\>={C\|C is closed and A\<subseteq\>C}> then
<math|<wide|A|\<bar\>>=<big|cap><rsub|C\<in\>\<cal-A\>>C> is closed and
contains <math|A>. Further if <math|C> is closed and
<math|A\<subseteq\>C> then <math|<wide|A|\<bar\>>\<supseteq\>C>
</proof>
<\remark>
<label|characterization of closed sets 1>Let <math|X,\<cal-T\>> be a set
then <math|A> is closed iff <math|A=<wide|A|\<bar\>>>
</remark>
<\proof>
If <math|A=<wide|A|\<bar\>>> then <math|A> is closed as
<math|<wide|A|\<bar\>>> is closed. If <math|A> is closed then using that
<math|<wide|A|\<bar\>>> is the smallest closed set containing <math|A> we
have \ <math|A\<subseteq\><wide|A|\<bar\>>\<subseteq\>A\<Rightarrow\>A=<wide|A|\<bar\>>>
</proof>
<\definition>
<index|limit point><index|accumulation point><index|derived set>Let
<math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X> then
<math|x\<in\>X> is a limit point or accumulation point of <math|A> iff
<math|\<forall\>U\<in\>\<cal-T\>> with <math|x\<in\>U> we have
<math|\<emptyset\>\<neq\>(A<mid|\\>{x})<big|cap>U>. The set of all limit
points of <math|A> is called the derived set of <math|A> denoted by
<math|A<rprime|'>>.
</definition>
<\theorem>
<label|characterization of closure>Let <math|X,\<cal-T\>> be a
topological space and <math|A\<subseteq\>X> then
<math|<wide|A|\<bar\>>=A<big|cup>A<rprime|'>> (this means that
<math|x\<in\><wide|A|\<bar\>>\<Leftrightarrow\>\<forall\>U\<in\>\<cal-T\>\<vdash\>x\<in\>U>
we have <math|U<big|cap>A=\<emptyset\>>)
</theorem>
<\proof>
If <math|x\<in\><wide|A|\<bar\>>> then if
<math|x\<in\>A\<Rightarrow\>x\<in\>A<big|cup><wide|A|\<bar\>>>, if
<math|x\<nin\>A> then <math|x\<in\>A<rprime|'>> [for if
<math|x\<nin\>A<rprime|'>\<Rightarrow\>\<exists\>U\<in\>\<cal-T\>\<vdash\>x\<in\>U>
such that <math|A<big|cap>U=(A<mid|\\>{x})<big|cap>U=\<emptyset\>\<Rightarrow\>A\<subseteq\>(X<mid|\\>U)>
which is closed and thus <math|x\<in\><wide|A|\<bar\>>\<subseteq\>(X<mid|\\>U)\<Rightarrow\>x\<nin\>U>
a contradiction]
Assume now <math|x\<in\>A<big|cup>A<rprime|'>> then either
<math|x\<in\>A\<subseteq\><wide|A|\<bar\>>> or <math|x\<in\>A<rprime|'>>
and then if <math|x\<nin\><wide|A|\<bar\>>> we have
<math|x\<in\>X<mid|\\><wide|A|\<bar\>>> which is open and thus
<math|(X<mid|\\><wide|A|\<bar\>>)<big|cap>(A<mid|\\>{x})\<neq\>\<emptyset\>\<Rightarrow\>(A<mid|\\>{x})<mid|\\><wide|A|\<bar\>>\<neq\>\<emptyset\>>
which is a contradiction as <math|A\<subseteq\><wide|A|\<bar\>>> and so
we must have also in this case <math|x\<in\><wide|A|\<bar\>>>
</proof>
<\definition>
<index|finer topology>Let <math|X> be a set and
<math|\<cal-T\><rsub|1>,\<cal-T\><rsub|2>> two topologies for <math|X>
then <math|\<cal-T\><rsub|2>> is finer then <math|\<cal-T\><rsub|1>> iff
<math|\<cal-T\><rsub|1>\<subseteq\>\<cal-T\><rsub|1>>
</definition>
<\note>
If <math|\<cal-T\><rsub|1>> is finer then <math|\<cal-T\><rsub|2>> and
<math|\<cal-T\><rsub|2>> is finer then <math|\<cal-T\><rsub|1>> then
<math|\<cal-T\><rsub|1>=\<cal-T\><rsub|2>>
</note>
<\definition>
<index|basis (topology)>Let <math|X,\<cal-T\>> be a topological space
then <math|\<cal-B\>\<subseteq\>\<cal-T\>> is a basis for the topology
<math|\<cal-T\>> if every open set \ is the union of sets in
<math|\<cal-B\>>
</definition>
<\theorem>
<label|basis of suptopology>Let <math|X,\<cal-T\>> be a topological space
with a basis <math|\<cal-B\>> and <math|A\<subseteq\>X> then
<math|\<cal-B\><rsub|A>={B<big|cap>A\|B\<in\>\<cal-B\>}> is a basis for
the subspace topology <math|\<cal-T\><rsub|A>> on <math|A>
</theorem>
<\proof>
Let <math|U\<in\>\<cal-T\><rsub|A>\<Rightarrow\>\<exists\>W\<in\>\<cal-T\>\<succ\>U=W<big|cap>A>
now by the definition of a basis there exists a family of open sets
<math|{W<rsub|i>}<rsub|i\<in\>I>,\<forall\>i\<in\>I\<succ\>W<rsub|i>\<in\>\<cal-B\>>
such that <math|W=<big|cup><rsub|i\<in\>I>W<rsub|i>\<Rightarrow\>U=W<big|cap>A=<big|cup><rsub|i\<in\>I>(W<rsub|i><big|cap>A)>
where <math|\<forall\>i\<in\>I\<succ\>W<rsub|i><big|cap>A\<in\>\<cal-B\><rsub|A>>
</proof>
<\theorem>
<label|characterization of a basis in a topology>Let <math|X,\<cal-T\>>
be a topological space and let <math|\<cal-C\>\<subseteq\>\<cal-T\>> such
that <math|\<forall\>U\<in\>\<cal-T\>> and <math|\<forall\>x\<in\>U> we
have <math|\<exists\>C\<in\>\<cal-C\>\<succ\>x\<in\>C\<subseteq\>U> then
<math|\<cal-C\>> is a basis
</theorem>
<\proof>
Let <math|U\<in\>\<cal-T\>> then <math|\<forall\>x\<in\>U> there exists a
<math|C<rsub|x>\<in\>\<cal-C\>> such that
<math|x\<in\>\<cal-C\><rsub|x>\<subseteq\>U> then
<math|U=<big|cup><rsub|x\<in\>U>U<rsub|x>> and <math|\<cal-C\>> is thus a
basis
</proof>
<\theorem>
<label|conditions to make a basis>Let <math|X,\<cal-T\>> be a topological
space and let <math|\<cal-B\>> be a basis for the topology then\
<\enumerate>
<item><math|\<forall\>x\<in\>X> we have
<math|\<exists\>B\<in\>\<cal-B\>\<vdash\>x\<in\>B>
<item><math|\<forall\>B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>> if
<math|x\<in\>B<rsub|1><big|cap>B<rsub|2>> then
<math|\<exists\>B<rsub|3>\<in\>\<cal-B\>\<vdash\>x\<in\>B<rsub|3>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
</enumerate>
<\proof>
\;
<\enumerate>
<item>As <math|X\<in\>\<cal-T\>\<Rightarrow\>X=<big|cup><rsub|B\<in\>\<cal-I\>\<subseteq\>\<cal-B\>>B\<Rightarrow\>\<forall\>x\<in\>X>
<math|\<exists\>B\<in\>\<cal-I\>\<subseteq\>B> such that
<math|x\<in\>B>
<item><math|\<forall\>B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>> if
<math|x\<in\>B<rsub|1><big|cap>B<rsub|2>> \ then
<math|x\<in\>B<rsub|1><big|cap>B<rsub|2>=<big|cup><rsub|B\<in\>\<cal-I\>\<subseteq\>\<cal-B\>>B\<Rightarrow\>\<exists\>B\<in\>\<cal-I\>\<subseteq\>\<cal-B\>>
such that <math|x\<in\>B\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
</enumerate>
</proof>
</theorem>
<\theorem>
<label|generating basis in topology><index|generation basis>Let <math|X>
be a set and <math|\<cal-B\>\<subseteq\>2<rsup|X>> be a set of subsets of
<math|X> fulfilling
<\enumerate>
<item><math|\<forall\>x\<in\>X\<succ\>\<exists\>B\<in\>\<cal-B\>\<vdash\>x\<in\>B>
<item><math|\<forall\>B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>> if
<math|x\<in\>B<rsub|1><big|cap>B<rsub|2>> then
<math|\<exists\>B<rsub|3>\<vdash\>x\<in\>B<rsub|3>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
</enumerate>
then <math|\<cal-T\>={U\<subseteq\>X\|\<forall\>x\<in\>U\<succ\>\<exists\>B\<in\>\<cal-B\>>
such that <math|x\<in\>B\<subseteq\>U}> is a topology and has
<math|\<cal-B\>> as its basis. <math|\<cal-B\>> is called the generating
basis for <math|\<cal-T\>> and <math|\<cal-T\>> is called the topology
generated by <math|\<cal-B\>>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<emptyset\>\<in\>\<cal-T\>> because every element of the
empty set fulfills every condition
<item><math|X\<in\>\<cal-T\>> for if <math|x\<in\>X> then
<math|\<exists\>B\<in\>\<cal-B\>> such that
<math|x\<in\>B\<subseteq\>X>
<item>Let <math|\<cal-I\>\<subseteq\>\<cal-T\>> then
<math|\<forall\>x\<in\><big|cup><rsub|U\<in\>\<cal-I\>>U> there exists
a <math|U\<in\>\<cal-I\>\<subseteq\>\<cal-T\>> such that
<math|x\<in\>U> and thus <math|\<exists\>B\<in\>\<cal-B\>\<succ\>x\<in\>B\<subseteq\>U\<subseteq\><big|cup><rsub|U\<in\>\<cal-I\>>U\<Rightarrow\><big|cup><rsub|U\<in\>\<cal-I\>>U\<in\>\<cal-T\>>
<item>If <math|U<rsub|1>,U<rsub|2>\<in\>\<cal-T\>> and let
<math|x\<in\>U<rsub|1><big|cap>U<rsub|2>> then
<math|\<exists\>B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>> such that
<math|x\<in\>B<rsub|1>\<subseteq\>U<rsub|1>,x\<in\>B<rsub|2>\<subseteq\>U<rsub|2>\<Rightarrow\>x\<in\>B<rsub|1><big|cap>B<rsub|2>\<subseteq\>U<rsub|1><big|cap>U<rsub|2>>
and <math|\<exists\>B<rsub|3>\<in\>\<cal-B\>> such that
<math|x\<in\>B<rsub|3>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>\<subseteq\>U<rsub|1><big|cap>U<rsub|2>\<Rightarrow\>B<rsub|1><big|cap>B<rsub|2>\<in\>\<cal-T\>>
</enumerate>
This proves that <math|\<cal-T\>> is a topology. Now if
<math|U\<in\>\<cal-T\>> then <math|\<forall\>x\<in\>U> there exists a
<math|B<rsub|x>\<in\>\<cal-B\>> such that
<math|x\<in\>B<rsub|x>\<subseteq\>U> and thus
<math|U=<big|cup><rsub|x\<in\>U>B<rsub|x>> which proves that
<math|\<cal-B\>> is a basis of the generated topology
</proof>
<\theorem>
<label|characterization of closure>Let <math|X,\<cal-T\>> be a
topological space and <math|A\<subseteq\>X> then\
<\enumerate>
<item><math|x\<in\><wide|A|\<bar\>>\<Leftrightarrow\>\<forall\>U\<in\>\<cal-T\>\<vdash\>x\<in\>U>
we have <math|U<big|cap>A\<neq\>\<emptyset\>>
<item>If <math|\<cal-B\>> is a basis of the topology <math|\<cal-T\>>
then <math|x\<in\><wide|A|\<bar\>>\<Leftrightarrow\>\<forall\>U\<in\>\<cal-B\>>
with <math|x\<in\>U> we have <math|U<big|cap>A\<neq\>\<emptyset\>>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>If <math|x\<in\><wide|A|\<bar\>>> then assume that there exists a
<math|U\<in\>\<cal-T\>> such that <math|U<big|cap>A=\<emptyset\>> and
<math|x\<in\>U\<Rightarrow\>x\<nin\>X<mid|\\>U\<supseteq\>A> and as
<math|X<mid|\\>U> is closed we have
<math|X<mid|\\>A\<supseteq\><wide|A|\<bar\>>> and thus
<math|x\<in\><wide|A|\<bar\>>> a contradiction, so
<math|U<big|cap>A=\<emptyset\> <space|-0.2spc>
\<forall\>U\<in\>\<cal-T\>>. On the other hand let <math|x\<in\>X> such
that <math|\<forall\>U\<in\>\<cal-T\>\<vdash\>x\<in\>U> we have
<math|U<big|cap>A\<neq\>\<emptyset\>> then if
<math|x\<nin\><wide|A|\<bar\>>> we have
<math|x\<subset\>X<mid|\\><wide|A|\<bar\>>> which is open and thus
<math|\<emptyset\>\<neq\>(X<mid|\\><wide|A|\<bar\>>)<big|cap>A=(A<mid|\\><wide|A|\<bar\>>)=\<emptyset\>>
a contradiction
<item>Let <math|x\<in\><wide|A|\<bar\>>> then if
<math|U\<in\>\<cal-B\>\<subseteq\>\<cal-T\>\<vdash\>x\<in\>U> we have
by (1) that <math|U<big|cap>A\<neq\>\<emptyset\>>. On the other hand
assume that <math|x\<in\>X> has <math|\<forall\>U\<in\>\<cal-B\>\<vdash\>x\<in\>U>
we have <math|U<big|cap>A=\<emptyset\>> then if
<math|W\<in\>\<cal-T\>\<vdash\>x\<in\>W> we have the existence of a
<math|B\<in\>\<cal-B\>> such that <math|x\<in\>B\<subseteq\>W> and thus
<math|\<emptyset\>\<neq\>A<big|cap>B\<subseteq\>A<big|cap>W> and thus
using (1) we have that <math|x\<in\><wide|A|\<bar\>>>
</enumerate>
</proof>
<\theorem>
<label|characterization of closed sets 2>Let <math|X,\<cal-T\>> be a
topological space then if <math|A\<subset\>X> is closed we have\
<\enumerate>
<item><math|x\<in\>A\<Leftrightarrow\>\<forall\>U\<in\>\<cal-T\>\<vdash\>x\<in\>U>
we have <math|U<big|cap>A\<neq\>\<emptyset\>>
<item>If <math|\<cal-B\>> is a basis of the topology <math|\<cal-T\>>
then <math|x\<in\>A\<Leftrightarrow\>\<forall\>U\<in\>\<cal-B\>> with
<math|x\<in\>U> we have <math|U<big|cap>A\<neq\>\<emptyset\>>
</enumerate>
</theorem>
<\proof>
This is trivial using <reference|characterization of closed sets 1> and
<reference|characterization of closure>
</proof>
<\theorem>
<label|finer basis>Let be a set and <math|\<cal-T\><rsub|1>,\<cal-T\><rsub|2>>
two topologies on <math|X> with basis
<math|\<cal-B\><rsub|1>,\<cal-B\><rsub|2>> then the following are
equivalent
<\enumerate>
<item><math|\<cal-T\><rsub|2>> is finer then <math|\<cal-T\><rsub|1>>
<item><math|\<forall\>x\<in\>X,\<forall\>B\<in\>\<cal-B\><rsub|1>\<vdash\>x\<in\>B>
there <math|\<exists\>B<rprime|'>\<in\>\<cal-B\><rsub|2>> such that
<math|x\<in\>B<rprime|'>\<subseteq\>B>
</enumerate>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>If <math|x\<in\>X> and
<math|x\<in\>B\<in\>\<cal-B\><rsub|1>\<subseteq\>\<cal-T\><rsub|1>\<subseteq\>\<cal-T\><rsub|2>>
(<math|\<cal-T\><rsub|2>> is finer then
<math|\<cal-T\><rsub|1>)\<Rightarrow\>B\<in\>\<cal-B\><rsub|2>> so
<math|\<exists\>B<rprime|'>\<in\>\<cal-B\><rsub|2>> such that
<math|x\<in\>B<rprime|'>\<subseteq\>B>
<math|\<Leftarrow\>>Let <math|U\<in\>\<cal-T\><rsub|1>> then
<math|\<forall\>x\<in\>U\<succ\>\<exists\>B<rsub|x>\<in\>\<cal-B\><rsub|1>>
such that <math|x\<in\>B<rsub|x>\<subseteq\>U<rsub|>> and using (2)
<math|\<exists\>B<rsub|x><rprime|'>\<in\>\<cal-B\><rsub|2>> such that
<math|x\<in\>B<rprime|'><rsub|x>\<subseteq\>B<rsub|x>\<subseteq\>U\<Rightarrow\>U=<big|cup><rsub|x\<in\>U>B<rprime|'><rsub|x>>
</proof>
<\theorem>
<index|subbasis (topology)><label|sub basis topology>Let <math|X> be a
set and <math|\<cal-S\>\<subseteq\>2<rsup|X>> then the set
<math|\<cal-B\>={<big|cap><rsub|U\<in\>\<cal-I\>>U\|\<cal-I\>> a finite
subset of <math|\<cal-S\><big|cup>{X}}> forms a basis and generates thus
a topology. This topology is called the topology generated by the
subbasis <math|\<cal-S\>>
</theorem>
<\proof>
We have to proof that <reference|generating basis in topology> is
fulfilled
<\enumerate>
<item>Let <math|x\<in\>X> then <math|x\<in\>X=<big|cup><rsub|U\<in\>{X}>U\<in\>\<cal-B\>>
<item>Let <math|x\<in\>B<rsub|1><big|cap>B<rsub|2>> where
<math|B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>> then there exists finite
subsets <math|\<cal-I\><rsub|1>,\<cal-I\><rsub|2>\<subseteq\>\<cal-S\><big|cup>{x}>
such that <math|B<rsub|1>=<big|cap><rsub|U\<in\>\<cal-I\><rsub|1>>U,B<rsub|2>=<big|cap><rsub|U\<in\>\<cal-I\><rsub|2>>U>
then <math|B<rsub|1><big|cap>B<rsub|2>=<big|cap><rsub|U\<in\>\<cal-I\><rsub|1><big|cup>\<cal-I\><rsub|2>>U\<in\>\<cal-B\>>
(because <math|\<cal-I\><rsub|1><big|cup>\<cal-I\>> is a finite subset
of <math|S<big|cup>{x}>
</enumerate>
</proof>
<\definition>
<label|box topology><index|box topology>Let
<math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>> be a family of
topological spaces, then the box topology on the
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> is the topology generated by
the basis <math|\<cal-B\>={<big|prod><rsub|i\<in\>I>U<rsub|i>\|U<rsub|i>\<in\>\<cal-T\><rsub|i>}.>
we have of course to prove that <math|\<cal-B\>> fulfills the conditions
of <reference|generating basis in topology>. The topology is called the
box topology
</definition>
<\proof>
\;
<\enumerate>
<item>Let <math|x\<in\><big|prod><rsub|i\<in\>I>X<rsub|i>> now as
<math|X<rsub|i>\<in\>\<cal-T\><rsub|i>> we have
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>\<in\>\<cal-B\>>
<item>Let <math|U<rsub|1>=<big|prod><rsub|i>U<rsub|i,1>,U<rsub|2>=<big|prod><rsub|i\<in\>I>U<rsub|i,2>>
where <math|U<rsub|i,1>,U<rsub|i,2>\<in\>\<cal-T\><rsub|i>>. then as
<math|U<rsub|i,1><big|cap>U<rsub|i,2>\<in\>\<cal-T\><rsub|i>> we have
<math|U<rsub|1><big|cap>U<rsub|2>=(<big|prod><rsub|i\<in\>I>U<rsub|i,1>)<big|cap>(<big|prod><rsub|i\<in\>I>U<rsub|i,2>)\<equallim\><rsub|<reference|union
and intersection of a family>><big|prod><rsub|i\<in\>I>(U<rsub|i,1><big|cap>U<rsub|i,2>)\<in\>\<cal-B\>>
</enumerate>
</proof>
<\theorem>
<label|box topology and base>Let <math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>>
be a family of topological spaces with
<math|\<forall\>i\<in\>I\<succ\>\<cal-B\><rsub|i>> is a basis of
<math|\<cal-T\><rsub|i>>. Then <math|\<cal-B\>={<big|prod><rsub|i\<in\>I>B<rsub|i>\|B<rsub|i>\<in\>\<cal-B\><rsub|i>,\<forall\>i\<in\>I}>
is a basis for the box topology\
</theorem>
<\proof>
Let <math|x\<in\>U> open in the box topology then there exists a family
of <math|{U<rsub|i>}<rsub|i\<in\>I>> with
<math|\<forall\>i\<in\>I\<succ\>U<rsub|i>\<in\>\<cal-T\><rsub|i>> and
<math|x\<in\><big|prod><rsub|i\<in\>I>U<rsub|i>\<Rightarrow\>\<forall\>i\<in\>I\<succ\>x<rsub|i>=x(i)\<in\>U<rsub|i>>
and thus <math|\<forall\>i\<in\>I\<succ\>\<exists\>B<rsub|i>\<in\>\<cal-B\><rsub|i>>
such that <math|x(i)\<in\>B<rsub|i>\<subseteq\>U<rsub|i>\<Rightarrow\>x\<in\><big|prod><rsub|i\<in\>I>B<rsub|i>\<subseteq\><big|prod><rsub|i\<in\>I>U<rsub|i>=U>
which by the fact that <math|<big|prod><rsub|i\<in\>I>B<rsub|i>\<in\>\<cal-B\>>
and <reference|characterization of a basis in a topology> proves that
<math|\<cal-B\>> is a basis
</proof>
<\definition>
<label|product topology><index|product topology>Let
<math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>> be a family of
topological spaces then the product topology on
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> is defined by the subbasis
<math|\<cal-S\>={\<pi\><rsub|i><rsup|-1>(V)\|i\<in\>I,V\<in\>\<cal-T\><rsub|i>}>
</definition>
<\theorem>
<label|basis of product topology>Let <math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>>
be a family of topological spaces then the product topology on
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> has as its basis
<math|\<cal-B\>={<big|prod><rsub|i\<in\>I>U<rsub|i>\|\<exists\>finite
A\<subseteq\>I> such that <math|\<forall\>i\<in\>I<mid|\\>A\<succ\>U<rsub|i>=X<rsub|i>>
and <math|\<forall\>i\<in\>A\<succ\>U<rsub|i>\<in\>\<cal-T\><rsub|i}>>
</theorem>
<\proof>
Let <math|\<cal-S\>={\<pi\><rsub|i><rsup|-1>(V)\|i\<in\>I,V\<in\>\<cal-T\><rsub|i>}>
the subbasis of the product topology, then as
<math|\<forall\>i\<in\>I,\<forall\>x\<in\>X> we have
<math|\<pi\><rsub|i>(x)=x(i)\<in\>X<rsub|i>\<in\>\<cal-T\><rsub|i>> we
have <math|\<pi\><rsub|i><rsup|-1>(X<rsub|i>)=X\<Rightarrow\>X\<in\>\<cal-S\>>
and thus <math|\<cal-S\><big|cup>{X}=\<cal-S\>>. The basis of the product
topology is then by definition <math|\<cal-B\><rprime|'>={<big|cap><rsub|A\<in\>\<cal-A\>>A>
where <math|\<cal-A\>> is a finite subset of <math|\<cal-S\>}>. So if
<math|U\<in\>\<cal-B\><rprime|'>\<Rightarrow\>\<exists\>\<cal-A\>={A<rsub|1>,\<ldots\>,A<rsub|n>}\<subseteq\>\<cal-S\>>
such that <math|U=<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>A<rsub|i>> where
<math|A<rsub|i>=\<pi\><rsub|j<rsub|i>><rsup|-1>(U<rsub|i>)\<equallim\><rsub|<reference|inverse
of projection>><big|prod><rsub|k\<in\>I>B<rsub|i<rsub|i>,k>> where
<math|B<rsub|j<rsub|i>,k>=U<rsub|i>\<in\>\<cal-T\><rsub|j<rsub|i>>=\<cal-T\><rsub|k>>
if <math|j<rsub|i>=k> and <math|B<rsub|j<rsub|i>,k>=X<rsub|k>\<in\>\<cal-T\><rsub|k>>
if <math|j<rsub|i>\<neq\>k> then using <reference|union and intersection
of a family> we have <math|U=<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>(<big|prod><rsub|k\<in\>I>B<rsub|j<rsub|i>,k>)=<big|prod><rsub|k\<in\>I>(<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|j<rsub|i>,k>)>
and if <math|k\<in\>{j<rsub|1>,\<ldots\>,j<rsub|n>}<rsub|>> then
<math|<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|j<rsub|i>,k>><math|\<in\>\<cal-T\><rsub|k>>
and if <math|k\<nin\>{j<rsub|1>,\<ldots\>,j<rsub|n>}> then
<math|<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|j<rsub|i>,k>=X<rsub|k>>
so we have proved that <math|U\<in\>\<cal-B\>>. Now if
<math|U\<in\>\<cal-B\>> then <math|\<exists\>fA\<subseteq\>I,A> finite
such that <math|U=<big|prod><rsub|i\<in\>I>U<rsub|i>> where
<math|U<rsub|i>=X<rsub|i>> if <math|i\<in\>A> and
<math|U<rsub|i>\<in\>\<cal-T\><rsub|i>> if <math|i\<in\>I<mid|\\>A> then
if <math|x\<in\>U\<Rightarrow\>\<forall\>i\<in\>A\<succ\>\<pi\><rsub|i>(x)=x(i)\<in\>U<rsub|i>\<Rightarrow\>\<forall\>i\<in\>A\<succ\>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)\<Rightarrow\>x\<in\><big|cap><rsub|i\<in\>A>\<pi\><rsup|-1>(U<rsub|i>)>,
also if <math|x\<in\><big|cap><rsub|i\<in\>A>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)\<subseteq\><big|prod><rsub|i\<in\>I>X<rsub|i>\<Rightarrow\>(\<forall\>i\<in\>I\<succ\>x(i)\<in\>X<rsub|i>)\<wedge\>\<forall\>i\<in\>A\<succ\>x(i)=\<pi\><rsub|i>(x)\<in\>U<rsub|i>\<Rightarrow\>x\<in\>U>.
So we have proved that <math|U=<big|cap><rsub|i\<in\>A>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)\<in\>\<cal-B\><rprime|'>>
</proof>
<\theorem>
<label|product topology of subspace topolgies>Let
<math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>> be a family of
topological spaces and let <math|\<forall\>i\<in\>I,
\ A<rsub|i>\<subseteq\>X<rsub|i>> be equipped with its subspace topology
<math|\<cal-T\><rsub|A<rsub|i>>> then the subspace topology on
<math|\<Pi\><rsub|i\<in\>I>A<rsub|i>> considered as a subset of
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> \ is the same as the product
topology of <math|{X<rsub|,>\<cal-T\><rsub|A<rsub|i>>}<rsub|i\<in\>I>>
</theorem>
<\proof>
Let <math|\<cal-B\>> be the base of the product topology on
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> then the subspace topology on
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>> is defined by
<math|\<cal-B\><rsub|1>={B<big|cap>(<big|prod><rsub|i\<in\>I>A<rsub|i>)\|B\<in\>\<cal-B\>}>
and let <math|\<cal-B\><rsub|2>> be a basis for the product topology of
<math|<big|prod><rsub|i\<in\>I>A<rsub|i>>.\
<\enumerate>
<item>If <math|U\<in\>\<cal-B\><rsub|1>\<Rightarrowlim\><rsub|<reference|basis
of product topology>>U=(<big|prod><rsub|i\<in\>I>U<rsub|i>)<big|cap>(<big|prod><rsub|j\<in\>I>A<rsub|j>)>
where <math|U<rsub|i>=X<rsub|i>> except for a finite subset of <math|I>
where <math|U<rsub|i>\<in\>\<cal-T\><rsub|i>\<Rightarrow\>U=<big|prod><rsub|i\<in\>I>(U<rsub|i><big|cap>A<rsub|i>)>
where <math|(U<rsub|i><big|cap>A<rsub|i>)=A<rsub|i> (if
U<rsub|i>=X<rsub|i>)> except for a finite subset of <math|I> where
<math|(U<rsub|i><big|cap>A<rsub|i>)\<in\>\<cal-T\><rsub|A<rsub|i>>> and
thus <math|U\<in\>\<cal-B\><rsub|2>>
<item>If <math|U\<in\>\<cal-B\><rsub|2> > then
<math|B=<big|prod><rsub|i\<in\>I>U<rsub|i>> where
<math|U<rsub|i>=A<rsub|i>=A<rsub|i><big|cap>A<rsub|i>> except for a
finite subset of <math|I> where <math|U<rsub|i>=W<rsub|i><big|cap>A<rsub|i>>
(<math|W<rsub|i>\<in\>\<cal-T\><rsub|i>)> and thus if
<math|V<rsub|i>=X<rsub|i>> except for a finite subset where
<math|V<rsub|i>=W<rsub|i>> then <math|B=<big|prod><rsub|i\<in\>I>(V<rsub|i><big|cap>A<rsub|i>)=(<big|prod><rsub|i\<in\>I>V<rsub|i>)<big|cap>(<big|prod><rsub|i\<in\>I>A<rsub|i>)>
where <math|V<rsub|i>=A<rsub|i>> except for a finite subset of <math|I>
where <math|V<rsub|i>\<in\>\<cal-T\><rsub|i>\<Rightarrow\>B\<in\>\<cal-B\><rsub|1>>
</enumerate>
now as we just have proved that <math|\<cal-B\><rsub|1>=\<cal-B\><rsub|2>>
we use <reference|finer basis> to prove that both topologies are the
same.
</proof>
<\theorem>
<label|finite product topology>Let <math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>>
be a family of topological spaces then the box topology on
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> is finer then the product
topology on <math|<big|prod><rsub|i\<in\>I>X<rsub|i>>. If <math|I> is
finite then both topologies are the same.
</theorem>
<\proof>
Let <math|\<cal-B\><rsub|1>> be the basis for the product topology and
<math|\<cal-B\><rsub|2>> be the basis for the box topology. Then if
<math|B\<in\>\<cal-B\><rsub|1>> we using <reference|basis of product
topology> that <math|B=<big|prod><rsub|i\<in\>I>U<rsub|i>> where either
<math|U<rsub|i>=X<rsub|i>\<in\>\<cal-T\><rsub|i>> except for a finite
number of cases where \ <math|U<rsub|i>\<in\>\<cal-T\><rsub|i>\<Rightarrow\>B\<in\>\<cal-B\><rsub|1>>.
This proves that the box topology is finer then the product topology.
Also if <math|I> is finite then if <math|B\<in\>\<cal-B\><rsub|2>> then
<math|B=<big|prod><rsub|i\<in\>I>U<rsub|i>> where
<math|U<rsub|i>\<in\>\<cal-T\><rsub|i>> then as <math|I> if finite, we
have that <math|J={i\<in\>I\|U<rsub|i>\<neq\>X<rsub|i>}\<subseteq\>I> is
finite and thus <math|B\<in\>\<cal-B\><rsub|1>>
</proof>
<\definition>
<index|dense subset>Let <math|X,\<cal-T\>> be a topological space then
<math|A\<subseteq\>X> is a dense subset if <math|<wide|A|\<bar\>>=X>
</definition>
<\definition>
<index|Baire set>A topological space <math|X,\<cal-T\>> is a Baire if for
every <math|{A<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> with
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> we have that <math|A<rsub|i>>
is closed and <math|A<rsup|\<circ\>><rsub|i>=\<emptyset\>> then
<math|(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)<rsup|\<circ\>>=\<emptyset\>>
. In other words the union of any sequence of closed sets with empty
interior has also empty interior
</definition>
<\theorem>
Let <math|X,\<cal-T\>> be a topological space and <math|A\<subseteq\>X>
then <math|A<rsup|\<circ\>>=\<emptyset\>\<Leftrightarrow\>X<mid|\\>A> is
dense in <math|X>\
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>As <math|A<rsup|\<circ\>>=\<emptyset\>> then
if <math|x\<in\>X> and <math|U> open with <math|x\<in\>U> we have
<math|U\<nsubseteq\>A\<Rightarrow\>U\<cap\>(X<mid|\\>A)\<neq\>\<emptyset\>\<Rightarrow\><wide|X<mid|\\>A|\<bar\>>=X>
proving that <math|X<mid|\\>A> is dense in <math|X>
<item><math|\<Leftarrow\>>Assume <math|X<mid|\\>A> is dense in <math|X>
and let <math|U> open and <math|U\<subseteq\>A> and let
<math|x\<in\>U\<Rightarrow\>U<big|cap>(X<mid|\\>A)\<neq\>\<emptyset\>>
contradicting <math|U\<subseteq\>A>
</enumerate>
</proof>
<\theorem>
<label|alternative definition of Baire sets><math|X,\<cal-T\>> is a Baire
space if and only if for every <math|{U<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
with <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> we have that
<math|U<rsub|i>> is open and dense in <math|X> then
<math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>> is dense in
<math|X>. In other words the intersection of every sequence of dense sets
in <math|X> is dense in <math|X>.
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>Let <math|{U<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
be a set of open dense sets in <math|X> then
<math|{A<rsub|i>}<rsub|i\<in\>\<bbb-N\>>> where
<math|A<rsub|i>=X<mid|\\>U<rsub|i>> is a closed set and
<math|U<rsub|i>=X<mid|\\>(X<mid|\\>A<rsub|i>)> so that by the previous
theorem we have <math|A<rsub|i><rsup|0>=\<emptyset\>> and thus
<math|(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)<rsup|\<circ\>>=\<emptyset\>\<Rightarrow\>X<mid|\\>(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)>
is dense in <math|X> by the previous theorem. Also
<math|X<mid|\\>(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)=<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>(X<mid|\\>A<rsub|i>)=<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>>
proving that <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>><math|>
<item><math|\<Leftarrow\>>Let <math|{A<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
be a set of closed sets with <math|A<rsup|\<circ\>><rsub|i>=\<emptyset\>>
then <math|{U<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> where
<math|U<rsub|i>=X<mid|\\>A<rsub|i>> is open then by the previous
theorem <math|U<rsub|i>> is dense in <math|X> and thus
<math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>> is dense in
<math|X> now <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>=<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>(X<mid|\\>A<rsub|i>)=X<mid|\\>(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)>
and using the previous theorem we have then
<math|(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)<rsup|\<circ\>>=\<emptyset\>>
</enumerate>
</proof>
<section|Metric and normed spaces>
<\definition>
<index|pseudo metric space><index|metric space>A pseudo metric space is a
set together with a map <math|d> from
<math|X\<times\>X\<rightarrow\>\<bbb-R\>> (<math|d> is called the pseudo
metric) such that <math|\<forall\>x,y,z> we have\
<\enumerate>
<item><math|d(x,y)\<geqslant\>0>
<item><math|d(x,y)=d(y,x)>
<item><math|d(x,z)\<leqslant\>d(x,y)+d(y,z)>
</enumerate>
If also <math|d(x,y)\<Rightarrow\>x=y> then <math|d> is a metric and
<math|X,d> is a metric space
</definition>
<\definition>
<index|limit of a function><index|<math|lim<rsub|h\<rightarrow\>x>f(x)>>Let
<math|X,d<rsub|X>> and <math|Y,d<rsub|Y>> be metric spaces and
<math|f:X\<rightarrow\>Y> be a <with|font-series|bold|partial function>
(should not be a mapping or function, but could be a mapping or function
if <math|dom(f)=X>) then given <math|x\<in\>X> and <math|y\<in\>Y> we say
that <math|lim<rsub|h\<rightarrow\>x>f(x)=y\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0>
there exists a <math|\<delta\>\<gtr\>0> such that
<math|\<forall\>x<rprime|'>\<in\>dom(f)> with
<math|0\<less\>d<rsub|X>(x,x<rprime|'>)\<less\>\<delta\>> that
<math|d<rsub|Y>(y,f(x<rprime|'>))\<less\>\<varepsilon\>>
<\note>
Note that neither <math|x> should be in <math|dom(f)> or <math|y>
should be in <math|rng(f)>
</note>
</definition>
<\definition>
<index|open ball><index|<math|B<rsub|d>(x,\<varepsilon\>)>>Given a metric
space <math|X,d> and <math|\<varepsilon\>\<in\>\<bbb-R\><rsub|+>={r\<in\>\<bbb-R\>\|r\<gtr\>0}>
then <math|B<rsub|d>(x,\<varepsilon\>)={y\<in\>X\|d(x,y)\<less\>\<varepsilon\>}>
is called a open ball centered around x with radius <math|\<varepsilon\>>
</definition>
<\definition>
<index|closed ball><index|<math|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)>>Given
a metric space <math|X,d> and <math|\<varepsilon\>\<in\>\<bbb-R\><rsub|+>>
then <math|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)={y\<in\>X\|d(x,y)\<leqslant\>\<varepsilon\>}>
is called the closed ball centered around <math|x> with radius
<math|\<varepsilon\>>
</definition>
<\remark>
Until further notice we always assume that <math|\<varepsilon\>\<gtr\>0>
actual means <math|\<varepsilon\>\<in\>\<bbb-R\><rsub|+>>
</remark>
<\theorem>
<label|intersection of balls>Given a pseudo metric space <math|X,d> then
if <math|x\<in\>B<rsub|d>(x<rsub|1>,\<varepsilon\><rsub|1>)<big|cap>B<rsub|d>(x<rsub|2>,\<varepsilon\><rsub|2>)>
we have <math|\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d>(x,\<varepsilon\>)\<subseteq\>B<rsub|d>(x<rsub|1>,\<varepsilon\><rsub|1>)<big|cap>B<rsub|d>(x<rsub|2>,\<varepsilon\><rsub|2>)>
</theorem>
<\proof>
Let <math|x\<in\>B<rsub|d>(x<rsub|1>,\<varepsilon\><rsub|1>)<big|cap>B<rsub|d>(x<rsub|2>,\<varepsilon\><rsub|2>)\<Rightarrow\>d(x<rsub|1>,x)\<less\>\<varepsilon\><rsub|1>,d(x<rsub|2>,x)\<less\>\<varepsilon\><rsub|2>>
take then <math|\<varepsilon\>=min(\<varepsilon\><rsub|1>-d(x<rsub|1>,x),\<varepsilon\><rsub|2>-d(x<rsub|2>,x))\<gtr\>0>
then if <math|y\<in\>B<rsub|d>(x,\<varepsilon\>)\<Rightarrow\>d(x,y)\<less\>\<varepsilon\>>
so for <math|i=1,2> we have <math|d(x<rsub|i>,y)\<leqslant\>d(x<rsub|i>,x)+d(x,y)\<less\>d(x<rsub|i>,x)+\<varepsilon\>\<leqslant\>d(x<rsub|i>,x)+(\<varepsilon\><rsub|i>-d(x<rsub|i>,x))=\<varepsilon\><rsub|i><rsub|>\<Rightarrow\>y\<in\>B<rsub|d>(x<rsub|i>,\<varepsilon\><rsub|i>)>
</proof>
<\corollary>
<label|ball in balls>Given a pseudo metric space <math|X,d> and
<math|y\<in\>B<rsub|d>(x,\<varepsilon\>)> then
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|B<rsub|d>(y,\<delta\>)\<subseteq\>B<rsub|d>(x,\<varepsilon\>)>
</corollary>
<\proof>
<math|y\<in\>B<rsub|d>(x,\<varepsilon\>)=B<rsub|d>(x,\<varepsilon\>)<big|cap>B<rsub|d>(x,\<varepsilon\>)\<Rightarrowlim\><rsub|<reference|intersection
of balls>>\<exists\>\<delta\>\<gtr\>0\<succ\>y\<in\>B<rsub|d>(y,\<delta\>)\<subseteq\>B<rsub|d>(x,\<varepsilon\>)<big|cap>B<rsub|d>(x,\<varepsilon\>)=B<rsub|d>(x,\<varepsilon\>)>
</proof>
<\theorem>
<label|metric topology>Given a pseudo metric space <math|X,d> then
<math|\<cal-B\>={B<rsub|d>(x,\<varepsilon\>)\|\<varepsilon\>\<in\>\<bbb-R\><rsub|+>,x\<in\>X}>
is a generating basis, the generated topology of <math|X> is called the
topology generated by the pseudo metric. If not specified we will for the
rest that the topology of <math|X,d> is the generated topology.
</theorem>
<\proof>
We use <reference|generating basis in topology> in our proof, so\
<\enumerate>
<item><math|\<forall\>x\<in\>X> we have
<math|x\<in\>B<rsub|d>(x,1)\<in\>\<cal-B\>>
<item><math|\<forall\>B<rsub|1>,B<rsub|2>\<in\>\<cal-B\>\<vdash\>x\<in\>B<rsub|1><big|cap>B<rsub|2>>
we have <math|B<rsub|i>=B<rsub|d>(x<rsub|i>,e<rsub|i>)\<Rightarrowlim\><rsub|<reference|intersection
of balls>>\<exists\>\<delta\>\<gtr\>0\<succ\>x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
</enumerate>
</proof>
<\corollary>
<label|open sets in metric topology>Given a pseudo metric space
<math|X,d> then <math|U> is a open set in the metric topology iff
<math|\<forall\>x\<in\>U> there exists <math|\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d>(x,\<delta\>)>
</corollary>
<\proof>
\;
<\math>
\<Rightarrow\>
</math>
Let <math|x\<in\>U> open then by the definition of a basis we have
<math|\<exists\>y\<in\>X,\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d>(y,\<varepsilon\>)\<subseteq\>U\<Rightarrowlim\><rsub|<reference|ball
in balls>>\<exists\>\<delta\>\<gtr\>0\<succ\>x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>B<rsub|d>(y,\<varepsilon\>)\<subseteq\>U>
<math|\<Leftarrow\>>
Assume that <math|\<forall\>x\<in\>U\<succ\>\<exists\>\<delta\><rsub|x>\<gtr\>0\<succ\>x\<in\>B<rsub|d>(x,\<delta\><rsub|x>)\<Rightarrow\>U=<big|cup><rsub|x\<in\>U>B<rsub|d>(x,\<delta\><rsub|x>)>
which is open because <math|B<rsub|d>(x,\<delta\><rsub|x>)> are open by
the definition of a basis
</proof>
<\theorem>
<label|closed balls are closed>Let <math|X,d> be a pseudo metric space
then closed balls are indeed closed
</theorem>
<\proof>
Let <math|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)> be a closed ball
and let <math|y\<in\>X<mid|\\><wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)\<Rightarrow\>d(x,y)\<gtr\>\<varepsilon\>>
take then <math|\<delta\>=d(x,y)-\<varepsilon\>\<gtr\>0>, now let
<math|z\<in\>B<rsub|d>(y,\<delta\>)> then assume that
<math|d(z,x)\<leqslant\>\<varepsilon\>\<Rightarrow\>d(x,y)\<leqslant\>d(x,z)+d(z,y)\<leqslant\>\<varepsilon\>+d(z,y)\<less\>\<varepsilon\>+\<delta\>=\<varepsilon\>+d(x,y)-\<varepsilon\>=d(x,y)\<Rightarrow\>d(x,y)\<less\>d(x,y)>
a contradiction, so <math|d(z,x)\<gtr\>\<varepsilon\>\<Rightarrow\>z\<in\>X<mid|\\><wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)\<Rightarrow\>B<rsub|d>(y,\<delta\>)\<subseteq\>X<mid|\\><wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)\<Rightarrowlim\><rsub|<reference|open
sets in metric topology>>X<mid|\\><wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)>
is open in the metric topology and thus
<math|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)> is closed
</proof>
<\theorem>
<label|subspace topology of a metric space>Let <math|X,d> be a (pseudo)
metric space and <math|A\<subseteq\>X> then the subspace topology on
<math|A> is generated by the (pseudo) metric <math|d<rsub|\|A>>
</theorem>
<\proof>
First we prove that <math|d<rsub|\|A>> is a (pseudo) metric, so let
<math|x,y,z\<in\>A>
<\enumerate>
<item><math|d<rsub|\|A>(x,y)=d(x,y)\<geqslant\>0>
<item><math|d<rsub|\|A>(x,y)=d(x,y)=d(y,x)=d<rsub|\|A>(y,x)>
<item><math|d<rsub|\|A>(x,y)=d(x,y)\<leqslant\>d(x,z)+d(z,y)=d<rsub|\|A>(x,z)+d<rsub|\|A>(z,x)>
</enumerate>
Second <math|B<rsub|d<rsub|\|A>>(x,\<delta\>)={y\<in\>A\|d<rsub|\|A>(x,y)\<less\>\<delta\>}={y\<in\>A\|d(x,y)\<less\>\<delta\>}={y\<in\>X\|d(x,y)\<less\>\<delta\>}<big|cap>A=B<rsub|d>(x,\<delta\>)<big|cap>A>
so using <reference|basis of suptopology> we see that
<math|{B<rsub|d<rsub|\|A>>(x,\<delta\>)\|x\<in\>A,\<delta\>\<gtr\>0}={B<rsub|d>(x,\<delta\>)<big|cap>A\|x\<in\>A,\<delta\>\<gtr\>0}>
which is the basis for the subtopology and which is equal to the metric
topology generated by <math|><math|d<rsub|\|A>>.
</proof>
<\definition>
<index|equivalent metrics>Two pseudo metrics <math|d<rsub|1>,d<rsub|2>>
on a set <math|X> are equivalent iff they generate the same topology
</definition>
<\theorem>
<label|finer metric>Let <math|d<rsub|1>,d<rsub|2>> be two metrics on
<math|X> and let <math|\<cal-T\><rsub|1>,\<cal-T\><rsub|2>> be the
generated metric topologies then <math|\<cal-T\><rsub|2>> is finer then
<math|\<cal-T\><rsub|2>> iff <math|\<forall\>x\<in\>X,\<forall\>\<varepsilon\>\<gtr\>0>
there <math|\<exists\>\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d<rsub|2>>(x,\<delta\>)\<subseteq\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>><math|\<forall\>x\<in\>X,\<forall\>\<varepsilon\>\<gtr\>0>
we use <reference|finer basis> to find the existence of a
<math|y\<in\>X,\<tau\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d<rsub|2>>(y,\<tau\>)\<subseteq\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)\<Rightarrowlim\><rsub|<reference|ball
in balls>>\<exists\>\<delta\>\<gtr\>0\<succ\>x\<in\>B<rsub|d<rsub|2>>(x,\<delta\>)\<subseteq\>B<rsub|d<rsub|2>>(y,\<tau\>)\<subseteq\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)>
<math|\<Leftarrow\>>If <math|y\<in\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)>
then we have using <reference|ball in balls> to find the existence of a
<math|\<delta\>\<gtr\>0> such that <math|y\<in\>B<rsub|d<rsub|1>>(y,\<delta\>)\<subseteq\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)\<Rightarrowlim\><rsub|hypothese>\<exists\>\<tau\>\<gtr\>0\<succ\>y\<in\>B<rsub|d<rsub|2>>(y,\<tau\>)\<subseteq\>B<rsub|d<rsub|1>>(y,\<delta\>)\<subseteq\>B<rsub|d<rsub|1>>(x,\<varepsilon\>)\<Rightarrowlim\><rsub|<reference|finer
basis>>><math|\<cal-T\><rsub|2>> is finer then <math|\<cal-T\><rsub|1>>
</proof>
<\definition>
<index|isometry>Let <math|X,d<rsub|X>> and <math|Y,d<rsub|Y>> be two
pseudo metric spaces then a function <math|\<varphi\>:X\<rightarrow\>Y>
is called a isometry iff\
<\enumerate>
<item><math|\<varphi\>> is a bijective map
<item><math|\<forall\>x,y\<in\>X> we have
<math|d<rsub|X>(x,y)=d<rsub|Y>(\<varphi\>(x),\<varphi\>(y))>
</enumerate>
</definition>
<\theorem>
<label|inverse of isometry metric>Let <math|X,d<rsub|X>> and
<math|Y,d<rsub|Y>> be two pseudo metric spaces and
<math|\<varphi\>:X\<rightarrow\>Y> a isometry then
<math|\<varphi\><rsup|-1>> is a isometry
</theorem>
<\proof>
<math|d<rsub|X>(\<varphi\><rsup|-1>(x),\<varphi\><rsup|-1>(y))=d<rsub|Y>(\<varphi\>(\<varphi\><rsup|-1>(x)),\<varphi\>(\<varphi\><rsup|-1>(y)))=d<rsub|Y>(x,y)>
and <math|\<varphi\><rsup|-1>> is bijective because <reference|inverse of
bijective mapping>
</proof>
<\theorem>
<label|composition of isometries metric>Let <math|X,d<rsub|X> ,
Y,d<rsub|Y> , Z,d<rsub|Z>> then if <math|\<varphi\><rsub|1>:X\<rightarrow\>Y>
and <math|\<varphi\><rsub|2>:Y\<rightarrow\>Z> then
<math|\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>:X\<rightarrow\>Z> is a
isometry
</theorem>
<\proof>
<math|d<rsub|Z>((\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>)(x),(\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>)(y))=d<rsub|Z>(\<varphi\><rsub|2>(\<varphi\><rsub|1>(x)),\<varphi\><rsub|2>(\<varphi\><rsub|1>(y)))=d<rsub|Y>(\<varphi\><rsub|1>(x),\<varphi\><rsub|1>(y))=d<rsub|X>(x,y)>
and <math|\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>> is bijective
because of <reference|composition of bijective mappings>
</proof>
<\theorem>
<label|isometry and balls>Let <math|X,d<rsub|X>> and <math|Y,d<rsub|Y>>
be two metric spaces with a isometry <math|\<varphi\>> between <math|X>
and <math|Y> then \ <math|\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))=B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)>
and <math|\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>*(y,\<delta\>))=B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(y),\<delta\>)>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|y\<in\>\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))\<Rightarrow\>y=\<varphi\>(z),z\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<Rightarrow\>y=\<varphi\>(z),d<rsub|x>(z,x)\<less\>\<delta\>\<Rightarrow\>y=\<varphi\>(z),d<rsub|Y>(\<varphi\>(x),\<varphi\>(z)\<less\>\<delta\>\<Rightarrow\>d<rsub|Y>(\<varphi\>(x),y)\<less\>\<delta\>\<Rightarrow\>y\<in\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)>
so <math|\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))\<subseteq\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)>.
Secondly assume that <math|y\<in\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<Rightarrowlim\><rsub|\<varphi\>
is bijective>y=\<varphi\>(\<varphi\><rsup|-1>(y)),d<rsub|Y>(\<varphi\>(x),\<varphi\>(\<varphi\><rsup|-1>(y)))\<less\>\<delta\>\<Rightarrow\>d<rsub|x>(x,\<varphi\><rsup|-1>(y))\<less\>\<delta\>\<Rightarrow\>\<varphi\><rsup|-1>(y)\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<Rightarrow\>y=\<varphi\>(\<varphi\><rsup|-1>(y))\<in\>\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))\<Rightarrow\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<subseteq\>\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))>.
<item>Let <math|x\<in\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(y,\<delta\>))=\<varphi\>(x)\<in\>B<rsub|d<rsub|Y>>(y,\<delta\>)\<Rightarrow\>d<rsub|Y>(y,\<varphi\>(x))\<less\>\<delta\>)\<Rightarrowlim\><rsub|\<varphi\>
is bijective>d<rsub|Y>(\<varphi\>(\<varphi\><rsup|-1>(y)),\<varphi\>(x))\<less\>\<delta\>\<Rightarrow\>d<rsub|X>(\<varphi\><rsup|-1>(y),x)\<less\>\<delta\>\<Rightarrow\>x\<in\>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(y),\<delta\>)\<Rightarrow\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(y,\<delta\>))\<subseteq\>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(y),\<delta\>)>.
Secondly if <math|x\<in\>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(y),\<delta\>)\<Rightarrow\>d<rsub|X>(\<varphi\><rsup|-1>(y),x)\<less\>\<delta\>\<Rightarrow\>d<rsub|Y>(\<varphi\>(\<varphi\><rsup|-1>(y)),\<varphi\>(x))\<less\>\<delta\>\<Rightarrow\>\<varphi\>(x)\<in\>B<rsub|d<rsub|Y>>(y,\<delta\>)\<Rightarrow\>x\<in\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(y,\<delta\>))\<Rightarrow\>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(y),\<delta\>)\<subseteq\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(y,\<delta\>))>
</enumerate>
</proof>
<\theorem>
<label|topologies and isometries>Let <math|X,d<rsub|X>> and
<math|Y,d<rsub|Y>> be two metric spaces with a isometry <math|\<varphi\>>
between them. Then if <math|\<cal-T\><rsub|X>,\<cal-T\><rsub|Y>> be the
metric topologies on <math|X,Y> then <math|\<cal-T\><rsub|X>={\<varphi\><rsup|-1>(V)\|V\<in\>\<cal-T\><rsub|Y>}={U\<subseteq\>X\|\<varphi\>(U)\<in\>\<cal-T\><rsub|Y>}>
and <math|\<cal-T\><rsub|Y>={\<varphi\>(U)\|U\<in\>\<cal-T\><rsub|X>}={V\<subseteq\>Y\|\<varphi\><rsup|-1>(V)\<in\>\<cal-T\><rsub|X>}>.
If two topologies are related in the way the theorem state then we call
them equivalent
</theorem>
<\proof>
Let <math|\<cal-U\>={\<varphi\><rsup|-1>(V)\|V\<in\>\<cal-T\><rsub|Y>}>,
<math|\<cal-V\>={U\<subseteq\>X\|\<varphi\>(U)\<in\>\<cal-T\><rsub|Y>}>
<\enumerate>
<item>Let <math|U\<in\>\<cal-T\><rsub|X>> then
<math|U=\<varphi\><rsup|-1>(V)> where <math|V=\<varphi\>(U)> using the
fact that <math|\<varphi\>> is a bijective function. Then
<math|\<forall\>y\<in\>V> we have <math|y=\<varphi\>(x),x\<in\>U> then
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<subseteq\>U\<Rightarrow\>B<rsub|d<rsub|Y>>(y,\<delta\>)\<equallim\><rsub|<reference|isometry
and balls>>\<varphi\>(B<rsub|d<rsub|X>>(x,\<delta\>))\<subseteq\>\<varphi\>(U)=V\<Rightarrow\>V>
is open and thus <math|V\<in\>\<cal-T\><rsub|Y>\<Rightarrow\>U\<in\>\<cal-U\>>.
Now if <math|U\<in\>\<cal-U\>> then
<math|\<exists\>V\<in\>\<cal-T\><rsub|Y>> such that
<math|U=\<varphi\><rsup|-1>(V)> now if
<math|x\<in\>U=\<varphi\>(x)\<in\>V\<Rightarrow\>\<exists\>\<delta\>\<gtr\>0\<succ\>\<varphi\>(x)\<in\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<subseteq\>V\<Rightarrow\>x\<in\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>))\<equallim\><rsub|<reference|isometry
and balls>>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(\<varphi\>(x)),\<delta\>)=B<rsub|d<rsub|X>>(x,\<delta\>)\<subseteq\>\<varphi\><rsup|-1>(V)=U>
and thus <math|U> is open and thus <math|\<cal-U\>\<subseteq\>\<cal-T\><rsub|X>>.
<item>Let <math|U\<in\>\<cal-T\><rsub|X>> then
<math|\<forall\>x\<in\>U\<succ\>\<exists\>\<delta\>\<gtr\>0\<vdash\>x\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<subseteq\>U\<Rightarrowlim\>\<varphi\>(x)\<in\>\<varphi\>(B<rsub|d<rsub|X>>(x,y))\<equallim\><rsub|<reference|isometry
and balls>>\<varphi\>(x)\<in\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<subseteq\>\<varphi\>(U)>
so if <math|y\<in\>\<varphi\>(U)\<Rightarrow\>y=\<varphi\>(x),x\<in\>U\<Rightarrow\>y\<in\>B<rsub|d<rsub|Y>>(y,\<delta\>)\<subseteq\>\<varphi\>(U)\<Rightarrow\>\<varphi\>(U)\<in\>\<cal-T\><rsub|Y>\<Rightarrow\>\<cal-T\><rsub|X>\<subseteq\>\<cal-V\>>.
Also if <math|U\<in\>\<cal-V\>> then
<math|\<varphi\>(U)\<in\>\<cal-T\><rsub|Y>> now if <math|x\<in\>U> then
<math|\<varphi\>(x)\<in\>\<varphi\>(U)\<Rightarrow\>\<exists\>\<delta\>\<gtr\>0\<succ\>\<varphi\>(x)\<in\>B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<subseteq\>\<varphi\>(U)\<Rightarrow\>x\<in\>\<varphi\><rsup|-1>(B<rsub|d<rsub|Y>>(\<varphi\>(x),\<delta\>)\<subseteq\>\<varphi\><rsup|-1>(\<varphi\>(U))\<equallim\><rsub|\<varphi\>
s bijective>U\<Rightarrowlim\><rsub|<reference|isometry and
balls>>x\<in\>B<rsub|d<rsub|X>>(\<varphi\><rsup|-1>(\<varphi\>(x)),\<delta\>)=B<rsub|d<rsub|X>>(x,\<delta\>)\<subseteq\>U\<Rightarrow\>U>
is open and thus <math|\<cal-V\>\<subseteq\>\<cal-T\><rsub|X>>
<item>For the remaining of the proof, note that
<math|\<phi\>=\<varphi\><rsup|-1>> is also a symmetry between <math|Y>
and <math|X> so that by <math|(1)> and <math|(2)> we have
<math|\<cal-T\><rsub|Y>={\<phi\><rsup|-1>(U)\|U\<in\>\<cal-T\><rsub|X>}={V\<subseteq\>Y\|\<phi\>(V)\<in\>\<cal-T\><rsub|X>}>
then using <math|\<phi\><rsup|-1>(U)\<equallim\>\<varphi\>(U)
[x\<in\>\<phi\><rsup|-1>(U)\<Leftrightarrow\>\<phi\>(x)\<in\>U\<Leftrightarrowlim\><rsub|\<phi\>=\<varphi\><rsup|-1>>\<varphi\><rsup|-1>(x)\<in\>U\<Leftrightarrowlim\><rsub|\<varphi\>
is bijective>x\<in\>\<varphi\>(U)]> and
<math|\<phi\>(V)=\<varphi\><rsup|-1>(V)
[x\<in\>\<phi\>(V)\<Leftrightarrow\>x=\<phi\>(y),y\<in\>V\<Leftrightarrow\>x\<equallim\><rsub|\<varphi\><rsup|-1>=\<phi\>>\<varphi\><rsup|-1>(y),y\<in\>V\<Leftrightarrowlim\><rsub|\<varphi\>
is bijective>x\<in\>\<varphi\><rsup|-1>(V)]> and thus
<math|\<cal-T\><rsub|Y>={\<varphi\>(U)\|U\<in\>\<cal-T\><rsub|X>}={V\<subseteq\>Y\|\<varphi\><rsup|-1>(V)\<in\>\<cal-T\><rsub|X>}>
</enumerate>
</proof>
<\definition>
<label|bounded sets><index|bounded set><index|diameter>Let <math|X,d> be
a pseudo metric space, a subset <math|A\<subseteq\>X> is called bounded
iff <math|\<exists\>M\<in\>\<bbb-R\>> such that
<math|\<forall\>x,y\<in\>X> we have <math|d(x,y)\<leqslant\>M>. If
<math|A> is bounded then <math|sup({d(x,y)\|x,y\<in\>A})> exists and is
called the diameter of <math|A> noted as <math|diam(A)>
</definition>
<\theorem>
<label|product of metric spaces>Let <math|{X<rsub|i>,d<rsub|i>}<rsub|i\<in\>I>>
be a finite family (<math|I> is finite) \ of (pseudo) metric spaces then
<math|d:<big|prod><rsub|i\<in\>I>X<rsub|i>\<rightarrow\>\<bbb-R\>>
defined by <math|d(x,y)=max(d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\|i\<in\>I}>
(defined by <reference|maximum (minumu,) of finite sets>) is a (pseudo)
metric and the topology generated is the product topology. The metric
<math|d> is called the product metric.
</theorem>
<\proof>
First we proof that <math|d> is a metric
<\enumerate>
<item><math|d(x,y)=max{d<rsub|i>(\<pi\><rsub|i>(x).\<pi\><rsub|i>(y)\|i\<in\>I}\<geqslant\>0>
(as <math|d<rsub|i>> is positive)
<item><math|d(x,y)=max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y)\|i\<in\>I}=max{d<rsub|i>(\<pi\><rsub|i>(y),\<pi\><rsub|i>(x)\|i\<in\>I}=d(y,x)>
<item><math|d(x,z)=max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y)\|i\<in\>I}\<leqslant\>max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))+d<rsub|i>(\<pi\><rsub|i>(y),\<pi\><rsub|i>(z))\|i\<in\>I}\<leqslant\>max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))+max({d<rsub|i>(\<pi\><rsub|i>(y),\<pi\><rsub|i>(z))\|i\<in\>I}\|i\<in\>I}\<leqslant\>max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\|i\<in\>I}+max{d<rsub|i>(\<pi\><rsub|i>(y),\<pi\><rsub|i>(z))\|i\<in\>I}=d(x,y)+d(y,z)>
<item>If <math|d(x,y)=0\<Rightarrow\>max{d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\|i\<in\>I}=0\<Rightarrow\>\<forall\>i\<in\>I\<succ\>d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))=0\<Rightarrow\>\<forall\>i\<in\>I\<succ\>\<pi\><rsub|i>(x)=\<pi\><rsub|i>(y)\<Rightarrow\>x=y>
</enumerate>
\ Next as <math|I> is finite we have by <reference|finite product
topology> that the product topology and the box topology are the same and
thus that <math|\<cal-B\><rsub|1>={<big|prod><rsub|i\<in\>I>U<rsub|i>\|U<rsub|i>>
open in the metric topology of <math|X<rsub|i>>} and let
<math|\<cal-B\><rsub|2>={B<rsub|d>(x,\<delta\>)\|x\<in\><big|prod><rsub|i\<in\>I>X<rsub|i>,\<delta\>\<gtr\>0}>.
We now use <reference|finer basis> to prove that the topology generated
by <math|\<cal-B\><rsub|1>> and <math|\<cal-B\><rsub|2>> are the same. So
if <math|B\<in\>\<cal-B\><rsub|1>> then
<math|B=<big|prod><rsub|i\<in\>I>U<rsub|i>,U<rsub|i>\<in\>\<cal-T\><rsub|i>>
so if <math|x\<in\>B> then <math|\<forall\>i\<in\>I\<succ\>x(i)\<in\>U<rsub|i>\<Rightarrow\>\<exists\>\<delta\><rsub|i>\<gtr\>0>
such that <math|x(i)\<in\>B<rsub|d<rsub|i>>(x(i),\<delta\><rsub|i>)\<subseteq\>U<rsub|i>>
now let <math|\<delta\>=min(\<delta\><rsub|i>\|i\<in\>I)> (this exists
because <math|I> is finite) then if <math|y\<in\>B<rsub|d>(x,\<delta\>)\<in\>\<cal-B\><rsub|2>\<Rightarrow\>d(x,y)\<less\>\<delta\>\<Rightarrow\>d(x,y)=max(d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\|i\<in\>I}\<less\>\<delta\>\<Rightarrow\>\<forall\>i\<in\>I\<succ\>d<rsub|i>(x(i),y(i))=d<rsub|i>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\<less\>\<delta\>\<leqslant\>\<delta\><rsub|i>\<Rightarrow\>y(i)\<in\>B<rsub|d<rsub|i>>(x(i),\<delta\><rsub|i>)\<Rightarrow\>y\<in\><big|prod><rsub|i\<in\>I>B<rsub|d<rsub|i>>(x(i),\<delta\><rsub|i>)=B\<Rightarrow\>x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>B>.
On the other hand if <math|B\<in\>\<cal-B\><rsub|2>> then
<math|B=B<rsub|d>(x,\<delta\>)> now if
<math|z\<in\>B<rsub|d>(x,\<delta\>)> then if
<math|y\<in\><big|prod><rsub|i\<in\>I>B<rsub|d<rsub|i>>(z(i),\<delta\>)\<in\>\<cal-B\><rsub|1>\<Rightarrow\>\<forall\>i\<in\>I\<succ\>y(i)\<in\>B<rsub|d<rsub|i>>(z(i),\<delta\>)\<Rightarrow\>\<forall\>i\<in\>I\<succ\>d<rsub|i>(z(i),y(i))\<less\>\<delta\>\<Rightarrow\>max(d<rsub|i>(\<pi\><rsub|i>(z),\<pi\><rsub|i>(y))\|i\<in\>I}=max(d<rsub|i>(\<pi\><rsub|i>(z),\<pi\><rsub|i>(y))\|i\<in\>I)\<less\>\<delta\>\<Rightarrow\>d(z,y)\<less\>\<delta\>\<Rightarrow\>y\<in\>B<rsub|d>(z,\<delta\>)\<Rightarrow\>z\<in\><big|prod><rsub|i\<in\>I>B<rsub|d<rsub|i>>(z(i),\<delta\>)\<subseteq\>B<rsub|d>(z,\<delta\>)>
</proof>
<\definition>
<index|pseudo normed space><index|normed space>A pseudo normed space
<math|X,\<shortparallel\>\<shortparallel\>> is a (real or complex) vector
space <math|X> together with a function
<math|\<shortparallel\>\<shortparallel\>> from <math|X> to
<math|\<bbb-R\>> (where we write <math|\<shortparallel\>\<shortparallel\>(x)=\<shortparallel\>x\<shortparallel\>>)
such that\
<\enumerate>
<item><math|\<forall\>x\<in\>X> we have
<math|\<shortparallel\>x\<shortparallel\>\<geqslant\>0>
<item><math|\<forall\>x\<in\>X,\<forall\>\<alpha\>\<in\>\<bbb-R\>(\<bbb-C\>)>
we have <math|\<shortparallel\>\<alpha\>.x\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>x\<shortparallel\>>
<item><math|\<forall\>x,y\<in\>X> we have
<math|\<shortparallel\>x+y\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
</enumerate>
\ if also <math|\<shortparallel\>x\<shortparallel\>=0\<Rightarrow\>x=0>
then <math|X,\<shortparallel\>\<shortparallel\>> is a normed vector
space.
</definition>
<\theorem>
<label|norm of a finite sum>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a normed vector space then <math|\<shortparallel\><big|sum><rsub|i=1><rsup|n>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\>>
</theorem>
<\proof>
We prove this by induction\
<\enumerate>
<item><math|n=1> then <math|\<shortparallel\><big|sum><rsub|i=1><rsup|1>\<shortparallel\>x<rsub|i>\<shortparallel\>=\<shortparallel\>x<rsub|1>\<shortparallel\>=<big|sum><rsub|i=1><rsup|1>\<shortparallel\>x<rsub|i>\<shortparallel\>>
<item>If the theorem is true for <math|n> then
<math|\<shortparallel\><big|sum><rsub|i=1><rsup|n+1>\<shortparallel\>x<rsub|i>\<shortparallel\>=\<shortparallel\>(<big|sum><rsub|i=1><rsup|n>x<rsub|i>)+x<rsub|n+1>\<shortparallel\>\<leqslant\>\<shortparallel\><big|sum><rsub|i=1><rsup|n>x<rsub|i>\<shortparallel\>+\<shortparallel\>x<rsub|n+1>\<shortparallel\>\<leqslant\>(<big|sum><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\>)+\<shortparallel\>x<rsub|n+1>\<shortparallel\>=<big|sum><rsub|i=1><rsup|n+1>\<shortparallel\>x<rsub|i>\<shortparallel\>>
</enumerate>
</proof>
<\theorem>
<label|normed space properties>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a normed vector space then <math|\<forall\>\<alpha\>\<neq\>0,\<alpha\>\<in\>\<bbb-K\>>
and <math|\<forall\>x\<in\>X> we have\
<\enumerate>
<item><math|\<forall\>M\<subseteq\>X> we have
<math|<wide|\<alpha\>M|\<bar\>>=\<alpha\><wide|M|\<bar\>>>
<item><math|\<forall\>M> open in <math|X> we have that
\ <math|\<alpha\>M> is open
<item><math|\<forall\>M> open in <math|X >we have that \ <math|x+M> is
open
<item><math|\<forall\>N\<subseteq\>X> we have
<math|<wide|M|\<bar\>>-<wide|M|\<bar\>>\<subseteq\><wide|M-M|\<bar\>>>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|x\<in\><wide|\<alpha\>M|\<bar\>>> \ and take
<math|z\<equallim\><rsub|\<alpha\>\<neq\>0><frac|x|\<alpha\>>> then
<math|x=\<alpha\>.z>, now if <math|z\<in\>U, U> open then there exists
a <math|\<delta\>\<gtr\>0> such that
<math|z\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z,\<delta\>)\<subseteq\>U>.
Now as <math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<alpha\>\<delta\>)>
a open set and <math|x\<in\><wide|\<alpha\>M|\<bar\>>> there exists a
<math|m\<in\>M> such that <math|\<alpha\>m\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\|\<alpha\>\|\<delta\>)\<Rightarrow\>\<shortparallel\>\<alpha\>m-x\<shortparallel\>\<less\>\|\<alpha\>\|\<delta\>\<Rightarrow\>\<shortparallel\>\<alpha\>m-\<alpha\>z\<shortparallel\>\<less\>\|\<alpha\>\|\<delta\>\<Rightarrow\>\|\<alpha\>\|\<shortparallel\>m-z\<shortparallel\>\<less\>\|\<alpha\>\|\<delta\>\<Rightarrow\>\<shortparallel\>m-z\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>m\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z,\<delta\>)<big|cap>M\<subseteq\>U<big|cap>M>
and thus <math|z\<in\><wide|M|\<bar\>>><math|> and thus
<math|x\<in\>\<alpha\><wide|M|\<bar\>>. > If
<math|x\<in\>\<alpha\><wide|M|\<bar\>> > then \ if <math|x\<in\>U, U>
open there exists a <math|\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)\<subseteq\>U>,
now <math|x=\<alpha\>.n> where <math|n\<in\><wide|M|\<bar\>>>. Then
<math|n\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(n,<frac|\<delta\>|\|\<alpha\>\|>)>
which is open so that there exists a
<math|m\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(n,<frac|\<delta\>|\|\<alpha\>\|>)<big|cap>M>
and thus <math|\<alpha\>m\<in\>\<alpha\>M> also
<math|\<shortparallel\>n-m\<shortparallel\>\<less\><frac|\<delta\>|\|a\|>\<Rightarrow\>\|\<alpha\>\|\<shortparallel\>n-m\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>\<alpha\>n-\<alpha\>m\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>x-\<alpha\>m\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<alpha\>m\<in\>\<alpha\>M<big|cap>B<rsub|\<shortparallel\>>(x,\<delta\>)\<subseteq\>\<alpha\>M<big|cap>U>
proving that <math|x\<in\><wide|\<alpha\>M|\<bar\>>>
<item>If <math|x\<in\>\<alpha\>M> then <math|x=\<alpha\>m,m\<in\>M> and
as <math|M> is open there exists a <math|\<delta\>\<gtr\>0> such that
<math|m\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(m,\<delta\>)\<subseteq\>M>
then if <math|z\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x\<comma\>\|\<alpha\>\|\<delta\>)>
then we have <math|\<shortparallel\>z-x\<shortparallel\>\<less\>\|\<alpha\>\|\<delta\>\<Rightarrow\>\<shortparallel\><frac|z|\<alpha\>>-<frac|x|\<alpha\>>\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\><frac|z|\<alpha\>>-m\<shortparallel\>\<less\>\<delta\>\<Rightarrow\><frac|z|\<alpha\>>\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(m,\<delta\>)\<subseteq\>M\<Rightarrow\>z\<in\>\<alpha\>M\<Rightarrow\>x\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\|\<alpha\>\|\<delta\>)\<subseteq\>\<alpha\>M>
proving that <math|\<alpha\>M> is open
<item> Let <math|z\<in\>x+M> then <math|z-x\<in\>M> and there exist a
<math|\<delta\>\<gtr\>0> such that <math|(z-x)\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z-x,\<delta\>)\<subseteq\>M>.
Now if <math|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z,\<delta\>)\<Rightarrow\>\<shortparallel\>(y-x)-(z-x)\<shortparallel\>=\<shortparallel\>y-z\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>(y-x)\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z-x,\<delta\>)\<subseteq\>M\<Rightarrow\>y\<in\>x+M\<Rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\>>(z,\<delta\>)\<subseteq\>x+M\<Rightarrow\>z\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(z,\<delta\>)\<subseteq\>x+M>
proving that <math|x+M> is open
<item>If <math|x\<in\><wide|M|\<bar\>>-<wide|M|\<bar\>>> then
<math|x=y<rsub|1>-y<rsub|2>,y<rsub|1>\<in\><wide|M|\<bar\>>,
y<rsub|2>\<in\><wide|M|\<bar\>>> and suppose that <math|x\<in\>U,U>
open so there exists a <math|\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)\<subseteq\>U>
now there exists a <math|z<rsub|1>\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(y<rsub|1>,<frac|\<delta\>|2>)<big|cap>M,z<rsub|2>\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(y<rsub|2>,<frac|\<delta\>|2>)<big|cap>M>
take then <math|z=z<rsub|1>-z<rsub|2>\<in\>M-M> then
<math|\<shortparallel\>x-z\<shortparallel\>=\<shortparallel\>y<rsub|1>-y<rsub|2>-(z<rsub|1>-z<rsub|2>)\<shortparallel\>\<leqslant\>\<shortparallel\>y<rsub|1>-z<rsub|1>\<shortparallel\>+\<shortparallel\>y<rsub|2>-z<rsub|2>\<shortparallel\>\<less\><frac|\<delta\>|2>+<frac|\<delta\>|2>=\<delta\>\<Rightarrow\>z\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)<big|cap>(M-M)\<subseteq\>U<big|cap>(M-M)\<Rightarrow\>x\<in\><wide|M-M|\<bar\>>>
</enumerate>
</proof>
<\remark>
<math|\<shortparallel\>0\<shortparallel\>=\<shortparallel\>0.0\<shortparallel\>=0.\<shortparallel\>0\<shortparallel\>=0>
</remark>
<\theorem>
<label|metric from norm>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a (pseudo) normed space then <math|X,d<rsub|\<shortparallel\>\<shortparallel\>>>
with <math|d<rsub|\<shortparallel\>\<shortparallel\>>(x,y)=\<shortparallel\>x-y\<shortparallel\>>
is a (pseudo) metric space
</theorem>
<\proof>
\;
<\enumerate>
<item><math|d<rsub|\<shortparallel\>\<shortparallel\>>(x,y)=\<shortparallel\>x-y\<shortparallel\>\<geqslant\>0>
<item><math|d<rsub|\<shortparallel\>\<shortparallel\>>(x,y)=\<shortparallel\>x-y\<shortparallel\>=\|\<um\>1\|\<shortparallel\>y-x\<shortparallel\>=\<shortparallel\>y-x\<shortparallel\>=d<rsub|\<shortparallel\>\<shortparallel\>>(y,x)>
<item><math|d<rsub|\<shortparallel\>\<shortparallel\>>(x,z)=\<shortparallel\>x-z\<shortparallel\>\<leqslant\>\<shortparallel\>x-y+y-z\<shortparallel\>\<leqslant\>\<shortparallel\>x-y\<shortparallel\>+\<shortparallel\>y-z\<shortparallel\>=\<shortparallel\>x-y\<shortparallel\>+\<shortparallel\>z-y\<shortparallel\>=d(x,y)+d(y,z)>
<item>If <math|\<shortparallel\>\<shortparallel\>> is normed then
<math|d(x,y)=0\<Rightarrow\>\<shortparallel\>x-y\<shortparallel\>=0\<Rightarrow\>x-y=0\<Rightarrow\>x=y>
</enumerate>
</proof>
<\theorem>
<label|norm and absolute value>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a pseudo normed space then <math|\<forall\>x,y\<in\>X> we have
<math|\|\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>\|\<leqslant\>\<shortparallel\>x-y\<shortparallel\>>
</theorem>
<\proof>
<math|\<shortparallel\>x\<shortparallel\>=\<shortparallel\>x-y+y\<shortparallel\>\<leqslant\>\<shortparallel\>x-y\<shortparallel\>+\<shortparallel\>y\<shortparallel\>\<Rightarrow\>\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>\<leqslant\>\<shortparallel\>x-y\<shortparallel\>>
and <math|\<shortparallel\>y\<shortparallel\>=\<shortparallel\>y-x+x\<shortparallel\>\<leqslant\>\<shortparallel\>y-x\<shortparallel\>+\<shortparallel\>x\<shortparallel\>\<Rightarrow\>\<shortparallel\>y\<shortparallel\>-\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>x-y\<shortparallel\>>
so <math|\|\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>\|\<leqslant\>\<shortparallel\>x-y\<shortparallel\>>
</proof>
<\definition>
<index|normed topology>Let <math|X,\<shortparallel\>\<shortparallel\>> be
a (pseudo) normed space then the topology generated by the associated
(pseudo) metric <math|d<rsub|\<shortparallel\>\<shortparallel\>>> is
called the normed topology\
</definition>
<\example>
<label|topology of real numbers>The vectorspace of real numbers
<math|\<bbb-R\>> together with the absolute value <math|\| \|> is a
normed space and the basis generated by the norm is the set of open
intervals <math|\<cal-B\>={]a,b[\|a,b\<in\>\<bbb-R\>}>
</example>
<\proof>
First we prove that <math|\| \|> is a norm so
<math|\<forall\>\<alpha\>\<in\>\<bbb-R\>,x,y\<in\>\<bbb-R\>>
<\enumerate>
<item><math|><math|\|x\|=max(x,0)\<geqslant\>0>
<item>To prove <math|\|\<alpha\>.x\|=\|\<alpha\>\|.\|x\|> consider the
following possibilities
<\enumerate>
<item><math|0\<leqslant\>\<alpha\>,0\<leqslant\>x> then
<math|0\<leqslant\>\<alpha\>.x\<Rightarrow\>\|\<alpha\>.x\|=\<alpha\>.x=\|\<alpha\>\|.\|x\|>
<item><math|0\<leqslant\>\<alpha\>,x\<leqslant\>0> then
<math|\<alpha\>.x\<leqslant\>0\<Rightarrow\>\|\<alpha\>.x\|=\<um\>(\<alpha\>.x)=\<alpha\>.(\<um\>x)=\|\<alpha\>\|.\|x\|>
<item><math|\<alpha\>\<leqslant\>0,0\<leqslant\>x> then
<math|\<alpha\>.x\<leqslant\>0\<Rightarrow\>\|\<alpha\>.x\|=\<um\>(\<alpha\>.x)=(\<um\>\<alpha\>).x=\|\<alpha\>\|.\|x\|>
<item><math|\<alpha\>\<leqslant\>0,x\<leqslant\>0> then
<math|0\<leqslant\>\<alpha\>.x\<Rightarrow\>\|\<alpha\>.x\|=\<alpha\>.x=(\<um\>\<alpha\>).(\<um\>x)=\|\<alpha\>\|.\|x\|>
</enumerate>
<item>To prove that <math|><math|\|x+y\|\<less\>\|x\|+\|y\|> consider
the following possibilities
<\enumerate>
<item><math|0\<leqslant\>x,0\<leqslant\>y> then
<math|0\<leqslant\>x+0\<Rightarrow\>\|x+y\|=x+y=\|x\|+\|y\|\<Rightarrow\>\|x+y\|\<leqslant\>\|x\|+\|y\|>
<item><math|0\<leqslant\>x,y\<leqslant\>0> then we have the following
possibilities for <math|\|y\|=\<um\>y>
<\enumerate>
<item><math|x=\|y\|\<Rightarrow\>x=\<um\>y\<Rightarrow\>x+y=0\<Rightarrow\>\|x+y\|=0\<leqslant\>\|x\|+\|y\|>
(as <math|0\<leqslant\>\| .\|>) and thus
<math|\|x+y\|\<leqslant\>\|x\|+\|y\|>
<item><math|><math|x\<less\>\|y\|> then
<math|x\<less\>\<um\>y\<Rightarrow\>x+y\<less\>0\<Rightarrow\>\|x+y\|=\<um\>(x+y)=(\<um\>x)+(\<um\>y)\<equallim\><rsub|x\<geqslant\>0>(\<um\>x)+\|y\|\<equallim\><rsub|\<um\>x\<leqslant\>0>\|y\|\<leqslant\>\|x\|+\|y\|<rsub|>>
(as <math|0\<leqslant\>\|x\|>) and thus
<math|\|x+y\|\<leqslant\>\|x\|+\|y\|>
<item><math|\|y\|\<less\>x> then
<math|\<um\>y\<less\>x\<Rightarrow\>0\<less\>x+y\<Rightarrow\>\|x+y\|=x+y>
now as <math|y\<leqslant\>0\<Rightarrow\>x+y\<leqslant\>x> and thus
<math|x+y\<leqslant\>x=\|x\|\<leqslant\>\|x\|+\|y\|> (as
<math|0\<leqslant\>\|y\|>) so we have proved that
<math|\|x+y\|\<leqslant\>\|x\|+\|y\|>
</enumerate>
<item><math|x\<leqslant\>0,0\<leqslant\>y> then by setting
<math|x<rprime|'>=y,y<rprime|'>=x> we have
<math|0\<leqslant\>x<rprime|'>,0\<leqslant\>y<rprime|'>> and use (b)
to prove that <math|\|x+y\|=\|y+x\|=\|x<rprime|'>+y<rprime|'>\|\<leqslant\>\|x<rprime|'>\|+\|y<rprime|'>\|=\|y<rprime|'>\|+\|x<rprime|'>\|=\|x\|+\|y\|>
<item><math|x\<leqslant\>0,y\<leqslant\>0> then
<math|x+y\<leqslant\>y\<leqslant\>0\<Rightarrow\>\|x+y\|=\<um\>(x+y)=(\<um\>x)+(\<um\>y)=\|x\|+\|y\|\<leqslant\>\|x\|+\|y\|>
and thus <math|\|x+y\|\<leqslant\>\|x\|+\|y\|>
</enumerate>
<item>Assume now that <math|\|x\|=0> then we have the following
possibilities for <math|x>
<\enumerate>
<item><math|x\<less\>0> then <math|0\<less\>(\<um\>x)=\|x\|=0> a
contradiction
<item><math|0\<less\>x> then <math|0\<less\>\|x\|=0> again a
contradiction
<item><math|x=0> this is what we have to proof
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|finer norm>Let <math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
be two norms on <math|X> and <math|\<cal-T\><rsub|1>,\<cal-T\><rsub|2>>
the generated topologies by these norms. Then <math|\<cal-T\><rsub|2>> is
finer then <math|\<cal-T\><rsub|1>> iff there exists a <math|M\<gtr\>0>
such that <math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>M.\<shortparallel\>x\<shortparallel\><rsub|2>>
</theorem>
<\proof>
We use <reference|finer metric> to prove this theorem\
<\enumerate>
<item><math|\<Rightarrow\>> Take then
<math|0\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|1>>>(0,1)\<Rightarrowlim\><rsub|<reference|finer
metric>>\<exists\>\<delta\>\<gtr\>0\<succ\>0\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|2>>>(0,\<delta\>)\<subseteq\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|1>>>(0,1)\<Rightarrow\>\<forall\>z\<vdash\>\<shortparallel\>z\<shortparallel\><rsub|2>\<less\>\<delta\>>
we have <math|z\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|2>>>(0,\<delta\>)\<Rightarrow\>\<shortparallel\>z\<shortparallel\><rsub|1>\<less\>1>
take then <math|M=<frac|2|\<delta\>>> for <math|x\<in\>X> we have now
two cases:
<\enumerate>
<item><math|x=0> then <math|><math|0=\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>0=M.0=M.\<shortparallel\>x\<shortparallel\><rsub|2>>
<item><math|x\<neq\>0> then for <math|z=<frac|1|M.\<shortparallel\>x\<shortparallel\><rsub|2>>.x>
we have <math|\<shortparallel\>z\<shortparallel\><rsub|2>=<frac|1|M.\<shortparallel\>x\<shortparallel\><rsub|2>>.\<shortparallel\>x\<shortparallel\><rsub|2>=<frac|\<delta\>|2>\<less\>\<delta\>\<Rightarrow\><frac|1|M.\<shortparallel\>x\<shortparallel\><rsub|2>>.\<shortparallel\>x\<shortparallel\><rsub|1>=\<shortparallel\>z\<shortparallel\><rsub|1>\<less\>1\<Rightarrow\>\<shortparallel\>x\<\|\|\><rsub|1>\<less\>M.\<shortparallel\>x\<shortparallel\><rsub|2>>
</enumerate>
<item><math|\<Leftarrow\>> <math|\<forall\>x\<in\>X,\<varepsilon\>\<gtr\>0>
let <math|x\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|1>>>(x,\<varepsilon\>)>
and using the fact that <math|M\<gtr\>0\<Rightarrow\>\<delta\>=<frac|\<varepsilon\>|M>\<gtr\>0>
let <math|y\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|2>>>(x,\<delta\>)\<Rightarrow\>\<shortparallel\>x-y\<shortparallel\><rsub|2>\<less\><frac|\<varepsilon\>|M>>
and then <math|\<shortparallel\>x-y\<shortparallel\><rsub|1>\<leqslant\>M.\<shortparallel\>x-y\<shortparallel\><rsub|2>\<less\>M.<frac|\<varepsilon\>|M>=\<varepsilon\>\<Rightarrow\>y\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|1>>>(x,\<varepsilon\>)>
and thus <math|x\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|2>>>(x,0)\<subseteq\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\><rsub|1>>>(x,\<varepsilon\>)>
and we use then <reference|finer metric> to prove the theorem
</enumerate>
</proof>
<\lemma>
<label|maximum of products>Let <math|A\<subseteq\>\<bbb-R\>> and \ given
<math|\<alpha\>\<in\>\<bbb-R\><rsub|+,0>> then if
<math|\<alpha\>.A={\<alpha\>.a\|a\<in\>A}> we have if <math|max(A)>
exists that <math|max(\<alpha\>.A)> exists and
<math|max(\<alpha\>.A)=\<alpha\>.max(A)><math|><math|>
</lemma>
<\proof>
As <math|max(A)> exists <math|\<exists\>m\<in\>A\<succ\>\<forall\>a\<in\>A>
we have <math|a\<leqslant\>m> then if <math|b\<in\>\<alpha\>.A> there
exists a <math|a\<in\>A> such that <math|b=\<alpha\>.a> and as
<math|\<alpha\>\<geqslant\>0> we have from <math|a\<leqslant\>m> that
<math|b=\<alpha\>.a\<leqslant\>\<alpha\>.m> which together with the fact
that <math|\<alpha\>.m\<in\>a.A> proves that <math|a.m> is the maximum of
<math|\<alpha\>.A>
</proof>
<\theorem>
<label|norm of finite product of normed spaces>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>I>,I>
finite, be a finite family of normed spaces then
<math|\<shortparallel\>\<shortparallel\>> defined by
<math|\<shortparallel\>x\<shortparallel\>=max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)>
is a norm and its generated topology coincidences with the product
topology (<reference|product topology>) (using the norm topologies of
<math|X<rsub|i>>). The norm <math|\<shortparallel\>\<shortparallel\>> is
called the maximum norm
</theorem>
<\proof>
First we prove that <math|\<shortparallel\>\<shortparallel\>> is a norm,
(note that the maximum is defined by <reference|maximum minimum of finite
sets>)\
<\enumerate>
<item><math|\<shortparallel\>x\<shortparallel\>=max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)\<geqslant\>0>
as <math|\<shortparallel\>\<shortparallel\><rsub|i>\<geqslant\>0>
<item><math|><math|\<shortparallel\>\<alpha\>.x\<shortparallel\>=max(\<shortparallel\>\<pi\><rsub|i>(\<alpha\>.x)\<shortparallel\><rsub|i>\|i\<in\>I)\<equallim\><rsub|\<pi\><rsub|i>(a.x)=\<alpha\>.x(i)=\<alpha\>.\<pi\><rsub|i>(x)>max(\<shortparallel\>\<alpha\>.\<pi\><rsub|i>(x)\<shortparallel\>\|i\<in\>I)=max(\|\<alpha\>\|.\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)\<equallim\><rsub|<reference|maximum
of products>>\|\<alpha\>\|.max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)=\|\<alpha\>\|.\<shortparallel\>x\<shortparallel\>>
<item><math|\<shortparallel\>x+y\<shortparallel\>=max(\<shortparallel\>\<pi\><rsub|i>(x+y)\<shortparallel\><rsub|i>\|i\<in\>I)\<equallim\><rsub|\<pi\><rsub|i>(x+y)=(x+y)(i)=x(i)+y(i)=\<pi\><rsub|i>(x)+\<pi\><rsub|i>(y)>max(\<shortparallel\>\<pi\><rsub|i>(x)+\<pi\><rsub|i>(y)\<shortparallel\><rsub|i>\|i\<in\>I)\<leqslant\><rsub|<reference|maximum
of two sets>,<reference|maximum minimum of finite
sets>>max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>+\<shortparallel\>\<pi\><rsub|i>(y)\<shortparallel\><rsub|i>\|i\<in\>I)\<leqslant\><rsub|<reference|maximum
of two sets>>max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>+max(\<shortparallel\>\<pi\><rsub|i>(y)\<shortparallel\><rsub|i>\|i\<in\>I)\<leqslant\>max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)+max(\<shortparallel\>\<pi\><rsub|i>(y)\<shortparallel\><rsub|i>\|i\<in\>I)=\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
<item>If <math|\<shortparallel\>x\<shortparallel\>=0\<Rightarrow\>max(\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>\|i\<in\>I)=0\<Rightarrow\>\<forall\>i\<in\>I>
we have <math|\<shortparallel\>\<pi\><rsub|i>(x)\<shortparallel\><rsub|i>=0\<Rightarrow\>\<pi\><rsub|i>(x)=0\<Rightarrow\>x(i)=0\<Rightarrow\>x=0>
</enumerate>
Next using <reference|product of metric spaces> we find that the product
topology is generated by <math|d(x,y)=max(d<rsub|\<shortparallel\>\<shortparallel\><rsub|i>>(\<pi\><rsub|i>(x),\<pi\><rsub|i>(y))\|i\<in\>I)=max(\<shortparallel\>\<pi\><rsub|i>(x)-\<pi\><rsub|i>(y)\<\|\|\><rsub|i>\|i\<in\>I)=max(\<shortparallel\>\<pi\><rsub|i>(x-y)\<shortparallel\><rsub|i>\|i\<in\>I)=\<\|\|\>x-y\<\|\|\>>
so <math|\<shortparallel\>\<shortparallel\>> defines indeed the metric
that generates the product topology.
</proof>
<\example>
The product topology of <math|\<bbb-R\><rsup|n>=\<Pi\><rsub|i\<in\>{1,\<ldots\>,n}>\<bbb-R\>>
is generated by the maximum norm <math|\<shortparallel\>\<shortparallel\>>
with <math|\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=max(\|x<rsub|i>\|
\|i\<in\>I)> (<math|\<shortparallel\>\<shortparallel\><rsub|i>=\| \|>)
</example>
<\definition>
<index|equivalent norms>Two norms on a vector space are equivalent iff
they generate the same normed topology
</definition>
<\lemma>
<label|equivalence of norms>Let <math|X> be a vector space and
<math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>two
norms on <math|X> then these norms are equivalent iff there exists
<math|M<rsub|1>,M<rsub|2>\<gtr\>0> such that
<math|M<rsub|1>\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|2>\<leqslant\>M<rsub|2>\<shortparallel\>x\<shortparallel\><rsub|1>>
</lemma>
<\proof>
By <reference|finer norm> we have the existence of
<math|M<rprime|'><rsub|1>,M<rprime|'><rsub|2>\<gtr\>0> such that
<math|\<forall\>x\<in\>X> we have <math|\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>M<rprime|'><rsub|1>\<shortparallel\>x\<shortparallel\><rsub|2>>
and <math|\<shortparallel\>x\<shortparallel\><rsub|2>\<leqslant\>M<rprime|'><rsub|2>\<shortparallel\>x\<shortparallel\><rsub|1>>
take then <math|M<rsub|1>=<frac|1|M<rsub|1><rprime|'>>,M<rsub|2>=M<rprime|'><rsub|2>>
then <math|\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\><frac|1|M<rsub|1>>\<shortparallel\>x\<shortparallel\><rsub|2>\<Rightarrow\>M<rsub|1>\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|2>\<leqslant\>M<rsub|2>\<shortparallel\>x\<shortparallel\><rsub|1>>
</proof>
<\definition>
<index|isometry>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>
and <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be two normed
spaces then a isometry is a bijective function <math|\<varphi\>> between
<math|X> and <math|Y> such that <math|\<forall\>x\<in\>X\<succ\>\<shortparallel\>\<varphi\>(x)\<shortparallel\><rsub|Y>=\<shortparallel\>x\<shortparallel\><rsub|X>>
</definition>
<\theorem>
<label|inverse of isomorphisme norm>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y> be two normed spaces
and > <math|\<varphi\>> a isometry between <math|X> and <math|Y> then
<math|\<varphi\><rsup|-1>> is a isometry\
</theorem>
<\proof>
First because <reference|inverse of bijective mapping>
<math|\<varphi\><rsup|-1>> is bijective. Second
<math|\<shortparallel\>\<varphi\><rsup|-1>(x)\<shortparallel\><rsub|X>=\<shortparallel\>\<varphi\>(\<varphi\><rsup|-1>(x))\<shortparallel\><rsub|Y>=\<shortparallel\>x\<shortparallel\><rsub|Y>>
</proof>
<\theorem>
<label|composition of isomorphisme norm>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> and
<math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be three normed
spaces and <math|\<varphi\><rsub|1>:X\<rightarrow\>Y,\<varphi\><rsub|2>:Y\<rightarrow\>Z>
be isometries then <math|\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>:X\<rightarrow\>Z>
is a isometry
</theorem>
<\proof>
First because of <reference|composition of bijective mappings>
<math|\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1>> is a bijective
function. Secondly <math|\<shortparallel\>(\<varphi\><rsub|2>\<circ\>\<varphi\>)(x)\<shortparallel\><rsub|Z>=\<shortparallel\>\<varphi\><rsub|2>(\<varphi\><rsub|1>(x))\<shortparallel\><rsub|Z>=\<shortparallel\>\<varphi\><rsub|1>(x)\<shortparallel\><rsub|Y>=\<shortparallel\>x\<shortparallel\><rsub|X>>
</proof>
<\theorem>
<label|isometric normed spaces>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be two normed spaces with a isometry <math|\<varphi\>> between them, then
if <math|\<cal-T\><rsub|X>,\<cal-T\><rsub|Y>> are the topologies generate
by <math|\<shortparallel\>\<shortparallel\><rsub|X>,\<shortparallel\>\<shortparallel\><rsub|Y>>
then <math|\<cal-T\><rsub|X>={\<varphi\><rsup|-1><rsub|>(V)\|V\<in\>\<cal-T\><rsub|Y>}={U\<subseteq\>X\|\<varphi\>(U)\<in\>\<cal-T\><rsub|Y>}>
and <math|\<cal-T\><rsub|Y>={V\<subseteq\>Y\|\<varphi\><rsup|-1>(V)\<in\>\<cal-T\><rsub|X>}={\<varphi\>(U)\|U\<in\>\<cal-T\><rsub|X>}>
or in other words the topologies are equivalent.
</theorem>
<\proof>
The normed topologies are generated by the metric
<math|d<rsub|X>(x,y)=\<shortparallel\>x-y\<shortparallel\><rsub|X>,d<rsub|Y>(x,y)=\<shortparallel\>x-y\<shortparallel\><rsub|Y>>
and then we have <math|d<rsub|Y>(\<varphi\>(x).\<varphi\>(y))=\<shortparallel\>\<varphi\>(x)-\<varphi\>(y)\<shortparallel\><rsub|Y>=\<shortparallel\>x-y\<shortparallel\><rsub|X>=d<rsub|X>(x,y)>
so using <reference|topologies and isometries> we prove our theorem.
</proof>
<\definition>
<index|convergent sequence>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a normed space then a sequence <math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
is convergent to <math|x\<in\>X> if it is convergent to <math|x> in the
associated metric space or if we note that the associated metric is
defined by <math|d(x,y)=\<shortparallel\>x-y\<shortparallel\>>,
<math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>,n\<geqslant\>0> we have
<math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\>\<varepsilon\>>
</definition>
<\definition>
<index|Cauchy sequence>Let <math|X,\<shortparallel\>\<shortparallel\>> be
a normed space then a sequence <math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
is Cauchy to <math|x\<in\>X> if it is convergent to <math|x> in the
associated metric space or if we note that the associated metric is
defined by <math|d(x,y)=\<shortparallel\>x-y\<shortparallel\>>,
<math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>m,n\<in\>\<bbb-N\><rsub|0>,n,m\<geqslant\>0> we have
<math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\>\<less\>\<varepsilon\>>
</definition>
<section|Inner product spaces>
<\definition>
<index|real inner product space><index|inner product space>A real inner
product space is a real vector space <math|V> with a map
<math|\<langle\>\<rangle\>:V\<times\>V\<rightarrow\>\<bbb-R\>> (the inner
product) satisfying the following:
<\enumerate>
<item><math|\<forall\>x,y\<in\>V> we have
<math|\<langle\>x,y\<rangle\>=\<langle\>y,x\<rangle\>>
<item><math|\<forall\>x,y,z\<in\>V> and
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>> we have
<math|\<langle\>\<alpha\>.x+\<beta\>.y,z\<rangle\>=\<alpha\>\<langle\>x,z\<rangle\>+\<beta\>\<langle\>y,z\<rangle\>>
<\note>
Because of (1) we have also <math|\<langle\>z,\<alpha\>.x+\<beta\>.y\<rangle\>=\<alpha\>\<langle\>z,x\<rangle\>+\<beta\>\<langle\>z,y\<rangle\>>
<item><math|\<forall\>x\<in\>V> we have
<math|\<langle\>x,x\<rangle\>\<geqslant\>0>
<item><math|\<forall\>x\<in\>V> if we have
<math|\<langle\>x,x\<rangle\>=0\<Rightarrow\>x=0>
</note>
</enumerate>
</definition>
<\definition>
<index|complex inner product space><index|inner product space>A complex
inner product space is a complex vector space <math|V> with a map
<math|\<langle\>\<rangle\>:V\<times\>V\<rightarrow\>\<bbb-C\>> (the inner
product) satisfying the following:
<\enumerate>
<item><math|\<forall\>x,y\<in\>V> we have
<math|\<langle\>x,y\<rangle\>=<wide|\<langle\>y,x\<rangle\>|\<bar\>>>\
<\note>
This implies <math|\<langle\>x,x\<rangle\>=<wide|\<langle\>x,x\<rangle\>|\<bar\>>>
or <math|\<langle\>x,x\<rangle\>> is real
<item><math|\<forall\>x,y,z\<in\>V> and
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-C\>> we have
<math|\<langle\>\<alpha\>.x+\<beta\>.y,z\<rangle\>=\<alpha\>\<langle\>x,z\<rangle\>+\<beta\>\<langle\>y,z\<rangle\>>\
<\note>
Because of (1) we have <math|\<langle\>z,\<alpha\>.x+\<beta\>.y\<rangle\>=<wide|\<langle\>\<alpha\>.x+\<beta\>.y,z\<rangle\>|\<bar\>>=<wide|\<alpha\>\<langle\>x,z\<rangle\>+\<beta\>\<langle\>y,z\<rangle\>|\<bar\>>=<wide|\<alpha\>|\<bar\>>\<langle\>z,x\<rangle\>+<wide|\<beta\>|\<bar\>>\<langle\>z,y\<rangle\>>
<item><math|\<forall\>x\<in\>V> we have
<math|\<langle\>x,x\<rangle\>\<geqslant\>0> (make sense since
<math|\<langle\>x,x\<rangle\>> is real so)
<item><math|\<forall\>x\<in\>V> if
<math|\<langle\>x,x\<rangle\>=0\<Rightarrow\>x=0>
</note>
</note>
</enumerate>
</definition>
<\theorem>
<label|inner product of neutral elements>Let
<math|V,\<langle\>\<rangle\>> be a real (complex) inner product space
then <math|\<forall\>x\<in\>V> we have
<math|\<langle\>x,0\<rangle\>=\<langle\>0,x\<rangle\>=0>
</theorem>
<\proof>
<math|\<langle\>0,x\<rangle\>=\<langle\>0.x,x\<rangle\>=\|0\|\<langle\>x,x\<rangle\>=0>
and for the real case we have <math|\<langle\>0,x\<rangle\>=\<langle\>x,0\<rangle\>=0>
and in the complex case <math|\<langle\>0,x\<rangle\>=<wide|\<langle\>x,0\<rangle\>|\<bar\>>=<wide|0|\<bar\>>=0>
</proof>
<\theorem>
<label|sum of positive numbers is positive>Let <math|A={1,\<ldots\>,n}>
and <math|s:A\<rightarrow\>\<bbb-R\>(\<bbb-C\><rsub|\<bbb-R\>>)> be a
function so that <math|\<forall\>i\<in\>A\<succ\>s(i)\<geqslant\>0> then
<math|<big|sum><rsub|i=1><rsup|n>s(i)=0\<Rightarrow\>\<forall\>i\<in\>A\<succ\>s(i)=0>
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|><math|0=<big|sum><rsub|i=1><rsup|1>s(i)=s(1)>
<item>Assume that the theorem is true for <math|n> then
<math|<big|sum><rsub|i=1><rsup|n+1>s(i)=(<big|sum><rsub|i=1><rsup|n>s(i))+s(n+1)=0\<Rightarrowlim\><rsub|<reference|zero
sum>>(<big|sum><rsub|i=1><rsup|n>s(i))=0,s(n+1)=0> and the induction
hypothesis proves that also <math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>s(i)=0>
and this together with <math|s(n+1)=0> proves the theorem for
<math|n+1>
</enumerate>
</proof>
<\example>
Take <math|V=\<bbb-R\><rsup|n>> and with
<math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(y<rsub|1>,\<ldots\>,y<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.y<rsub|i>>
forms a real inner product space.
</example>
<\proof>
\;
<\enumerate>
<item><math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(y<rsub|1>,\<ldots\>,y<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.y<rsub|i>=<big|sum><rsub|i\<in\>1><rsup|n>y<rsub|i>.x<rsub|i>=\<langle\>(y<rsub|1>,\<ldots\>,y<rsub|n>),(x<rsub|1>,\<ldots\>,x<rsub|n>)\<rangle\>>
<item><math|><math|\<langle\>\<alpha\>.(x<rsub|1>,\<ldots\>,x<rsub|n>)+\<beta\>.(y<rsub|1>,\<ldots\>,y<rsub|n>),(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>=\<langle\>(\<alpha\>.x<rsub|1>+\<beta\>.y<rsub|<rsub|1>>,\<ldots\>,\<alpha\>.x<rsub|n>+\<beta\>.y<rsub|n>,(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>=<big|sum><rsub|1><rsup|n>(\<alpha\>x<rsub|i>+\<beta\>y<rsub|i>).z<rsub|i>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i<rsub|>>.z<rsub|i>+\<beta\>.y<rsub|i>.z<rsub|i>)\<equallim\><rsub|<reference|sum
of sums>,<reference|general sum is finite sum in the finite
case>>(<big|sum><rsub|i=1><rsup|n>\<alpha\>.<rsub|>x<rsub|i>.z<rsub|i>)+(<big|sum><rsub|i=1><rsup|n>\<beta\><rsub|i>.y<rsub|i>.z<rsub|i>)\<equallim\><rsub|<reference|scalair
product and sum>,<reference|general sum is finite sum in the finite
case>>(\<alpha\>.<big|sum><rsub|i=1><rsup|n>x<rsub|i>.z<rsub|i>)+(\<beta\>.<big|sum><rsub|i=1><rsup|n>y<rsub|i>.z<rsub|i>)=\<alpha\>\<langle\>x,z\<rangle\>+\<beta\>\<langle\>y,z\<rangle\>>
<item><math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(x<rsub|1>,\<ldots\>,x<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i><rsup|2>\<geqslant\>0>
(because of <reference|inequality and sum>)
<item>Let <math|0=\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(x<rsub|1>,\<ldots\>,m)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i<rsup|>><rsup|2>\<Rightarrowlim\><rsub|<reference|sum
of positive numbers is positive>>\<forall\>i\<in\>I\<succ\>x<rsub|i><rsup|2>=0\<Rightarrowlim\><rsub|<reference|kwadrat
is positive>>\<forall\>i\<in\>I\<succ\>x<rsub|i>=0\<Rightarrow\>x=0>
</enumerate>
</proof>
<\example>
Take <math|V=\<bbb-C\><rsup|n>> and define
<math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(y<rsub|1>,\<ldots\>,y<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|y<rsub|i>|\<bar\>>>
this forms a complex inner product space
</example>
<\proof>
\;
<\enumerate>
<item><math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(y<rsub|1>,\<ldots\>,y<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|y<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n><wide|<wide|x<rsub|i>|\<bar\>>.y<rsub|i>|\<bar\>>\<equallim\><rsub|<reference|conjugate
of sum>><wide|<big|sum><rsub|i=1><rsup|n>y<rsub|i>.<wide|x<rsub|i>|\<bar\>>|\<bar\>>=<wide|\<langle\>(y<rsub|1>,\<ldots\>,y<rsub|n>),(x<rsub|1>,\<ldots\>,x<rsub|n>)\<rangle\>|\<bar\>>>
<item><math|\<langle\>\<alpha\>.(x<rsub|1>,\<ldots\>,x<rsub|n>)+\<beta\>.(y<rsub|1>,\<ldots\>,y<rsub|n>),(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>=\<langle\>(\<alpha\>.x<rsub|1>+\<beta\>.y<rsub|1>,\<ldots\>,\<alpha\>.x<rsub|n>+\<beta\>.y<rsub|n>),(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>+\<beta\>.y<rsub|i>).<wide|z<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>.<wide|z<rsub|i>|\<bar\>>+\<beta\>.x<rsub|i>.<wide|z<rsub|i>|\<bar\>>)\<equallim\><rsub|<reference|sum
of sums>>(<big|sum><rsub|i=1><rsup|n>\<alpha\>.x<rsub|i>.<wide|z<rsub|i>|\<bar\>>)+<big|sum><rsub|i=1><rsup|n>\<beta\>.y<rsub|i>.<wide|z<rsub|i>|\<bar\>>\<equallim\><rsub|<reference|scalair
product and sum>>\<alpha\>.(<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|z<rsub|i>|\<bar\>>)+\<beta\>.(<big|sum><rsub|i=1><rsup|n>y<rsub|i>.<wide|z<rsub|i>|\<bar\>>)=\<alpha\>.\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>+\<beta\>.\<langle\>(y<rsub|1>,\<ldots\>,y<rsub|n>),(z<rsub|1>,\<ldots\>,z<rsub|n>)\<rangle\>>
<item><math|\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(x<rsub|1>,\<ldots\>,x<rsub|n>)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|x<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|<rsup|2>\<geqslant\>0>
(using <reference|sum and inequalities> and
<math|\|x<rsub|i>\|<rsup|2>\<geqslant\>0>)
<item>Let <math|0=\<langle\>(x<rsub|1>,\<ldots\>,x<rsub|n>),(x<rsub|1>,\<ldots\>,m)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i<rsup|>><rsup|2>\<Rightarrowlim\><rsub|<reference|sum
of positive numbers is positive>>\<forall\>i\<in\>I\<succ\>x<rsub|i><rsup|2>=0\<Rightarrowlim\><rsub|<reference|kwadrat
is positive>>\<forall\>i\<in\>I\<succ\>x<rsub|i>=0\<Rightarrow\>x=0><math|>
</enumerate>
</proof>
<\definition>
<index|inner product norm>Let <math|V,\<langle\>\<rangle\>> be a real
(complex) inner product space then the inner product norm is defined as
<math|\<shortparallel\>x\<shortparallel\>=<sqrt|\<langle\>x,x\<rangle\>>>
(in the case of a complex inner we define as
<math|\<langle\>x,x\<rangle\>> is real and thus
<math|\<langle\>x,x\<rangle\>=(z,0)> (uniquely defined) we define
<math|<sqrt|\<langle\>x,x\<rangle\>>=<sqrt|z>>)
</definition>
<\note>
If <math|x=0\<Rightarrow\>\<shortparallel\>x\<shortparallel\>=<sqrt|\<langle\>0,0\<rangle\>>=<sqrt|0>=0>
</note>
<\theorem>
<label|schwartz inequality>Let <math|V,\<langle\>\<rangle\>> be a real
(complex) inner product space and <math|\<shortparallel\>\<shortparallel\>>
the inner product norm then <math|\<forall\>x,y\<in\>V> then we have
<math|\|\<langle\>x,y\<rangle\>\|\<leqslant\>\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>,
equality holds if and only <math|x> and <math|y> are linearly dependent
</theorem>
<\proof>
We have the following cases
<\enumerate>
<item><math|x=0> then <math|x,y> are linearly dependent and using
<reference|inner product of neutral elements> we have
<math|\|\<langle\>x,y\<rangle\>\|=\|0\|=0.\<shortparallel\>y\<shortparallel\>=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
<item><math|y=0> then <math|x,y> are linearly dependent and using
<reference|inner product of neutral elements> we have
<math|\|\<langle\>x,y\<rangle\>\|=\|0\|=\<shortparallel\>x\<shortparallel\>.0=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
<item><math|x\<neq\>0,y\<neq\>0\<Rightarrow\>\<langle\>y,y\<rangle\>\<neq\>0>
take then <math|\<alpha\>=<frac|\<um\>\<langle\>x,y\<rangle\>|\<langle\>y,y\<rangle\>>>
then we can consider a the real and complex cases:
<\enumerate>
<item>(real case) we have <math|0\<leqslant\>\<langle\>x+\<alpha\>.y,x+\<alpha\>.y\<rangle\>=\<langle\>x,x\<rangle\>+2\<alpha\>\<langle\>x,y\<rangle\>+\<alpha\><rsup|2>\<langle\>y,y\<rangle\>=\<langle\>x,x\<rangle\>-2<frac|\<langle\>x,y\<rangle\><rsup|2>|\<langle\>y,y\<rangle\>>+<frac|\<langle\>x,y\<rangle\><rsup|2>|\<langle\>y,y\<rangle\><rsup|2>>\<langle\>y,y\<rangle\>=\<langle\>x,x\<rangle\>-<frac|\<langle\>x,y\<rangle\><rsup|2>|\<langle\>y,y\<rangle\>>\<Rightarrow\><frac|\<langle\>x,y\<rangle\><rsup|2>|\<langle\>y,y\<rangle\>>\<leqslant\>\<langle\>x,x\<rangle\>\<Rightarrow\>\<langle\>x,y\<rangle\><rsup|2>\<leqslant\>\<langle\>x,x\<rangle\>.\<langle\>y,y\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|\<leqslant\><sqrt|\<langle\>x,x\<rangle\>>+<sqrt|\<langle\>y,y\<rangle\>>=\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
proving the inequality. Now assume that <math|x> and <math|y> are
linearly dependent then with either :
<\enumerate>
<item><math|x=\<beta\>y> then <math|\<langle\>x,y\<rangle\><rsup|2>=\<beta\>\<langle\>y,y\<rangle\>\<langle\>x,y\<rangle\>=\<langle\>y,y\<rangle\>.\<langle\>x,x\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
<item><math|y=\<beta\>x> then <math|\<langle\>x,y\<rangle\><rsup|2>=\<beta\>\<langle\>x,x\<rangle\>\<langle\>x,y\<rangle\>=\<langle\>x,x\<rangle\>\<langle\>y,y\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>.
</enumerate>
Assume now that <math|\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
then <math|\<langle\>\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y,\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y\<rangle\>=\<langle\>y,y\<rangle\><rsup|2>.\<langle\>x,x\<rangle\>-2.\<langle\>y,y\<rangle\>.\<langle\>x,y\<rangle\><rsup|2>+\<langle\>x,y\<rangle\><rsup|2>.\<langle\>y,y\<rangle\>=\<shortparallel\>y\<shortparallel\><rsup|4>\<shortparallel\>x\<shortparallel\><rsup|2>-2.\<shortparallel\>y\<shortparallel\><rsup|2>\<shortparallel\>x\<shortparallel\><rsup|2>\<shortparallel\>y\<shortparallel\><rsup|2>+\<shortparallel\>x\<shortparallel\><rsup|2>\<shortparallel\>y\<shortparallel\><rsup|2>\<shortparallel\>y\<shortparallel\><rsup|2>=0\<Rightarrow\>\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y=0>
proving that <math|x,y> are linear dependent as
<math|\<langle\>y,y\<rangle\>\<neq\>0>\
<item>(complex case) we have <math|0\<leqslant\>\<langle\>x+\<alpha\>.y,x+\<alpha\>.y\<rangle\>=\<langle\>x,x\<rangle\>+<wide|\<alpha\>|\<bar\>>.\<langle\>x,y\<rangle\>+\<alpha\>.<wide|\<langle\>x,y\<rangle\>|\<bar\>>+\|\<alpha\>\|<rsup|2>\<langle\>y,y\<rangle\>\<equallim\>\<langle\>x,x\<rangle\>-<frac|<wide|\<langle\>x,y\<rangle\>|\<bar\>>|\<langle\>y,y\<rangle\>>.\<langle\>x,y\<rangle\>-<frac|\<langle\>x,y\<rangle\>|\<langle\>y,y\<rangle\>>.<wide|\<langle\>x,y\<rangle\>|\<bar\>>+<frac|\|\<langle\>x,y\<rangle\>\|<rsup|2>|\<langle\>y,y\<rangle\><rsup|2>>\<langle\>y,y\<rangle\>=\<langle\>x,x\<rangle\>-<frac|\|\<langle\>x,y\<rangle\>\|<rsup|2>|\<langle\>y,y\<rangle\>>\<Rightarrow\><frac|\|\<langle\>x,y\<rangle\>\|<rsup|2>|\<langle\>y,y\<rangle\>>\<leqslant\>\<langle\>x,x\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|\<leqslant\><sqrt|\<langle\>x,x\<rangle\>>+<sqrt|\<langle\>y,y\<rangle\>>=\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>.
Now assume that <math|x> and <math|y> are linearly dependent then
with either <math|\<beta\>\<neq\>0> :
<\enumerate>
<item><math|x=\<beta\>.y\<Rightarrow\>>then
<math|\|\<langle\>x,y\<rangle\>\|<rsup|2>=\<langle\>x,y\<rangle\>.\<langle\>y,x\<rangle\>=\<beta\>\<langle\>y,y\<rangle\>.\<langle\>y,x\<rangle\>=\<langle\>y,y\<rangle\>\<langle\>x,x\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
<item><math|y=\<beta\>.x > then
<math|\|\<langle\>x,y\<rangle\>\|<rsup|2>=\<langle\>x,y\<rangle\>.\<langle\>y,x\<rangle\>=\<beta\>\<langle\>x,y\<rangle\>.\<langle\>x,x\<rangle\>=\<langle\>y,y\<rangle\>.\<langle\>x,x\<rangle\>\<Rightarrow\>\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>.
</enumerate>
\ Assume now that <math|\|\<langle\>x,y\<rangle\>\|=\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>>
then <math|\<langle\>\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y,\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y\<rangle\>=\<langle\>y,y\<rangle\><rsup|2>.\<langle\>x,x\<rangle\>-\<langle\>y,y\<rangle\><wide|\<langle\>x,y\<rangle\>|\<bar\>>.\<langle\>x,y\<rangle\>-\<langle\>y,y\<rangle\>\<langle\>x,y\<rangle\><wide|\<langle\>x,y\<rangle\>|\<bar\>>+\<langle\>x,y\<rangle\><wide|\<langle\>x,y\<rangle\>|\<bar\>>\<langle\>y,y\<rangle\>=\<langle\>y,y\<rangle\><rsup|2>.\<langle\>x,x\<rangle\>-\<langle\>y,y\<rangle\>\|\<langle\>x,y\<rangle\>\|<rsup|2>=\<shortparallel\>y\<shortparallel\><rsup|4>\<shortparallel\>x\<shortparallel\><rsup|2>-\<shortparallel\>y\<shortparallel\><rsup|2>\<shortparallel\>x\<shortparallel\><rsup|2>\<shortparallel\>y\<shortparallel\><rsup|2>=0\<Rightarrow\>\<langle\>y,y\<rangle\>.x-\<langle\>x,y\<rangle\>.y=0>
proving that <math|x,y> are linear independent as
<math|\<langle\>y,y\<rangle\>\<neq\>0>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|triangle inequality>Let <math|V,\<langle\>\<rangle\>> be a inner
product space and <math|\<shortparallel\>\<shortparallel\>> be the inner
product norm then <math|\<forall\>x,y\<in\>V> then
<math|\<shortparallel\>x+y\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
</theorem>
<\proof>
\;
(Real Case)
<math|\<shortparallel\>x+y\<shortparallel\><rsup|2>=\<langle\>x+y,x+y\<rangle\>=\<langle\>x,x\<rangle\>+2.\<langle\>x,y\<rangle\>+\<langle\>y,y\<rangle\>\<leqslant\>\<langle\>x,x\<rangle\>+2.\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>+\<langle\>y,y\<rangle\>=\<shortparallel\>x\<shortparallel\><rsup|2>+2.\<shortparallel\>x\<shortparallel\>.\<shortparallel\>y\<shortparallel\>+\<shortparallel\>y\<shortparallel\><rsup|2>=(\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>)<rsup|2>\<Rightarrow\>\<shortparallel\>x+y\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
(Complex Case)
<math|\<shortparallel\>x+y\<shortparallel\><rsup|2>=\<langle\>x+y,x+y\<rangle\>=\<langle\>x,x\<rangle\>+\<langle\>x,y\<rangle\>+<wide|\<langle\>x,y\<rangle\>|\<bar\>>+\<langle\>y,y\<rangle\>=\<langle\>x,x\<rangle\>+2Re(\<langle\>x,y\<rangle\>)+\<langle\>y,y\<rangle\>\<leqslant\>\<langle\>x,x\<rangle\>+2\|\<langle\>x,y\<rangle\>\|+\<langle\>y,y\<rangle\>\<leqslant\>\<langle\>x.x\<rangle\>+2\<shortparallel\>x\<shortparallel\>\<shortparallel\>y\<shortparallel\>+\<langle\>y,y\<rangle\>=\<shortparallel\>x\<shortparallel\><rsup|2>+2\<shortparallel\>x\<shortparallel\>\<shortparallel\>y\<shortparallel\>+\<shortparallel\>y\<shortparallel\><rsup|2>=(\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>)<rsup|2>\<Rightarrow\>\<shortparallel\>x+y\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
</proof>
<\corollary>
<label|inner product norm is a norm>Let <math|V,\<langle\>\<rangle\>> be
a inner product space then the inner space norm is a norm
</corollary>
<\proof>
\;
<\enumerate>
<item><math|\<shortparallel\>x\<shortparallel\>=<sqrt|\<langle\>x,x\<rangle\>>\<geqslant\>0>
<item><math|\<shortparallel\>\<alpha\>.x\<shortparallel\>=<sqrt|\<langle\>\<alpha\>.x,\<alpha\>.x\<rangle\>>=<sqrt|\|\<alpha\>\|<rsup|2>\<langle\>x,x\<rangle\>>=\|\<alpha\>\|<sqrt|\<langle\>x,x\<rangle\>>=\|\<alpha\>\|.\<shortparallel\>x\<shortparallel\>>
<item><math|\<shortparallel\>x+y\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\>+\<shortparallel\>y\<shortparallel\>>
(<reference|triangle inequality>)
<item>If <math|\<shortparallel\>x\<shortparallel\>=0\<Rightarrow\><sqrt|\<langle\>x,x\<rangle\>>=0\<Rightarrow\>\<langle\>x,x\<rangle\>=0\<Rightarrow\>x=0>
</enumerate>
</proof>
<\example>
<label|finite dimensional space norm and inner product>Let <math|V> be a
finite dimensional real (complex) vector space and let
<math|e<rsub|1>,\<ldots\>,e<rsub|n>> be a basis then we define the inner
product as follows. Let <math|x,y\<in\>V> then there exists a unique
<math|x<rsub|1>,\<ldots\>,x<rsub|n>> and
<math|y<rsub|1>,\<ldots\>,y<rsub|n>> such that
<math|x=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>,y=<big|sum><rsub|i=1><rsup|n>y<rsub|i>.e<rsub|i>>
then <math|\<langle\>x,y\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|y<rsub|i>|\<bar\>>>
(or <math|<big|sum><rsub|i=1><rsup|n>x<rsub|i>.y<rsub|i>> in the real
case). The inner space norm is then defined by
<math|\<shortparallel\>v\<\|\|\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|x<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|<rsup|2>>
(or <math|\<shortparallel\>v\<shortparallel\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.x<rsub|i>=<big|sum><rsub|i=1><rsup|n>(x<rsub|i>)<rsup|2>>
in the real case)<math|>
</example>
<\proof>
As the coordinates are unique (definition of a basis) we only have to
prove that <math|\<langle\>.,.\<rangle\><rsub|V>> is indeed a inner
product:
<\enumerate>
<item>(Complex case)\
<\enumerate>
<item><math|\<langle\>x,y\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i><wide|y<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n><wide|<wide|x<rsub|i>|\<bar\>>.y<rsub|i>|\<bar\>>=<wide|<big|sum><rsub|i=1><rsup|n>y<rsub|i>.<wide|x<rsub|i>|\<bar\>>|\<bar\>>=\<langle\>y,x\<rangle\><rsub|V>>
<item><math|\<langle\>\<alpha\>.x+\<beta\>.y,z\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>+\<beta\>.y<rsub|i>).<wide|z<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>.<wide|z<rsub|i>|\<bar\>>+\<beta\>.y<rsub|i>.<wide|z<rsub|i>|\<bar\>>)=(<big|sum><rsub|i=1><rsup|n>\<alpha\>.x<rsub|i>.<wide|z<rsub|i>|\<bar\>>)+(<big|sum><rsub|i=1><rsup|n>\<beta\>.y<rsub|i>.<wide|z<rsub|i>|\<bar\>>)=(\<alpha\>.<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|z<rsub|i>|\<bar\>>)+(\<beta\>.<big|sum><rsub|i=1><rsup|n>y<rsub|i>.<wide|z<rsub|i>|\<bar\>>)=\<alpha\>.\<langle\>x,z\<rangle\><rsub|V>+\<beta\>.\<langle\>y,z\<rangle\><rsub|V>>
<item><math|\<langle\>x,x\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|x<rsub|i>|\<bar\>>=<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|<rsup|2>\<geqslant\><rsub|<reference|inequality
and sum>>0>
<item>If <math|\<langle\>x,x\<rangle\><rsub|V>=0\<Rightarrow\>0=<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|<rsup|2>\<Rightarrowlim\><rsub|<reference|sum
of positive numbers is positive>>\<forall\>i\<in\>I\<succ\>\|x<rsub|i>\|<rsup|2>=0\<Rightarrow\>\<forall\>i\<in\>I\<succ\>x<rsub|i>=0\<Rightarrow\>x=<big|sum><rsub|i=1><rsup|n>0.e<rsub|i>\<equallim\><rsub|<reference|all
zeroes in sum>>0>
</enumerate>
<item>(Real case)
<\enumerate>
<item><math|\<langle\>x,y\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>y<rsub|i>=<big|sum><rsub|i=1><rsup|n>y<rsub|i>.x<rsub|i>=\<langle\>y,x\<rangle\><rsub|V>>
<item><math|\<langle\>\<alpha\>.x+\<beta\>.y,z\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>+\<beta\>.y<rsub|i>).z<rsub|i>=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>.z<rsub|i>+\<beta\>.y<rsub|i>.z<rsub|i>)=(<big|sum><rsub|i=1><rsup|n>\<alpha\>.x<rsub|i>.z<rsub|i>)+(<big|sum><rsub|i=1><rsup|n>\<beta\>.y<rsub|i>.z<rsub|i>)=(\<alpha\>.<big|sum><rsub|i=1><rsup|n>x<rsub|i>.z<rsub|i>)+(\<beta\>.<big|sum><rsub|i=1><rsup|n>y<rsub|i>.z<rsub|i>)=\<alpha\>.\<langle\>x,z\<rangle\><rsub|V>+\<beta\>.\<langle\>y,z\<rangle\><rsub|V>>
<item><math|\<langle\>x,x\<rangle\><rsub|V>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.x<rsub|i>=<big|sum><rsub|i=1><rsup|n>(x<rsub|i>)<rsup|2>\<geqslant\><rsub|<reference|inequality
and sum>>0>
<item>If <math|\<langle\>x,x\<rangle\><rsub|V>=0\<Rightarrow\>0=<big|sum><rsub|i=1><rsup|n>(x<rsub|i>)<rsup|2>\<Rightarrowlim\><rsub|<reference|sum
of positive numbers is positive>>\<forall\>i\<in\>I\<succ\>(x<rsub|i>)<rsup|2>=0\<Rightarrow\>\<forall\>i\<in\>I\<succ\>x<rsub|i>=0\<Rightarrow\>x=<big|sum><rsub|i=1><rsup|n>0.e<rsub|i>\<equallim\><rsub|<reference|all
zeroes in sum>>0>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|finite dimensional spaces are isometric>Let <math|V> be a finite
dim<math|>ensional real (complex) vector space and let
<math|e<rsub|1>,,,,e<rsub|n>> be its basis then
<math|\<varphi\>:V\<rightarrow\>\<bbb-R\><rsup|n>(\<bbb-C\><rsup|n>)>
defined by <math|\<varphi\>(v)=(v<rsub|1>,\<ldots\>,v<rsub|n>)> where
<math|v=<big|sum><rsub|i=1><rsup|n>v<rsub|i>.e<rsub|i>> (which is defined
because <math|e<rsub|1>,\<ldots\>,e<rsub|n>)> is a basis) is a
isomorphism, <math|\<langle\>\<varphi\>(x),\<varphi\>(y)\<rangle\>=\<langle\>x,y\<rangle\><rsub|V>>
and <math|\<varphi\>> is isometric.
</theorem>
<\proof>
\;
First we prove that <math|\<varphi\>> is a bijective function
<\enumerate>
<item>Injectivity, if <math|\<varphi\>(x)=\<varphi\>(y)\<Rightarrow\>(x<rsub|1>,\<ldots\>,x<rsub|n>)=(y<rsub|1>,\<ldots\>,y<rsub|n>)\<Rightarrow\>\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>x<rsub|i>=y<rsub|i>\<Rightarrow\>x=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>=<big|sum><rsub|i=1><rsup|n>y<rsub|i>.e<rsub|i>=y>
<item>If <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<bbb-R\><rsup|n>(\<bbb-C\><rsup|n>)\<Rightarrow\>\<varphi\>(<big|sum><rsub|i=1><rsup|n>x<rsub|i>)=(x<rsub|1>,\<ldots\>,x<rsub|n>)>
</enumerate>
Second because if <math|\<varphi\>(x)=(x<rsub|1>,\<ldots\>,x<rsub|n>),\<varphi\>(y)=(y<rsub|1>,\<ldots\>,y<rsub|n>)>
then <math|x=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>,y=<big|sum><rsub|i=1><rsup|n>y<rsub|i>.e<rsub|i>>
and <math|\<alpha\>.x+\<beta\>.y=(\<alpha\>.<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>)+(\<beta\>.<big|sum><rsub|i=1><rsup|n>y<rsub|i>.e<rsub|i>)=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>.e<rsub|i>+\<beta\>.x<rsub|i>.e<rsub|i>)=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.x<rsub|i>+\<beta\>.y<rsub|i>).e<rsub|i>>
and thus <math|\<varphi\>(\<alpha\>.x+\<beta\>.y)=(\<alpha\>.x<rsub|1>+\<beta\>.y<rsub|1>,\<ldots\>,\<alpha\>.x<rsub|n>+\<beta\>.y<rsub|n>)=\<alpha\>.(x<rsub|1>,\<ldots\>,x<rsub|n>)+\<beta\>.(y<rsub|1>\<ldots\>,y<rsub|n>)=\<alpha\>.\<varphi\>(x)+\<beta\>.\<varphi\>(y)>
so <math|\<varphi\>> is a isomorphism.
Thirdly if <math|\<varphi\>(x)=(x<rsub|1>,\<ldots\>,x<rsub|n>),\<varphi\>(y)=(y<rsub|1>,\<ldots\>,y<rsub|n>)>
then <math|\<langle\>\<varphi\>(x),\<varphi\>(y)\<rangle\>=<big|sum><rsub|i=1><rsup|n>x<rsub|i>y<rsub|i>(real
case) (or <big|sum><rsub|i=1><rsup|n>x<rsub|i>.<wide|y<rsub|i>|\<bar\>>
compex case)=\<langle\>x,y\<rangle\><rsub|V>>
Finally as <math|\<shortparallel\>x\<shortparallel\>=<sqrt|x<rsub|i><rsup|2>>
(real case) (<sqrt|\|x<rsub|i>\|<rsup|2>> complex
case)=<sqrt|\<langle\>x,x\<rangle\><rsub|V>>=\<shortparallel\>x\<shortparallel\><rsub|V>>
</proof>
<\corollary>
Let <math|X,Y> be two \ n-dimensional real (complex) vector spaces then
these spaces are isomorphic and isometric, so that the topologies
generated by there inner product norms are equivalent (see
<reference|isometric normed spaces>)
</corollary>
<\proof>
By <reference|finite dimensional spaces are isometric> there exists \ two
isomorphic,isometric functions <math|\<varphi\><rsub|1>:X\<rightarrow\>\<bbb-R\><rsup|n>(\<bbb-C\><rsup|n>),\<varphi\><rsub|2>:Y\<rightarrow\>\<bbb-R\><rsup|n>(\<bbb-C\><rsup|n>)>
which is by <reference|inverse of isomorphisme norm>, <reference|inverse
of isomorphisme>, <reference|composition of isomorphisme> and
<reference|composition of isomorphisme norm> indeed isomorphic and
isometric
</proof>
<section|Continuity>
<\definition>
<index|continuous mapping>Let <math|X,\<cal-T\><rsub|X>> and
<math|Y,\<cal-T\><rsub|Y>> be topological spaces and let
<math|f:X\<rightarrow\>Y> be a function between <math|X> and <math|Y>
then <math|f> is continuous at x iff <math|\<forall\>V\<in\>\<cal-T\><rsub|Y>\<vdash\>f(x)\<in\>V>
we have <math|\<exists\>U\<in\>\<cal-T\><rsub|X>\<vdash\>x\<in\>U\<wedge\>f(U)\<subseteq\>V>
</definition>
<\theorem>
<label|continuity of restricted maps>Let <math|X,\<cal-T\><rsub|X>> and
<math|Y,\<cal-T\><rsub|Y>> be topological spaces and let
<math|f:X\<rightarrow\>Y> be a continuous function at <math|x\<in\>X>,
then if <math|x\<in\>A\<subseteq\>X> we have that
<math|f<rsub|\|A>:A\<rightarrow\>Y> is continuous at <math|x> where we
have used the subspace topology <math|\<cal-T\><rsub|A> >on <math|A>
</theorem>
<\proof>
<math|\<forall\>V\<in\>\<cal-T\><rsub|Y>\<vdash\>f<rsub|A>(x)\<in\>V\<Rightarrowlim\><rsub|definition
of restriction>f(x)\<in\>V\<Rightarrowlim\><rsub|continuity of f at
x>\<exists\>U\<in\>\<cal-T\><rsub|X>\<vdash\>x\<in\>U\<wedge\>f(U)\<subseteq\>V\<Rightarrow\>x\<in\>A<big|cap>U\<in\>\<cal-T\><rsub|A>\<wedge\>f(A<big|cap>U)\<subseteq\>f(U)\<subseteq\>V>
</proof>
<\definition>
Let <math|X,\<cal-T\><rsub|X>> and <math|Y,\<cal-T\><rsub|Y>> be
topological spaces and let <math|f> be a function between <math|X> and
<math|Y> then <math|f> is continuous if and only
<math|\<forall\>x\<in\>X> <math|f> is continuous at <math|x>
</definition>
<\theorem>
<label|continuity characterization>Let <math|X,\<cal-T\><rsub|X>> and
<math|Y,\<cal-T\><rsub|Y>> be topological spaces and let <math|f> be a
function between <math|X> and <math|Y> then the following are equivalent
<\enumerate>
<item><math|f> is continuous
<item><math|\<forall\>V\<in\>\<cal-T\><rsub|Y>> we have
<math|f<rsup|-1>(V)\<in\>\<cal-T\><rsub|X>>
<item><math|\<forall\>A\<subseteq\>X> we have
<math|f(<wide|A|\<bar\>>)\<subseteq\><wide|f(A)|\<bar\>>>
<item><math|\<forall\>F\<subseteq\>Y,F> closed we have
<math|f<rsup|-1>(F)> is closed
</enumerate>
</theorem>
<\proof>
\;
<math|1\<Rightarrow\>2>
Let <math|V\<in\>\<cal-T\><rsub|Y>> then either
<math|f<rsup|-1>(V)=\<emptyset\>\<in\>\<cal-T\><rsub|X>> or
<math|\<forall\>x\<in\>f<rsup|-1>(V)> we have by continuity
<math|\<exists\>U<rsub|x>\<in\>\<cal-T\><rsub|X>\<vdash\>x\<in\>U<rsub|x>
<with|mode|text|and >f(U<rsub|x>)\<subseteq\>V\<Rightarrowlim\><rsub|<reference|image
and preimage>>x\<in\>U<rsub|x>\<subseteq\>f<rsup|-1>(V)\<Rightarrow\>f<rsup|-1>(V)=<big|cup><rsub|x\<in\>U>U<rsub|x>\<in\>\<cal-T\><rsub|X>>
<math|2\<Rightarrow\>1>
<math|\<forall\>x\<in\>X> if <math|f(x)\<in\>V\<in\>\<cal-T\><rsub|Y>>
then <math|x\<in\>U=f<rsup|-1>(V)\<in\>\<cal-T\><rsub|X>> and
<math|f(U)=f(f<rsup|-1>(V))\<subseteq\>V> (see <reference|image of
preimage>)
<math|1\<Rightarrow\>3>
Let <math|y\<in\>f(<wide|A|\<bar\>>>) then
<math|\<exists\>x\<in\><wide|A|\<bar\>>> such that <math|f(x)=y> now if
<math|V\<in\>\<cal-T\><rsub|Y>> such that <math|y\<in\>V> then by
continuity <math|\<exists\>U\<in\>\<cal-T\><rsub|X>\<vdash\>x\<in\>U\<wedge\>f(U)\<subseteq\>V>
now as <math|x\<in\><wide|A|\<bar\>>> we have by
<reference|characterization of closure>
<math|><math|U<big|cap>A\<neq\>\<emptyset\>\<Rightarrow\>\<emptyset\>\<neq\>f(U<big|cap>A)\<subseteq\><rsub|<reference|image
of union or intersection>>f(U)<big|cap>f(A)\<subseteq\>V<big|cap>f(A)\<Rightarrow\>y\<in\><wide|V|\<bar\>>>
<math|3\<Rightarrow\>2> (and thus <math|1)>
Let <math|A> be a set in <math|Y> then as
<math|f(<wide|A|\<bar\>>)\<subseteq\><wide|f(A)|\<bar\>>> we have
<reference|image and preimage> <math|<wide|A|\<bar\>>\<subseteq\>f<rsup|-1>(<wide|f(A))|\<bar\>>>
so if <math|B\<subseteq\>Y> is closed then <math|<wide|B|\<bar\>>=B> and
let <math|A=f<rsup|-1>(B)> then <math|<wide|f<rsup|-1>(B)|\<bar\>>\<subseteq\>f<rsup|-1>(<wide|f(A))|\<bar\>>\<subseteq\><rsub|<reference|image
of preimage>>f<rsup|-1>(<wide|B|\<bar\>>)><math|=f<rsup|-1>(B)> and as
<math|f<rsup|-1>(B)\<subseteq\><wide|f<rsup|-1>(B)|\<bar\>>> we have
<math|f<rsup|-1>(B)=<wide|f<rsup|-1>(B)|\<bar\>>> and thus
<math|f<rsup|-1>(B)> is closed (<reference|characterization of closed
sets 1>). Now if <math|V\<in\>\<cal-T\><rsub|Y>> then <math|Y<mid|\\>V>
is closed and <math|X<mid|\\>f<rsup|-1>(V)\<equallim\><rsub|<reference|preimage
of difference>>f<rsup|-1>(Y<mid|\\>V)> is closed then
<math|f<rsup|-1>(V)> is open
<\math>
2\<Rightarrow\>4
</math>
Let <math|F> be closed then <math|Y<mid|\\>F> is open
<math|f<rsup|-1>(Y<mid|\\>F)> is open by <math|(2)> then as by
<reference|preimage of difference> we have
<math|X<mid|\\>f<rsup|-1>(Y<mid|\\>F)=f<rsup|-1>(Y<mid|\\>(Y<mid|\\>F))=f<rsup|-1>(F)>
we have that <math|f<rsup|-1>(F)> is closed.
<math|4\<Rightarrow\>2>
Let <math|U> be open in <math|Y> then <math|Y<mid|\\>U> is closed and by
(4) we have <math|f<rsup|-1>(Y<mid|\\>U)> is closed and then
<math|X<mid|\\>(f<rsup|-1>(Y<mid|\\>U))\<equallim\><rsub|<reference|preimage
of difference>>f<rsup|-1>(Y<mid|\\>(Y<mid|\\>U))=f<rsup|-1>(U)> is open
proving <math|(2)>
</proof>
<\theorem>
<label|continuity is local>Let <math|X,\<cal-T\><rsub|X>> and
<math|Y,\<cal-T\><rsub|Y>> be topological spaces and let <math|f> be a
function between <math|X> and <math|Y> then <math|f> is continue if
<math|\<forall\>x\<in\>X> there exists a <math|U> open with
<math|x\<in\>U> such that <math|f<rsub|\|U>:U\<rightarrow\>Y> is
continuous (using the subspace topology on <math|U)>\
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>This is trivial using <reference|continuity of
restricted maps>
<math|\<Leftarrow\>>Let <math|x\<in\>X> and <math|U> open such that
<math|x\<in\>U> and <math|f<rsub|\|U>> is continuous then if <math|V> is
open with <math|f(x)=f<rsub|\|U>(x)> there exists a <math|W<rprime|'>>
open in <math|U> such that <math|f(W<rprime|'>)=f<rsub|\|U>(W<rprime|'>)\<subseteq\>V>
which, as <math|W<rprime|'>=W<big|cap>U,W> open in <math|X> and thus
<math|W<rprime|'>> is open, means that <math|f> is continuous\
\;
</proof>
<\definition>
<index|open mapping>Let <math|X,\<cal-T\><rsub|X>> and
<math|Y,\<cal-T\><rsub|Y>> be topological spaces then a function
<math|f:X\<rightarrow\>Y> is open iff
<math|\<forall\>U\<in\>\<cal-T\><rsub|X>> we have
<math|f(U)\<in\>\<cal-T\><rsub|Y>> (every image of a open set is open).
</definition>
<\theorem>
<label|inverse of open map is continuous>Let <math|X,\<cal-T\><rsub|X>>
and <math|Y,\<cal-T\><rsub|Y>> be topological spaces then if a injective
function <math|f:X\<rightarrow\>Y> is open then the function
<math|f<rsup|-1>:f(X)\<rightarrow\>X> is continuous
</theorem>
<\proof>
Let <math|><math|U\<in\>X> be open then <math|f(U)> is open and as
<math|(f<rsup|-1>)<rsup|-1>(U)\<equallim\><rsub|<reference|preimage of
injective mapping>>f(U)> which is open we have that <math|f<rsup|-1>> is
continuous
</proof>
<\theorem>
Let <math|X,\<cal-T\><rsub|X>,> <math|Y,\<cal-T\><rsub|Y>> and
<math|Z,\<cal-T\><rsub|Z>> be topological spaces then if
<math|f:X\<rightarrow\>Y> and <math|g:Y\<rightarrow\>Z> are open
functions then <math|g\<circ\>f:X\<rightarrow\>Z> is a open function
</theorem>
<\proof>
The proof is trivial for if <math|U\<subseteq\>X> is open then
<math|f(U)> is open and thus <math|(g\<circ\>f)(U)=g(f(U))> is open.
</proof>
<\theorem>
<label|continuitity of composition>Let <math|X,\<cal-T\><rsub|X>>,
<math|Y,\<cal-T\><rsub|Y>> and <math|Z,\<cal-T\><rsub|Z>> be topological
spaces and <math|f:X\<rightarrow\>Y> and <math|g:Y\<rightarrow\>Z>
continuous functions then <math|g\<circ\>f:X\<rightarrow\>Z> is a
continuous function
</theorem>
<\proof>
\;
Given <math|W\<in\>\<cal-T\><rsub|Z>> we have
<math|(g\<circ\>f)<rsup|-1>(W)\<equallim\><rsub|<reference|preimage of a
composed mapping>>><math|f<rsup|-1>(g<rsup|-1>(W))> which is open because
<math|g<rsup|-1>(W)> is open by continuity of <math|g> and then
<math|f<rsup|-1>(g<rsup|-1>(W))> is open by continuity of <math|f>
</proof>
<\theorem>
<label|projection map is continuous>Let
<math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>> be a family of
topological spaces and let <math|<big|prod><rsub|i\<in\>I>X<rsub|i>> be
equipped with the product topology then the projection map
<math|\<pi\><rsub|i>:X\<rightarrow\>X<rsub|i>> is continuous and open.
</theorem>
<\proof>
\;
Continuity is trivial because if <math|U\<in\>\<cal-T\><rsub|i> then by
the definition of the product topology ><math|\<pi\><rsub|i><rsup|-1>(U)>
is a element of the generating subbasis and thus the basis of the product
topology and thus open.
To prove that <math|\<pi\><rsub|i>> is open, let <math|U> be open in the
product topology and let <math|t\<in\>\<pi\><rsub|i>(U)> then
<math|\<exists\>x\<in\>U> such that <math|t=\<pi\><rsub|i>(x)=x(i)=x<rsub|i>>
and as <math|x\<in\>U> open so using (<reference|basis of product
topology>) there exists a finite <math|A\<subseteq\>I> such that
<math|x\<in\><big|prod><rsub|i\<in\>I>A<rsub|i>=V\<subseteq\>U> where
<math|A<rsub|j>> if <math|j\<in\>A> and <math|A<rsub|j>=X<rsub|j>> if
<math|j\<in\>A> (or <math|A<rsub|j>> is open in all case) so
\ <math|t=\<pi\><rsub|j>(x)\<subseteq\>\<pi\><rsub|j>(V)\<subseteq\>\<pi\><rsub|j>(U)>
\ and as <math|\<pi\><rsub|j>(V)=A<rsub|j>> is open we have proved that
<math|\<pi\><rsub|j>(U)> is a union of open sets and thus open. \
</proof>
<\theorem>
<label|continuous mappings to a product space>Let
<math|X,\<cal-T\><rsub|X>> be a topological space
<math|{X<rsub|i>,\<cal-T\><rsub|i>}<rsub|i\<in\>I>> be a family of
topological spaces and <math|<big|prod><rsub|i\<in\>I>X<rsub|i>> equipped
with the product topology then a function
<math|f:X\<rightarrow\><big|prod><rsub|i\<in\>I>X<rsub|I>> is continuous
<math|\<Leftrightarrow\>> <math|\<forall\>i\<in\>I>
<math|f<rsub|i>=\<pi\><rsub|i>\<circ\>f> is continuous
</theorem>
<\proof>
\;
<math|\<Rightarrow\>> This is trivial and follows from
<reference|continuitity of composition> (composition of continuous
functions) and <reference|projection map is continuous>
<math|\<Leftarrow\>> Let <math|x\<in\>X> and<math|f(x)\<in\>V> which is
open in the product topology then by the definition of the subspace
topology <math|\<exists\>\<cal-I\>> finite where
<math|\<forall\>B\<in\>\<cal-I\>\<succ\>B=\<pi\><rsub|i><rsup|-1>(U<rsub|B>),U<rsub|B>\<in\>\<cal-T\><rsub|i>>
and <math|f(x)\<in\>W=<big|cap><rsub|B\<in\>\<cal-I\>>B\<subseteq\>V>
then as <math|\<pi\><rsub|i>\<circ\>f> is continuous we have
<math|U=f<rsup|-1>(W)\<equallim\><rsub|<reference|preimage image of union
intersection of a family of sets>><big|cap><rsub|B\<in\>\<cal-I\>>f<rsup|-1>(B)>
which is open because <math|\<forall\>B\<in\>I> we have
<math|f<rsup|-1>(B)=f<rsup|-1>(\<pi\><rsub|i><rsup|-1>(U<rsub|B>))\<equallim\><rsub|<reference|preimage
of a composed mapping>>(\<pi\><rsub|i>\<circ\>f<rsub|i>)<rsup|-1>(U<rsub|B>)\<in\>\<cal-T\><rsub|X>>.
Now <math|f(U)=f(f<rsup|-1>(W))\<subseteq\>W\<subseteq\>V> and <math|f>
is continuous<math|>
</proof>
<\theorem>
<label|continuity of function on a product>Let
<math|X<rsub|1>,X<rsub|2>,Y> be topological spaces, consider then
<math|X<rsub|1>\<times\>X<rsub|2>> with the product topology and
<math|f:X<rsub|1>\<times\>X<rsub|2>\<rightarrow\>Y> a continuous function
then the following function are also continuous
<\enumerate>
<item><math|\<forall\>x\<in\>X<rsub|1>
\ \ f<rsub|1>(x):X<rsub|2>\<rightarrow\>Y:t\<rightarrow\>f<rsub|1>(x)(t)=f(x,t)>
<item><math|\<forall\>x\<in\>X<rsub|2>
f<rsub|2>(x):X<rsub|1>\<rightarrow\>Y:t\<rightarrow\>f<rsub|2>(x)(t)=(f(t,x)>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|V\<subseteq\>Y> be a open set then by continuity of
<math|f> we have that <math|f<rsup|-1>(V)> is open in
<math|X<rsub|1>\<times\>X<rsub|2>>. Now let
<math|y\<in\>f<rsub|1>(x)<rsup|-1>(V)> then
<math|f(x,y)=f<rsub|1>(x)(y)\<in\>V\<Rightarrow\>(x,y)\<in\>f<rsup|-1>(V)\<Rightarrowlim\><rsub|f<rsup|-1>(V)
is open>\<exists\>U\<in\>X<rsub|1>>
open,<math|\<exists\>W\<in\>X<rsub|2>> open such that
<math|(x,y)\<in\>U\<times\>W\<subseteq\>f<rsup|-1>(V)\<Rightarrow\>y\<in\>W,x\<in\>U>.
Now if <math|t\<in\>W\<Rightarrow\>(x,t)\<in\>U\<times\>W\<subseteq\>f<rsup|-1>(V)\<Rightarrow\>f<rsub|1>(x)(t)=f(x,t)\<in\>V\<Rightarrow\>t\<in\>f<rsub|1>(x)<rsup|-1>(V)\<Rightarrow\>W\<subseteq\>f<rsub|1>(x)<rsup|-1>(V)>.
So every point in <math|f<rsub|1>(x)<rsup|-1>(V)> has a open
neighborhood also in <math|f<rsub|1>(x)<rsup|-1>(V)> proving that
<math|f<rsub|1>(x)<rsup|-1>(V)> is open from which continuity of
<math|f<rsub|1>(x)> follows.
<item>Let <math|V\<subseteq\>Y> be a open set then by continuity of
<math|f> we have that <math|f<rsup|-1>(V)> is open in
<math|X<rsub|1>\<times\>X<rsub|2>>. Now let
<math|y\<in\>f<rsub|2>(x)<rsup|-1>(V)> then
<math|f(y,x)=f<rsub|2>(x)(y)\<in\>V\<Rightarrow\>(y,x)\<in\>f<rsup|-1>(V)\<Rightarrowlim\><rsub|f<rsup|-1>(V)
is open>\<exists\>U\<in\>X<rsub|1>>
open,<math|\<exists\>W\<in\>X<rsub|2>> open such that
<math|(y,x)\<in\>U\<times\>W\<subseteq\>f<rsup|-1>(V)\<Rightarrow\>y\<in\>U,x\<in\>W>.
Now if <math|t\<in\>W\<Rightarrow\>(t,x)\<in\>U\<times\>W\<subseteq\>f<rsup|-1>(V)\<Rightarrow\>f<rsub|2>(x)(t)=f(t,x)\<in\>V\<Rightarrow\>t\<in\>f<rsub|2>(x)<rsup|-1>(V)\<Rightarrow\>W\<subseteq\>f<rsub|1>(x)<rsup|-1>(V)>.
So every point in <math|f<rsub|2>(x)<rsup|-1>(V)> has a open
neighborhood also in <math|f<rsub|12>(x)<rsup|-1>(V)> proving that
<math|f<rsub|2>(x)<rsup|-1>(V)> is open from which continuity of
<math|f<rsub|2>(x)> follows.
</enumerate>
</proof>
\;
<\theorem>
<label|continuity on metric spaces>Let <math|X,d<rsub|X>> ,
<math|Y,d<rsub|Y>> be metric spaces and <math|f:X\<rightarrow\>Y> be a
function then <math|f> is continuous (in the metric topology) at
<math|x\<in\>X\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0\<succ\>\<exists\>\<delta\>\<gtr\>0>
such that <math|\<forall\>y\<in\>X\<succ\>d<rsub|X>(x,y)\<less\>\<delta\>>
we have <math|d<rsub|Y>(f(x),f(y))\<less\>\<varepsilon\>>
</theorem>
<\proof>
\;
<\math>
\<Rightarrow\>
</math>
Let <math|f> be continuous at <math|x> then if
<math|\<varepsilon\>\<gtr\>0> we have
<math|f(x)\<in\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)> which is open so
<math|\<exists\>U> open in <math|X> such that <math|x\<in\>U> and
<math|f(U)\<subseteq\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)> then
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<subseteq\>U> and
<math|f(B<rsub|d<rsub|X>>(x,\<delta\>))\<subseteq\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)>
so if <math|d<rsub|X>(x,y)\<less\>\<delta\>\<Rightarrow\>y\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)\<Rightarrow\>f(y)\<in\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)\<Rightarrow\>d<rsub|Y>(f(x),f(y))\<less\>\<varepsilon\>>
<math|\<Leftarrow\>>
Let <math|x\<in\>X> then if we have <math|\<forall\>\<varepsilon\>\<gtr\>0>
there <math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>y\<in\>X> with <math|d<rsub|X>(x,y)\<less\>\<delta\>> we
have <math|d<rsub|Y>(f(x),f(y))\<less\>\<varepsilon\>> now if
<math|f(x)\<in\>V,V> open then <math|\<exists\>\<varepsilon\>\<gtr\>0>
such that <math|f(x)\<in\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)> now let
<math|\<delta\>\<gtr\>0> such that <math|d<rsub|X>(x,y)\<less\>\<delta\>\<Rightarrow\>d<rsub|Y>(f(x),f(y))\<less\>\<varepsilon\>\<Rightarrow\>x\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)>
which is open and <math|\<forall\>y\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)>
we have <math|d<rsub|X>(x,y)\<less\>\<delta\>\<Rightarrow\>f(y)\<in\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)\<Rightarrow\>f(B<rsub|d<rsub|X>>(x,\<delta\>))\<subseteq\>B<rsub|d<rsub|Y>>(f(x),\<varepsilon\>)\<subseteq\>V>
and thus <math|f> is continue at <math|x>
</proof>
<\theorem>
<label|open mapping in metric spaces>Let <math|X,d<rsub|X>> and
<math|Y,d<rsub|Y>> be metric spaces and <math|f:X\<rightarrow\>Y> a
function then <math|f> is open if and only if
<math|\<forall\>x\<in\>X,\<forall\>\<delta\>\<gtr\>0> we have that there
exists a <math|\<delta\><rprime|'>\<gtr\>0> such that
<math|f(x)\<in\>B<rsub|d<rsub|Y>>(f(x),\<delta\><rprime|'>)\<subseteq\>f(B<rsub|d<rsub|X>>(x,\<delta\>))>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>Because <math|f> is open we have that if
<math|x\<in\>B<rsub|d<rsub|X>>(x,\<delta\>)> then
<math|f(B<rsub|d<rsub|X>>(x,\<delta\>))> is open and thus as
<math|f(x)\<in\>f(B<rsub|d<rsub|X>>(x,\<delta\>))> there exists a
<math|\<delta\><rprime|'>> such that
<math|f(x)\<in\>B<rsub|d<rsub|Y>>(f(x),\<delta\><rprime|'>)\<subseteq\>f(B<rsub|d<rsub|X>>(x,\<delta\>))>
<item><math|\<Leftarrow\>>Let <math|U\<subset\>X,> <math|U> open and
let <math|y\<in\>f(U)> then there exists a <math|x\<in\>U> such that
<math|f(x)=y> then by the hypothesis there exists a
<math|\<delta\><rsub|y>> with <math|y=f(x)\<in\>B<rsub|d<rsub|Y>>(f(x),\<delta\><rsub|y>)=B<rsub|d<rsub|Y>>(y,\<delta\><rsub|y>)\<subseteq\>f(B<rsub|d<rsub|X>>(x,\<delta\>))\<subseteq\>f(U)\<Rightarrow\>f(U)=<big|cup><rsub|y\<in\>f(U)>B<rsub|d<rsub|Y>>(y,\<delta\><rsub|y>0>
and thus <math|f(U)> is open\
</enumerate>
</proof>
<\definition>
<index|uniform continuous mapping>Let <math|X,d<rsub|X>> and
<math|Y,d<rsub|Y>> be metric spaces then a function
<math|f:X\<rightarrow\>Y> is uniform continuous in <math|K\<subseteq\>X>
iff <math|\<forall\>\<varepsilon\>\<gtr\>0>
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>x,y\<in\>K\<vdash\>d<rsub|X>(x,y)\<less\>\<delta\>\<Rightarrow\>d<rsub|Y>(f(x),f(y))\<less\>\<varepsilon\>>
</definition>
<\theorem>
<label|continuity in normed spaces>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces then
a function <math|f:X\<rightarrow\>Y> is continuous at x in the norm
topologies <math|\<Leftrightarrow\>><math|\<forall\>\<varepsilon\>\<gtr\>0>
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>y\<in\>X\<vdash\>\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>\<delta\>>
we have <math|\<shortparallel\>f(x)-f(y)\<shortparallel\><rsub|Y>\<less\>\<varepsilon\>>
</theorem>
<\proof>
This follows form <reference|continuity on metric spaces> and the fact
that <math|d<rsub|X>(x,y)=\<shortparallel\>x-y\<shortparallel\><rsub|X>,d<rsub|Y>(f(x),f(y))=\<shortparallel\>f(x)-f(y)\<shortparallel\><rsub|Y>>
</proof>
<\theorem>
<label|continuity of vectorspace operations>Let
<math|X,+,.,\<shortparallel\>\<shortparallel\>> be a normed vector space
then <math|+:X\<times\>X\<rightarrow\>X>,
<math|.:X\<times\>\<bbb-R\>(\<bbb-C\>)\<rightarrow\>X> and
<math|\<shortparallel\>\<shortparallel\>:X\<rightarrow\>\<bbb-R\>> are
continuous in the norm topology and product topologies of
<math|X\<times\>X> and <math|X\<times\>\<bbb-R\>(\<bbb-C\>)>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|x+y\<in\>U> (open in the norm topology) then
<math|\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|x+y\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x+y,\<varepsilon\>)={z\|
\<shortparallel\>x+y-z\<shortparallel\>\<less\>\<varepsilon\>}\<subseteq\>U>
take now <math|z<rprime|'>=(x<rprime|'>,y<rprime|'>)\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,<frac|\<varepsilon\>|2>)\<times\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(y,<frac|\<varepsilon\>|2>)>
(which is open in <math|X\<times\>X>) then
<math|\<shortparallel\>x+y-(x<rprime|'>+y<rprime|'>)\<shortparallel\>\<leqslant\>\<shortparallel\>x-x<rprime|'>\<shortparallel\>+\<shortparallel\>y-y<rprime|'>\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>\<Rightarrow\>+(B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,<frac|\<varepsilon\>|2>)\<times\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(y,<frac|\<varepsilon\>|2>))\<subseteq\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x+y,\<varepsilon\>)>
and thus <math|+(.,.)> is continuous at <math|(x,y)>.
<item>Let <math|\<lambda\>.x\<in\>U> (open in <math|X>) then there
exists a <math|\<varepsilon\>\<gtr\>0> such that
<math|\<lambda\>.x\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(\<lambda\>.x,\<varepsilon\>)={z\|
\<shortparallel\>\<lambda\>.x-z\<shortparallel\>\<less\>\<varepsilon\>}\<subseteq\>U.>
Let now <math|z<rprime|'>=(x<rprime|'>,y<rprime|'>)\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,\<delta\><rsub|1>)\<times\>B<rsub|d<rsub|\|
\|>>(\<lambda\>,\<delta\><rsub|2>)> which is open in
<math|X\<times\>\<bbb-R\>(\<bbb-C\>)> then
<math|\<shortparallel\>\<lambda\>.x-\<lambda\><rprime|'>.x<rprime|'>\<shortparallel\>\<leqslant\>\<shortparallel\>\<lambda\>.x-\<lambda\><rprime|'>.x+\<lambda\><rprime|'>.x-\<lambda\><rprime|'>.x<rprime|'>\<shortparallel\>\<leqslant\>\<shortparallel\>\<lambda\>.x-\<lambda\><rprime|'>.x\<shortparallel\>+\<shortparallel\>\<lambda\><rprime|'>.x-\<lambda\><rprime|'>.x<rprime|'>\<shortparallel\>=\|(\<lambda\>-\<lambda\><rprime|'>)\|.\<shortparallel\>x\<shortparallel\>+\|\<lambda\><rprime|'>\|.\<shortparallel\>x-x<rprime|'>\<shortparallel\>\<less\>\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+\|\<lambda\><rprime|'>\|.\<delta\><rsub|1>=\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+\|\<lambda\><rprime|'>-\<lambda\>+\<lambda\>\|.\<delta\><rsub|1>\<leqslant\>\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+(\|\<lambda\>\|+\|\<lambda\>-\<lambda\><rprime|'>\|).\<delta\><rsub|1>\<less\>\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+(\|\<lambda\>\|+\<delta\><rsub|2>).\<delta\><rsub|1>>
we have now the following possible cases
<\enumerate>
<item><math|\<shortparallel\>x\<shortparallel\>=0> take then
<math|\<delta\><rsub|2>=1> and <math|\<delta\><rsub|1>=<frac|\<varepsilon\>|1+\|\<lambda\>\|>>
then <math|\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+(\|\<lambda\>\|+\<delta\><rsub|2>).\<delta\><rsub|1>\<less\>(\|\<lambda\>\|+1).<frac|e|1+\|\<lambda\>\|>=\<varepsilon\>>
<item><math|\<shortparallel\>x\<shortparallel\>\<neq\>0\<Rightarrow\>\<shortparallel\>x\<shortparallel\>\<gtr\>0>
take then <math|\<delta\><rsub|2>=min(1,<frac|\<varepsilon\>|2.\<shortparallel\>x\<shortparallel\>>),\<delta\><rsub|1><frac|\<varepsilon\>|2(1+\|\<lambda\>\|)>>
then <math|\<shortparallel\>x\<shortparallel\>.\<delta\><rsub|2>+(\|\<lambda\>\|+\<delta\><rsub|2>).\<delta\><rsub|1>\<leqslant\>\<shortparallel\>x\<shortparallel\>.<frac|\<varepsilon\>|2.\<shortparallel\>x\<shortparallel\>>+(\|\<lambda\>\|+1).<frac|\<varepsilon\>|2.(1+\|\<lambda\>\|)>=<frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
</enumerate>
\ so with the correct choice of <math|\<delta\><rsub|1>,\<delta\><rsub|2>>
we have <math|\<shortparallel\>\<lambda\>.x-\<lambda\><rprime|'>.x<rprime|'>\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\>\<lambda\><rprime|'>.x<rprime|'>\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(\<lambda\>.x,\<varepsilon\>)>
and thus <math|><math|\<lambda\>(.,.)> is continuous at
<math|(x,\<lambda\>)>
<item>Let <math|x\<in\>X> and <math|\<shortparallel\>x\<shortparallel\>\<in\>U>
open in <math|\<bbb-R\>(\<bbb-C\>)> then
<math|\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|\<shortparallel\>x\<shortparallel\>\<in\>B<rsub|d<rsub|\|
\|>>(\<shortparallel\>x\<shortparallel\>,\<varepsilon\>)\<subseteq\>U>
then if <math|y\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,\<varepsilon\>)>
we have <math|\<shortparallel\>x-y\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\>\|\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>\|\<leqslant\>\<shortparallel\>x-y\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\>\<shortparallel\>y\<shortparallel\>\<in\>B<rsub|d<rsub|\|
\|>>(\<shortparallel\>x\<shortparallel\>,\<varepsilon\>)> and thus
<math|\<shortparallel\>.\<shortparallel\>> is continuous at <math|x>
</enumerate>
</proof>
<\theorem>
<label|norms mappings are continuous in the product of reals>When
<math|\<bbb-R\><rsup|n>> is equipped with the product topology (generated
by the maximum norm <math|\<shortparallel\>\<shortparallel\>>(see
<reference|norm of finite product of normed spaces>) and let
<math|\<shortparallel\>\<shortparallel\><rsup|*\<ast\>>> be another norm
on <math|\<bbb-R\><rsup|n>> then the map
<math|\<shortparallel\>.\<shortparallel\><rsup|\<ast\>>:\<bbb-R\><rsup|n>\<rightarrow\>\<bbb-R\>>
defined by <math|\<shortparallel\>.\<shortparallel\><rsup|\<ast\>>(x)=\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>>
is continuous.
</theorem>
<\proof>
Let <math|{e<rsub|1>,\<ldots\>,e<rsub|n>}> be the canonical basis on
<math|\<bbb-R\><rsup|n>> (see <reference|the canonical basis of product
of reals>) then <math|\<forall\>x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<bbb-R\><rsup|n>>
we have <math|<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>\<Rightarrowlim\><rsub|<reference|norm
of a finite sum>>\<shortparallel\>x\<shortparallel\>=\<shortparallel\><big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>\<shortparallel\><rsup|\<ast\>>\<leqslant\><big|sum><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>,e<rsub|i>\<shortparallel\><rsup|\<ast\>>=<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|.\<shortparallel\>e<rsub|i>\<shortparallel\><rsup|\<ast\>>\<leqslant\><rsub|<reference|sum
of majorants>><big|sum><rsub|i=1><rsup|n>max(\|x<rsub|i>\|i\<in\>{1,\<ldots\>,n}).\<shortparallel\>e<rsub|i>\<shortparallel\><rsup|\<ast\>>=max(\|x<rsub|i>\|i\<in\>{1,\<ldots\>,n}).<big|sum><rsub|i=1><rsup|n>\<shortparallel\>e<rsub|i>\<shortparallel\><rsup|\<ast\>>=\<shortparallel\>x\<shortparallel\>.A>
(where <math|0\<leqslant\>A=<big|sum><rsub|i=1><rsup|n>\<shortparallel\>e<rsub|i>\<shortparallel\><rsup|\<ast\>>>).
So if <math|x\<in\>\<bbb-R\><rsup|n>> \ then if
<math|\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<in\>V> (open in
<math|\<bbb-R\>>) \ so <math|\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<in\>B<rsub|d<rsub|\|
\|>>(\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>,\<varepsilon\>)>
and let <math|\<delta\>=<frac|\<varepsilon\>|A+1>\<gtr\>0> then if
<math|y\<in\>B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,\<delta\>)\<Rightarrow\>\<shortparallel\>x-y\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\|\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>-\<shortparallel\>y\<shortparallel\><rsup|\<ast\>>\|\<leqslant\>\<shortparallel\>x-y\<shortparallel\><rsup|\<ast\>>\<leqslant\>\<shortparallel\>x-y\<shortparallel\>.A\<less\><frac|\<varepsilon\>|A+1>.A\<less\>\<varepsilon\>>
proving that <math|\<shortparallel\>.\<shortparallel\><rsup|\<ast\>>(B<rsub|d<rsub|\<shortparallel\>\<shortparallel\>>>(x,\<delta\>)\<subseteq\>B<rsub|d<rsub|\|
\|>>(\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>,\<varepsilon\>)>
and thus <math|\<shortparallel\>.\<shortparallel\><rsup|\<ast\>>> is
continuous at <math|x>.<math|>
</proof>
\;
<\definition>
<label|definition of a homeomorphism><index|homeomorphism>Let
<math|X,\<cal-T\><rsub|X>> \ and <math|Y,\<cal-T\><rsub|Y>> be
topological spaces then a <math|f:X\<rightarrow\>Y> is a homeomorphism if
and only if <math|f> is a bijection and <math|f,f<rsup|-1>> are
continuous. <math|X and >Y are called homeomorphic.
</definition>
<\theorem>
<label|composition of homeomorphisms is a homemorphism>Let
<math|X,\<cal-T\><rsub|X>>, <math|Y,\<cal-T\><rsub|Y>> and
<math|Z,\<cal-T\><rsub|Z>> be topological spaces and
<math|f:X\<rightarrow\>Y> and <math|g:Y\<rightarrow\>Z> homeomorphisms
then <math|g\<circ\>f:X\<rightarrow\>Z> is a homeomorphism
</theorem>
<\proof>
Using <reference|composition of bijective mappings> we have that
<math|><math|g\<circ\>f> and <math|f<rsup|-1>\<circ\>g<rsup|-1>> are
bijective functions, this together with the fact that by the definition
of homeomorphism <math|f,g,f<rsup|-1>,g<rsup|-1>> are homeomorphism's and
<reference|continuitity of composition> proves that <math|g\<circ\>f> and
<math|(g\<circ\>f)<rsup|-1>=f<rsup|-1>\<circ\>g<rsup|-1>> are continuous.
</proof>
<\theorem>
<index|induced topology><label|induced topology>Let
<math|X,\<cal-T\><rsub|X>> be a topological space, <math|Y> a set and
<math|f:X\<rightarrow\>Y> a bijection then
<math|\<cal-T\><rsub|f,X>={f(U)\|U\<in\>\<cal-T\><rsub|X>}> forms a
topology on <math|Y> (called the induced topology on <math|Y>) so that
<math|f> is a homeomorphism. Further if <math|f,g:X\<rightarrow\>Y> are
such that <math|f=g\<circ\>h> where <math|h> is a homeomorphism (our
equivalent <math|g<rsup|-1>\<circ\>f> is a homeomorphism) then
<math|\<cal-T\><rsub|f,X>=\<cal-T\><rsub|g,X>>
</theorem>
<\proof>
First we proof that <math|\<cal-T\><rsub|f,X>> is a topology making use
of the fact that <math|f> is a bijection
<\enumerate>
<item><math|\<emptyset\>=f(\<emptyset\>)\<in\>\<cal-T\><rsub|f,X>>
<item><math|Y=f(X)\<in\>\<cal-T\><rsub|f,X>>
<item>If <math|{V<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>A>\<in\>\<cal-T\><rsub|f,X>\<Rightarrow\>V<rsub|\<alpha\>>=f(U<rsub|\<alpha\>>),U<rsub|\<alpha\>>\<in\>\<cal-T\><rsub|X>\<Rightarrow\><big|cup><rsub|\<alpha\>\<in\>A>V<rsub|\<alpha\>>=<big|cup><rsub|\<alpha\>\<in\>A>f(U<rsub|\<alpha\>>)=f(<big|cup><rsub|\<alpha\>\<in\>A>U<rsub|\<alpha\>>)\<in\>\<cal-T\><rsub|f,X>>
(because <math|<big|cup><rsub|\<alpha\>\<in\>A>U<rsub|\<alpha\>>\<in\>\<cal-T\><rsub|X>>
<item>Let <math|V<rsub|1>=f(U<rsub|1>),V<rsub|2>=f(U<rsub|2>)\<in\>\<cal-T\><rsub|f,X>\<Rightarrow\>V<rsub|1><big|cap>V<rsub|2>=f(U<rsub|1>)<big|cap>f(U<rsub|2>)=f(U<rsub|1><big|cap>U<rsub|2>)\<in\>\<cal-T\><rsub|f,X>>
as <math|U<rsub|1><big|cap>U<rsub|2>\<in\>\<cal-T\><rsub|X>>
</enumerate>
To prove that <math|f> is a homeomorphism. Note first that if
<math|U\<in\>\<cal-T\><rsub|X>\<Rightarrow\>f(U)\<in\>\<cal-T\><rsub|f,X>\<Rightarrow\>f>
is open. Secondly let <math|V\<in\>\<cal-T\><rsub|f,X>\<Rightarrow\>\<exists\>U\<in\>\<cal-T\><rsub|X>\<vdash\>V=f(U)\<Rightarrow\>f<rsup|-1>(V)=f<rsup|-1>(f(U))\<Rightarrowlim\><rsub|f
is a bijection>f<rsup|-1>(V)=U\<in\>\<cal-T\><rsub|X>\<Rightarrow\>f> is
continuous.
Next if <math|h=g<rsup|-1>\<circ\>f:X\<rightarrow\>X> is a homeomorphism
we have that\
<\enumerate>
<item><math|V\<in\>\<cal-T\><rsub|f,X>\<Rightarrow\>V=f(U),U\<in\>\<cal-T\><rsub|X>\<Rightarrow\>V=f(U)=f(h<rsup|-1>(h(U)))\<equallim\><rsub|W=h(U)\<in\>\<cal-T\><rsub|X>>f(h<rsup|-1>(W))=f(f<rsup|-1>(g(W)))=g(W)\<in\>\<cal-T\><rsub|g,X>>
<item><math|V\<in\>\<cal-T\><rsub|g,X>\<Rightarrow\>V=g(U),U\<in\>\<cal-T\><rsub|X>\<Rightarrow\>V=g(U)=g(h(h<rsup|-1>(U)))\<equallim\><rsub|W=h<rsup|-1>(U)\<in\>\<cal-T\><rsub|X>>g(h(W))=g(g<rsup|-1>(f(W)))=f(W)\<in\>\<cal-T\><rsub|f,X>>
</enumerate>
</proof>
<\theorem>
<label|inverse induced topology><index|inverse induced
topology><index|<math|\<cal-T\><rsup|-1><rsub|f,Y>>>Let <math|X> be a
set, <math|Y,\<cal-T\><rsub|Y>> a topological space and
<math|f:X\<rightarrow\>Y> a function then
<math|\<cal-T\><rsup|-1><rsub|f,Y>={f<rsup|-1>(U)\|U\<in\>\<cal-T\><rsub|Y>}>
is a topology called the inverse induced topology. Furthermore <math|f>
is a continuous function between <math|X> and <math|Y>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<emptyset\>=f<rsup|-1>(\<emptyset\>)=\<cal-T\><rsup|-1><rsub|f,X>>
<item><math|X=f<rsup|-1>(Y)\<in\>\<cal-T\><rsup|-1><rsub|f,Y>>
<item>Let <math|(V<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>I>> be a family
of sets with <math|V<rsub|\<alpha\>>\<in\>\<cal-T\><rsup|-1><rsub|f,Y>>
then <math|<big|cup><rsub|\<alpha\>\<in\>I>f<rsup|-1>(V<rsub|\<alpha\>>)\<equallim\><rsub|<reference|preimage
of union intersection of a family of
sets>>f<rsup|-1>(<big|cup><rsub|\<alpha\>\<in\>I>V<rsub|\<alpha\>>)\<in\>\<cal-T\><rsup|-1><rsub|f,Y>>
<item>Let <math|V<rsub|1>,V<rsub|2>\<in\>\<cal-T\><rsup|-1><rsub|f,Y>>
then <math|f<rsup|-1>(V<rsub|1>)<big|cap>f<rsup|-1>(V<rsub|2>)\<equallim\><rsub|<reference|preimage
image of union intersection of a family of
sets>>f<rsup|-1>(V<rsub|1><big|cap>V<rsub|2>)\<in\>\<cal-T\><rsup|-1><rsub|f,Y>>
</enumerate>
As by definition <math|f<rsup|-1>(U)\<in\>\<cal-T\><rsup|-1><rsub|f,X>>
if <math|U\<in\>\<cal-T\><rsub|Y>> so <math|f> is continue
\
</proof>
<\theorem>
<label|a isometry between nomred vectorspaces is a homeomorphisme>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed
vectorspaces over <math|\<bbb-R\>> or <math|\<bbb-C\>> and let
<math|f:X\<rightarrow\>Y> be a isometry then <math|f> is a homeomorphism
</theorem>
<\proof>
As <math|f> is a isometry (and also <math|f<rsup|-1>> is a isometry (see
<reference|inverse of isomorphisme norm>)) it is a bijection and
<math|>as <math|\<shortparallel\>x\<shortparallel\><rsub|X>=\<shortparallel\>f(x)\<shortparallel\><rsub|Y>>and
<math|> <math|\<shortparallel\>y\<shortparallel\><rsub|Y>=\<shortparallel\>f<rsup|-1>(y)\<shortparallel\><rsub|X>>
we can use <reference|continuity in normed spaces> to prove that
<math|f,f<rsup|-1>> is continuous.\
</proof>
<\theorem>
<label|there exists a isometry between a finite dimensional normed space
and the product of reals with a induced norm>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a finite dimensional
normed space over <math|\<bbb-R\> (\<bbb-C\>)> then there exists a norm
<math|\<shortparallel\>\<shortparallel\><rsub|X>> on
<math|\<bbb-R\><rsup|n> (\<bbb-C\><rsup|n>)> such that the function
<math|\<varphi\>:X\<rightarrow\>\<bbb-R\><rsup|n> (\<bbb-C\><rsup|n>)>
defined by <math|\<varphi\>(x)=(x<rsub|1>,\<ldots\>,x<rsub|n>)> where
<math|x=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>>
(<math|{e<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> being the base of
<math|X)> is a isometry. Further <math|\<varphi\>> is also a
homeomorphism and isomorphism.
</theorem>
<\proof>
First note that <math|\<varphi\>> is uniquely defined, injective,
surjective and a isomorphism (see the proof of <reference|finite
dimensional spaces are isometric>. Second define
<math|\<shortparallel\>\<shortparallel\><rsub|X>> on
<math|\<bbb-R\><rsup|n>(\<bbb-C\><rsup|n>)> by
<math|\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>\<shortparallel\>>
then <math|\<shortparallel\>\<shortparallel\><rsub|X>> is a norm for:
<\enumerate>
<item><math|\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>\<shortparallel\>\<geqslant\>0>
<item><math|\<shortparallel\>\<alpha\>.(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\>(\<alpha\>.x<rsub|1>,\<ldots\>,\<alpha\>.x<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(\<alpha\>.x<rsub|i>).e<rsub|i>\<shortparallel\>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,}>\<alpha\>.(x<rsub|i>.e<rsub|i>)\<shortparallel\>\<equallim\><rsub|<reference|scalair
product and sum>>\<shortparallel\>\<alpha\>.<big|sum><rsub|i\<in\>{1,\<ldots\>n}>x<rsub|i>.e<rsub|i>\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\><rsub|X>>
<item><math|\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)+(y<rsub|1>,\<ldots\>,y<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\>(x<rsub|1>+y<rsub|1>,\<ldots\>,x<rsub|n>+y<rsub|n>)\<shortparallel\><rsub|X>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(x<rsub|i>+y<rsub|i>).e<rsub|i>\<shortparallel\>\<equallim\><rsub|<reference|sum
of sums>>\<shortparallel\>(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>)+(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>y<rsub|i>.e<rsub|i>)\<shortparallel\>\<leqslant\>\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>\<shortparallel\>+\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>y<rsub|i>.e<rsub|i>\<shortparallel\>=\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>+\<shortparallel\>(y<rsub|1>,\<ldots\>,y<rsub|n>)\<shortparallel\>>
</enumerate>
Also by the definition of now if <math|x\<in\>X> then as
<math|{e<rsub|1>,\<ldots\>,e<rsub|n>}> is a base we find a
<math|(x<rsub|1>,\<ldots\>,x<rsub|n>)> such that
<math|x=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>> and thus
<math|\<varphi\>(x)=(x<rsub|1>,\<ldots\>,x<rsub|n>)> and we have by the
definition of <math|\<shortparallel\>\<shortparallel\><rsub|X>> that
<math|\<shortparallel\>\<varphi\>(x)\<shortparallel\><rsub|X>=\<shortparallel\><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>.e<rsub|i>\<shortparallel\>=\<shortparallel\>x\<shortparallel\>>
so <math|\<varphi\>> is a isometry. We use then the previous theorem
(<reference|a isometry between nomred vectorspaces is a homeomorphisme>)
to prove that <math|\<varphi\>> is indeed a homeomorphism.
</proof>
<\theorem>
<label|sum of continuous functions is continuous>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> ,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-R\>(\<bbb-C\>)> then if <math|f:X\<rightarrow\>Y,g:X\<rightarrow\>Y>
are continuous functions on <math|x> then <math|f+g:X\<rightarrow\>Y> is
continuous on <math|x>. From this it follows that if <math|f,g> are
continuous on <math|X> then <math|f+g> is continuous on <math|X>
</theorem>
<\proof>
Given <math|\<varepsilon\>\<gtr\>0> there exists by continuity of
<math|f,g> a <math|\<delta\><rsub|1>\<gtr\>0,\<delta\><rsub|2>\<gtr\>0>
such that if <math|\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>>
then <math|\<shortparallel\>f(x)-f(y)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>,
if <math|\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|2>>
then <math|\<shortparallel\>g(x)-g(y)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>
and thus if we take <math|\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>)>
then if <math|\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>(f+g)(x)-(f+g)(y)\<shortparallel\><rsub|Y>=\<shortparallel\>f(x)-f(y)+g(x)-g(y)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>f(x)-f(y)\<shortparallel\><rsub|Y>+\<shortparallel\>g(x)-g(y)\<shortparallel\><rsub|Y>\<less\>
\ \ \ <frac|\<varepsilon\>|2>+ <frac|\<varepsilon\>|2>=\<varepsilon\>>
</proof>
<\theorem>
<label|product of continuous functions is continuous>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> ,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-R\>(\<bbb-C\>)> then if <math|f:X\<rightarrow\>Y,g:X\<rightarrow\>Y>
are continuous functions on <math|x> then <math|f.g:X\<rightarrow\>Y> is
continuous on <math|x>. From this it follows that if <math|f,g> are
continuous on <math|X> then <math|f.g> is continuous on <math|X>
</theorem>
<\proof>
Let <math|x\<in\>X> be given and <math|\<varepsilon\>\<gtr\>0>. First by
continuity of <math|f> there exists a <math|\<delta\><rsub|1>> such that
if <math|\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>\<Rightarrow\>\<shortparallel\>f(x<rprime|'>)-f(x)\<shortparallel\><rsub|Y>\<less\>1\<Rightarrow\>\<shortparallel\>f(x<rprime|'>)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>f(x<rprime|'>)-f(x)\<shortparallel\><rsub|Y>+\<shortparallel\>f(x)\<less\>1+\<shortparallel\>f(x)\<shortparallel\><rsub|Y>>.
Now choose <math|\<delta\><rsub|2>\<gtr\>0> such that if
<math|\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|2>\<Rightarrow\>\<shortparallel\>g(x<rprime|'>)-g(x)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2(1+\<shortparallel\>f(x)\<shortparallel\><rsub|Y>)>>,
choose <math|\<delta\><rsub|3>\<gtr\>0> so that
<math|\<shortparallel\>f(x<rprime|'>)-f(x)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2(\<shortparallel\>g(x)\<shortparallel\><rsub|Y>+1)>>.
Take now <math|\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>,\<delta\><rsub|3>)>
then if <math|\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|X>\<less\>\<delta\>>
we have
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>f(x<rprime|'>)g(x<rprime|'>)-f(x)g(x)\<shortparallel\><rsub|Y>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>f(x<rprime|'>)g(x<rprime|'>)-f(x<rprime|'>)g(x)\<shortparallel\><rsub|Y>+\<shortparallel\>f(x<rprime|'>)g(x)-f(x)g(x)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>f(x<rprime|'>)\<shortparallel\><rsub|Y>\<shortparallel\>g(x<rprime|'>)-g(x)\<shortparallel\><rsub|Y>+\<shortparallel\>g(x)\<shortparallel\><rsub|Y>\<shortparallel\>f(x<rprime|'>)-f(x)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<less\>>|<cell|(1+\<shortparallel\>f(x)\<shortparallel\><rsub|Y>)<frac|\<varepsilon\>|2(1+\<shortparallel\>f\<shortparallel\><rsub|Y>)>+\<shortparallel\>g(x)\<shortparallel\><rsub|Y><frac|\<varepsilon\>|2(1+\<shortparallel\>g(x)\<shortparallel\><rsub|Y>)>>>|<row|<cell|>|<cell|\<less\>>|<cell|<frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>>>>
</eqnarray*>
proving continuity at <math|x>.
\;
</proof>
<\theorem>
<label|power function is continuous>The mapping
<math|x<rsup|n>:\<bbb-R\>\<rightarrow\>\<bbb-R\>> defined by
<math|x\<rightarrow\>x<rsup|n>> is continuous where
<math|n\<in\>\<bbb-N\>>
</theorem>
<\proof>
The case <math|n=0> is trivial, for the case
<math|n\<in\>\<bbb-N\><rsub|0>> we proceed by induction
<\enumerate>
<item>Case <math|n=1> this is again trivial as <math|x<rsup|1>=x>
<item>Assume the theory to be true for <math|n> then as
<math|x<rsup|n+1>=x.x<rsup|n>> is the product of two continuous
functions hence it is continuous by the previous theorem
(<reference|product of continuous functions is continuous>)
</enumerate>
</proof>
<subsection|Linear maps and continuity>
<\definition>
<index|linear mapping>Let <math|X,Y> be vector spaces over a field
<math|F> then a function <math|L:X\<rightarrow\>Y> is a linear map iff
<math|\<forall\>\<alpha\>,\<beta\>\<in\>F> and
<math|\<forall\>x,y\<in\>X> we have <math|L(\<alpha\>.x+\<beta\>.y)=\<alpha\>.L(x)+\<beta\>.L(y)>
</definition>
<\theorem>
Let <math|X,Y,Z> be vector spaces over a field <math|F> and
<math|L<rsub|1>:X\<rightarrow\>Y>, <math|L<rsub|2>:Y\<rightarrow\>Z> be
linear maps then <math|L<rsub|2>\<circ\>L<rsub|1>> is also linear
</theorem>
<\proof>
<math|(L<rsub|2>\<circ\>L<rsub|1>(\<alpha\>.x+\<beta\>.y)=L<rsub|2>(L<rsub|1>(\<alpha\>.x+\<beta\>.y))\<equallim\><rsub|L<rsub|2>
is linear>L<rsub|2>(\<alpha\>.L<rsub|1>(x)+\<beta\>.L<rsub|1>(y))\<equallim\><rsub|L<rsub|2>
is linear>\<alpha\>.L<rsub|2>(L<rsub|1>(x))+\<beta\>.L<rsub|2>(L<rsub|1>(y))=\<alpha\>.(L<rsub|2>\<circ\>L<rsub|1>)(x)+\<beta\>.(L<rsub|2>\<circ\>L<rsub|1>)(y)
>
</proof>
<\theorem>
<label|hom(X,Y)><index|<math|Hom(X,Y)>>Let <math|X,Y> be vector spaces
over a field over a field and <math|Hom(X,Y)={L\|L:X\<rightarrow\>Y, L is
a linear mapping}\<subseteq\>M(X,Y)> together with the operations induced
by <math|M(X,Y)> (see <reference|vector space of mappings>) forms a
vectorspace.
</theorem>
<\proof>
Using <reference|vector space of mappings> and <reference|subspace> and
the fact that <math|<wide|0|\<bar\>>\<in\>Hom(X,Y)\<Rightarrow\>\<emptyset\>\<neq\>Hom(X,Y)\<subseteq\>M(X,Y)>
we only have to prove that <math|\<forall\>\<alpha\>,\<beta\>\<in\>F> and
<math|L<rsub|1>,L<rsub|2>\<in\>Hom(X,Y)> that
<math|(\<alpha\>.L<rsub|1>)+(\<beta\>.L<rsub|2>)> is linear and thus
<math|\<in\>Hom(X,Y)>. So let <math|\<alpha\><rprime|'>,\<beta\><rprime|'>\<in\>F,x,y\<in\>X>
then \ <math|(\<alpha\>.L<rsub|1>)+(\<beta\>.L<rsub|2>)((\<alpha\><rprime|'>.x)+(\<beta\><rprime|'>.y))=\<alpha\>.L<rsub|1>((\<alpha\><rprime|'>.x)+(\<beta\><rprime|'>.y))+\<beta\>.L<rsub|2>((\<alpha\><rprime|'>.x)+(\<beta\><rprime|'>.y))=\<alpha\>.(\<alpha\><rprime|'>.L<rsub|1>(x)+\<beta\><rprime|'>.L<rsub|1>(y))+\<beta\>.(\<alpha\><rprime|'>.L<rsub|2>(x)+\<beta\><rprime|'>.L<rsub|2>(y))=\<alpha\>.\<alpha\><rprime|'>.L<rsub|1>(x)+\<alpha\>.\<beta\><rprime|'>.L<rsub|1>(y)=\<beta\>.\<alpha\><rprime|'>L<rsub|2>(x)+\<beta\>.\<beta\><rprime|'>.L<rsub|2>(y)=\<alpha\><rprime|'>.(\<alpha\>.L<rsub|1>(x)+\<beta\>.L<rsub|2>(x))+\<beta\><rprime|'>.(\<alpha\>.L<rsub|1>(y)+\<beta\>.L<rsub|2>(y))=\<alpha\><rprime|'>.((\<alpha\>.L<rsub|1>)+(\<beta\>.L<rsub|2>))(x)+\<beta\><rprime|'>((\<alpha\>.L<rsub|1>)+(\<beta\>.L<rsub|2>))(y)\<Rightarrow\>\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>
is linear>
</proof>
<\theorem>
<label|linear map applied to neutral element yields the neutral
element>Let X,Y be vectorspaces over a field <math|F> and
<math|L\<in\>Hom(X,Y)> then <math|L(0)=0>
</theorem>
<\proof>
<math|L(0)=L(0.1)=0.L(1)\<equallim\><rsub|<reference|product of zero>>0>
</proof>
<\definition>
<label|invertible linear transformation><index|invertible linear
transformation>If <math|X,+,.> is a vector space over a field <math|F>
then a invertible linear transformation is a linear bijection (or
isomorphism) from <math|X> to <math|X>
</definition>
<\theorem>
<label|general linear group><index|general linear group>Let <math|X,+,.>
be a vector space then the set of all invertible linear transformations
on <math|X> noted by <math|GL(X)={f\<in\>Hom(X,X):f is a bijection}>
forms a group with the group operation <math|\<circ\>> (the composition
operator) and neutral element <math|i<rsub|X>>. <math|GL(X),\<circ\>> is
called the general linear group. Furthermore if we define
<math|\<rightarrow\> : GL(X)\<rightarrow\>X> by
<math|\<vartriangleright\>(f,x)=f\<vartriangleright\>x=f(x)> then this is
a faithful left action.\
</theorem>
<\proof>
To proof that <math|GL(X),\<circ\>> is a group note that
<math|\<forall\>f,g,h\<in\>GL(X)> we have that
<\enumerate>
<item><math|f\<circ\>g\<in\>GL(X)> as the composition of bijection's is
again bijective and from <math|(f\<circ\>g)(\<alpha\>.x+\<beta\>.y)=f(g(\<alpha\>.x+\<beta\>.y))=f(\<alpha\>.g(x)+\<beta\>.g(y))=\<alpha\>.f(g(x))+\<beta\>.f(g(y))>
it follows also that <math|f\<circ\>g> is linear. <math|\<circ\>> is
thus indeed a function from <math|GL(X)\<times\>GL(X)\<rightarrow\>GL(X)>
<item>(Associativity) is trivial as the composition of functions is
associative
<item>(Neutral element) <math|f\<circ\>i<rsub|X>=f=i<rsub|X>\<circ\>f>
and <math|i<rsub|X>> is trivially bijective and linear so that
<math|i<rsub|X>\<in\>GL(X)>
<item>(Inverse element) Using <reference|inverse of isomorphisme> we
have that <math|\<forall\>f\<in\>GL(X)\<Rightarrow\>f<rsup|-1>\<in\>GL(X)>
and then <math|f\<circ\>f<rsup|-1>=i<rsub|X>=f<rsup|-1>\<circ\>f>
</enumerate>
Now to prove that <math|\<vartriangleright\>> is a left action\
<\enumerate>
<item><math|\<forall\>x\<in\>X\<succ\>i<rsub|X>\<vartriangleright\>x=i<rsub|X>(X)=x>
<item><math|\<forall\>f,g\<in\>GL(X),\<forall\>x\<in\>X\<succ\>(f\<circ\>g)\<vartriangleright\>x=(f\<circ\>g)(x)=f(g(x))=(f\<vartriangleright\>(g\<vartriangleright\>x))>
</enumerate>
\ Next <math|\<vartriangleright\>> is faithful because if for
<math|f\<in\>GL(X)> we define <math|\<vartriangleright\><rsub|f>:X\<rightarrow\>X>
to be <math|\<vartriangleright\><rsub|f>(x)=f\<vartriangleright\>x> then
if <math|\<vartriangleright\><rsub|f>=i<rsub|X>> we must have
<math|\<forall\>x\<in\>X> that <math|\<vartriangleright\><rsub|f>(x)=f\<vartriangleright\>x=f(x)=i<rsub|X>(x)=x>
or <math|f=i<rsub|X>>.
</proof>
<\theorem>
<label|continuous linear maps in normed spaces>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over the real (complex) field <math|F> and <math|L\<in\>Hom(X,Y)>
(<math|L> is a linear function from <math|X> to <math|Y>) then the
following are equivalent
<\enumerate>
<item><math|L> is continuous
<item><math|L> is continuous at <math|0\<in\>X>
<item><math|\<exists\>k\<gtr\>0> such that
<math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k>
<item><math|\<exists\>k\<gtr\>0> such that <math|\<forall\>x\<in\>X> we
have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>>
</enumerate>
</theorem>
<\proof>
\;
<math|1\<Rightarrow\>2>
This is trivial as L is continuous on X then L is continuous on
<math|0\<in\>X>
<math|2\<Rightarrow\>3>
Take <math|1\<gtr\>0> then using <reference|continuity in normed spaces>
<math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=\<shortparallel\>x-0\<shortparallel\><rsub|X>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x-0)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x)-L(0)\<shortparallel\><rsub|Y>\<less\>1>.
Now <math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1>
take then <math|0\<less\>\<delta\><rprime|'>\<less\>\<delta\>> (possible
see <reference|number between two different numbers>) then
<math|\<shortparallel\>\<delta\><rprime|'>.x\<shortparallel\><rsub|X>=\<delta\><rprime|'>.\<shortparallel\>x\<shortparallel\><rsub|X>=\<delta\><rprime|'>.1=\<delta\><rprime|'>\<less\>\<delta\>\<Rightarrow\>\|\<delta\><rprime|'>\|.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>\<delta\><rprime|'>.L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(\<delta\><rprime|'>.x)\<shortparallel\><rsub|Y>\<less\>1\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\><frac|1|\<delta\><rprime|'>>>
which yields the theorem if we take <math|0\<less\>k=<frac|1|\<delta\><rprime|'>>>
<math|3\<Rightarrow\>4>
Let <math|k> be such that <math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k>.
Take <math|x\<in\>X> then if <math|\<shortparallel\>x\<shortparallel\><rsub|X>=0\<Rightarrow\>x=0
and L(x)=0> so that <math|\<shortparallel\>x\<shortparallel\><rsub|X>=0\<leqslant\>k=k.\<shortparallel\>0\<shortparallel\><rsub|Y>=k.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>>,
so assume that <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<neq\>1>
then for <math|<frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>.y> we
have <math|\<shortparallel\>y\<shortparallel\><rsub|X>=\|<frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>\|.\<shortparallel\>x\<shortparallel\><rsub|X>=1\<Rightarrow\><frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\><frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>.L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(<frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(y)\<shortparallel\><rsub|Y>\<leqslant\>k\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>>
<math|4\<Rightarrow\>1>
Let <math|k\<gtr\>0> such that <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>>
then <math|\<forall\>x\<in\>X> if <math|\<varepsilon\>\<gtr\>0> take then
<math|\<delta\>=<frac|\<varepsilon\>|k+1>> then if
<math|\<forall\>y\<in\>X\<vdash\>\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>L(x)-L(y)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x-y)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x-y\<shortparallel\><rsub|X>\<less\>k.\<delta\>=<frac|\<varepsilon\>|k+1>\<less\>\<varepsilon\>>
and thus using <reference|continuity in normed spaces> we have proved
that <math|L> is continuous
</proof>
<\corollary>
<label|linear maps between products of the real spaces are continuous>Let
<math|L:\<bbb-R\><rsup|n>,\<shortparallel\>\<shortparallel\><rsub|n>\<rightarrow\>\<bbb-R\><rsup|m>,\<shortparallel\>\<shortparallel\><rsub|m>>
be a linear map between <math|\<bbb-R\><rsup|n>> and
<math|\<bbb-R\><rsup|m>> equipped with the maximum norm then <math|L> is
continuous
</corollary>
<\proof>
Let<math|{e<rsub|i>}<rsub|i\<in\>{1,.,,n}>> be the canonical base for
<math|\<bbb-R\><rsup|n>> then <math|\<forall\>x\<in\>\<bbb-R\><rsup|n>>
we have <math|x=<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>> (where
<math|x<rsub|i>=x(i)=\<pi\><rsub|i>(x)> and thus
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|m>=\<shortparallel\>L(<big|sum><rsub|i=1><rsup|n>x<rsub|i>.e<rsub|i>\<shortparallel\><rsub|m>\<equallim\><rsub|linearity>\<shortparallel\><big|sum><rsub|i=1><rsup|n>x<rsub|i>.L(e<rsub|i>)\<shortparallel\><rsub|m>\<leqslant\><big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|.\<shortparallel\>L(e<rsub|i>)\<shortparallel\><rsub|m>\<leqslant\>k.<big|sum><rsub|i=1><rsup|n>\|x<rsub|i>\|>
(where <math|k=max(\<shortparallel\>L(e<rsub|i>)\<shortparallel\><rsub|m>\|i\<in\>{1,\<ldots\>,n})>
and as <math|\<shortparallel\>x\<shortparallel\><rsub|n>=max(\|x<rsub|i>\|
\| i\<in\>{1,\<ldots\>,m})\<Rightarrow\>\|x<rsub|i>\|\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|n>>
and thus <math|\<shortparallel\>L(x)\<shortparallel\><rsub|m>\<leqslant\>n.k.\<shortparallel\>x\<shortparallel\><rsub|n>>
so <math|L> is continuous
</proof>
<\lemma>
<label|definition of the norm of a linear mapping>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be vector spaces and
let <math|L:X\<rightarrow\>Y> be a continuous linear function then if we
define <math|A<rsub|s>={k\<in\>\<bbb-R\><rsub|+>\|
\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k>
<math|\<forall\>x\<in\>X> with <math|\<shortparallel\>x\<shortparallel\><rsub|X>=1}>,
<math|A<rsub|r>={k\<in\>\<bbb-R\><rsub|+>\|
\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>
\<forall\>x\<in\>X}> then <math|A<rsub|s>=A<rsub|r>> and
<math|\<shortparallel\>L\<shortparallel\>=inf(A<rsub|s>)=inf(A<rsub|r>)>
exists (is finite) and <math|\<geqslant\>0>
</lemma>
<\proof>
First we prove that <math|A<rsub|s>=A<rsub|r>> then\
<\enumerate>
<item><math|k\<in\>A<rsub|s>> then let <math|x\<in\>X>, if
<math|\<shortparallel\>x\<shortparallel\><rsub|X>=0\<Rightarrow\>x=0\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(0)\<shortparallel\><rsub|Y>=\<shortparallel\>0\<shortparallel\><rsub|Y>=0\<leqslant\>k.0=k.\<shortparallel\>x\<shortparallel\><rsub|X>>
and if <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<neq\>0> then
<math|\<shortparallel\><frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>x\<\|\|\>=<frac|\<shortparallel\>x\<shortparallel\><rsub|X>|\<shortparallel\>x\<shortparallel\><rsub|X>>=1>
and thus <math|<frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\><frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(<frac|1|\<shortparallel\>x\<shortparallel\><rsub|X>>x)\<shortparallel\><rsub|Y>\<leqslant\>k\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>>
and thus we have proved that <math|k\<in\>A<rsub|r>> so
<math|A<rsub|s>\<subseteq\>A<rsub|r>>
<item><math|k\<in\>A<rsub|r>> then <math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.\<shortparallel\>x\<shortparallel\><rsub|X>=k\<Rightarrow\>k\<in\>A<rsub|r>>
<item>Now by continuity of <math|L> we use <reference|continuous linear
maps>(4) to prove the existence of a <math|k\<in\>A<rsub|r>> so that
<math|\<emptyset\>\<neq\>A<rsub|r>=A<rsub|s>> and as
<math|\<forall\>k\<in\>A<rsub|r>=A<rsub|s>> we have <math|k\<gtr\>0> we
find that the nonempty set <math|A<rsub|r>(A<rsub|s>)> is bounded below
by 0 and then use the upper bound property of the real's
(<reference|real numbers> and <reference|upper bound property implies
lower bound property>) to prove the existence of
<math|inf(A<rsub|s>)=inf(A<rsub|r>)>
</enumerate>
</proof>
<\lemma>
<label|norm properties of the norm of a linear map>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over the real (complex) field <math|F> and
<math|L,L<rsub|1>,L<rsub|2>> be linear continuous functions between
<math|X> and <math|Y> then
<\enumerate>
<item><math|\<forall\>x\<in\>X> we have
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>>
<item><math|L<rsub|1>+L<rsub|2>> is continuous and
<math|\<shortparallel\>L<rsub|1>+L<rsub|2>\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>>
<item>If <math|L=<wide|0|\<bar\>>> (the <math|0> function) then
<math|\<shortparallel\>L\<shortparallel\>=0>
<item><math|\<forall\>\<alpha\>\<in\>F> we have <math|\<alpha\>.L> is
continuous and <math|\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>=\<shortparallel\>\<alpha\>.L\<shortparallel\>>
</enumerate>
</lemma>
<\proof>
\;
<\enumerate>
<item>If <math|x=0> then we have trivially
<math|\<shortparallel\>L(0)\<shortparallel\><rsub|Y>=\<shortparallel\>0\<shortparallel\><rsub|Y>=0\<leqslant\>0=\<shortparallel\>L\<shortparallel\>.\<shortparallel\>0\<shortparallel\><rsub|X>>
and thus <math|\<shortparallel\>L(0)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>.\<shortparallel\>0\<shortparallel\>>.
If <math|x\<neq\>0> (an thus <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<neq\>0>)
then if <math|\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\>\<less\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<Rightarrow\>\<shortparallel\>L\<shortparallel\>\<less\><frac|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>|\<shortparallel\>x\<shortparallel\><rsub|X>>>
and as <math|\<shortparallel\>L\<shortparallel\>=inf(A<rsub|r>)>
<math|\<exists\>k\<in\>A<rsub|r>> such that
<math|\<shortparallel\>L\<shortparallel\>\<leqslant\>k\<less\><frac|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>|\<shortparallel\>x\<shortparallel\><rsub|X>>\<Rightarrow\>k.\<shortparallel\>x\<shortparallel\><rsub|X>\<less\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>>
contradicting the fact that <math|k\<in\>A<rsub|r>> so we must have
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>>
<item>Because of <math|(1)> we have <math|\<forall\>x\<in\>X>
<math|\<shortparallel\>L<rsub|1>(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>>
and <math|\<shortparallel\>L<rsub|2>(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L<rsub|2>\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>>
and thus <math|\<shortparallel\>(L<rsub|1>+L<rsub|2>)(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L<rsub|1>(x)+L<rsub|2>(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L<rsub|1>(x)\<shortparallel\><rsub|Y>+\<shortparallel\>L<rsub|2>(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>+\<shortparallel\>L<rsub|2>\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>=(\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>).\<shortparallel\>x\<shortparallel\><rsub|X>>
and thus using <reference|continuous linear maps in normed spaces> we
find that <math|L<rsub|1>+L<rsub|2>> is continuous and also using the
definition of <math|\<shortparallel\>.\<shortparallel\>> (see
<reference|definition of the norm of a linear mapping>) we have
<math|\<shortparallel\>L<rsub|1>+L<rsub|2>\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>>
<item>If <math|L=<wide|0|\<bar\>>> then <math|\<forall\>x\<in\>X> we
have <math|L(x)=0> and thus <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>0\<shortparallel\><rsub|Y>=0\<leqslant\>0.\<shortparallel\>x\<shortparallel\><rsub|X>\<Rightarrow\>0\<in\>A<rsub|r>\<Rightarrow\>0\<leqslant\>\<shortparallel\>L\<shortparallel\>\<leqslant\>0\<Rightarrow\>\<shortparallel\>L\<shortparallel\>=0>
<item>Using (1) we have<math| \<forall\>x\<in\>X>
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>\<Rightarrow\>\<shortparallel\>(\<alpha\>.L)(x)\<shortparallel\><rsub|Y>=\<shortparallel\>\<alpha\>.L(x)\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>>
so by <reference|continuous linear maps> we find that
<math|\<alpha\>.L> is continuous and by <math|<reference|definition of
the norm of a linear mapping>> that
<math|\<shortparallel\>\<alpha\>.L\<shortparallel\>\<leqslant\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>.>
Now if <math|\<alpha\>=0> then <math|\<alpha\>.L=<op|<wide|0|\<bar\>>>><math|>
and <math|\<shortparallel\>0.L\<shortparallel\>=0=0.\<shortparallel\>L\<shortparallel\>>
assume now that <math|\<alpha\>\<neq\>0> then as
<math|\<shortparallel\>(\<alpha\>.L)(x)\<shortparallel\><rsub|Y>=\|\<alpha\>\|.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=<frac|\<shortparallel\>(\<alpha\>.L)(x)\<shortparallel\><rsub|Y>|\|\<alpha\>\|>\<leqslant\><frac|\<shortparallel\>\<alpha\>.L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>|\|\<alpha\>\|>=<frac|\<shortparallel\>\<alpha\>.L\<shortparallel\><rsub|Y>|\|\<alpha\>\|>.\<shortparallel\>x\<shortparallel\><rsub|X>\<Rightarrow\>\<shortparallel\>L\<shortparallel\>\<leqslant\><frac|\<shortparallel\>\<alpha\>.L\<shortparallel\>|\|\<alpha\>\|>\<Rightarrow\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>\<leqslant\>\<shortparallel\>\<alpha\>.L\<shortparallel\>\<Rightarrow\>\<shortparallel\>\<alpha\>.L\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>>
</enumerate>
</proof>
<\lemma>
<label|alternative definition of norm of linear map>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over the real (complex) field <math|F> with <math|X> containing at
least a non zero element and <math|L> be a linear continuous function
between <math|X> and <math|Y> then\
<\enumerate>
<item><math|\<shortparallel\>L\<shortparallel\>=sup{\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\|x\<in\>X>
with <math|\<shortparallel\>x\<shortparallel\><rsub|X>=1}>
<item><math|\<shortparallel\>L\<shortparallel\>=sup{\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\|x\<in\>X>
with <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>1}>
</enumerate>
</lemma>
<\proof>
Let <math|A<rsub|t>={><math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\|x\<in\>X>
with <math|\<shortparallel\>x\<shortparallel\><rsub|X>=1}> and
<math|A<rsub|u>={\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\|x\<in\>X>
with <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>1}>
then\
<\enumerate>
<item>If <math|k\<in\>A<rsub|t> then ><math|\<exists\>x\<in\>X> with
<math|\<shortparallel\>x\<shortparallel\><rsub|X>=1\<leqslant\>1> such
that <math|k=\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<Rightarrow\>k\<in\>A<rsub|u>
and thus ><math|A<rsub|t>\<subseteq\>A<rsub|u>>
<item>Now because <math|X> contains a non zero element <math|x<rsub|0>>
we have <math|\<shortparallel\>x<rsub|0>\<shortparallel\><rsub|X>\<neq\>0>
and <math|x<rsub|1>=<frac|1|\<shortparallel\>x<rsub|0>\<shortparallel\><rsub|X>>.x>
has <math|\<shortparallel\>x<rsub|1>\<shortparallel\>=1> and thus
<math|\<emptyset\>\<neq\>A<rsub|t>\<subseteq\>A<rsub|u>> so
<math|A<rsub|t>,A<rsub|u>> are not empty. Also by <reference|continuous
linear maps> <math|\<exists\>k\<gtr\>0> such that
<math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k\<Rightarrow\>k>
is a upper bound of <math|A<rsub|t>> and thus by the upper bound
property of the real numbers we have <math|sup(A<rsub|t>)> exists and
<math|t=sup(A<rsub|t>)\<leqslant\>k>. Now if <math|l\<in\>A<rsub|s>>
then <math|\<forall\>x\<in\>X> such that
<math|\<shortparallel\>x\<shortparallel\><rsub|X>=1> we have
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>l> and
thus <math|sup(A<rsub|t>)\<leqslant\>l\<Rightarrow\>t=sup(A<rsub|t>)\<leqslant\>inf(A<rsub|s>)=\<shortparallel\>L\<shortparallel\>>
<item>Now <math|\<forall\>m\<in\>A<rsub|u>>
<math|\<exists\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>1\<succ\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=m>
then <math|\<forall\>l\<in\>A<rsub|r>> we have
<math|m=\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>l.\<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>l>
and thus <math|m\<leqslant\>inf(A<rsub|r>)=\<shortparallel\>L\<shortparallel\>>
so <math|A<rsub|u>> is bounded above by
<math|\<shortparallel\>L\<shortparallel\>> and thus
<math|u=sup(A<rsub|u>)\<leqslant\>\<shortparallel\>L\<shortparallel\>><math|>
also because <math|A<rsub|t>\<subseteq\>A<rsub|u>> we have
<math|t=sup(A<rsub|t>)\<leqslant\>sup(A<rsub|u>)=u\<leqslant\>\<shortparallel\>L\<shortparallel\>><math|>
<item><math|\<forall\>x\<in\>X\<vdash\>\<shortparallel\>x\<shortparallel\><rsub|X>=1>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<in\>A<rsub|t>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>sup(A<rsub|t>)=t\<Rightarrow\>t\<in\>A<rsub|s>\<Rightarrow\>\<shortparallel\>L\<shortparallel\>=inf(A<rsub|s>)\<leqslant\>>t
<item>Finally using (3) we have that
<math|\<shortparallel\>L\<shortparallel\>\<leqslant\>t\<leqslant\>u\<leqslant\>\<shortparallel\>L\<shortparallel\>\<Rightarrow\>\<shortparallel\>L\<shortparallel\>=t=u>
</enumerate>
</proof>
<\theorem>
<label|composition of continuous linear map is continuous>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,
Y,\<shortparallel\>\<shortparallel\><rsub|Y>> and
<math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be normed spaces and
<math|L<rsub|1>:X\<rightarrow\>Y>, <math|L<rsub|2>:X\<rightarrow\>Y> be
continuous linear functions then <math|L<rsub|2>\<circ\>L<rsub|1>:X-Z> is
a continuous linear function and <math|\<shortparallel\>L<rsub|2>\<circ\>L<rsub|1>\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|2>\<shortparallel\>\<shortparallel\>L<rsub|1>\<shortparallel\>>
</theorem>
<\proof>
First we prove linearity\
<\eqnarray*>
<tformat|<table|<row|<cell|(L<rsub|2>\<circ\>L<rsub|1>)(\<alpha\>x+\<beta\>y)>|<cell|=>|<cell|L<rsub|2>(L<rsub|1>(\<alpha\>x+\<beta\>y))>>|<row|<cell|>|<cell|=>|<cell|L<rsub|2>(\<alpha\>L<rsub|1>(x)+\<beta\>L<rsub|1>(y))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>L<rsub|2>(L<rsub|1>(x))+\<beta\>L<rsub|2>(L<rsub|1>(y))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(L<rsub|2>\<circ\>L<rsub|1>)(x)+\<beta\>(L<rsub|2>\<circ\>L<rsub|1>)(y)>>>>
</eqnarray*>
Secondly we prove continuity, take <math|x\<in\>X> then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>(L<rsub|2>\<circ\>L<rsub|1>)(x)\<shortparallel\><rsub|Z>>|<cell|=>|<cell|\<shortparallel\>L<rsub|2>(L<rsub|1>(x))\<shortparallel\><rsub|Z>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>L<rsub|2>\<shortparallel\>\<shortparallel\>L<rsub|1>(x)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>L<rsub|2>\<shortparallel\>\<shortparallel\>L<rsub|1>\<shortparallel\>\<shortparallel\>x\<shortparallel\><rsub|X>>>>>
</eqnarray*>
proving continuity and also that <math|\<shortparallel\>L<rsub|2>\<circ\>L<rsub|1>\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|2>\<shortparallel\>\<shortparallel\>L<rsub|1>\<shortparallel\>>
\;
</proof>
<\theorem>
<label|normed vector space of continuous linear
mappings><index|L(X,Y)><index|operator norm>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be vector spaces over
<math|F> (real or complex field) then
<math|L(X,Y)={L\|L:X\<rightarrow\>Y>, <math|L> is a linear continuous
function between <math|X> and <math|Y}\<subseteq\>Hom(X,Y)> is a subspace
of <math|Hom(X,Y)> (and thus a vectorspace). Further given
<math|L\<in\>L(X,Y)> then <math|\<shortparallel\>L\<shortparallel\>>
forms a norm on <math|L(X,Y)> and <math|L(X,Y)> is thus a normed
vectorspace. <math|\<shortparallel\>\<shortparallel\>> is called the
operator norm
</theorem>
<\proof>
First we prove that <math|L(X,Y)> is a subspace. For this using
<reference|hom(X,Y)> and <reference|subspace> we only have to prove that
for <math|L<rsub|1>,L<rsub|2>\<in\>L(X,Y)> and
<math|\<alpha\>,\<beta\>\<in\>F> \ <math|\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>\<in\>L(X,Y)>
which means that it must be continuous, and this follows from
<reference|norm properties of the norm of a linear map>. Secondly to
prove that <math|\<shortparallel\>\<shortparallel\>> is a norm we can use
again <reference|norm properties of the norm of a linear map> so that the
only thing still must be proved is that from
<math|\<shortparallel\>L\<shortparallel\>=0> if follows that <math|L=0>.
Now if <math|\<shortparallel\>L\<shortparallel\>=0\<Rightarrow\>\<forall\>x\<in\>X\<succ\>\<shortparallel\>L(x)\<shortparallel\>\<leqslant\>\<shortparallel\>L\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X>=0\<Rightarrow\>L(x)=0\<Rightarrow\>L=0>
</proof>
<\notation>
<index|<math|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>>Given
<math|n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>We note
<math|{1,\<ldots\>,n}<mid|\\>{i}> as <math|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>
(or <math|{1,\<ldots\>,n-1}> if <math|i=n> or <math|{2,\<ldots\>,n}> if
<math|i=1> and given <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we note <math|x<rsup|[i]>={x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=x<rsub|\|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>>
as the restriction of <math|x> as a function of
<math|{1,\<ldots\>,n}\<rightarrow\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}><rsup|>X<rsub|i>>
to <math|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>. If <math|i=1> then sometimes
we note <math|(x<rsub|2>,\<ldots\>,x<rsub|n>)> and if <math|i=n> then we
note <math|(x<rsub|1>,\<ldots\>,x<rsub|n-1>)>
</notation>
<\definition>
<label|function of ith variable>Let <math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
be a family of sets given <math|i\<in\>{1,\<ldots\>,n}> and
<math|(y<rsub|1>,\<ldots\>,y<rsub|n>)\<in\><big|prod><rsub|j\<in\>{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>X<rsub|j>>
then we define <math|(y<rsub|1>,\<ldots\>,\<ast\>,\<ldots\>,y<rsub|n>):X<rsub|i>\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
by\
<\eqnarray*>
<tformat|<table|<row|<cell|((y<rsub|1>,\<ldots\>,\<ast\>,\<ldots\>,y<rsub|n>)(t))<rsub|j>=>|<cell|y<rsub|j>>|<cell|j\<in\>{1,\<ldots\>,n}<mid|\\>{i}>>|<row|<cell|>|<cell|
t>|<cell|j=i>>>>
</eqnarray*>
</definition>
As a easy notation we use <math|(y<rsub|1>,\<ldots\>,y<rsub|i-1>,\<ast\>,y<rsub|i+1>,\<ldots\>,y<rsub|n>)(t)\<equallim\><rsub|notation>(y<rsub|1>,\<ldots\>,y<rsub|i-1>,t,y<rsub|i+1>,\<ldots\>,y<rsub|n>)>,
also if <math|i=n> (=1) we note this by
<math|(y<rsub|1>,\<ldots\>,y<rsub|n-1>,\<ast\>)> (or
<math|(\<ast\>,y<rsub|2>,\<ldots\>,y<rsub|n>)>)
<\note>
Given <math|y\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> then
<math|(y<rsup|[i]><rsub|1>,\<ldots\>,y<rsup|[i]><rsub|i-1>,\<ast\>,y<rsup|[i]><rsub|i+1>,\<ldots\>,y<rsup|[i]><rsub|n>)(y<rsub|i>)=y>
</note>
<\definition>
<label|multilinear mapping><index|multilinear mapping>Let
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of vector spaces over a field <math|F> and <math|Y> a vector
space over <math|F> then a function <math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
is multilinear if in case of <math|n=1> it is linear and if
<math|n\<gtr\>1> we have \ <math|\<forall\>i\<in\>{1,\<ldots\>,n}> and
<math|\<forall\>y\<in\><big|prod><rsub|j\<in\>{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>X<rsub|i>>
\ that <math|L\<circ\>(y<rsub|1>,\<ldots\>,y<rsub|i-1>,\<ast\>,y<rsub|i+1>,\<ldots\>,y<rsub|n>)>
is linear. As a easy notation we use <math|L(y<rsub|1>,\<ldots\>,y<rsub|i-1>,\<ast\>,y<rsub|i+1>,\<ldots\>,y<rsub|n>)\<equallim\><rsub|notation>L\<circ\>(y<rsub|1>,\<ldots\>,y<rsub|i-1>,\<ast\>,y<rsub|i+1>,\<ldots\>,y<rsub|n>)>
and <math|L(y<rsub|1>,\<ldots\>,y<rsub|i-1>,t,y<rsub|i+1>,\<ldots\>,y<rsub|n>)=<with|mode|text|<math|L\<circ\>(y<rsub|1>,\<ldots\>,y<rsub|i-1>,\<ast\>,y<rsub|i+1>,\<ldots\>,y<rsub|n>)>(t)>>
This means that \ <math|L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<alpha\>.x+\<beta\>.y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=\<alpha\>.L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,y,x<rsub|i+1>,\<ldots\>,x<rsub|n>),\<forall\>i\<in\>{1,\<ldots\>,n}>
</definition>
<\notation>
We note <math|L\<circ\>(y<rsub|1>,\<ldots\>,y<rsub|n-1>,\<ast\>)> as
<math|L(y<rsub|1>,\<ldots\>,y<rsub|n-1>,\<ast\>)> and
<math|L\<circ\>(\<ast\>,y<rsub|2>,\<ldots\>,y<rsub|n>)=L(\<ast\>,y<rsub|2>,\<ldots\>,y<rsub|n>)>
</notation>
<\definition>
Let <math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
be a family of sets then if <math|x\<in\>X<rsub|n>> we define
<math|(*\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x):<big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>X<rsub|i>\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
by\
<\eqnarray*>
<tformat|<table|<row|<cell|((*\<ast\><rsub|1>,\<ldots\>,*\<ast\><rsub|n-1>,x)(y))<rsub|j>=>|<cell|y<rsub|j>>|<cell|j\<in\>{1,\<ldots\>,n-1}>>|<row|<cell|>|<cell|x>|<cell|j=n>>>>
</eqnarray*>
In easy notation we have <math|(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)(y<rsub|1>,\<ldots\>,y<rsub|n-1>)=(y<rsub|1>,\<ldots\>,y<rsub|n-1>,x)>
</definition>
<\note>
Given <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>I>X<rsub|i>>
then <math|(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x<rsub|n>)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)={x<rsub|1>,.\<ldots\>,x<rsub|n>)>
</note>
<\definition>
Let <math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
be a family of sets and <math|Y> be a set and
<math|f:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
a function and <math|x\<in\>X<rsub|n>> then
<math|f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x):<rsub|><big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>X<rsub|i>\<rightarrow\>Y>
is defined by <math|f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)=f\<circ\>(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)>
Or in easy notation <math|f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)=f(x<rsub|1>,\<ldots\>,x<rsub|n>,x)>
</definition>
<\lemma>
<label|n multilinear mapping yields n-1 multilinear map>Let
<math|{X<rsub|i>}<rsub|i\<in\>(1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>2>
be a family of vector spaces over a field <math|F>, <math|Y> a
vectorspace over <math|F> and <math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
be a multilinear function then given <math|x\<in\>X<rsub|n>> we have that
<math|L(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)> is multilinear
<\proof>
Let <math|i\<in\>{1,\<ldots\>,n-1}\<Rightarrow\>i\<neq\>n> and if
<math|t\<in\>X<rsub|i>,s\<in\>X<rsub|n>> and
<math|(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i+1>,\<ldots\>,x<rsub|n-1>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,i-1,i+1,\<ldots\>,n-1}>X<rsub|i>>
then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|((\<ast\><rsub|1>,\<ldots\>,*\<ast\><rsub|n-1>,s)((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n-1>)(t)))<rsub|j>=>|<cell|x<rsub|j><rsub|>>|<cell|j\<neq\>i,n>>|<row|<cell|>|<cell|t>|<cell|j=i\<neq\>n>>|<row|<cell|>|<cell|s>|<cell|j=n>>>>
</eqnarray*>
so if we define <math|(u<rsub|1>,\<ldots\>,u<rsub|n>)> by\
<\eqnarray*>
<tformat|<table|<row|<cell|u<rsub|i>=>|<cell|x<rsub|i>>|<cell|i\<neq\>n>>|<row|<cell|>|<cell|s>|<cell|i=n>>>>
</eqnarray*>
then for <math|i\<in\>{1,\<ldots\>,n-1}>
<\eqnarray*>
<tformat|<table|<row|<cell|((u<rsub|1>,\<ldots\>,u<rsub|i-1>,\<ast\>,u<rsub|i+1>,\<ldots\>,u<rsub|n>)(t))<rsub|j>=>|<cell|u<rsub|j>=x<rsub|j>>|<cell|j\<neq\>i>>|<row|<cell|>|<cell|t>|<cell|j=i>>|<row|<cell|>|<cell|s>|<cell|j=n>>>>
</eqnarray*>
So <math|(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,s)\<circ\>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=(u<rsub|1>,\<ldots\>,u<rsub|i-1>,\<ast\>,u<rsub|i+1>,\<ldots\>,u<rsub|n>)>
and thus <math|L(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,s)\<circ\>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=L\<circ\>(u<rsub|1>,\<ldots\>,u<rsub|i-1>,\<ast\>,u<rsub|i+1>,\<ldots\>,u<rsub|n>)>
which is linear because <math|L> is multilinear.
</proof>
</lemma>
<\theorem>
<label|mulilinearity and coefficients>Let
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n+1}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of vector spaces over a field F and let <math|Y> also be a
vector space over F. Then for every multilinear function
<math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>X<rsub|i>\<rightarrow\>Y>
we have <math|\<forall\>x\<in\><big|prod><rsub|i\<subset\>{1,\<ldots\>,n+1}>X<rsub|i>>
and <math|\<forall\>f<rsup|>\<in\>F<rsup|n>> that
<math|L(f.x)=(<big|prod><rsub|i=1><rsup|n+1>f<rsub|i>).L(x)> (where
<math|f.x> is defined by <math|\<forall\>i\<in\>{1,\<ldots\>,n+1}>
<math|(f.x)<rsub|i>=f<rsub|i>.x<rsub|i>>)
</theorem>
<\proof>
We proof this by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|L> is linear by definition and we have
<math|L(f<rsub|1>.x<rsub|1>)=f<rsub|1>.L(x<rsub|1>)=(<big|prod><rsub|i=1><rsup|1>f<rsub|1>).L(x<rsub|1>)>
<item><math|n=2> then <math|L(f<rsub|1>.x<rsub|1>,f<rsub|2>.x<rsub|2>)=L(\<ast\>,f<rsub|2>.x<rsub|2>)(f<rsub|1>.x<rsub|1>)\<equallim\><rsub|multilinearity>f<rsub|1>\<ast\>L(\<ast\>,f<rsub|2>.x<rsub|2>)=f<rsub|1>.L(x<rsub|1>,f<rsub|2>.x<rsub|2>)=f<rsub|1>.L(x<rsub|1>,\<ast\>)(f<rsub|2>.x)=f<rsub|1>.f<rsub|2>.L(x<rsub|1>,\<ast\>)(x<rsub|2>)=f<rsub|1>.f<rsub|2>.L(x<rsub|1>,x<rsub|2>)=(<big|prod><rsub|i=1><rsup|2>f<rsub|i>).L(x<rsub|1>,x<rsub|2>)>
<item>Assume the theorem is true for <math|n> then let
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n+1>X<rsub|i>,f\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n+1}>F<rsub|i>>
then <math|L(f<rsub|1>.x<rsub|1>,\<ldots\>,f<rsub|n+1>.x<rsub|n+1>)=L(f<rsub|1>.x<rsub|1>,\<ldots\>,f<rsub|n>.x<rsub|n>,\<ast\>)(f<rsub|n+1>.x<rsub|n+1>)\<equallim\><rsub|multilinearity>f<rsub|n+1>.L(f<rsub|1>.x<rsub|1>,\<ldots\>,f<rsub|n>.x<rsub|n>,\<ast\>)(x<rsub|n+1>)=f<rsub|n+1>.L(f<rsub|1>.x<rsub|1>,\<ldots\>,f<rsub|n>.x<rsub|n>,x<rsub|n+1>)=f<rsub|n+1>.L(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,x<rsub|n+1>)(f<rsub|1>.x<rsub|1>,\<ldots\>,f<rsub|n>.x<rsub|n>)\<equallim\><rsub|induction
hypothese and previous lemma>=f<rsub|n+1>.(<big|prod><rsub|i=1><rsup|n>f<rsub|i>).L(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,x<rsub|n+1>)(x<rsub|1>,\<ldots\>,x<rsub|n>)=(<big|prod><rsub|i=1><rsup|n+1>f<rsub|i>).L(x<rsub|1>,\<ldots\>,x<rsub|n+1>)>
</enumerate>
</proof>
<\theorem>
<label|vector space of multilinear mappings><index|<math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>>Let
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of vector spaces over a field <math|F>, <math|Y> a vector
spaces over the field <math|F> and <math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)={L\|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
is a multilinear function<math|}\<subseteq\>M(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>,Y)>
(see <reference|vector space of mappings>) then
<math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> forms a subspace of
<math|M(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>,Y)> and is thus
is a vectorspace using the induced operations from
<math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>, note that if <math|n=1> then
we have <math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)=Hom(X<rsub|1>,Y)>
which is already proved to be a vector subspace from
<math|M(X<rsub|1>,Y)>
</theorem>
<\proof>
By <reference|subspace> and <reference|vector space of mappings> we must
only proof that <math|\<forall\>\<alpha\>,\<beta\>\<in\>F,\<forall\>L<rsub|1>,L<rsub|2>\<in\>Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
we have <math|\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|1>\<in\>Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>.
So choose <math|i\<in\>{1,\<ldots\>,n}> and
<math|x\<in\><big|prod><rsub|j\<in\>{1,\<ldots\>,n}<mid|\\>{1}>X<rsub|j>>,
and <math|s,t\<in\>X<rsub|i>,\<delta\>,\<gamma\>\<in\>F> then
<math|(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>)((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(\<delta\>.s+\<gamma\>.t))=\<alpha\>.L<rsub|1>((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(\<delta\>.s+\<gamma\>.t))+\<beta\>.L<rsub|2>((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(\<delta\>.s+\<gamma\>.t))\<equallim\><rsub|multilinearity
of L<rsub|1>,L<rsub|2>>><math|\<alpha\>.(\<delta\>.L<rsub|1>(x<rsub|1,\<ldots\>,>x<rsub|i-1>,s,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<gamma\>.L<rsub|1>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>,x<rsub|n>))+\<beta\>.(\<delta\>.L<rsub|1>(x<rsub|1,\<ldots\>,>x<rsub|i-1>,s,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<gamma\>.L<rsub|1>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>,x<rsub|n>))\<equallim\>\<delta\>.(\<alpha\>.L<rsub|1>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,s,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.L<rsub|1>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,s,x<rsub|i+1>,\<ldots\>,x<rsub|n>))+\<gamma\>.(\<alpha\>.L<rsub|1>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>x<rsub|n>)+\<beta\>.L<rsub|2>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>,x<rsub|n>))=\<delta\>.(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>)(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<gamma\>.(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>)(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=\<delta\>.(\<alpha\>.L<rsub|1>+\<beta\>.l<rsub|2>)(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(s)+\<gamma\>.(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>)(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
proving multilinearity of <math|\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>>\
</proof>
<\theorem>
<label|multilinear mappings and neutral element>Let
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of vector spaces over a field <math|F> an <math|Y> also a
vector space over <math|F>. Then for a multilinear function
<math|L:<big|prod><rsub|i\<in\>I>X<rsub|i>\<rightarrow\>Y> we have
<math|L(x)=0> if <math|\<exists\>i\<in\>{1,\<ldots\>,n}> such that
<math|x(i)=0>
</theorem>
<\proof>
If <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> is such
that <math|\<exists\>i\<in\>{1,\<ldots\>,n}\<vdash\>x(i)=0> then
<math|x=f.x> where <math|f> is defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|f(j)=>|<cell|1>|<cell|j\<neq\>i>>|<row|<cell|>|<cell|0>|<cell|j=i>>>>
</eqnarray*>
and this <math|L(x)=L(f.x)=(<big|prod><rsub|i=1><rsup|n>f(i)).L(x)\<equallim\><rsub|<reference|general
product with a zero>>0.L(x)=0>\
</proof>
\;
<\theorem>
<label|continuitity and multilinearity>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of normed vector spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be a vector space
over <math|\<bbb-R\>(\<bbb-C\>)> then for a multilinear function
<math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
the following are equivalent in the product topology on
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> (generated by the
normed topologies on <math|X<rsub|i>>) and the normed topology on
<math|Y>
<\enumerate>
<item><math|L> is continuous
<item><math|L> is continuous at <math|0\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>n}>X<rsub|i>>
<item><math|\<exists\>k\<gtr\>0> such that
<math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>
with \<forall\>i\<in\>I\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<equallim\><rsub|definition>\<shortparallel\>x(i)\<shortparallel\><rsub|i>=1>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k>
<item><math|\<exists\>k\<gtr\>0> such that
<math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
</enumerate>
</theorem>
<\proof>
The product topology on <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
is by <reference|norm of finite product of normed spaces> generated by
the norm <math|\<shortparallel\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=max(\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\|i\<in\>{1,\<ldots\>,n})>
(here <math|x<rsub|i>=x(i)=\<pi\><rsub|i>(x)<rsub|>>), now to prove our
theorem
<math|1\<Rightarrow\>2> This is trivial
<math|2\<Rightarrow\>3> Using <reference|continuity in normed spaces> we
find that <math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
with <math|\<shortparallel\>x\<shortparallel\>=\<shortparallel\>x-0\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x-0)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x)-L(0)\<shortparallel\><rsub|Y>\<less\>1>.
Now take <math|0\<less\>\<delta\><rprime|'>\<less\>\<delta\>> and form
<math|k=<frac|1|\<delta\><rprime|'><rsup|n>>> then if
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<vdash\>\<forall\>i\<in\>{1,\<ldots\>,n}>
we have <math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1> then
<math|\<shortparallel\>\<delta\><rprime|'>.x\<shortparallel\>=\<delta\><rprime|'>.\<shortparallel\>x\<shortparallel\>=\<delta\><rprime|'>.max{\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\|i\<in\>{1,\<ldots\>,n})\<leqslant\>\<delta\><rprime|'>\<less\>\<delta\>\<Rightarrow\>\<delta\><rprime|'><rsup|n>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=(<big|prod><rsub|i=1><rsup|n>\<delta\><rprime|'>).\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>\<delta\><rprime|'>.L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(\<delta\><rprime|'>.x)\<shortparallel\><rsub|Y>\<less\>1\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\>\<leqslant\><frac|1|\<delta\><rprime|'><rsup|n>>=k>
<math|3\<Rightarrow\>4> Let <math|k> be as in (3) \ then if
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>> such that
<math|\<exists\>i\<in\>{1,\<ldots\>,n}> with <math|x<rsub|i>=0> then
<math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=0\<Rightarrow\><big|prod><rsub|j=1><rsup|n>\<shortparallel\>x<rsub|j>\<shortparallel\><rsub|j>=0>
and <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<equallim\><rsub|<reference|multilinear
mappings and neutral element>>0\<leqslant\>k.0=k.<big|prod><rsub|j=1><rsup|n>\<shortparallel\>x<rsub|j>\<shortparallel\><rsub|j>>,
so assume now that <math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>x<rsub|i>\<neq\>0\<Rightarrow\>\<shortparallel\>x<rsub|i>\<\|\|\><rsub|i>\<neq\>0>
define then <math|y> by <math|y<rsub|i>=<frac|x<rsub|i>|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>>
so that <math|\<shortparallel\>y<rsub|i>\<shortparallel\><rsub|i>=<frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1>
and thus <math|<frac|1|\<Pi\><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>\<shortparallel\>L(x)\<shortparallel\><rsub|i>\<equallim\><rsub|<reference|inverse
of product of many elements in a field>>(<big|prod><rsub|i=1><rsup|n><frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>).\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L(y)\<shortparallel\><rsub|Y>\<leqslant\>k\<Rightarrow\>\<shortparallel\>L(y)\<shortparallel\><rsub|Y>\<leqslant\>k.(<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)><math|><math|>
<math|4\<Rightarrow\>1> Let <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
and take <math|\<varepsilon\>\<gtr\>0> then if we choose
<math|\<delta\>\<gtr\>0> such that <math|0\<less\>\<delta\>\<less\>min(1,<frac|\<varepsilon\>|k>)>
then if <math|\<shortparallel\>x-y\<shortparallel\>\<less\>\<delta\>>
then <math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|\<shortparallel\>x<rsub|i>-y<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>(x-y)<rsub|i>\<shortparallel\><rsub|i>\<less\>\<delta\>>
and thus <math|\<shortparallel\>L(x)-L(y)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x-y)\<shortparallel\><rsub|Y>\<leqslant\>k.(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>-y<rsub|i>\<shortparallel\><rsub|i>\<leqslant\>k.\<delta\><rsup|n>>
(and as we can prove by induction from <math|0\<less\>\<delta\>\<less\>1>
it follows that <math|0\<less\>\<delta\><rsup|n>\<less\>1> we have that
<math|\<shortparallel\>L(x)-L(y)\<shortparallel\><rsub|Y>\<leqslant\>k.\<delta\>\<less\>\<varepsilon\>>
which proves by <reference|continuity in normed spaces> that <math|L> is
continuous.
</proof>
<\lemma>
<label|introduction of the multilinear norm>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>>a normed vector space
over <math|\<bbb-R\>(\<bbb-C\>)> and a continuous multilinear function
<math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
then if <math|A<rsub|ms>={k\<in\>\<bbb-R\><rsub|+>\|
\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<vdash\>
\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1
\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k}>
and <math|A<rsub|mr>={k\<in\>\<bbb-R\><rsub|+>\|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<succ\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>}>
then <math|A<rsub|mr>=A<rsub|ms>> and
<math|\<shortparallel\>L\<shortparallel\>=inf(A<rsub|ms>)=inf(A<rsub|mr>)>
exists and is finite (and of course <math|\<geqslant\>0>)
</lemma>
<\proof>
As the theorem is already proved in the linear case <math|(n=1)> we
proceed proving the theorem for <math|n\<gtr\>1.> First we prove that
<math|A<rsub|mr>=A<rsub|ms>> so\
<\enumerate>
<item>If <math|k\<in\>A<rsub|ms>> and
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> then we
have two cases\
<\enumerate>
<item><math|\<exists\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=0>
then <math|x<rsub|i>=0> and thus using <reference|multilinear
mappings and neutral element> <math|L(x)=0\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>0\<shortparallel\><rsub|Y>=0=k.0\<equallim\><rsub|<reference|general
product with a zero>>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x\<shortparallel\><rsub|i>>
<item><math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<neq\>0>
then if we define <math|y\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
by <math|y<rsub|i>=<frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>.x<rsub|i>>
then <math|\<shortparallel\>y\<shortparallel\><rsub|i>=><math|\<shortparallel\><frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>.x<rsub|i>\<shortparallel\><rsub|i>=1>
and thus <math|<frac|1|<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<equallim\><rsub|<reference|inverse
of product of many elements in a field>>(<big|prod><rsub|i=1><rsup|n><frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>)\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>(<big|prod><rsub|i=1><rsup|n><frac|1|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>).L(x)\<shortparallel\><rsub|Y>\<equallim\><rsub|<reference|mulilinearity
and coefficients>>\<shortparallel\>L(y)\<shortparallel\><rsub|Y>\<leqslant\>k\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<Rightarrow\>k\<in\>A<rsub|mr>>
</enumerate>
<item>If <math|k\<in\>A<rsub|mr>> and let
<math|x\<in\><big|prod><rsub|i={1,\<ldots\>,n}>X<rsub|i>\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1
\ \<forall\>i\<in\>I> then <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=k\<Rightarrow\>k\<in\>A<rsub|ms>>
</enumerate>
\ Now by <reference|continuitity and multilinearity>
<math|\<exists\>k\<gtr\>0> such that <math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
and thus <math|k\<in\>A<rsub|ms>\<neq\>\<emptyset\>> and
<math|A<rsub|ms>> is bounded below by <math|0> then by the lower bound
property of the real's we have the existence of
<math|inf(A<rsub|ms>)=inf(A<rsub|mr>)\<geqslant\>0>
</proof>
<\theorem>
<label|norm of a multilinear mapping>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of normed vector spaces over <math|\<bbb-R\>(\<bbb-C\>)>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> a normed vector space
over <math|\<bbb-R\>(\<bbb-C\>)> and continuous multilinear
<math|>function <math|L,L<rsub|1>,L<rsub|2>:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
then we have the following properties for
<math|\<shortparallel\>.\<shortparallel\>>
<\enumerate>
<item><math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<succ\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\><big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
<item><math|L<rsub|1>+L<rsub|2>> is continuous and
<math|\<shortparallel\>L<rsub|1>+L<rsub|2>\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>>
<item>If <math|L=<wide|0|\<bar\>>> (the <math|0> function) then
<math|\<shortparallel\>L\<shortparallel\>=0>
<item>If <math|\<shortparallel\>L\<shortparallel\>=0\<Rightarrow\>L=<wide|0|\<bar\>>>
<item><math|\<forall\>\<alpha\>\<in\>\<bbb-R\>(\<bbb-C\>)> we have
<math|\<alpha\>.L> \ is continuous and
<math|\|\<alpha\>\|\<shortparallel\>L\<shortparallel\>=\<shortparallel\>\<alpha\>.L\<shortparallel\>><math|>
</enumerate>
</theorem>
<\proof>
As this is already proved in the linear case (<math|n=1>) all that left
is proving the multilinear case <math|n\<gtr\>1>
<\enumerate>
<item>Let <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
then we have the following possibilities
<\enumerate>
<item><math|\<exists\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=0>
then <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>0\<shortparallel\><rsub|Y>=0=\<shortparallel\>L\<shortparallel\>.0=\<shortparallel\>L\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
<item><math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<neq\>0>
then assume that <math|\<shortparallel\>L\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<less\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>>
and as <math|<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<neq\>0>
we have <math|\<shortparallel\>L\<shortparallel\>\<less\><frac|1|<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>>
and by the definition of infimum there exists
<math|k\<in\>A<rsub|mr>> such that
<math|\<shortparallel\>L\<shortparallel\>\<leqslant\>k\<less\><frac|1|<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>k.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<less\><frac|1|<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<less\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>>
which is a contradiction so we must have
<math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
</enumerate>
<item>Let <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
then <math|\<shortparallel\>(L<rsub|1>+L<rsub|2>)(x)\<shortparallel\><rsub|Y>=\<shortparallel\>L<rsub|1>(x)+L<rsub|2>(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L<rsub|1>(x)\<shortparallel\><rsub|Y>+\<shortparallel\>L<rsub|2>(x)\<shortparallel\><rsub|Y>\<leqslant\><rsub|continuity
of L<rsub|1>,L<rsub|2>>(\<shortparallel\>L<rsub|1>\<shortparallel\><rsub|>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)+(\<shortparallel\>L<rsub|2>\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)=(\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>)<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
which by <reference|continuitity and multilinearity> means that
<math|L<rsub|1>+L<rsub|2>> is continuous and that
<math|\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>\<in\>A<rsub|mr>>
(the one for <math|L<rsub|1>+L<rsub|2>>) and thus
<math|\<shortparallel\>L<rsub|1>+L<rsub|2>\<shortparallel\>=inf(A<rsub|mr>)\<leqslant\>\<shortparallel\>L<rsub|1>\<shortparallel\>+\<shortparallel\>L<rsub|2>\<shortparallel\>>
<item>If <math|L=<wide|0|\<bar\>>> then
<math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we have <math|L(x)=0\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=0=0.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<Rightarrow\>0\<in\>A<rsub|mr>\<Rightarrow\>inf(A<rsub|mr>)=0\<Rightarrow\>\<shortparallel\>L\<shortparallel\>=0>
<item>If <math|\<shortparallel\>L\<shortparallel\>=0> then
<math|\<forall\>x\<in\><big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>0.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=0\<Rightarrow\>\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=0\<Rightarrow\>L(x)=0\<Rightarrow\>L=<wide|0|\<bar\>>>
<item>Let <math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
and <math|\<alpha\>\<in\>\<bbb-R\>(\<bbb-C\>)> we have
<math|\<shortparallel\>\<alpha\>.L(x)\<\|\|\><rsub|Y>=\|\<alpha\>\|.\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<leqslant\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
which by <reference|continuitity and multilinearity> proves that
<math|\<alpha\>.L> is continuous and also that
<math|\<shortparallel\>\<alpha\>.L\<shortparallel\>\<leqslant\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>>.\
Now assume that <math|\<shortparallel\>\<alpha\>.L\<shortparallel\>\<less\>\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>>
then we have the following possibilities\
<\enumerate>
<item><math|\<alpha\>=0> then <math|\<alpha\>.L=<wide|0|\<bar\>> and
thus ><math|\<shortparallel\>\<alpha\>.L\<shortparallel\><rsub|Y>=\<shortparallel\><wide|0|\<bar\>>\<shortparallel\>=0=0.\<shortparallel\>L\<shortparallel\>=\<alpha\>.\<shortparallel\>L\<shortparallel\><rsub|Y>>
a contradiction so <math|\<shortparallel\>\<alpha\>.L\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>>
<item><math|\<alpha\>\<neq\>0> then <math|\|\<alpha\>\<nvDash\>0> and
thus <math|<frac|1|\|\<alpha\>\|>\<shortparallel\>\<alpha\>.L\<shortparallel\><rsub|Y>\<less\>\<shortparallel\>L\<shortparallel\>>
then by the definition of the infimum
<math|\<exists\>k\<in\>A<rsub|mr>> such that
<math|<frac|1|\|\<alpha\>\|>\<shortparallel\>\<alpha\>.L\<shortparallel\>\<leqslant\>k\<less\>\<shortparallel\>L\<shortparallel\>>
and <math|\<forall\>x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we have <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>=<frac|\<shortparallel\>\<alpha\>.L(x)\<shortparallel\><rsub|Y>|\|\<alpha\>\|>\<leqslant\><frac|k|\|\<alpha\>\|>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<less\><frac|\<shortparallel\>L\<shortparallel\>|\|a\|>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<Rightarrow\>\<shortparallel\>L\<shortparallel\>\<leqslant\><frac|\<shortparallel\>\<alpha\>.L\<shortparallel\>|\|\<alpha\>\|>\<less\>\<shortparallel\>L\<shortparallel\>>
which is a contradiction so <math|\<shortparallel\>\<alpha\>.L\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>L\<shortparallel\>>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|vector space of multilinear continuous
mappings><index|<math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>><index|multilinear
continuous mapping>Let <math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>>
be a family of normed vector spaces over <math|\<bbb-R\>(\<bbb-C\>)>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> a normed vector space
over <math|\<bbb-R\>,\<bbb-C\>> then if
<math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)={L\|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
a multilinear continuous map<math|}\<subseteq\>Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
then <math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> is a subspace of
<math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> and is thus a vector space.
Furthermore <math|\<shortparallel\>\<shortparallel\>> (see
<reference|introduction of the multilinear norm>) forms a norm on
<math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> and is this a normed vector
space.
</theorem>
<\proof>
If <math|n=1> then <math|L(X<rsub|1>,\<ldots\>,X<rsub|1>;Y)=L(X<rsub|1>,Y)>
which is a normed space. If <math|n\<gtr\>1> then we proceed as follows.
Using <reference|vector space of multilinear mappings> and
<reference|subspace> we must prove that
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>(\<bbb-C\>),\<forall\>L<rsub|1>,L<rsub|2>\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
we have <math|\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
and this is trivially using <reference|norm of a multilinear mapping>.
Further using again <reference|norm of a multilinear mapping> we have
that <math|\<shortparallel\>\<shortparallel\>> has the property of a
norm.
</proof>
<\theorem>
<label|composition of continuous linear maps is continuous>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
and <math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be normed spaces
over <math|\<bbb-R\>(\<bbb-C\>)> then the composition function
<math|C<rsub|\<circ\>>:L(Y,Z)\<times\>L(X,Y)\<rightarrow\>L(X,Z)> defined
by <math|C<rsub|\<circ\>>(f,g)=f\<circ\>g> is multilinear and continuous
(if we consider <math|L(Y,Z),L(X,Y)> as normed spaces with the operator
norm, and the product topology on <math|L(Y,Z)\<times\>L(X,Y))>(so
<math|C<rsub|\<circ\>>\<in\>L(L(Y,Z)\<times\>L(X,Y),L(X,Z))> and
<math|\<shortparallel\>C<rsub|\<circ\>>\<shortparallel\>\<leqslant\>1>
</theorem>
<\proof>
\;
First note that the composition of continuous linear maps is again linear
and continuous, so <math|C<rsub|\<circ\>>(f,g)\<in\>L(X,Z)>\
Second multilinearity
<math|\<forall\>x\<in\>X\<succ\>(C<rsub|\<circ\>>(\<alpha\>.f<rsub|1>+\<beta\>.f<rsub|1>,g))(x)=(\<alpha\><rsub|>.f<rsub|1>+\<beta\>.f<rsub|2>)(g(x))=\<alpha\>.f<rsub|1>(g(x))+\<beta\>.f<rsub|2>(g(x))=(\<alpha\>.\<circ\>(f<rsub|1>,g))(x)+(\<beta\>.\<circ\>(f<rsub|2>,g))(x)=(\<alpha\>.\<circ\>(f<rsub|1>,g)+\<beta\>.\<circ\>(f<rsub|2>,g))(x)\<Rightarrow\>C<rsub|\<circ\>>(\<alpha\>.f<rsub|1>+\<beta\>.f<rsub|2>,g)=\<alpha\>.C<rsub|\<circ\>>(f<rsub|1>,g)+\<beta\>.C<rsub|\<circ\>>(f<rsub|2>,g)>
<math|\<forall\>x\<in\>X\<succ\>(C<rsub|\<circ\>>(f,\<alpha\>.g<rsub|1>+\<beta\>.g<rsub|2>))(x)=f((\<alpha\>.g<rsub|1>+\<beta\>.g<rsub|2>)(x))\<equallim\>f(\<alpha\>.g<rsub|1>(x)+\<beta\>.g<rsub|2>(x))\<equallim\><rsub|f
is linear>\<alpha\>.f(g<rsub|1>(x))+\<beta\>.f(g<rsub|1>(x))=\<alpha\>.C<rsub|\<circ\>>(f,g<rsub|1>)(x)+\<beta\>.C<rsub|\<circ\>>(f,g<rsub|2>)(x)=(\<alpha\>.C<rsub|\<circ\>>(f,g<rsub|1>)+\<beta\>.C<rsub|\<circ\>>(f,g<rsub|2>))(x)\<Rightarrow\>C<rsub|\<circ\>>(f,\<alpha\>.g<rsub|1>+\<beta\>.g<rsub|2>)=\<alpha\>.C<rsub|\<circ\>>(f,g<rsub|1>)+\<beta\>.(f,g<rsub|2>)>
Third to prove continuity note that <math|\<forall\>x\<in\>X> we have
<math|\<shortparallel\>(f\<circ\>g)(x)\<shortparallel\>f(g(x))\<shortparallel\><rsub|Z>\<leqslant\><rsub|continuitity
of f>\<shortparallel\>f\<shortparallel\>.\<shortparallel\>g(x)\<shortparallel\><rsub|Y>\<leqslant\><rsub|continuity
of g>\<shortparallel\>f\<shortparallel\>.\<shortparallel\>g\<shortparallel\>.\<shortparallel\>x\<shortparallel\><rsub|X><rsub|>>
and thus <math|\<shortparallel\>C<rsub|\<circ\>>(f,g)\<shortparallel\>=\<shortparallel\>f\<circ\>g\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>.\<shortparallel\>g\<shortparallel\>>
proving continuity of <math|C<rsub|\<circ\>>> and also that
<math|\<shortparallel\>C<rsub|\<circ\>>\<shortparallel\>\<leqslant\>1>
</proof>
<\definition>
<index|<math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y))>>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n>,n\<in\>\<bbb-N\><rsub|0>>
be a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\>(\<bbb-C\>)> then we define
<math|L(X<rsub|1>,\<ldots\>,L(X<rsub|n>,Y)\<ldots\>)> recursively as
follows
<\enumerate>
<item><math|n=1> then <math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|1>,Y)>
<item><math|n\<gtr\>1> then <math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|n>,L(X<rsub|n-1><rsub|>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>))>
</enumerate>
</definition>
<\example>
\;
<\enumerate>
<item><math|L(X<rsub|2>,L(X<rsub|1>,Y))=L(X<rsub|2>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
<item><math|L(X<rsub|3>,L(X<rsub|2>,L(X<rsub|1>,Y))=L(X<rsub|3>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
</enumerate>
</example>
<\theorem>
<label|L(X1,..L(Xn,Y)..) is Banach if Y is Banach>If
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
is a family of nnormed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> a Banach space then
<math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y),\<ldots\>)> is a Banach space
</theorem>
<\proof>
We prove this by induction
<\enumerate>
<item>Case <math|n=1> this follows at once from <reference|space of
linear continuous maps to a Banach space is Banach>
<item>Assume the theorem true for <math|n> then prove it to be true for
<math|n+1>. As <math|L(X<rsub|n+1>,\<ldots\>,L(X<rsub|1>,Y),\<ldots\>)=L(X<rsub|n+1>,L(X<rsub|n>,\<ldots\>,L(X,Y),\<ldots\>)>
which is Banach because again <reference|space of linear continuous
maps to a Banach space is Banach> and the induction hypothese.
</enumerate>
</proof>
<\definition>
<index|<math|f(x<rsub|m>:\<ldots\>:x<rsub|n>)>>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n>,n\<in\>\<bbb-N\><rsub|0>,n\<geqslant\>2,>be
a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\>(\<bbb-C\>)> then if <math|f\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
and <math|x=(x<rsub|1>,\<ldots\>,x<rsub|n>)=(x<rsub|i>)<rsub|i\<in\>{1,\<ldots\>,n}>\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
then given <math|1\<less\>m\<leqslant\>n> we define
<math|f(x<rsub|m>:\<ldots\>:x<rsub|n>)\<in\>L(X<rsub|m-1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
by recursion on <math|k=n-m+1\<Rightarrow\>m=n-k+1> as follows\
<\enumerate>
<item><math|k=1> (or <math|m=n>) then
<math|f(x<rsub|n-1+1>:\<ldots\>:x<rsub|n>)=f(x<rsub|n>:\<ldots\>:x<rsub|n>)\<equallim\><rsub|def>f(x<rsub|n>)\<in\>L(X<rsub|n-1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)<rsub|>=L(X<rsub|n-k,\<ldots\>,>L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>\
<item><math|n\<gtr\>k\<gtr\>1\<Rightarrow\>> then as we have already
defined <math|f(x<rsub|n-(k-1)+1>:\<ldots\>:x<rsub|n>)\<in\>L(X<rsub|n-(k-1)>,\<ldots\>,L(X<rsub|n>,Y)\<ldots\>)=L(X<rsub|n-k+1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)\<Rightarrow\>f(x<rsub|n-(k-1)+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k+1>)\<in\>L(X<rsub|n-k>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)
and we define then ><math|f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)\<equallim\><rsub|def>f(x<rsub|n-(k-1)+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k+1>)\<in\>L(X<rsub|n-k>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
or in terms of <math|m> we have <math|f(x<rsub|m>:\<ldots\>:x<rsub|n>)=f(x<rsub|m+1>:\<ldots\>:x<rsub|n>)(x<rsub|m>)\<in\>L(X<rsub|m-1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
<item><math|n=k> then <math|f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)=f(x<rsub|1>:\<ldots\>x<rsub|n>)\<equallim\><rsub|def>f(x<rsub|2>:\<ldots\>x<rsub|n>)(x<rsub|1>)\<in\>Y<rsub|>>
because by <math|1> and <math|2> we have
<math|f(x<rsub|2>:\<ldots\>:x<rsub|n>)\<in\>L(X<rsub|1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|1>,Y)>
</enumerate>
</definition>
<\remark>
The definition of <math|(x<rsub|n>:\<ldots\>:x<rsub|1>)> seems difficult
but the following example illustrates the definition let
<math|f\<in\>L(X<rsub|4>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|4>,L(X<rsub|3>,L(X<rsub|2>,L(X<rsub|1>,Y))))>
then we have if <math|(x<rsub|1>,x<rsub|2>,x<rsub|3>,x<rsub|4>)\<in\>X<rsub|1>\<times\>X<rsub|2>*\<times\>X<rsub|3>\<times\>X<rsub|4>>\
<\itemize>
<item><math|f(x<rsub|4>:\<ldots\>:x<rsub|4>)=f(x<rsub|4>)\<in\>L(X<rsub|3>,L(X<rsub|2>,L(X<rsub|1>,Y)))>
<item><math|f(x<rsub|3>:\<ldots\>:x<rsub|4>)=f(x<rsub|4>)(x<rsub|3>)\<in\>L(X<rsub|2>,L(X<rsub|1>,Y))>
<item><math|f(x<rsub|2>:\<ldots\>:x<rsub|4>)=f(x<rsub|4>)(x<rsub|3>)(x<rsub|2>)\<in\>L(X<rsub|1>,Y)>
<item><math|f(x<rsub|1>:\<ldots\>:x<rsub|4>)=f(x<rsub|4>)(x<rsub|3>)(x<rsub|2>)(x<rsub|1>)\<in\>Y>
</itemize>
</remark>
<\note>
<label|symmetry of multiple appliance>For <math|1\<less\>m\<leqslant\>n>
and <math|1\<leqslant\>k\<less\>m\<leqslant\>n> we have
<math|f(x<rsub|k>:\<ldots\>:x<rsub|n>)=f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|k>:\<ldots\>:x<rsub|m-1>)>
(more specifically if we take <math|k=1> then
<math|f(x<rsub|1>:\<ldots\>:x<rsub|n>)=f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|1>:\<ldots\>:x<rsub|m-1>)>
Another special case is by taking <math|m=n> then
<math|f(x<rsub|1>:\<ldots\>:x<rsub|n>)=f(x<rsub|n>)(x<rsub|1>:\<ldots\>:x<rsub|n-1>)>
</note>
<\proof>
We proof this by induction on <math|l=m-k> or <math|k=m-l>
<\enumerate>
<item><math|l=1> then <math|k=m-1> then
<math|f(x<rsub|k>:\<ldots\>:x<rsub|n>)=f(x<rsub|m-1>:\<ldots\>:x<rsub|n>)=f(x<rsub|m>:\<ldots\>x<rsub|n>)(x<rsub|m-1>)=f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|m-1>:\<ldots\>:x<rsub|m-1>)=f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|k>:\<ldots\>:x<rsub|m>-1)>
<item>Assume that the theorem is true for <math|l> or <math|k> then
prove it for <math|l+1> if <math|1\<leqslant\>l+1\<less\>m> or for
<math|k-1> if <math|1\<leqslant\>k-1\<less\>m> now by the induction
hypothesis we have <math|f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|k>:\<ldots\>:x<rsub|m-1>)=f(x<rsub|k>:\<ldots\>:x<rsub|n>)\<Rightarrow\>f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|k>:\<ldots\>:x<rsub|m-1>)(x<rsub|k-1>)=f(x<rsub|k>:\<ldots\>x<rsub|n>)(x<rsub|k-1>)\<Rightarrow\>f(x<rsub|m>:\<ldots\>:x<rsub|n>)(x<rsub|k-1>:\<ldots\>:x<rsub|m-2>)=f(x<rsub|k-1>:\<ldots\>:x<rsub|n>)>
proving the theorem for <math|k-1 (or l+1)><math|>
</enumerate>
</proof>
<\lemma>
<label|equality of multiple appliance>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n>,n\<in\>\<bbb-N\><rsub|0>
>be a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\>(\<bbb-C\>)> then if <math|f,g\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
is such that <math|\<forall\>(x<rsub|1>,\<ldots\>,x<rsub|n>)> we have
<math|f(x<rsub|1>:\<ldots\>x<rsub|n>)=g(x<rsub|1>:\<ldots\>:x<rsub|n>)>
then <math|f=g>
</lemma>
<\proof>
We prove by induction on if <math|1\<leqslant\>k\<leqslant\>n> it follows
from <math|f(x<rsub|1>:\<ldots\>x<rsub|n>)=g(x<rsub|1>:\<ldots\>x<rsub|n>)>
that <math|f(x<rsub|k>:\<ldots\>:x<rsub|n>)=g(x<rsub|k>:\<ldots\>:x<rsub|n>)>
<\enumerate>
<item>If <math|k=1> then <math|f(x<rsub|1>:\<ldots\>:x<rsub|n>)=g(x<rsub|1>:\<ldots\>:x<rsub|n>)>
which is trivially true
<item>Assume it is true for <math|k> and prove it for
<math|0\<leqslant\>k+1\<leqslant\>n> then by the induction hypothesis
\ <math|f(x<rsub|k>,\<ldots\>,x<rsub|n>)=g(x<rsub|k>,\<ldots\>,x<rsub|n>)\<Rightarrow\>f(x<rsub|k+1>,\<ldots\>,x<rsub|n>)(x<rsub|k>)=g(x<rsub|k+1>,\<ldots\>,x<rsub|n>)(x<rsub|k>)><math|>
which means because <math|x<rsub|k>> is chosen arbitrarily, that
<math|f(x<rsub|k+1>,\<ldots\>,x<rsub|n>)=g(x<rsub|k+1>,\<ldots\>,x<rsub|n>)>
proving the recursive case
</enumerate>
We conclude the proof by noting that for <math|>k=n we have proved that
<math|f(x<rsub|n>)=f(x<rsub|n>:\<ldots\>:x<rsub|n>)=g(x<rsub|n>:\<ldots\>:x<rsub|n>)=g(x<rsub|n>)>
which means that <math|f=g>
</proof>
<\lemma>
<label|sum, scalair product of multiple appliance>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n>,n\<in\>\<bbb-N\><rsub|0>
>be a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\>(\<bbb-C\>)> then <math|\<forall\>f,g\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
and <math|\<forall\>\<alpha\>\<in\>\<bbb-R\>(\<bbb-C\>)> we have
<math|(f+g)(x<rsub|1>:\<ldots\>:x<rsub|n>)=f(x<rsub|1>:\<ldots\>:x<rsub|n>)+g(x<rsub|1>:\<ldots\>:x<rsub|n>)>
and <math|(\<alpha\>.f)(x<rsub|1>:\<ldots\>:x<rsub|n>)=\<alpha\>.f(x<rsub|1>:\<ldots\>:x<rsub|n>)>
</lemma>
<\proof>
We prove this by proving first that <math|(f+g)(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)=f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)+g(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)>
and <math|(\<alpha\>.f)(x<rsub|n-k+1>:\<ldots\>x<rsub|n>)> we do this by
induction on <math|m=n-k+1>
<\enumerate>
<item><math|k=1> then <math|(f+g)(x<rsub|n>:\<ldots\>:x<rsub|n>)=(f+g)(x<rsub|n>)=f(x<rsub|n>)+g(x<rsub|n>)=f(x<rsub|n>:\<ldots\>:x<rsub|n>)+g(x<rsub|n>:\<ldots\>:x<rsub|n>)>
and also <math|(\<alpha\>.f)(x<rsub|n>:\<ldots\>:x<rsub|n>)=(\<alpha\>.f)(x<rsub|n>)=\<alpha\>.f(x<rsub|n>)=\<alpha\>.f(x<rsub|n>:\<ldots\>:x<rsub|n>)>
<item>Assume the lemma true for <math|k> then prove it for
<math|1\<leqslant\>k+1\<leqslant\>n> then
<math|(f+g)(x<rsub|n-(k+1)+1>:\<ldots\>:x<rsub|n>)=(f+g)(x<rsub|n-k>:\<ldots\>x<rsub|n>)=(f+g)(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>)=(f(x<rsub|n-k+1>:\<ldots\>x<rsub|n>)+g(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>)=f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>)+g(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>)=f(x<rsub|n-k>:\<ldots\>x<rsub|n>)+g(x<rsub|n-k>:\<ldots\>:x<rsub|n>)=f(x<rsub|n-(k+1)+1>:\<ldots\>x<rsub|n>)+g(x<rsub|n-(k+1)+1>:\<ldots\>:x<rsub|n>)>
and likewise <math|(\<alpha\>.f)(x<rsub|n-(k+1)>:\<ldots\>:x<rsub|n>)=(\<alpha\>.f)(x<rsub|n-k>:\<ldots\>:x<rsub|n>)=(\<alpha\>.f)(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>)=(\<alpha\>.(f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)))(x<rsub|n-k>)=\<alpha\>.(f(x<rsub|n-k+1>:\<ldots\>:x<rsub|n>)(x<rsub|n-k>))=\<alpha\>.f(x<rsub|n-k>:\<ldots\>:x<rsub|n>)=\<alpha\>.f(x<rsub|n-(k+1)+1>:\<ldots\>:x<rsub|n>)>
</enumerate>
The lemma follows then from the case <math|k=n>
\
</proof>
<\theorem>
Given <math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\> (\<bbb-C\>)> then we define the function
<math|\<cal-P\>:L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)\<rightarrow\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
as follows , given a <math|f\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
define then <math|\<cal-P\>(f)\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>)> by
<math|\<cal-P\>(f)(x)=f(x<rsub|1>:\<ldots\>:x<rsub|m>)> where
<math|x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>,
also <math|\<shortparallel\>\<cal-P\>(f)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>>
</theorem>
<\proof>
We must of course prove that <math|\<cal-P\>(f)> is indeed multilinear
and continuous. To prove this let <math|y<rsub|i>=x<rsup|(m,n)>(t)(i)>
and note that for <math|1\<less\>m\<leqslant\>n>
<\enumerate>
<item><math|\<cal-P\>(f)(x<rsup|(m,n)>(t))=\<cal-P\>(f(y<rsub|1>,\<ldots\>,y<rsub|n>))=f(y<rsub|1>:\<ldots\>:y<rsub|n>)=f(y<rsub|m>:\<ldots\>:y<rsub|n>)(y<rsub|1>:\<ldots\>:y<rsub|m-1>)<rsub|>>
so that we have the following cases\
<\enumerate>
<item><math|m=n\<Rightarrow\>f(y<rsub|1>:\<ldots\>:y<rsub|n>)=f(y<rsub|n>:\<ldots\>:y<rsub|n>)(y<rsub|1>:\<ldots\>:y<rsub|m-1>)=f(y<rsub|n>)(y<rsub|1>:\<ldots\>:y<rsub|n-1>)=f(t)(x<rsub|1>:\<ldots\>:x<rsub|n-1>)>
which is linear in <math|t> because
<math|f\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|n>,L(x<rsub|n-1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
<item><math|1\<less\>m\<less\>n\<Rightarrow\>f(y<rsub|1>:\<ldots\>:y<rsub|n>)=f(y<rsub|m>:\<ldots\>:y<rsub|n>)(y<rsub|1>:\<ldots\>:y<rsub|m-1>)=f(y<rsub|m+1>:\<ldots\>:y<rsub|n>)(y<rsub|m>)(y<rsub|1>:\<ldots\>:y<rsub|m-1>)=f(x<rsub|m+1>:\<ldots\>:x<rsub|n>)(t)(x<rsub|1>:\<ldots\>:x<rsub|m>)>
which is linear in <math|t> because
<math|f(x<rsub|m+1>:\<ldots\>:x<rsub|n>)\<in\>L(X<rsub|m>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|m>,L(X<rsub|m-1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
</enumerate>
<item>Finally if <math|m=1> then <math|\<cal-P\>(f)(x<rsup|(m,n)>(t))=\<cal-P\>(f(y<rsub|1>,\<ldots\>,y<rsub|n>))=f(y<rsub|1>:\<ldots\>:y<rsub|n>)=f(y<rsub|2>\<ldots\>y<rsub|n>)(y<rsub|1>)=f(x<rsub|2>:\<ldots\>:x<rsub|n>)(t)>
which is linear in <math|t> because
<math|f(x<rsub|2>:\<ldots\>:x<rsub|n>)\<in\>L(X<rsub|1>,Y)>
</enumerate>
which proves that <math|\<cal-P\>(f)\<circ\>x<rsup|(m,n)>> is linear and
thus that <math|\<cal-P\>(f)> is multilinear.
Now to prove that <math|\<cal-P\>(f)> is continuous let assume that
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> and prove
that for <math|1\<less\>m\<leqslant\>n>
<math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|m>:\<ldots\>:x<rsub|n>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,m-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
we prove this by induction on <math|m>
<\enumerate>
<item>Let <math|m=1> then <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>=\<shortparallel\>f(x<rsub|2>:\<ldots\>:x<rsub|n>)(x<rsub|1>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|2>:\<ldots\>:x<rsub|n>)\<shortparallel\>.\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|1>>
<item>Assume the theorem proved for <math|m> and prove it for
<math|m+1\<leqslant\>n> then <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|m>:\<ldots\>:x<rsub|n>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,m-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>f(x<rsub|m+1>:\<ldots\>:x<rsub|n>)(x<rsub|m>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,m-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<leqslant\>\<shortparallel\>f(x<rsub|m+1>:\<ldots\>:x<rsub|n>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,m}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
</enumerate>
So by taking <math|m=n> we get <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|n>:\<ldots\>:x<rsub|n>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>f(x<rsub|n>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<leqslant\>\<shortparallel\>f\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\>>
proving that <math|\<shortparallel\>\<cal-P\>(f)(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>.<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\>>
and thus that <math|\<cal-P\>(f)> is continuous and also that
<math|\<shortparallel\>\<cal-P\>(f)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>>
</proof>
<\lemma>
<label|norm of muliple appliance>Given
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>},
Y,\<shortparallel\>\<shortparallel\>> normed spaces over
<math|\<bbb-R\>(\<bbb-C\>)> and <math|f\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>Y)>
then <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\><big|prod><rsub|i=1><rsup|n>\<shortparallel\>x\<shortparallel\><rsub|i>>
</lemma>
<\proof>
First if <math|n=1> then the lemma is trivial true
(<math|f\<in\>L(X<rsub|1>,Y))> so lets take <math|1\<less\>n> and prove
first that if <math|1\<leqslant\>k\<less\>n> then
<math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|k+1>:\<ldots\>:x<rsub|n>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|k>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>,
this is easily done by recursion of <math|k>
<\enumerate>
<item><math|k=1> then <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>=\<shortparallel\>f(x<rsub|2>,\<ldots\>,x<rsub|n>)(x<rsub|1>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|2><rprime|'>:\<ldots\>:x<rsub|n>)\<shortparallel\>.\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|1>=\<shortparallel\>f(x<rsub|2>:\<ldots\>:x<rsub|n>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|1>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i><rsub|>>
<item>Assume now trueness for <math|k> and prove it is true for
<math|1\<leqslant\>k+1\<less\>n> then
<math|\<shortparallel\>f(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|k+1>:\<ldots\>:x<rsub|n>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|k>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>f(x<rsub|k+2>:\<ldots\>:x<rsub|n>)(x<rsub|k+1>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|k>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<leqslant\>\<shortparallel\>f(x<rsub|k+1>:\<ldots\>:x<rsub|n>)\<shortparallel\>.\<shortparallel\>x<rsub|k+1>\<shortparallel\>.<big|prod><rsub|i=1><rsup|k>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>f(x<rsub|(k+1)+1>:\<ldots\>:x<rsub|n>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|k>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
</enumerate>
</proof>
So using <math|k=n-1> we find <math|\<shortparallel\>f(x<rsub|1>:\<ldots\>x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|n>)\<shortparallel\>.<big|prod><rsub|i=1><rsup|n-1>\<leqslant\>\<shortparallel\>f\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
proving the lemma
<\lemma>
<label|n-multilinear map becomes linear map to n-1 multilinear map>Given
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,
Y,\<shortparallel\>\<shortparallel\>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
be normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> then for
<math|x\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> and
<math|f\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> then
<math|f<rsup|\<ast\>>:X<rsub|n>\<rightarrow\>L(X<rsub|1>,\<ldots\>,X<rsub|n-1>;Y)>
defined by <math|f<rsup|\<ast\>>(x)=f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)>
is a element of <math|L(X<rsub|n>,L(X<rsub|1>,\<ldots\>,X<rsub|n-1>;Y)>
and <math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>=\<shortparallel\>f\<shortparallel\>>
</lemma>
<\proof>
\;
First we prove that <math|f<rsup|\<ast\>>(x)> is indeed a element of
<math|L(X<rsub|1>,\<ldots\>,X<rsub|n>)>, by <reference|n multilinear
mapping yields n-1 multilinear map> we have that
<math|f<rsup|\<ast\>>(x)> is multilinear and
<math|\<shortparallel\>f<rsup|\<ast\>>(x)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)\<shortparallel\>=\<shortparallel\>f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,x)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)\<shortparallel\>=\<shortparallel\>f(x<rsub|1>,\<ldots\>,x<rsub|n-1>,x)\<shortparallel\>\<leqslant\>(\<shortparallel\>f\<shortparallel\>.\<shortparallel\>x\<shortparallel\>).<big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i><rsub|>>so
<math|f*<rsup|\<ast\>>(x)> is indeed continuous and
<math|\<shortparallel\>f<rsup|\<ast\>>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>.\<shortparallel\>x\<shortparallel\>>
Second we prove linearity of <math|f<rsup|\<ast\>>> let
<math|(x<rsub|1>,\<ldots\>,x<rsub|n-1>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>X<rsub|i>>
then <math|f*<rsup|\<ast\>>(\<alpha\>.s+\<beta\>.t)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)=f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,\<alpha\>.s+\<beta\>.t)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)=f(x<rsub|1>,\<ldots\>,x<rsub|n-1>,\<alpha\>.s+\<beta\>.t)=f(x<rsub|1>,\<ldots\>,x<rsub|n-1>,\<ast\>)(\<alpha\>.s+\<beta\>.t)\<equallim\><rsub|multilinearity
of f>\<alpha\>.f(x<rsub|1>,\<ldots\>,x<rsub|n-1>,s)+\<beta\>.f(x<rsub|1>,\<ldots\>,x<rsub|n-1>,t)=\<alpha\>.f(*\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,s)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)+\<beta\>.f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n-1>,t)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)=(\<alpha\>.f<rsup|\<ast\>>(s)+\<beta\>.f<rsup|\<ast\>>(t))(x<rsub|1>,\<ldots\>,x<rsub|n-1>)\<Rightarrow\>f<rsup|\<ast\>>(\<alpha\>.s+\<beta\>.t)=\<alpha\>.f<rsup|\<ast\>>(s)+\<beta\>.f<rsup|\<ast\>>(t)>
proving linearity. Now for continuity as we have already proved that
<math|\<shortparallel\>f<rsup|\<ast\>>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>.\<shortparallel\>x\<shortparallel\>>
we find that <math|f<rsup|\<ast\>>> is indeed continuous and that <math|>
<math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>>.\
Now to prove that <math|\<shortparallel\>f\<shortparallel\>=\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>>,
let <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)> be such that
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> with
<math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1> we have
<math|\<shortparallel\>f(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=\<shortparallel\>f<rsup|\<ast\>>(x<rsub|n>)(x<rsub|1>,\<ldots\>,x<rsub|n-1>)\<shortparallel\>\<leqslant\>\<shortparallel\>f<rsup|\<ast\>>(x<rsub|n><rsup|>)\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,n-1}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>f<rsup|\<ast\>>(x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>\<shortparallel\>x<rsub|n>\<shortparallel\><rsub|n>=\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>>
and thus <math|\<shortparallel\>f\<shortparallel\>\<leqslant\>\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>>
and this together with <math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>>
gives <math|\<shortparallel\>f\<shortparallel\>=\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>>
</proof>
<\theorem>
<label|equivalence of multilinear and linear continuous
mappings><index|<math|\<cal-P\>>>Given
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,n\<in\>\<bbb-N\><rsub|0>,n\<gtr\>1>
a family of normed spaces over <math|\<bbb-R\>(\<bbb-C\>)> and
<math|Y,\<shortparallel\>\<shortparallel\>> also a normed space over
<math|\<bbb-R\> (\<bbb-C\>)> then the function
<math|\<cal-P\>:L(X<rsub|n>,\<ldots\>.,L(X<rsub|1>,Y),\<ldots\>)\<rightarrow\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>
is a norm preserving isomorphism
</theorem>
<\proof>
First we prove that <math|\<cal-P\>> is bijective\
<\enumerate>
<item>injectivity if <math|\<cal-P\>(f)=\<cal-P\>(g)\<Rightarrow\>\<forall\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
we have then <math|f(x<rsub|1>:\<ldots\>:x<rsub|n>)=g(x<rsub|1>:\<ldots\>:x<rsub|n>)>
this follows from one of the previous lemmas (<reference|equality of
multiple appliance>)
<item>surjectivity, let <math|f\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>:Y)>
we prove surjectivity (the existence of a <math|f<rprime|'>> such that
<math|\<cal-P\>(f<rprime|'>)=f> and
<math|\<shortparallel\>f\<shortparallel\>=\<shortparallel\>f<rprime|'>\<shortparallel\>>)
by induction as follows
<\enumerate>
<item><math|n=2> then we use a previous lemma
(<reference|n-multilinear map becomes linear map to n-1 multilinear
map>) to prove the existence of a
<math|f<rsup|\<ast\>>\<in\>L(X<rsub|2>,L(X<rsub|1>;Y))=L(X<rsub|2>,L(X<rsub|1>,Y))>
with <math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>=\<shortparallel\>f\<shortparallel\>>
and <math|f<rsup|*\<ast\>>(x)=f(\<ast\><rsub|1>,x)> so that
<math|\<cal-P\>(f<rsup|\<ast\>>)(x<rsub|1>,x<rsub|2>)=f<rsup|\<ast\>>(x<rsub|2>)(x<rsub|1>)=f(\<ast\><rsub|1>,x<rsub|2>)(x<rsub|1>)=f(x<rsub|1>,x<rsub|2>)\<Rightarrow\>\<cal-P\>(f<rsup|\<ast\>><rsup|*>)=f>
and thus <math|f<rsup|\<ast\>>> is the searched for
<math|f<rprime|'>>, note also that
<math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>=\<shortparallel\>f<rprime|'>\<shortparallel\>=\<shortparallel\>f\<shortparallel\>=\<shortparallel\>\<cal-P\>(f)\<shortparallel\>>
<item>Assume the theorem is true for <math|n> and prove it
<math|n=n+1> then given a <math|f\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n+1>)>
we use the previous lemma (<reference|n-multilinear map becomes
linear map to n-1 multilinear map>) to find a
<math|f<rsup|\<ast\>>\<in\>L(X<rsub|n+1>,L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y))>
such that given <math|x\<in\>X<rsub|n+1>>
<math|f<rsup|\<ast\>>(x)=f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,x)>
and <math|\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>=\<shortparallel\>f\<shortparallel\>>.
Now using our induction hypothesis on
<math|f<rsup|\<ast\>>(x)\<in\>L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> we
find a <math|f<rsup|<rprime|'>>(x)=(f<rsup|\<ast\>>(x))<rprime|'>\<in\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
with <math|\<shortparallel\>f<rsup|<rprime|'>>(x)\<shortparallel\>=\<shortparallel\>(f<rsup|\<ast\>>(x))<rprime|'>\<shortparallel\>=\<shortparallel\>f<rsup|\<ast\>>(x)\<shortparallel\>>
and <math|\<cal-P\>(f<rsup|\<upl\>>(x))=f<rsup|\<ast\>>(x)> or
<math|><math|f<rsup|\<ast\>>(x)(x<rsub|1>,\<ldots\>,x<rsub|n>)=f<rsup|<rprime|'>>(x)(x<rsub|1>:\<ldots\>:x<rsub|n>)>.
Then define <math|f<rprime|'>:X<rsub|n+1>\<rightarrow\>L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
by <math|x\<rightarrow\>f<rprime|'>(x)> then
<math|f<rprime|'>(\<alpha\>.x+\<beta\>.y)(x<rsub|1>:\<ldots\>:x<rsub|n>)=f<rsup|\<ast\>>(a.x+\<beta\>.y)(x<rsub|1>,\<ldots\>,x<rsub|n>)=f(x<rsub|1>,\<ldots\>,x<rsub|n>,\<alpha\>.x+\<beta\>.y)=f(x<rsub|1>,\<ldots\>,x<rsub|n>,\<ast\>)(\<alpha\>.x+\<beta\>.y)\<equallim\><rsub|multilinearity
of f>\<alpha\>.f(x<rsub|1>,\<ldots\>,x<rsub|n>,\<ast\>)(x)+\<beta\>.f(x<rsub|1>,\<ldots\>,x<rsub|n>,\<ast\>)(y)=\<alpha\>.f(x<rsub|1>,\<ldots\>,x<rsub|n>,x)+\<beta\>.f(x<rsub|1,>\<ldots\>,x<rsub|n>,y)=\<alpha\>.f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,x)(x<rsub|1>:\<ldots\>:x<rsub|n>)+\<beta\>.f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,y)(x<rsub|1>:\<ldots\>:x<rsub|n>)=\<alpha\>.f<rprime|'>(x)(x<rsub|1><rsup|>:\<ldots\>:x<rsub|n>)+\<beta\>.f<rprime|'>(y)(x<rsub|1>:\<ldots\>:x<rsub|n>)\<equallim\><rsub|<reference|sum,
scalair product of multiple appliance>>(\<alpha\>.f<rprime|'>(x)+\<beta\>.f<rprime|'>(y))(x<rsub|1>:\<ldots\>:x<rsub|n>)\<Rightarrowlim\><rsub|<reference|equality
of multiple appliance>>f<rprime|'>(\<alpha\>.x+\<beta\>.y)=\<alpha\>.f<rprime|'>(x)+\<beta\>.f<rprime|'>(y)>
proving the linearity of <math|f<rprime|'>>. To prove continuity of
<math|f<rprime|'>> note that <math|\<shortparallel\>f<rprime|'>(x)\<shortparallel\>=\<shortparallel\>f<rsup|\<ast\>>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>f<rsup|\<ast\>>\<shortparallel\>\<shortparallel\>x\<shortparallel\>=\<shortparallel\>f\<shortparallel\>.\<shortparallel\>x\<shortparallel\>>
proving that <math|f<rprime|'>> is continuous and thus together with
linearity we have that <math|f<rprime|'>\<in\>L(X<rsub|n+1>,L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>))=L(X<rsub|n+1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>.
Further we have also that <math|\<shortparallel\>f<rprime|'>\<shortparallel\>\<leqslant\>\<shortparallel\>f\<shortparallel\>.>
Now note that<math|f(x<rsub|1>,\<ldots\>,x<rsub|n+1>)=f(\<ast\><rsub|1>,\<ldots\>,\<ast\><rsub|n>,x<rsub|n+1>)(x<rsub|1>,\<ldots\>,x<rsub|n>)=f<rsup|\<ast\>>(x<rsub|n+1>)(x<rsub|1>,\<ldots\>,x<rsub|n>)=f<rprime|'>(x<rsub|n+1>)(x<rsub|1>:\<ldots\>:x<rsub|n>)\<equallim\><rsub|<reference|symmetry
of multiple appliance>>f<rprime|'>(x<rsub|1>:\<ldots\>:x<rsub|n+1>)>
proving that <math|\<cal-P\>(f<rprime|'>)=f>. Finally to prove that
<math|\<shortparallel\>f<rprime|'>\<shortparallel\>=\<shortparallel\>f\<shortparallel\>>
note if we take <math|(x<rsub|1>,\<ldots\>,x<rsub|n+1>)> with
<math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1>
(<math|i\<in\>{1,\<ldots\>,n+1})> then
<math|\<shortparallel\>f(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=\<shortparallel\>f<rprime|'>(x<rsub|1>:\<ldots\>:x<rsub|n>)\<shortparallel\>\<leqslant\><rsub|<reference|norm
of muliple appliance>>\<shortparallel\>f<rprime|'>\<shortparallel\>.<big|prod><rsub|i=1><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<Rightarrow\>\<shortparallel\>f\<shortparallel\>\<leqslant\>\<shortparallel\>f(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>f<rprime|'>\<shortparallel\>\<Rightarrow\>\<shortparallel\>f\<shortparallel\>=\<shortparallel\>f<rprime|'>\<shortparallel\>>
</enumerate>
</enumerate>
The fact that <math|\<cal-P\>> is norm preserving follows from (2) for if
<math|\<cal-P\>(f)=g> then there exists a <math|f<rprime|'>> such that
<math|\<cal-P\>(f<rprime|'>)=g> and <math|\<shortparallel\>f<rprime|'>\<shortparallel\>=\<shortparallel\>g\<\|\|\>>
but because of injectivity we have <math|f<rprime|'>=g> so
<math|\<shortparallel\>g\<shortparallel\>=\<shortparallel\>f\<shortparallel\>>
or <math|\<shortparallel\>\<cal-P\>(f)\<shortparallel\>=\<shortparallel\>g\<shortparallel\>>
Now to prove that <math|\<cal-P\>> is a linear and thus a isomorphism,
take <math|\<cal-P\>(\<alpha\>.f+\<beta\>.g)(x<rsub|1>,\<ldots\>,x<rsub|n>)=(\<alpha\>.f+\<beta\>.g)(x<rsub|1>:\<ldots\>:x<rsub|n>)\<equallim\><rsub|<reference|sum,
scalair product of multiple appliance>>\<alpha\>.f(x<rsub|1>:\<ldots\>:x<rsub|n>)+\<beta\>.g(x<rsub|1>:\<ldots\>:x<rsub|n>)=\<alpha\>.\<cal-P\>(f)(x<rsub|1>,\<ldots\>,x<rsub|n>)+\<beta\>.\<cal-P\>(g)(x<rsub|1>,\<ldots\>,x<rsub|n>)=(\<alpha\>.\<cal-P\>(f)+\<beta\>.\<cal-P\>(g))(x<rsub|1>,\<ldots\>,x<rsub|n>)\<Rightarrow\>\<cal-P\>(\<alpha\>.f+\<beta\>.g)=\<alpha\>.\<cal-P\>(f)+\<beta\>.\<cal-P\>(g)>
</proof>
<\theorem>
<label|linear open mappings>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be metric vector
spaces on <math|\<bbb-K\>> then a linear function
<math|L:X\<rightarrow\>Y> is open if and only if \ there exists a
<math|\<delta\>\<gtr\>0> such that <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<delta\>)\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,1))>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>This is trivial using <reference|open
mapping in metric spaces>
<item><math|\<Leftarrow\>>Using <reference|open mapping in metric
spaces> we only have to prove that <math|\<forall\>x\<in\>X,\<delta\>\<gtr\>0>
we have the existence of a <math|\<delta\><rprime|'>\<gtr\>0> such that
<math|L(x)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(L(x),\<delta\><rprime|'>)\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(x,\<delta\>))>.
Now as we now by the hypothesis that there exists a
<math|\<delta\><rprime|''>\<gtr\>0> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<delta\><rprime|''>)\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,1))>
then\
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(L(x),<frac|\<delta\><rprime|''>|\<delta\>>)>|<cell|\<Rightarrow\>>|<cell|\<shortparallel\>y-L(x)\<shortparallel\><rsub|Y>\<less\><frac|\<delta\><rprime|''>|\<delta\>>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<shortparallel\>\<delta\>(y-L(x)\<shortparallel\><rsub|Y>\<less\>\<delta\><rprime|''>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<delta\>(y-L(x))\<in\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,1))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x<rsub|0>\<vdash\>\<shortparallel\>x<rsub|0>\<shortparallel\><rsub|X>\<less\>1
and L(x<rsub|0>)=\<delta\>(y-L(x))>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x<rsub|0>\<vdash\>\<shortparallel\>x<rsub|0>\<shortparallel\><rsub|X>\<less\>1
and y=<frac|L(x<rsub|0>)|\<delta\>>+L(x), take now
x<rsub|1>=<frac|x<rsub|0>|\<delta\>>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x<rsub|1>\<vdash\>\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|X>\<less\>\<delta\>
and y=L(x<rsub|1>)+L(x)=L(x<rsub|1>+x) take now
x<rsub|2>=x<rsub|1>+x>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<exists\>x<rsub|2>\<vdash\>\<shortparallel\>x<rsub|2>-x\<shortparallel\>\<less\>\<delta\>
and y=L(x<rsub|2>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|y\<in\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(x,\<delta\>))>>>>
</eqnarray*>
Proving our theorem
</enumerate>
</proof>
\;
<subsection|Separation>
<\definition>
<index|Hausdorff space>A topological space <math|X,\<cal-T\>> is called
Hausdorff if <math|\<forall\>x,y\<in\>X\<succ\>x\<neq\>y> we have
<math|\<exists\>U,V\<in\>\<cal-T\>> such that <math|x\<in\>U,y\<in\>V>
and <math|U<big|cap>V=\<emptyset\>>
</definition>
<\example>
<label|metric space is hausdorf>A metric space <math|X,d> is Hausdorff in
the metric topology
</example>
<\proof>
Let <math|x,y\<in\>X> such that <math|x\<neq\>y\<Rightarrow\>\<varepsilon\>=d(x,y)\<gtr\>0>
take then the open sets <math|x\<in\>B<rsub|d>(x\<comma\><frac|\<varepsilon\>|2>),y\<in\>B<rsub|d>(y,<frac|\<varepsilon\>|2>)
and let ><math|z\<in\>B<rsub|d>(x,<frac|\<varepsilon\>|2>)<big|cap>B<rsub|d>(y,<frac|\<varepsilon\>|2>)\<Rightarrow\>\<varepsilon\>=d(x,y)\<leqslant\>d(x,z)+d(y,z)\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
a contradiction so <math|B<rsub|d>(x,<frac|\<varepsilon\>|2>)<big|cap>B<rsub|d>(y,<frac|\<varepsilon\>|2>)=\<emptyset\>>
</proof>
<\theorem>
<label|every finite subset of a hausdorf space is closed>Every finite
subset of a Hausdorff space is closed
</theorem>
<\proof>
First we prove that <math|{x}> is closed or equivalent that
<math|X<mid|\\>{x}> is open, so let <math|y\<in\>X<mid|\\>{x}\<Rightarrow\>y\<neq\>x\<Rightarrow\>\<exists\>U<rsub|x>,U<rsub|y>>
open with <math|x\<in\>U<rsub|x>,y\<in\>U<rsub|y>> and
<math|U<rsub|x><big|cap>U<rsub|y>=\<emptyset\>\<Rightarrow\>x\<nin\>U<rsub|y>\<Rightarrow\>U<rsub|y>\<in\>X<mid|\\>{x}\<Rightarrow\>X<mid|\\>{x}>
is open. The rest follows from the fact that a finite union of closed
sets is closed. <math|>
</proof>
<\definition>
<index|regular space>A topological space <math|X,\<cal-T\>> is called
regular if for every closed set <math|A> and
<math|\<forall\>x\<in\>X\<vdash\>x\<nin\>A> there exists disjoint sets
<math|U,V\<in\>\<cal-T\>> with <math|x\<in\>U,A\<subseteq\>V> and
<math|U<big|cap>V=\<emptyset\>>
</definition>
<\theorem>
<label|a metric space is regular>A metric space <math|X,d> is regular
</theorem>
<\proof>
Let <math|A> be a closed set and <math|x\<nin\>A> then there exists a
<math|U> open <math|x\<in\>U> with <math|U<big|cap>A=\<emptyset\>> and
thus there exists a <math|\<delta\>\<gtr\>0> such that
<math|B<rsub|d>(x,\<delta\>)\<subseteq\>U> and thus
<math|B<rsub|d>(x,\<delta\>)<big|cap>A=\<emptyset\>>. Then if
<math|a\<in\>a> we have if <math|z\<in\>B<rsub|d>(x,<frac|\<delta\>|2>)<big|cap>B<rsub|d>(a,<frac|\<delta\>|2>)\<Rightarrow\>d(x,z)\<less\><frac|\<delta\>|2>,d(z,a)\<less\><frac|\<delta\>|2>\<Rightarrow\>d(x,a)\<leqslant\>d(x,z)+d(z,a)\<less\><frac|\<delta\>|2>+<frac|\<delta\>|2>=\<delta\>\<Rightarrow\>a\<in\>B<rsub|d>(x,\<delta\>)\<Rightarrow\>B<rsub|d>(x,\<delta\>)<big|cap>A\<neq\>\<emptyset\>>
giving a contradiction so <math|B<rsub|d>(x,<frac|\<delta\>|2>)<big|cap>B<rsub|\<delta\>>(a,<frac|\<delta\>|2>)=\<emptyset\>>
and thus <math|A\<subseteq\><big|cup><rsub|a\<in\>A>B<rsub|d>(a,<frac|\<delta\>|2>)>
a open set that does not intersect the open set
<math|B<rsub|d>(x,<frac|\<delta\>|2>)>
</proof>
<\definition>
<index|normal space>A topological space <math|X,\<cal-T\>> is called
normal if for every pair of disjoint closed sets <math|A,B> there exists
a pair of disjoint open sets <math|U,V> with
<math|A\<subseteq\>U,B\<subseteq\>V>
</definition>
<\theorem>
A normal space where every singleton is closed is regular, a regular
space where every singleton is closed is Hausdorff
</theorem>
<\proof>
\;
Let <math|X> be normal and let <math|x\<in\>X> and let <math|A> be closed
and <math|x\<in\>A\<Rightarrow\>{x}> and <math|A> are closed sets and
thus there exists disjoint open sets <math|U,V> with
<math|x\<in\>{x}\<subseteq\>U,A\<subseteq\>V>.
Let <math|X> be regular and <math|x,y\<in\>X> with
<math|x\<neq\>y\<Rightarrow\>y\<nin\>{x}> which is closed
<math|\<Rightarrow\>> there exists disjoint open sets <math|U,V> with
<math|x\<in\>{x}\<subseteq\>U> and <math|y\<in\>V>
</proof>
<\definition>
<index|neighborhood><index|open neighborhood>Let <math|X> be a
topological space and <math|x\<in\>X> then a set <math|A> is a
neighborhood of <math|x> iff there exists a open set <math|U> with
<math|x\<in\>U\<subseteq\>A>. If <math|A> is open then it is called a
open neighborhood
</definition>
<\definition>
<index|fundamentally system of neighborhoods>Let <math|X> be a
topological space and <math|x\<in\>X> then a fundamentally system of
neighborhoods of <math|x> is a set <math|\<cal-N\>> of neighborhoods of
<math|x> such that for every neighborhood <math|A> of <math|x> we have a
<math|N\<in\>\<cal-N\>> such that <math|x\<in\>N\<subseteq\>A>
</definition>
<\definition>
<index|first countable set>A topological space <math|X> is first
countable if every point has a countable fundamental system of
neighborhoods
</definition>
<\example>
A metric space is first countable
</example>
<\proof>
Let <math|X,d> be a metric space and <math|x\<in\>X> then
<math|\<cal-N\><rsub|x>={B<rsub|d>(x,<frac|1|n>)\|n\<in\>\<bbb-N\>}>
(which is countable) is a fundamental system of neighborhoods of
<math|x>. For if <math|N> is a neighborhood of <math|x> then
<math|\<exists\>U> open with <math|x\<in\>U\<subseteq\>N> and then
<math|\<exists\>\<varepsilon\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d>(x,\<varepsilon\>)\<subseteq\>U\<subseteq\>N> now
by <reference|Archimedean ordering property> there exists
<math|n\<in\>\<bbb-N\>> such that <math|<frac|1|\<varepsilon\>>\<less\>n\<Rightarrow\>0\<less\><frac|1|n>\<less\>\<varepsilon\>\<Rightarrow\>x\<in\>B<rsub|d>(x,<frac|1|n>)\<subseteq\>B<rsub|d>(x,\<varepsilon\>)\<subseteq\>U\<subseteq\>N>
</proof>
<\definition>
<index|second countable space>A topological space is second countable if
it has a countable basis
</definition>
<\remark>
It is trivial to prove that a second countable space is first countable.
</remark>
<\definition>
<index|limit of sequence>Let <math|X> be a topological space and let
<math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\>>> be a sequence then
<math|x\<in\>X> is a limit of this sequence iff <math|\<forall\>U> open
with <math|x\<in\>U> there exists a <math|N\<in\>\<bbb-N\><rsub|0>> such
that <math|\<forall\>n\<in\>\<bbb-N\><rsub|0> <with|mode|text|with >
n\<geqslant\>N> we have <math|x<rsub|n>\<in\>U>
</definition>
<\theorem>
Let <math|X> be a Hausdorff topological space then every sequence has
only one limit.
</theorem>
<\proof>
Suppose that the sequence <math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
has two limits <math|x,y> with <math|x\<neq\>y> then there exists
<math|U,V> open with <math|x\<in\>U,y\<in\>V> and
<math|U<big|cap>V=\<emptyset\>> and <math|\<exists\>N<rsub|1>,N<rsub|2>\<in\>\<bbb-N\><rsub|0>\<vdash\>N<rsub|1>,N<rsub|2>\<geqslant\>0>
such that <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<vdash\>n\<geqslant\>N<rsub|1>\<succ\>x<rsub|n>\<in\>U<rsub|>>
and <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<vdash\>n\<geqslant\>N<rsub|2>\<succ\>x<rsub|n>\<in\>V>
then if <math|n=max(N<rsub|1>,N<rsub|2>)\<geqslant\>N<rsub|1>,N<rsub|2>>
and <math|x<rsub|n>\<in\>U> and <math|x<rsub|n>\<in\>V\<Rightarrow\>U<big|cap>V\<neq\>\<emptyset\>>
a contradiction so we must have <math|x=y>
</proof>
<\theorem>
If <math|X,d> is a metric space (then by <reference|metric space is
hausdorf> it is Hausdorff) then a sequence
<math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> has a limit
<math|x\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0\<succ\>\<exists\>N\<in\>\<bbb-N\><rsub|0>>
such that <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<geqslant\>N\<succ\>d(x,x<rsub|i>)\<less\>\<varepsilon\>>.
(Note that because of Hausdorff the limit if it exists is unique)
</theorem>
<\proof>
\;
<math|\<Rightarrow\>\<forall\>\<varepsilon\>\<gtr\>0> we have
<math|x\<in\>B<rsub|d>(x,\<varepsilon\>)> (a open set by definition) so
there exists <math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<geqslant\>N> we have
<math|x<rsub|i>\<in\>B<rsub|d>(x,\<varepsilon\>)> or
<math|d(x,x<rsub|i>)\<less\>\<varepsilon\>>
<math|\<Leftarrow\>>Assume that <math|x\<in\>U,U> open then there exists
<math|\<varepsilon\>\<gtr\>0> such that
<math|x\<in\>B<rsub|d>(x,\<varepsilon\>)\<subseteq\>U> and by the
hypothesis <math|\<exists\>N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<geqslant\>N\<succ\>d(x,x<rsub|i>)\<less\>\<varepsilon\>\<Rightarrow\>x<rsub|i>\<in\>B<rsub|d>(x,\<varepsilon\>)\<subseteq\>U>
</proof>
<subsection|Topological Vector Spaces>
<\definition>
<label|topological vector space><index|topological vector space>A
topological vector space is a vector space <math|X,+,.> over the field of
the real numbers (<math|\<bbb-R\>>) such that\
<\enumerate>
<item><math|+:X\<times\>X\<rightarrow\>X> is continuous
<item><math|.:\<bbb-R\>\<times\>X\<rightarrow\>X> is continuous
</enumerate>
</definition>
<\example>
Using <reference|continuity of vectorspace operations> we have that a
normed vector space is a topological vector space
</example>
<\definition>
<label|toplinear isomorphism><index|toplinear isomorphism>Let <math|X,Y>
be topological vector spaces then a function <math|L:X\<rightarrow\>Y> is
a toplinear isomorphism iff
<\enumerate>
<item>L is a linear isomorphism, meaning that\
<\enumerate>
<item>L is a bijection
<item><math|L(x+y)=L(x)+L(y)>
<item><math|L(\<alpha\>.x)=\<alpha\>.L(x)>
</enumerate>
<item><math|L> and <math|L<rsup|-1>> are continuous
</enumerate>
</definition>
<\theorem>
<label|continuous mappings in a topological space>Let <math|X,+,.> be a
topological vector space then the functions defined by\
<\enumerate>
<item><math|+<rsub|x>:X\<rightarrow\>X> where <math|+<rsub|x>(y)=x+y>
<item><math|.<rsub|\<alpha\>>:X\<rightarrow\>X> where
<math|.<rsub|\<alpha\>>(x)=\<alpha\>.x>
</enumerate>
are continuous.
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|U> open and take <math|y\<in\>(+<rsub|x>)<rsup|-1>(U)={y\<in\>X\|x+y\<in\>U}>
then <math|(x,y)\<in\>+<rsup|-1>(U)> and by continuity of <math|+>
there exists and the definition of the product topology there exists
open <math|V<rsub|1>,V<rsub|2>> with
<math|x\<in\>V<rsub|1>,y\<in\>V<rsub|2>> such that
<math|+(V<rsub|1>,V<rsub|2>)\<subseteq\>U> thus
<math|+<rsub|x>(V<rsub|2>)={x}+V<rsub|2>\<subseteq\>V<rsub|1>+V<rsub|2>=+(V<rsub|1>,V<rsub|2>)\<subseteq\>U>
proving continuity of <math|+<rsub|x>>
<item>Let <math|U> open and take <math|x\<in\>.<rsub|\<alpha\>>(U)={x\<in\>X:\<alpha\>.x\<in\>U}>
then by continuity of <math|.> there exists a <math|V<rsub|1>> open in
<math|\<bbb-R\>> and a <math|V<rsub|2>> open in <math|X> with
<math|\<alpha\>\<in\>V<rsub|1>,x\<in\>V<rsub|2>> such that
<math|.(V<rsub|1>\<times\>V<rsub|2>)\<subseteq\>U> and thus
<math|.<rsub|\<alpha\>>(V<rsub|2>)={\<alpha\>}.V<rsub|2>\<subseteq\>.(V<rsub|1>\<times\>V<rsub|2>)\<subseteq\>U>
proving continuity of <math|.<rsub|\<alpha\>>>
</enumerate>
</proof>
<subsection|Bundles>
<\definition>
<index|topological group>A topological group is a group <math|X,+>
together with a topology <math|\<cal-T\>> (so that <math|X,\<cal-T\>> is
a topological space such that\
<\enumerate>
<item><math|+:X\<times\>X\<rightarrow\>X> definied by <math|+(x,y)=x+y>
<item><math|-:X\<rightarrow\>X> defined by <math|<rsup|->(x)=-x> (the
inverse function)
</enumerate>
are continuous functions.
</definition>
<\example>
A topological vector space is a topological group\
</example>
<\proof>
(1) Follows directly from the definition of a topological vector space
(see <reference|topological vector space>). To proof (2) let <math|U> be
a open set then as <math|\<minusassign\>.<rsub|-1>> we use
<math|><reference|continuous mappings in a topological space> to prove
continuity of the <math|-> mapping.
</proof>
<\definition>
<label|fiber bundle><index|fiber bundle>A fiber bundle consists of the
tuple <math|E,B,\<pi\>,F,\<cal-T\>> where <math|E,B> and <math|F> are
topological spaces, <math|\<pi\>:E\<rightarrow\>B> is a continuous
surjection, <math|\<cal-T\>> is a family
<math|{U<rsub|i>,\<varphi\><rsub|i>)<rsub|i\<in\>I>> where
<math|U<rsub|i>> is a open set in <math|B> such that
<math|B=<big|cup><rsub|i\<in\>I>U<rsub|i>> and
<math|\<varphi\><rsub|i>:\<pi\><rsup|-1>(U<rsub|i>)\<rightarrow\>U<rsub|i>\<times\>F>
is a homeomorphism. The continuous surjection must fulfil
<math|\<pi\><rsub|1>\<circ\>\<varphi\><rsub|i>=\<pi\>> [where
<math|\<pi\><rsub|1>> is the first projection function on
<math|U<rsub|i>\<times\>F> (or <math|\<pi\><rsub|1>((u,f))=u>]. In other
way the following diagram must commute:
<math|<tabular|<tformat|<table|<row|<cell|\<pi\><rsup|-1>(U<rsub|i>)>|<cell|<huge|\<rightarrowlim\><rsup|<normal-size|\<varphi\><rsub|i>>>>>|<cell|U\<times\>F>>|<row|<cell|>|<cell|<huge|\<searrow\>>>|<cell|<huge|\<downarrow\><normal-size|\<pi\><rsub|1>>>>>|<row|<cell|>|<cell|\<pi\>>|<cell|U<rsub|i>>>>>>>
\;
We also introduce the following names
<\itemize>
<item><math|E> is called the total space of the fiber bundle
<item><math|B> is called the base space of the bundle
<item><math|\<pi\>> is called the projection map of the bundle
<item><math|\<cal-T\>> is called a trivialization and
<math|(U<rsub|i>,\<varphi\><rsub|i>)> is called a trivializing
neighborhood
</itemize>
</definition>
<\definition>
<index|trivial fiber bundle><label|trival fiber bundle>If in the fiber
bundle <math|E,B,\<pi\>,F,\<cal-T\>> we have that the total space
<math|E=B\<times\>F> and the projection map <math|\<pi\>=\<pi\><rsub|1>>
(so that <math|\<pi\><rsub|1>\<circ\>\<varphi\><rsub|i>=\<pi\><rsub|1>>)
then this fiber bundle is called trivial.
</definition>
<\definition>
<index|vector bundle><label|vector bundle>A vector bundle is a tuple
<math|E,B,\<pi\>,F,\<cal-T\>> where <math|E,B> are topological spaces,
\ <math|\<pi\>:E\<rightarrow\>B> a continuous surjection, <math|F> a
normed space, <math|\<cal-T\>> is a family
<math|{U<rsub|i>,\<varphi\><rsub|i>}<rsub|i\<in\>I>> of homeomorphism
<math|\<varphi\><rsub|i>:U<rsub|i>\<times\>F\<rightarrow\>\<pi\><rsup|-1>(U<rsub|i>)>
with <math|B=<big|cup><rsub|i\<in\>I>U<rsub|i>> such that
<\itemize-dot>
<item><math|\<forall\>b\<in\>B\<succ\>\<pi\><rsup|-1>({b})> has the
structure of a vectorspace
<item><math|\<forall\>i\<in\>I> we have
<math|\<forall\>x\<in\>U<rsub|i>> and <math|\<forall\>v\<in\>F> that
<math|\<pi\>(\<varphi\><rsub|i>(x,v))=x>
<\note>
From this we have <math|\<varphi\><rsub|i\|{x}\<times\>F>> is a
bijection from <math|{x}\<times\>F\<rightarrow\>\<pi\><rsup|-1>({x})>
</note>
<item><math|\<forall\>i\<in\>I,x\<in\>U<rsub|i>> the map
<math|\<varphi\><rsub|i><rsup|(x)>:F\<rightarrow\>\<pi\><rsup|-1>({x})>
defined by <math|\<varphi\><rsub|i><rsup|(x)>(v)=\<varphi\><rsub|i>(x,v)>
is a linear function between the vector spaces <math|F> and
<math|\<pi\><rsup|-1>({x})>\
<\note>
From the previous note it already follows that
<math|\<varphi\><rsub|i><rsup|(x)>> is already a bijection, so we
have that <math|\<varphi\><rsub|i><rsup|(x)>> is a isomorphism.
</note>
</itemize-dot>
We call\
<\itemize>
<item><math|E> the total space of the vector bundle
<item><math|B> the base space of the vector bundle
<item><math|\<pi\>> is the projection map of the bundle\
<item><math|\<cal-T\>> is called a trivialization and
<math|(U<rsub|i>,\<varphi\><rsub|i>)> is called a trivializing
neighborhood.
</itemize>
</definition>
<\theorem>
Given a vector bundle <math|E,B,\<pi\>,F,\<cal-T\>={(U<rsub|i>,\<varphi\><rsub|i>)}<rsub|i\<in\>I>>
then if <math|\<cal-T\><rsup|-1>={(U<rsub|i>,\<varphi\><rsub|i><rsup|-1>)}<rsub|i\<in\>I>>
then <math|E,B,\<pi\>,F,\<cal-T\><rsup|-1>> is a fiber bundle.
</theorem>
<\proof>
By definition <math|\<pi\>> is a continuous surject and as
<math|B=<big|cup><rsub|i\<in\>I>U<rsub|i>> and
<math|\<varphi\><rsub|i><rsup|-1>:\<pi\><rsup|-1>(U<rsub|i>)\<rightarrow\>U<rsub|i>\<times\>F>
is a homeomorphism (as <math|\<varphi\><rsub|i>> is a homeomorphism), so
we must only proof that <math|\<pi\><rsub|1>\<circ\>\<varphi\><rsup|-1><rsub|i>=\<pi\>>.
Now if <math|x\<in\>\<pi\><rsup|-1>(U<rsub|i>)\<Rightarrowlim\><rsub|\<varphi\><rsub|i>
is a homeomorphism>\<exists\>(y,v)\<in\>U<rsub|i>\<times\>F\<vdash\>x=\<varphi\><rsub|i>(y,v)\<Rightarrow\>(y,v)=\<varphi\><rsub|i><rsup|-1>(x)>
and thus <math|\<pi\><rsub|1>(\<varphi\><rsup|-1><rsub|i>(x))=v=\<pi\>(\<varphi\><rsub|i>(y,v))=\<pi\>(x)>
</proof>
\;
<section|Compact Spaces>
<\definition>
<index|compact space>A topological space <math|X> is compact if for every
family <math|{U<rsub|i>}<rsub|i\<in\>A>> of open sets such that
<math|X=<big|cup><rsub|i\<in\>A>U<rsub|i>> (covers <math|X>) there exists
a finite subset <math|B\<subseteq\>A> that covers X
(<math|X=<big|cup><rsub|i\<in\>B>U<rsub|i>>)
</definition>
<\definition>
<index|compact set>Let <math|X> be a topological space and <math|C> a
subset of <math|X> then C is compact in X iff it is compact in the
subspace of topology
</definition>
<\theorem>
Let <math|X> be a topological space and <math|C> a subset of <math|X>
then <math|C> is compact (meaning compact in the subspace topology) if
and only if for every collection <math|{U<rsub|i>}<rsub|i\<in\>A>> of
open sets in <math|X> that covers <math|C>
(<math|C\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>>) there exists a
finite subset <math|B\<subseteq\>A> that covers <math|C>
(<math|C\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>>)
</theorem>
<\proof>
\;
<\math>
\<Rightarrow\>
</math>
If <math|{U<rsub|i>}<rsub|i\<in\>A>> is a set of open sets in X such that
<math|C\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>\<Rightarrow\>C=<big|cup><rsub|i\<in\>A>(U<rsub|i><big|cap>C)\<Rightarrowlim\><rsub|C
is compact>\<exists\> B\<subseteq\>A> where <math|B> is finite and
<math|C=<big|cup><rsub|i\<in\>B>(U<rsub|i><big|cap>C)\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>>
<math|\<Leftarrow\>>
If <math|{V<rsub|i>}<rsub|i\<in\>A>> is a collection of open sets in
<math|C> such that <math|C=<big|cup><rsub|i\<in\>A>V<rsub|i>\<Rightarrowlim\><rsub|definition
of subspace topology>>there exists a collection
<math|{U<rsub|i>}<rsub|i\<in\>A>> of open sets in <math|X> such that
<math|\<forall\>i\<in\>A\<succ\>V<rsub|i>=U<rsub|i><big|cap>C><math|\<Rightarrow\>C\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>\<Rightarrowlim\><rsub|hypothese>\<exists\>B\<subseteq\>A,B>
finite and <math|C\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>\<Rightarrow\>C=<big|cup><rsub|i\<in\>B>(U<rsub|i><big|cap>C)=<big|cup><rsub|i\<in\>B>V<rsub|i>\<Rightarrow\>C>
is compact in the subspace topology and thus <math|C> is compact in X
</proof>
<\theorem>
<label|compact subsets of subsets>Let <math|X> be a topological space and
let <math|A> be a subset of <math|X> then a set <math|C\<subseteq\>A>
that is compact in the subspace topology of <math|A> is also compact in
the topology of <math|X>
</theorem>
<\proof>
Let <math|{U<rsub|i>}<rsub|i\<in\>A>> be a open cover of
<math|C\<Rightarrow\>C\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>\<Rightarrow\>C=C<big|cap>A\<subseteq\><big|cup><rsub|i\<in\>A>(U<rsub|i><big|cap>A)\<Rightarrowlim\><rsub|C
compact in A, subspace topology>\<exists\>B\<subseteq\>A,B> finite and
<math|C\<subseteq\><big|cup><rsub|i\<in\>B>(U<rsub|i><big|cap>C)\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>>
and <math|C> is compact in <math|X>
</proof>
<\theorem>
<label|compact space in a metric space is bounded>Let <math|X,d> be a
metric space and let <math|C> be a compact subset of <math|X> then
<math|C> is bounded (see <reference|bounded sets>)
</theorem>
<\proof>
Let <math|C> be a compact subset of <math|X> and
<math|\<forall\>c\<in\>C> take <math|c\<in\>B<rsub|d>(c,1)> which is open
and then <math|C\<subseteq\><big|cup><rsub|c\<in\>C>B<rsub|d>(c,1)> and
by compactness there exists a finite set
<math|{c<rsub|1>,\<ldots\>,c<rsub|n>}\<succ\>C\<subseteq\><big|cup><rsub|c\<in\>{1,\<ldots\>,n}>B<rsub|d>(c<rsub|i>)>
take then <math|M=2+max(d(c<rsub|i>,c<rsub|j>)\|(i,j)\<in\>{1,\<ldots\>,n}\<times\>{1,\<ldots\>,n})>
(which exists because of <reference|maximum minimum of finite sets> and
<reference|{1,..n}*{1,..,m} is finite>). Then if
<math|x,y\<in\>C\<Rightarrow\>\<exists\>i,j\<in\>{1,\<ldots\>,n}> such
that <math|x\<in\>B<rsub|d>(c<rsub|i>,1),y\<in\>B<rsub|d>(c<rsub|j>,1)>
and thus <math|d(x,y)\<leqslant\>d(x,c<rsub|i>)+d(c<rsub|i>,c<rsub|j>)+d(y,c<rsub|j>)\<leqslant\>1+d(c<rsub|i>,c<rsub|j>)+1\<leqslant\>M>
proving that <math|C> is bounded
</proof>
<\theorem>
<label|continuous image of compact sets>Let <math|X> be a topological
space, <math|Y> a topological space, <math|C> a compact subset of
<math|X> and let <math|f:X\<rightarrow\>Y> be a continuous function then
<math|f(C)> is compact
</theorem>
<\proof>
Let <math|{V<rsub|i>}<rsub|i\<in\>B>> be a family of open sets so that
<math|f(C)\<subseteq\><big|cup><rsub|i\<in\>B>V<rsub|i>> \ then
<math|\<forall\>x\<in\>C> we have <math|f(x)\<in\>f(C)> and
<math|\<exists\>i\<in\>B\<succ\>f(x)=y\<in\>V<rsub|i>> so by continuity
<math|\<exists\>U<rsub|i>> open in <math|X> with <math|x\<in\>U<rsub|i>>
and <math|f(x)\<in\>f(U<rsub|i>)\<subseteq\>V<rsub|i>> and thus
<math|C\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>> and by compactness
we have <math|\<exists\>finite A\<subseteq\>B> such that
<math|C\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>> and thus if
<math|y\<in\>f(C)\<Rightarrow\>\<exists\>x\<in\>C\<vdash\>f(x)=y\<in\>f(C)\<Rightarrow\>\<exists\>i\<in\>B\<vdash\>x\<in\>U<rsub|i>>
and <math|f(x)=y\<subseteq\>f(U<rsub|i>)\<subseteq\>V<rsub|i>\<Rightarrow\>f(C)\<subseteq\><big|cup><rsub|i\<in\>B>V<rsub|i>>
proving compactness
</proof>
<\theorem>
<label|compactness and closed sets>Let <math|X> be a Hausdorff
topological space then every compact subspace is closed.
</theorem>
<\proof>
Let <math|C> be a compact subset of <math|X> then if
<math|y\<in\>X<mid|\\>C> then <math|\<forall\>x\<in\>C> we have
<math|\<exists\>U<rsub|x>> open with <math|x\<in\>U<rsub|x>> and a
<math|V<rsub|x>> open with <math|y\<in\>V<rsub|x>> such that
<math|U<rsub|x><big|cap>V<rsub|x>=\<emptyset\>>. So
<math|C\<subseteq\><big|cup><rsub|x\<in\>C>U<rsub|x>> then there exists a
finite set <math|{U<rsub|x<rsub|i>>,V<rsub|x<rsub|i>>}<rsub|i\<in\>{1,\<ldots\>,n}>>
such that <math|C\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|x<rsub|i>>>
and <math|y\<in\><big|cap><rsub|i\<in\>{1,.,,n}>V<rsub|x<rsub|i>>> which
is open (see <reference|intersection of finite open sets is open>, so
<math|X<mid|\\>C> is open and thus <math|C> is closed.
</proof>
<\theorem>
<label|closed subsets are compact>Let <math|X> be a topological space
then every closed subset of a compact subset of <math|X> is compact\
</theorem>
<\proof>
Let <math|C> be compact, <math|A\<subseteq\>C> and let
<math|{U<rsub|i>}<rsub|i\<in\>I>> be a family of open sets then
<math|W={U<rsub|i>}<rsub|i\<in\>A><big|cup>{X<mid|\\>A}> covers <math|C>
and so a finite subset of <math|W> must cover <math|C> and thus <math|A>.
And as <math|a\<in\>A\<Rightarrow\>a\<nin\>X<mid|\\>A> so we must have a
finite subset of <math|{U<rsub|i>}<rsub|i\<in\>A>> covering <math|A>.
</proof>
<\theorem>
<label|compact Hausdorf space is regular and normal>Let <math|X> be a
Hausdorff compact topological vector space then it is regular and normal
</theorem>
<\proof>
\;
<\enumerate>
<item>Regularity Let <math|A> be closed set and
<math|x\<in\>X,x\<nin\>A> then because of Hausdorff we have
<math|\<forall\>a\<in\>A> we have the existence of disjoint open sets
<math|V<rsub|a>,U<rsub|a>> with <math|x\<in\>V<rsub|a>,a\<in\>U<rsub|a>\<Rightarrow\>V<rsub|a><big|cap>U<rsub|a>=\<emptyset\>>
then <math|A\<subseteq\><big|cup><rsub|a\<in\>A>U<rsub|a>> and because
<math|A> is compact (previous theorem) we have
<math|{a<rsub|1>,\<ldots\>,a<rsub|n>}\<subseteq\>A> with
<math|A\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|a<rsub|i>>=U>
a open set and <math|x\<in\><big|cap><rsub|i\<in\>[1,\<ldots\>,n}>V<rsub|a<rsub|i>>>
also a open set and <math|U<big|cap>V=(<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>V<rsub|i>)<big|cap>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|i>)=<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>((<big|cap><rsub|j\<in\>{1,\<ldots\>,n}>V<rsub|j>)<big|cap>U<rsub|i>)\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>(V<rsub|i><big|cap>U<rsub|i>)=\<emptyset\>>\
<item>Normality Let <math|A,B> be two disjoint closed sets then for
every <math|a\<in\>A> we have using (1) the existence of disjoint open
sets <math|V<rsub|a>,U<rsub|a>> such that
<math|a\<in\>V<rsub|a>,B\<subseteq\>U<rsub|a>> and
<math|V<rsub|a><big|cap>U<rsub|a>=\<emptyset\>> then
<math|A\<subseteq\><big|cup><rsub|a\<in\>A>U<rsub|a>> and as <math|A>
is compact there exists a finite set
<math|{a<rsub|1>,\<ldots\>,a<rsub|n>}> such that
<math|A\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>V<rsub|a<rsub|i>>-U>
a open set and <math|B\<subseteq\><big|cap><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|a<rsub|i>>=V>
also a open set and <math|U<big|cap>V=(<big|cap><rsub|i\<in\>{1,\<ldots\>,n}>V<rsub|i>)<big|cap>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|i>)=<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>((<big|cap><rsub|j\<in\>{1,\<ldots\>,n}>V<rsub|j>)<big|cap>U<rsub|i>)\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>(V<rsub|i><big|cap>U<rsub|i>)=\<emptyset\>>
</enumerate>
</proof>
<\theorem>
<label|compactness and base>Let <math|X> be a topological space with a
basis <math|\<cal-B\>> then a subset <math|C\<subseteq\>X> is compact in
the subspace topology iff every cover of <math|C> by sets in the base
contains a finite covering
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
This is trivial as <math|U\<in\>\<cal-B\>\<Rightarrow\>U> is open
<math|\<Leftarrow\>>
Let <math|{U<rsub|i>}<rsub|i\<in\>I>> be a family of open sets such that
<math|><math|C\<subseteq\><big|cup><rsub|i\<in\>I>U<rsub|i>> then
<math|\<forall\>x\<in\>C> there exists
<math|i<rsub|x>\<in\>I\<succ\>x\<in\>U<rsub|i<rsub|x>>> and thus
<math|\<exists\>B<rsub|i<rsub|x>>\<in\>\<cal-B\>> such that
<math|x\<in\>B<rsub|i<rsub|x>>\<subseteq\>U<rsub|i<rsub|x>>> and then
<math|C\<subseteq\><big|cup><rsub|x\<in\>C>B<rsub|i<rsub|x>>> and thus
there exists a finite subset <math|i<rsub|x<rsub|1>>,\<ldots\>,i<rsub|x<rsub|n>>\<Rightarrow\>C\<subseteq\><big|cup><rsub|j\<in\>{1,\<ldots\>,n}>B<rsub|i<rsub|x<rsub|j>>>\<subseteq\><big|cup><rsub|j\<in\>{1,\<ldots\>,n}>U<rsub|i<rsub|x<rsub|j>>>>
and thus a finite subsets of <math|{U<rsub|i>}<rsub|i\<in\>I>> covers
<math|C>.
</proof>
<\definition>
<index|sequential compact set>Let <math|X> be a topological space then a
subset <math|C> is called sequential compact if every infinite
<math|B\<subseteq\>C> has a limit point that is in <math|C>
</definition>
<\theorem>
<label|compact subspaces are sequential compact>A compact subset of a
topological space <math|X> is sequential compact.
</theorem>
<\proof>
Let <math|C> be a compact subset and let <math|A\<subseteq\>C> be a
infinite subset that does not have a limit point in <math|C>. Then
<math|\<forall\>x\<in\>C> we can find a <math|U<rsub|x>> open with
<math|U<rsub|x><big|cap>(A<mid|\\>{x})=\<emptyset\>>. So if
<math|y\<in\>U<rsub|x><big|cap>A> then we have two cases
<math|y=x\<Rightarrow\>y\<in\>{x}> or
<math|y\<neq\>x\<Rightarrow\>y\<in\>U<rsub|x><big|cap>(A<mid|\\>{x})=\<emptyset\>>
a contradiction. So <math|U<rsub|x><big|cap>A\<subseteq\>{x}>. Now
<math|C\<subseteq\><big|cup><rsub|x\<in\>C>U<rsub|x>> and as <math|C> is
compact <math|\<exists\>{x<rsub|1>,\<ldots\>,x<rsub|n>}\<subseteq\>C>
such that <math|C\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|x<rsub|i>>>
and thus <math|A\<equallim\><rsub|A\<subseteq\>C>A<big|cap>C\<subseteq\>A<big|cap>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|x<rsub|i>>)=<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>(U<rsub|x<rsub|i>><big|cap>A)=<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>{x<rsub|i>}={x<rsub|1>,\<ldots\>,x<rsub|n>}<rsub|>>
which is finite, and thus <math|A> is finite a contradiction
</proof>
<\theorem>
<label|ball cover of a metric space>Let <math|X,d> be a metric space
equipped with the metric topology. Let <math|K\<subseteq\>X> be a
sequential compact subset and <math|{U<rsub|i>}<rsub|i\<in\>A>> be a open
cover of <math|K> then <math|\<exists\>\<delta\>\<gtr\>0> such that
<math|\<forall\>x\<in\>K> there <math|\<exists\>i\<in\>A> so that
<math|x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>U<rsub|i>>
</theorem>
<\proof>
We prove this by contradiction, so assume that
<math|\<forall\>\<delta\>\<in\>\<bbb-R\><rsub|+>> there
<math|\<exists\>x\<in\>K> such that <math|\<forall\>i\<in\>A> we have
<math|B<rsub|d>(x,\<delta\>)\<nsubseteq\>U<rsub|i>>. In particular this
means that <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<succ\>\<exists\>x<rsub|n>\<in\>K>
such that <math|\<forall\>i\<in\>A> we have
<math|B<rsub|d>(x<rsub|n>,<frac|1|n>)>. This forms a sequence
<math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> of elements of <math|K>
such that <math|B<rsub|d>(x<rsub|n>,<frac|1|n>)\<nsubseteq\>U<rsub|i>,\<forall\>i\<in\>A,\<forall\>n\<in\>\<bbb-N\><rsub|0>>.
Define now the set <math|O={x<rsub|n>\|n\<in\>\<bbb-N\><rsub|0>}\<subseteq\>K>.
If this set was finite then \ using <reference|mapping of N into a finite
set> <math|\<exists\>x<rsub|N>\<in\>O> such that
<math|\<forall\>n\<in\>\<bbb-N\><rsub|0>\<succ\>\<exists\>m\<in\>\<bbb-N\><rsub|0>\<vdash\>m\<geqslant\>n\<wedge\>x<rsub|m>=x<rsub|N>>.
Now as <math|x<rsub|N>\<in\>O\<subseteq\>K\<Rightarrow\>\<exists\>i\<in\>A\<vdash\>x<rsub|N>\<in\>U<rsub|i>\<Rightarrowlim\><rsub|U<rsub|i>is
open>\<exists\>\<varepsilon\>\<gtr\>0\<vdash\>x<rsub|N>\<in\>B<rsub|d>(x<rsub|N>,\<varepsilon\>)\<subseteq\>U<rsub|i>>.
So let <math|M> be such that <math|0\<less\><frac|1|M>\<less\>\<varepsilon\>>
(see <reference|Archimedean ordering property>) then
<math|\<exists\>m\<in\>\<bbb-N\><rsub|0>,m\<geqslant\>M\<Rightarrow\><frac|1|M>\<geqslant\><frac|1|m>>
such that <math|x<rsub|N>=x<rsub|m>> and
<math|x<rsub|m>\<in\>B<rsub|d>(x<rsub|m>,<frac|1|m>)=B<rsub|d>(x<rsub|N>,<frac|1|m>)\<subseteq\>B<rsub|d>(x<rsub|N>,<frac|1|M>)\<subseteq\>B<rsub|d>(x<rsub|N>,\<varepsilon\>)\<subseteq\>U<rsub|i>>,
contradicting the fact that <math|B<rsub|d>(x<rsub|m>,<frac|1|m>)\<subseteq\>U<rsub|i>,\<forall\>i\<in\>A>.
We conclude thus that <math|O> is infinite . Now as <math|K> is
sequential compact we have a limit point <math|k\<in\>K> for <math|O> and
as <math|K> is covered by <math|{U<rsub|i>}<rsub|i\<in\>I>> we have
<math|\<exists\>i<rsub|0>\<in\>I\<vdash\>k\<in\>U<rsub|i>\<Rightarrowlim\><rsub|U<rsub|i>
is open>\<exists\>\<varepsilon\><rsub|0>\<gtr\>0\<vdash\>k\<in\>B<rsub|d>(k,\<varepsilon\><rsub|0>)\<subseteq\>U<rsub|i>>.
The set <math|P={n\<in\>\<bbb-N\><rsub|0>\|x<rsub|n>\<in\>B<rsub|d>(k,<frac|e<rsub|0>|2>)}>
is a infinite set set of natural numbers, for if it was finite we could
take <math|\<varepsilon\>=min{<frac|\<varepsilon\><rsub|0>|2>,min
(d(k,x<rsub|i>)\|i\<in\>P))> then <math|k\<in\>B<rsub|d>(k,\<varepsilon\>)\<subseteq\>B<rsub|d>(k,<frac|\<varepsilon\><rsub|0>|2>)>
and <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>,\<forall\>y\<in\>B<rsub|d>(k,\<varepsilon\>)>
we have either <math|i\<nin\>P> and then
<math|x<rsub|i>\<nin\>B<rsub|d>(k,<frac|\<varepsilon\><rsub|0>|2>)\<supseteq\>B<rsub|d>(k,\<varepsilon\>)>
or <math|i\<in\>P> and then as <math|d(k,x<rsub|i>)\<geqslant\>\<varepsilon\>\<Rightarrow\>x<rsub|i>\<nin\>B<rsub|d>(k,\<varepsilon\>)>
so in all cases we have <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> that
<math|x<rsub|i>\<nin\>B<rsub|d>(k,\<varepsilon\>)\<Rightarrow\>O<big|cap>B<rsub|d>(k,\<varepsilon\>)=\<emptyset\>>
contradicting the fact that <math|k> is the limit point of <math|O>. So
we have indeed that <math|P> is infinite. Then as <math|P> is infinite we
can chose a <math|m\<in\>P> such that
<math|<frac|1|m>\<less\><frac|\<varepsilon\><rsub|0>|2>> and as
<math|m\<in\>P> we have <math|x<rsub|m>\<in\>B<rsub|d>(k,<frac|\<varepsilon\><rsub|0>|2>)>
now if <math|z\<in\>B<rsub|d>(x<rsub|m>,<frac|1|m>)\<Rightarrow\>d(x<rsub|m>,z)\<less\><frac|1|m>>
and thus <math|d(k,z)\<leqslant\>d(k,x<rsub|m>)+d(x<rsub|m>,z)\<less\><frac|\<varepsilon\><rsub|0>|2>+<frac|1|m>\<less\><frac|\<varepsilon\><rsub|0>|2>+<frac|\<varepsilon\><rsub|0>|2>=\<varepsilon\><rsub|0>\<Rightarrow\>z\<in\>B<rsub|d>(k,\<varepsilon\><rsub|0>)\<Rightarrow\>B<rsub|d>(x<rsub|m>,<frac|1|m>)\<subseteq\>B<rsub|d>(k,\<varepsilon\><rsub|0>)\<subseteq\>U<rsub|i>>
contradicting the fact that <math|B<rsub|d>(x<rsub|n>,<frac|1|n>)\<nsubseteq\>U<rsub|i>,
\<forall\>i\<in\>A,\<forall\>n\<in\>\<bbb-N\><rsub|0>>.
</proof>
<\theorem>
<label|compactness and sequenctial compactness>Let <math|X,d> be a metric
space with the metric topology and <math|K\<subseteq\>X> then the
following are equivalent
<\enumerate>
<item><math|K> is compact
<item><math|K> is sequential compact
</enumerate>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
This is already proved in <reference|compact subspaces are sequential
compact>
<math|\<Leftarrow\>>
Let <math|K> be sequential compact and <math|{U<rsub|i>}<rsub|i\<in\>A>>
a cover of <math|K>. If <math|K=\<emptyset\>> then it is trivially
compact so lets assume that <math|K=\<emptyset\>>. Then by
<reference|ball cover of a metric space>
<math|\<exists\>\<delta\>\<gtr\>0> such that <math|\<forall\>x\<in\>K>
there <math|\<exists\>i\<in\>A> such that
<math|x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>U<rsub|i>>. Suppose now
that we can not find a finite set of <math|{y<rsub|1>,\<ldots\>,y<rsub|n>}\<subseteq\>K>
such that <math|K\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|d>(y<rsub|i>,\<delta\>)>
(1). Then if <math|x<rsub|1>\<in\>K> we can not have
<math|K\<subseteq\>B<rsub|d>(x<rsub|1>,\<delta\>)> by the assumption (1)
so there exists a <math|x<rsub|2>\<in\>K<mid|\\>B<rsub|d>(x<rsub|1>,\<delta\>)>
now proceed recursively and assume that we have constructed a
<math|{x<rsub|1>,\<ldots\>,x<rsub|n>}> such that
<math|x<rsub|i>\<in\>K<mid|\\>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n-1}>B<rsub|d>(x<rsub|i>,\<delta\>))<rsub|>>
then as we can not have <math|K\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>U<rsub|i>>
there exists a <math|x<rsub|n+1>\<in\>K<mid|\\>(<big|cup><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|d>(x<rsub|i>,\<delta\>)>.
So we have constructed a sequence <math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
such that <math|\<forall\>i\<in\>\<bbb-N\><rsub|0><mid|\\>{1}\<succ\>x<rsub|i>\<in\>K<mid|\\>(<big|cup><rsub|i\<in\>{1,\<ldots\>,i-1}>B<rsub|d>(x<rsub|i>,\<delta\>))>.
Then if <math|i\<neq\>j> for example <math|i\<gtr\>j> (for the other case
just switch <math|i> and <math|j>) <math|x\<nin\>B<rsub|d>(x<rsub|j>,\<delta\>)>
so that <math|d(x<rsub|i>,x<rsub|j>)\<geqslant\>\<delta\>\<Rightarrow\>x<rsub|i>\<neq\>x<rsub|j>>
making the function <math|x:A\<rightarrow\>K:i\<rightarrow\>x<rsub|i>=x(i)>
injective. Then using <reference|injection from infinite set> we have
that <math|O={x<rsub|i>\|i\<in\>\<bbb-N\><rsub|0>}\<subseteq\>K> is
infinite so it has a accumulation point <math|x\<in\>K> (because of
sequential compactness). Take then <math|B<rsub|d>(x,\<delta\>)> then we
have a <math|x<rsub|i>\<neq\>x> such that
<math|x<rsub|i>\<in\>B<rsub|d>(x,<frac|\<delta\>|2>)> take then
<math|\<varepsilon\>=min(<frac|\<delta\>|4>,<frac|d(x,x<rsub|i>)|2>)>
then <math|d(x,x<rsub|i>)\<gtr\>\<varepsilon\>\<Rightarrow\>x<rsub|i>\<nin\>B<rsub|d>(x,\<varepsilon\>)>
and if <math|m\<neq\>i> then if <math|x<rsub|m>\<in\>B<rsub|d>(x,\<varepsilon\>)\<Rightarrow\>\<delta\>\<leqslant\>d(x<rsub|m>,x<rsub|i>)\<leqslant\>d(x<rsub|m>,x)+d(x,x<rsub|i>)\<less\>\<varepsilon\>+<frac|\<delta\>|2>\<less\><frac|\<delta\>|2>+<frac|\<delta\>|2>=\<delta\>>
so <math|\<delta\>\<less\>\<delta\>> a contradiction and thus
<math|x<rsub|m>\<nin\>B<rsub|d>(x,\<varepsilon\>)\<Rightarrow\>O<big|cap>B<rsub|d>(x,\<varepsilon\>)=\<emptyset\>>
contradicting the fact that <math|x> is a accumulation point.
</proof>
<\theorem>
<label|principle of nested intervals>Let <math|[a<rsub|i>,b<rsub|i>],
a<rsub|i>,b<rsub|i>\<in\>\<bbb-R\>,i\<in\>\<bbb-N\><rsub|0>,a<rsub|i>\<leqslant\>b<rsub|i>>
be a decreasing sequence (<math|[a<rsub|i+1>,b<rsub|i+1>]\<subseteq\>[a<rsub|i>,b<rsub|i>])>
of closed bounded intervals <math|\<bbb-R\>> then\
<\enumerate>
<item>Then <math|\<exists\>a,b\<in\>\<bbb-R\>,a\<leqslant\>b> then
<math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]=[a,b]>
<item>If the sequence <math|{b<rsub|i>-a<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
has the limit <math|0> then there exists a number <math|a> such that
<math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]={a}>
</enumerate>
</theorem>
<\proof>
First we prove that <math|\<forall\>i,j\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<leqslant\>j\<Rightarrow\>a<rsub|i>\<leqslant\>a<rsub|j>\<leqslant\>b<rsub|j>\<leqslant\>b<rsub|i>>
by induction on <math|j>\
<\enumerate>
<item><math|j=1\<Rightarrow\>i=1> and of course we have
<math|a<rsub|1>\<leqslant\>a<rsub|1>\<leqslant\>b<rsub|1>\<leqslant\>b<rsub|1>>
<item>Assume that it is true for <math|j> and prove it for <math|j+1>
then <math|a<rsub|j>\<leqslant\>a<rsub|j+1>\<leqslant\>b<rsub|j+1>\<leqslant\>b<rsub|j>>
and so it is true for <math|j,j+1> if <math|i\<less\>j> then by the
induction hypothesis <math|a<rsub|i>\<leqslant\>a<rsub|j>\<leqslant\>a<rsub|j+1>\<leqslant\>b<rsub|j+1>\<leqslant\>b<rsub|j>\<leqslant\>b<rsub|i>><math|<rsub|>>
</enumerate>
So more specially we have <math|a<rsub|1>\<leqslant\>a<rsub|i>\<leqslant\>b<rsub|i>\<leqslant\>b<rsub|1>>
and then\
<\enumerate>
<item><math|A={a<rsub|i>\|i\<in\>\<bbb-N\><rsub|0>}> is bounded above
by <math|b<rsub|1>> and <math|B={b<rsub|i>\|i\<in\>\<bbb-N\><rsub|0>}>
is bounded below by <math|a<rsub|1>> and by the upper(lower) bound
property of <math|\<bbb-R\>> there exists a <math|>a=sup(A),b=inf(A)
and as <math|\<forall\>i,j\<in\>\<bbb-N\><rsub|0>> we have if
(<math|i\<leqslant\>j>) that <math|a<rsub|i>\<leqslant\>a<rsub|j>\<leqslant\>b<rsub|j>>
or if <math|(j\<leqslant\>i)> that <math|a<rsub|i>\<leqslant\>b<rsub|i>\<leqslant\>b<rsub|j>>
so in all cases we have <math|a<rsub|i>\<leqslant\>b<rsub|j>> and thus
<math|a<rsub|i>\<leqslant\>inf(B)=b> and
<math|a=sup(A)\<leqslant\>b\<Rightarrow\>a\<leqslant\>b> and
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> we have by the definition of
inf and sup that <math|a<rsub|i>\<leqslant\>a\<leqslant\>b\<leqslant\>b<rsub|i>>
so that <math|[a,b]\<subseteq\><big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]><math|>.
Now assume that <math|x\<in\><big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]>
then <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> we have
<math|a<rsub|i>\<leqslant\>x\<leqslant\>b<rsub|i>> so <math|x> is a
upper bound for <math|A> and a lower bound for <math|B> so
<math|a\<leqslant\>x\<leqslant\>b\<Rightarrow\>x\<in\>[a,b]>
<item>By (1) we have <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]=[a,b]>
suppose now that <math|a\<less\>b\<Rightarrow\>\<forall\>i\<in\>\<bbb-N\><rsub|0>>
we have <math|a<rsub|i>\<leqslant\>a\<less\>b\<leqslant\>b<rsub|i>\<Rightarrow\>0\<less\>b-a\<leqslant\>b<rsub|i>-a\<leqslant\>b<rsub|i>-a<rsub|i>>
so if <math|{b<rsub|i>-a<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>> has a
limit >0 then given <math|\<varepsilon\>=b-a\<gtr\>0> choose
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>\<vdash\>i\<geqslant\>N
<with|mode|text|we have >b<rsub|i>-a<rsub|i>=\|b<rsub|i>-a<rsub|i>\|\<less\>\<varepsilon\>\<Rightarrow\>b<rsub|N>-a<rsub|N>\<less\>b-a\<leqslant\>b<rsub|N>-a<rsub|N>>
and we reach a contradiction so we must have
<math|a=b\<Rightarrow\><big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]={a}>
</enumerate>
</proof>
<\theorem>
<label|[a,b] is compact>Let <math|[a,b],a\<leqslant\>b,a,b\<in\>\<bbb-R\>>
be a closed and bounded interval in <math|\<bbb-R\>> then <math|[a,b]> is
a compact subset in <math|\<bbb-R\>> (using the normed topology generated
by the norm <math|\| \|>)
</theorem>
<\proof>
\;
If <math|a=b> then <math|[a,b]={a}> so every covering of open sets
contains a open <math|U> such that <math|a\<in\>U\<Rightarrow\>{a}\<subseteq\>U>
and <math|[a,b]> is compact. So assume that <math|a\<less\>b>. We proceed
now by contradiction so assume that <math|[a,b]> is not compact, then
using \ <math|<reference|topology of real numbers>> and
<reference|compactness and base> assume that there exists a covering of
<math|[a,b]> by open intervals <math|U<rsub|i>=]x<rsub|i>,y<rsub|i>[,i\<in\>I>
that does not contains a finite sub covering. First take
<math|a<rsub|1>=a,b<rsub|1>=b> and <math|b<rsub|1>-a<rsub|1>=<frac|b-a|2<rsup|0>>>.
Take then <math|c=a+<frac|b-a|2>=<frac|a+b|2>\<Rightarrow\>c-a=<frac|a+b|2>=<frac|b-a|2<rsup|1>>>
then <math|a\<less\>c=<frac|a+b|2>\<less\><frac|b+b|2>=b> and
<math|b-c=b-<frac|a+b|2>=<frac|b-a|2>=<frac|b-a|2<rsup|1>>> and also
<math|[a,c]\<subseteq\>[a,b]> and <math|[c,b]\<subseteq\>[a,b]>. Now if
there exists two finite sets <math|A,B\<subseteq\>I> with
<math|[a,c]\<subseteq\><big|cup><rsub|i\<in\>A>U<rsub|i>,[c,b]\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>>
then <math|[a,b]\<subseteq\><big|cup><rsub|i\<in\>A<big|cup>B>U<rsub|i>>
which because <math|A<big|cup>B><math|> is finite contradicts our
assumption, so either <math|[a,c]> or <math|[c,b]> is not covered by a
finite number of <math|U<rsub|i><rsup|>> 's, let this interval be
<math|[a<rsub|2>,b<rsub|2>]> and <math|b<rsub|2>-a<rsub|2>=<frac|b-a|2<rsup|1>>>
and <math|[a<rsub|2>,b<rsub|2>]\<subseteq\>[a<rsub|1>,b<rsub|2>]>, we
proceed the construction now by recursion so assume that we have for
<math|i=1,\<ldots\>,n> intervals <math|[a<rsub|i>,b<rsub|i>]> not covered
by a finite number of <math|U<rsub|i><rprime|'>s> such that
<math|[a<rsub|i>,b<rsub|i>]\<supseteq\>[a<rsub|i+1>,b<rsub|i+1>],b<rsub|i>-a<rsub|i>=<frac|b-a|2<rsup|i-1>>,i=1,\<ldots\>,n>
then we can repeat the previous reasoning using
<math|a<rsub|n>\<less\>c=a<rsub|n>+<frac|b<rsub|n>-a<rsub|n>|2<rsup|>>=<frac|a<rsub|n>+b<rsub|n>|2>\<less\><frac|b<rsub|n>+b<rsub|n>|2>=b<rsub|n>>
and <math|c-a<rsub|n>=<frac|b<rsub|n>-a<rsub|n>|2<rsup|n>>\<less\><frac|b-a|2<rsup|n>>>
and <math|b<rsub|n>-c=b<rsub|n>-a<rsub|n>-<frac|b<rsub|n>-a<rsub|n>|2<rsup|>>=<frac|b<rsub|n>-a<rsub|n>|2>=<frac|b-a|2<rsup|n>>>
so that <math|[a<rsub|n>,c]\<subseteq\>[a<rsub|n>,b<rsub|n>]> and
<math|[c,b<rsub|n>]\<subseteq\>[a<rsub|n>,b<rsub|n>]>, now if there
exists two finite sets <math|A,B\<subseteq\>I> with
<math|[a<rsub|n>,c]\<in\><big|cup><rsub|i\<in\>A>U<rsub|i>,[c,b<rsub|n>]\<subseteq\><big|cup><rsub|i\<in\>B>U<rsub|i>>
then as <math|[a<rsub|n>,b<rsub|n>]\<subseteq\><big|cup><rsub|i\<in\>A<big|cup>B>U<rsub|i>>
would be a finite cover, so one of the <math|[a<rsub|n>,c],[c,b<rsub|n>]>
is not covered by a finite set of <math|U<rsub|i>> 's, lets label this as
<math|[a<rsub|n+1>,b<rsub|n+1>]>. So we have constructed a sequence
<math|[a<rsub|i>,b<rsub|i>]<rsub|i\<in\>\<bbb-N\><rsub|0>>> of closed
intervals where <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> we have that
<math|[a<rsub|i>,b<rsub|i>]> is not covered by a finite number of
<math|U<rsub|i>>'s, <math|[a<rsub|i+1>,b<rsub|i+1>]\<subseteq\>[a<rsub|i>,b<rsub|i>]>
and <math|b<rsub|i>-a<rsub|i>=<frac|b-a|2<rsup|i-1>>> (so each limit is
0) and then by <reference|principle of nested intervals> there exists a
<math|c\<in\>\<bbb-R\>> such that <math|{c}=<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>[a<rsub|i>,b<rsub|i>]\<subseteq\>[a<rsub|1>,b<rsub|1>]=[a,b]>.
Since <math|c\<in\>[a,b]\<Rightarrow\>\<exists\>i\<in\>I> such that
<math|c\<in\>U<rsub|i>=]x<rsub|i>,y<rsub|i>[\<Rightarrow\>x<rsub|i>\<less\>c\<less\>y<rsub|i>>,
now take <math|\<varepsilon\>=<frac|1|2>.min(c-x<rsub|i>,y<rsub|i>-c)\<gtr\>0>
and choose a <math|j> such that <math|b<rsub|j>-a<rsub|j>=<frac|b-a|2<rsup|j-1>>\<less\>\<varepsilon\>>
then <math|c\<in\>[a<rsub|j>,b<rsub|j>]> and thus
<math|a<rsub|j>\<less\>c\<less\>b<rsub|j>\<Rightarrow\>0\<less\>c-a<rsub|j>\<less\>b<rsub|j>-a<rsub|j>\<less\>\<varepsilon\>\<less\>c-x<rsub|i>\<Rightarrow\>-a<rsub|j>\<less\>-x<rsub|i>\<Rightarrow\>x<rsub|i>\<less\>a<rsub|j>>
also <math|b<rsub|j>-c\<less\>b<rsub|j>-a<rsub|j>=\<varepsilon\>\<less\>y<rsub|i>-c\<Rightarrow\>b<rsub|j>\<leqslant\>y<rsub|i>\<Rightarrow\>[a<rsub|j>,b<rsub|j>]\<subseteq\>[x<rsub|i>,y<rsub|i>]=U<rsub|i>
> so <math|[a<rsub|j>,b<rsub|j>]> is covered by a finite number of
<math|U<rsub|i>>'s contrary to its definition. So the assumption that
<math|[a,b]> is not compact leads to a contradiction and so <math|[a,b]>
is compact.
</proof>
<\theorem>
<label|continuous functions on a compact set are uniform continuous>Let
<math|X,d<rsub|X>> and <math|Y,d<rsub|Y>> be metric spaces,
<math|K\<subseteq\>X> a compact subset and <math|f:X\<rightarrow\>Y> a
continuous functions then <math|f> is uniformly continuous on <math|K>
</theorem>
<\proof>
Let <math|\<varepsilon\>\<gtr\>0> then by continuity of <math|f>
<math|\<forall\>x\<in\>K> there exists a <math|\<delta\>(x)> such that
<math|\<forall\>y\<in\>X> such that <math|d<rsub|X>(x,y)\<less\>\<delta\>(x)>
we have <math|d<rsub|Y>(f(x),f(y))\<less\><frac|\<varepsilon\>|2>> the
set of open balls <math|B<rsub|d<rsub|X>>(x,\<delta\>(x)/2)> covers
<math|K> and by compactness there exists a
<math|{x<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> such that
<math|K\<subseteq\><big|cup><rsub|i\<in\>{1,\<ldots\>,n}>B<rsub|d<rsub|X>>(x<rsub|i>,\<delta\>(x<rsub|i>)/2)>
take now <math|\<delta\>=min(<frac|\<delta\>(x<rsub|i>)|2>\|i\<in\>{1,\<ldots\>,n})>.
Let now <math|x,y\<in\>K> be such that
<math|d<rsub|X>(x,y)\<less\>\<delta\>> then there exists a
<math|x<rsub|i<rsub|> > such that ><math|x\<in\>B<rsub|d<rsub|X>>(x<rsub|i>,<frac|\<delta\>(x<rsub|i>)|2>)>
then <math|d<rsub|X>(x<rsub|i>,y)\<leqslant\>d<rsub|X>(x<rsub|>,y)+d<rsub|X>(x<rsub|i>,x)\<less\>\<delta\>+<frac|\<delta\>(x<rsub|i>)|2>\<leqslant\><frac|\<delta\>(x<rsub|i>)|2>+<frac|\<delta\>(x<rsub|i>)|2>=\<delta\>(x<rsub|i>)\<Rightarrow\>d<rsub|Y>(f(x),f(y))\<leqslant\>d<rsub|Y>(f(x),f(x<rsub|i>))+d<rsub|Y>(f(x<rsub|i>),f(y))\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\><rsub|>>
</proof>
<subsection|Filter bases in Spaces>
\;
<\definition>
<index|filterbase>Let <math|X> be a topological space. A filter base
<math|\<frak-U\>> in <math|X> is a nonempty family
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> of
subset's of <math|Y> having the following 2 properties:
<\enumerate>
<item><math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have
<math|A<rsub|\<alpha\>>\<neq\>\<emptyset\>>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-U\>> we have
<math|\<exists\>\<gamma\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>>
(this is of course implies that the intersection is non empty)
</enumerate>
</definition>
<\theorem>
<label|finite subset of filterbase>Let <math|X> be a topological space
and <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
is a filter base then for each finite
<math|{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n>}\<subseteq\>\<bbb-U\>>
there exists a <math|\<gamma\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<gamma\>>\<subseteq\><big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,a<rsub|n>}>A<rsub|\<alpha\>>>
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1\<Rightarrow\>{\<alpha\><rsub|1>}\<in\>\<bbb-U\>>, then
<math|A<rsub|\<alpha\><rsub|1>>\<subseteq\>A<rsub|\<alpha\><rsub|1>>=<big|cap><rsub|\<alpha\>\<in\>{a<rsub|1>}>A<rsub|\<alpha\>>>
<item>Assume the theorem is true for <math|n>, then we prove it for
<math|n+1> then <math|<big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n+1>}>A<rsub|\<alpha\>>=(<big|cap><rsub|a\<in\>{a<rsub|1>,\<ldots\>,a<rsub|n>}>A<rsub|\<alpha\>>)<big|cap>A<rsub|\<alpha\><rsub|n+1>>>
and by the induction hypothesis <math|\<exists\>\<gamma\>\<in\>\<bbb-U\>>
such that <math|A<rsub|\<gamma\>>\<subseteq\><big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n>}>A<rsub|a>\<Rightarrow\>A<rsub|\<gamma\>><big|cap>A<rsub|\<alpha\><rsub|n+1>>\<subseteq\>(<big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n>}>A<rsub|\<alpha\>>)<big|cap>A<rsub|\<alpha\><rsub|n+1>>=<big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n+1>}>A<rsub|\<alpha\>>>
then there exists a <math|\<beta\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<beta\>>\<subseteq\>A<rsub|\<gamma\>><big|cap>A<rsub|\<alpha\><rsub|n+1>>=<big|cap><rsub|\<alpha\>\<in\>{\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n+1>}>A<rsub|\<alpha\>>>
</enumerate>
</proof>
<\example>
<label|neighborhood filterbase><index|neighborhood filterbase>Let
<math|X> be a topological space then if <math|\<frak-U\>(x)={U\|U> is
open and <math|x\<in\>U}> (the set of all the open neighborhoods of
<math|x\<in\>X>) then <math|\<frak-U\>(x)={U}<rsub|U\<in\>\<frak-U\>(x)>>
(here <math|\<frak-U\>(x)> is indexed by itself by the identity map) is a
filter base. This is filter base is called the neighborhood filter base.\
</example>
<\proof>
\;
<\enumerate>
<item>If <math|U\<in\>\<frak-U\>(x)\<Rightarrow\>x\<in\>U,U> is open so
obviously <math|U\<neq\>\<emptyset\>>
<item>If <math|U,V\<in\>\<frak-U\>(x)\<Rightarrow\>x\<in\>U,x\<in\>V>
and <math|U,V> is open then <math|x\<in\>U<big|cap>V\<neq\>\<emptyset\>>
and by the definition of a topology <math|U<big|cap>V> is open so
<math|U<big|cap>V\<in\>\<frak-U\>(x)> and
<math|U<big|cap>V\<subseteq\>U<big|cap>V>
</enumerate>
</proof>
<\example>
<label|trivial filterbase>Let <math|X> be a topological space and let
<math|\<emptyset\>\<neq\>A\<subseteq\>X> then <math|{A}> is a filter base
which is trivial as <math|A\<neq\>\<emptyset\>> and
<math|A\<subseteq\>A<big|cap>A>
</example>
<\theorem>
<label|construction of new filterbases>Let
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>,\<frak-W\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-W\>>>
be two filter bases in <math|X> then:
<\enumerate>
<item><math|\<frak-U\><big|cup>\<frak-W\>={A<rsub|\<alpha\>><big|cup>B<rsub|\<beta\>>}<rsub|(\<alpha\>,\<beta\>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>>
is a filter base
<item>If <math|\<forall\>\<alpha\>\<in\>\<bbb-U\>,\<forall\>\<beta\>\<in\>\<bbb-W\>>
we have that <math|A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>\<neq\>\<emptyset\>>
then <math|\<frak-U\><big|cap>\<frak-W\>={A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>}<rsub|(\<alpha\>,\<beta\>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>>
is a filter base
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<frak-U\><big|cap>\<frak-W\>> is a filter base
<\enumerate>
<item><math|\<forall\>(\<alpha\>,\<beta\>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>
we have <math|A<rsub|\<alpha\>>\<neq\>\<emptyset\>,B<rsub|\<beta\>>\<neq\>\<emptyset\>\<Rightarrow\>A<rsub|\<alpha\>><big|cup>B<rsub|\<beta\>>\<neq\>\<emptyset\>>
<item><math|\<forall\>(\<alpha\><rsub|1>,\<beta\><rsub|1>),(\<alpha\><rsub|2>,\<beta\><rsub|2>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>
then there exists <math|\<gamma\><rsub|1>\<in\>\<bbb-U\>,\<gamma\><rsub|2>\<in\>\<bbb-W\>>
such that <math|A<rsub|\<gamma\><rsub|1>>\<subseteq\>A<rsub|\<alpha\><rsub|1>><big|cap>A<rsub|\<alpha\><rsub|2>>,B<rsub|\<gamma\><rsub|2>>\<subseteq\>B<rsub|\<beta\><rsub|1>><big|cap>B<rsub|\<beta\><rsub|2>>>
then <math|(\<gamma\><rsub|1>,\<gamma\><rsub|2>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>
and <math|A<rsub|\<gamma\><rsub|1>><big|cup>B<rsub|\<gamma\><rsub|2>>\<subseteq\>(A<rsub|\<alpha\><rsub|1>><big|cap>A<rsub|\<alpha\><rsub|2>>)<big|cup>(B<rsub|\<beta\><rsub|1>><big|cap>B<rsub|\<beta\><rsub|2>>)\<subseteq\>A<rsub|\<alpha\><rsub|1>><big|cup>B<rsub|\<beta\><rsub|1>>>
(and <math|\<subseteq\>A<rsub|\<alpha\><rsub|2>><big|cup>B<rsub|\<beta\><rsub|2>>>)
and thus <math|A<rsub|\<gamma\><rsub|1>><big|cup>A<rsub|\<gamma\><rsub|2>>\<subseteq\>(A<rsub|\<alpha\><rsub|1>><big|cup>B<rsub|\<beta\><rsub|1>>)<big|cap>(A<rsub|\<alpha\><rsub|2>><big|cup>B<rsub|\<beta\><rsub|2>>)>
</enumerate>
<item><math|\<frak-U\><big|cap>\<frak-W\>> is a filter base
<\enumerate>
<item><math|\<forall\>(\<alpha\>,\<beta\>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>
we have by the hypothesis that <math|A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>\<neq\>\<emptyset\>>
<item><math|\<forall\>(\<alpha\><rsub|1>,\<beta\><rsub|1>),(\<alpha\><rsub|2>,\<beta\><rsub|2>)\<in\>\<bbb-U\>*\<times\>\<bbb-W\>>
then <math|\<exists\>\<gamma\><rsub|1>\<in\>\<bbb-U\>,\<exists\>\<gamma\><rsub|2>\<in\>\<bbb-W\>>
such that <math|A<rsub|\<gamma\><rsub|1>>\<subseteq\>A<rsub|\<alpha\><rsub|1>><big|cap>A<rsub|\<alpha\><rsub|2>>,
B<rsub|\<gamma\><rsub|2>>\<subseteq\>B<rsub|\<beta\><rsub|1>><big|cap>B<rsub|\<beta\><rsub|2>><rsub|>>
Then <math|(\<gamma\><rsub|1>,\<gamma\><rsub|2>)\<in\>\<bbb-U\>\<times\>\<bbb-W\>>
and <math|A<rsub|\<gamma\><rsub|1>><big|cap>B<rsub|\<gamma\><rsub|2>>\<subseteq\>A<rsub|\<alpha\><rsub|1>><big|cap>A<rsub|\<alpha\><rsub|2>><big|cap>B<rsub|\<beta\><rsub|1>><big|cap>B<rsub|\<beta\><rsub|2>>=(A<rsub|\<alpha\><rsub|1>><big|cap>B<rsub|\<beta\><rsub|1>>)<big|cap>(A<rsub|\<alpha\><rsub|2>><big|cap>B<rsub|\<beta\><rsub|2>>)<rsub|>>
</enumerate>
</enumerate>
</proof>
<\theorem>
<label|filter base has the finite intersection property>Let <math|X> be a
topological space and <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
be a filter base then it has the finite intersection property (that is
<math|\<forall\>\<cal-B\>\<subseteq\>\<bbb-U\>,\<cal-B\>> is finite we
have <math|<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>\<neq\>\<emptyset\>>)
</theorem>
<\proof>
Using <reference|finite subset of filterbase>
<math|\<exists\>\<gamma\>\<in\>\<bbb-U\>> such that
<math|\<emptyset\>\<neq\>A<rsub|\<gamma\>>\<subseteq\><big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>>
</proof>
<\definition>
<index|convergent filterbase><index|accumulate (filterbase)>Let
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> be a
filter base in a topological space <math|X> then we define
<\enumerate>
<item><math|\<frak-U\>> converges to <math|y\<in\>X> written as
<math|\<frak-U\>\<rightarrow\>y> iff an only if <math|\<forall\>U> with
<math|y\<in\>U,U> open then <math|\<exists\>\<alpha\>\<in\>\<bbb-U\>>
with <math|A<rsub|\<alpha\>>\<subseteq\>U>
<item><math|\<frak-U\>> accumulates at y written as
<math|\<frak-U\>\<succ\>y> if and only if <math|\<forall\>U> with
<math|y\<in\>U,U> open then <math|\<forall\>\<alpha\>\<in\>\<bbb-U\>>
we have <math|U<big|cap>A<rsub|\<alpha\>>\<neq\>\<emptyset\>>
</enumerate>
</definition>
<\remark>
<label|accumulation and closure>If <math|\<frak-U\>\<succ\>y> then
<math|[\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have
[<math|\<forall\>U,y\<in\>U,U> open that
<math|A<rsub|\<alpha\>><big|cap>U\<neq\>\<emptyset\>]\<Leftrightarrowlim\><rsub|<reference|characterization
of closure>>y\<in\><wide|A<rsub|\<alpha\>>|\<bar\>>]\<Leftrightarrow\>y\<in\><big|cap><rsub|\<alpha\>\<in\>\<bbb-U\>><wide|A<rsub|\<alpha\>>|\<bar\>>>
so (2) really means <math|y\<in\><big|cap><rsub|\<alpha\>\<in\>\<bbb-U\>><wide|A<rsub|\<alpha\>>|\<bar\>>>
</remark>
<\example>
<label|neighborhood filterbase converts>Let <math|X> be a topological
space then <math|\<forall\>x\<in\>X> we have
<math|\<frak-U\>(x)\<rightarrow\>x>
</example>
<\proof>
If <math|x\<in\>U,U> open then <math|U\<in\>\<frak-U\>(x)> and
<math|U\<subseteq\>U>
</proof>
<\definition>
<index|subordinate filterbase>Let <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
and <math|\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>> be
two filter bases on a topological space <math|X> then <math|\<frak-W\>>
is subordinate to <math|\<frak-U\>> (noted by
<math|\<frak-W\>\<gg\>\<frak-U\>>) if and only of
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>\<succ\>\<exists\>\<beta\>\<in\>\<bbb-W\>>
such that <math|B<rsub|\<beta\>>\<subseteq\>A<rsub|\<alpha\>>>
</definition>
<\theorem>
<label|properties of two filterbases>Let <math|X> be a topological base
and <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>,\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>>be
two filter bases in a topological space then the following hold
<\enumerate>
<item>If <math|\<frak-U\>\<subseteq\>\<frak-W\>> (meaning that
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> there exists a
<math|\<beta\>\<in\>\<bbb-W\>> such that
<math|A<rsub|\<alpha\>>=B<rsub|\<beta\>>>, or in terms of a family that
<math|A(\<bbb-U\>)\<subseteq\>B(\<bbb-W\>)>) then
<math|\<frak-W\>\<gg\>\<frak-U\>>
<item>If <math|\<frak-W\>\<gg\>\<frak-U\>> then
<math|\<forall\>\<beta\>\<in\>\<bbb-W\>> we have
<math|\<forall\>\<alpha\>\<in\>\<frak-U\>> that
<math|A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>\<neq\>\<emptyset\>>
<item><math|\<frak-U\>\<rightarrow\>x> if and only if
<math|\<frak-U\>\<gg\>\<frak-U\>(x)>
<item>If <math|\<frak-W\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-W\>>,\<frak-W\><rprime|'>={A<rprime|'><rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-W\><rprime|'>>,\<frak-W\><rprime|''>={A<rprime|''><rsub|a>}<rsub|a\<in\>\<bbb-W\><rprime|''>>>
are filter bases such that <math|\<frak-W\>\<gg\>\<frak-W\><rprime|'>>
and <math|\<frak-W\><rprime|'>\<gg\>\<frak-W\><rprime|''>> then
<math|\<frak-W\>\<gg\>\<frak-W\><rprime|''>>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|\<alpha\>\<in\>\<bbb-U\>> then
<math|\<exists\>\<beta\>\<in\>\<bbb-W\>> such that
<math|A<rsub|\<alpha\>>=B<rsub|\<beta\>>> and thus
<math|B<rsub|\<beta\>>\<subseteq\>A<rsub|\<alpha\>>> so that
<math|\<frak-W\>\<gg\>\<frak-U\>>
<item>We prove this by contradiction so assume that
<math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> and
<math|\<exists\>\<beta\>\<in\>\<bbb-W\>> such that
<math|A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>=\<emptyset\>> then
since <math|\<frak-W\>\<gg\>\<frak-U\>>
<math|\<exists\>\<gamma\>\<in\>\<bbb-W\>> such that
<math|B<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>>> and then there
exists a <math|\<delta\>\<in\>\<bbb-W\>> such that
<math|B<rsub|\<delta\>>\<subseteq\>B<rsub|\<gamma\>><big|cap>B<rsub|\<beta\>>\<subseteq\>A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>\<Rightarrowlim\><rsub|B<rsub|\<delta\>>\<neq\>\<emptyset\>>A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>\<neq\>\<emptyset\>>
which is a contradiction.
<item><math|\<frak-U\>\<rightarrow\>x\<Leftrightarrow\>\<forall\>U>
open with <math|x\<in\>U> we have <math|\<exists\>\<alpha\>\<in\>\<bbb-U\>>
such that <math|A<rsub|\<alpha\>>\<subseteq\>U\<Leftrightarrow\>\<forall\>U\<in\>\<frak-U\>(x)>
then <math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U\<Leftrightarrow\>\<frak-U\>\<gg\>\<frak-U\>*(x)>
<item>Let <math|\<alpha\><rprime|''>\<in\>\<bbb-W\><rprime|''>> then
because of <math|\<frak-W\><rprime|'>\<gg\>\<frak-W\><rprime|''>> there
exists a <math|\<alpha\><rprime|'>\<in\>\<bbb-W\><rprime|'>> such that
<math|A<rprime|'><rsub|a<rprime|'>>\<subseteq\>A<rprime|''><rsub|a<rprime|''>>>
and then because of <math|\<frak-W\>\<gg\>\<frak-W\><rprime|'>> we have
<math|\<exists\>\<alpha\>\<in\>\<bbb-W\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>A<rprime|'><rsub|\<alpha\><rprime|'>>\<subseteq\>A<rprime|''><rsub|a<rprime|''>>\<Rightarrow\>A<rsub|\<alpha\>>\<subseteq\>A<rprime|''><rsub|\<alpha\><rprime|''>>\<Rightarrow\>\<frak-W\>\<gg\>\<frak-W\><rprime|''>>
</enumerate>
</proof>
<\theorem>
<label|each filterbase converges to exactly one point>Let <math|X> be a
topological space then <math|X> is Hausdorff <math|\<Leftrightarrow\>>
each filter base converges to exactly one point
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Assume that <math|X> is Hausdorff, <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
and let <math|\<frak-U\>\<rightarrow\>x> and
<math|\<frak-U\>\<rightarrow\>y> with <math|x\<neq\>y> then there exists
<math|U,V> open with <math|U<big|cap>V=\<emptyset\>> (Hausdorff property)
then <math|\<exists\>\<alpha\>,\<beta\>\<in\>\<bbb-U\>\<vdash\>A<rsub|\<alpha\>>\<subseteq\>U,A<rsub|\<beta\>>\<subseteq\>V\<Rightarrowlim\><rsub|definition
of filterbase>\<exists\>\<gamma\>\<in\>\<bbb-U\>\<vdash\>A<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>><rsub|>>
and thus <math|\<emptyset\>\<neq\>A<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>\<subset\>U<big|cap>V=\<emptyset\>>
<math|\<Leftarrow\>>
Assume that <math|X> is not Hausdorff then
<math|\<exists\>x,y\<in\>X,x\<neq\>y> such that <math|\<forall\>U,V> open
with <math|x\<in\>U,y\<in\>V> and <math|U<big|cap>V\<neq\>\<emptyset\>>
then by <reference|construction of new filterbases> we have that
<math|\<frak-W\>=\<frak-U\>(x)<big|cap>\<frak-U\>(y)> is a filter base
and if <math|x\<in\>U> (<math|U> open)<math|\<Rightarrow\>U\<in\>\<frak-U\>(x),X\<in\>\<frak-U\>(y)>
then <math|U=U<big|cap>X\<in\>\<frak-W\>> and thus
<math|\<frak-W\>\<rightarrow\>x,> in the same way we prove that
<math|\<frak-W\>\<rightarrow\>y> and by the hypothesis we have <math|x=y>
which contradicts our original assumption so we must have that <math|X>
is Hausdorff
</proof>
<\theorem>
<label|converging and accumulation>Let <math|X> be a topological space
and <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>,\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>>be
filter bases then\
<\enumerate>
<item>If <math|\<frak-U\>\<rightarrow\>x> then
<math|\<frak-U\>\<succ\>x> and if X is Hausdorff then if
<math|\<frak-U\>\<succ\>y\<Rightarrow\>x=y>
<item>Let <math|\<frak-W\>\<gg\>\<frak-U\>> then\
<\enumerate>
<item><math|\<frak-U\>\<rightarrow\>x\<Rightarrow\>\<frak-W\>\<rightarrow\>x>
<item><math|\<frak-W\>\<succ\>x\<Rightarrow\>\<frak-U\>\<succ\>x>
</enumerate>
</enumerate>
<\proof>
\;
<\enumerate>
<item>Let <math|U> be open with <math|x\<in\>U> then
<math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U> now given
<math|\<beta\>\<in\>\<bbb-U\>> then as (see <reference|finite
intersection property>) <math|\<emptyset\>\<neq\>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>\<subseteq\>U<big|cap>A<rsub|\<beta\>>\<Rightarrow\>\<frak-U\>\<succ\>x>.
Now if <math|X> is Hausdorff and let <math|\<frak-U\>\<succ\>y> with
<math|x\<neq\>y> then <math|\<exists\>U,V> open such that
<math|x\<in\>U,y\<in\>V> and <math|U<big|cap>V=\<emptyset\>> then
<math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U\<Rightarrow\>A<rsub|\<alpha\>><big|cap>V\<subseteq\>U<big|cap>V=\<emptyset\>>
contradicting the fact that <math|\<frak-U\>\<succ\>y>
<item>Assume that <math|\<frak-W\>\<gg\>\<frak-U\>> then\
<\enumerate>
<item>Assume that <math|\<frak-U\>\<rightarrow\>x> then if <math|U>
is open and <math|x\<in\>U> then
<math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U> and as
<math|\<frak-W\>\<gg\>\<frak-U\> >then
<math|\<exists\>\<beta\>\<in\>\<bbb-W\>> such that
<math|A<rsub|\<beta\>>\<subseteq\>A<rsub|\<alpha\>>\<subseteq\>U>
and thus <math|\<frak-W\>\<rightarrow\>x>
<item>Assume that <math|\<frak-W\>\<succ\>x> then if <math|U> is
open such that <math|x\<in\>U> then
<math|\<forall\>\<beta\>\<in\>\<bbb-W\>> we have
<math|U<big|cap>B<rsub|\<beta\>>\<neq\>\<emptyset\>>, now if
<math|\<alpha\>\<in\>\<bbb-U\>> then as
<math|\<frak-W\>\<gg\>\<frak-U\>>
<math|\<exists\>\<gamma\>\<in\>\<bbb-W\>> such that
<math|B<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>>\<Rightarrow\>\<emptyset\>\<neq\>U<big|cap>B<rsub|\<gamma\>>\<subseteq\>U<big|cap>A<rsub|\<alpha\>>\<Rightarrow\>\<frak-U\>\<succ\>x><math|>
</enumerate>
</enumerate>
</proof>
</theorem>
<\theorem>
Let <math|X> be a topological space and let
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> be a
filter base then
<\enumerate>
<item><math|\<frak-U\>\<rightarrow\>x\<Leftrightarrow\>\<forall\>\<frak-W\>,\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>>
a filter base with <math|\<frak-W\>\<gg\>\<frak-U\>> we have that there
exists a filter base <math|\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>>
with <math|\<frak-C\>\<gg\>\<frak-W\>> with
<math|\<frak-C\>\<rightarrow\>x>
<item><math|\<frak-U\>\<succ\>x\<Leftrightarrow\>\<exists\>\<frak-W\>,\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>>
a filter base with <math|\<frak-W\>\<gg\>\<frak-U\>> such that
<math|\<frak-W\>\<rightarrow\>x>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>
<\enumerate>
<item><math|\<Rightarrow\>> Assume <math|\<frak-U\>\<rightarrow\>x>
and <math|\<frak-W\>\<gg\>\<frak-U\>\<Rightarrowlim\><rsub|<reference|converging
and accumulation>>\<frak-W\>\<rightarrow\>x>, and as trivially
<math|\<frak-W\>\<gg\>\<frak-W\>> we can use
<math|\<frak-C\>=\<frak-W\>>
<item><math|\<Leftarrow\>>Assume that
<math|\<frak-U\>\<nrightarrow\>x> (<math|\<frak-U\>> does not
converge to x) then <math|\<exists\>U>open with <math|x\<in\>U> such
that <math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have
<math|A<rsub|\<alpha\>>\<nsubseteq\>U\<Rightarrow\>A<rsub|\<alpha\>><big|cap>(X<mid|\\>U)\<neq\>\<emptyset\>>
and then <math|\<frak-W\>={A<rsub|\<alpha\>><big|cap>(X
<mid|\\>U)}<rsub|\<alpha\>\<in\>\<bbb-U\>>> forms a filter base (see
<reference|trivial filterbase> and <reference|construction of new
filterbases>) and trivially <math|\<frak-W\>\<gg\>\<frak-U\>>
<math|>then <math|\<exists\>\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>>
such that <math|\<frak-C\>\<gg\>\<frak-W\>> and
<math|\<frak-C\>\<rightarrow\>x\<Rightarrowlim\><rsub|<reference|converging
and accumulation>>\<frak-W\>\<succ\>x> which means that
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> that
<math|(A<rsub|\<alpha\>><big|cap>(X<mid|\\>U))<big|cap>U\<neq\>\<emptyset\>>
which is a contradiction because <math|(A<rsub|\<alpha\>><big|cap>(X<mid|\\>U))<big|cap>U\<subseteq\>(X
<mid|\\>U)<big|cap>U=\<emptyset\>.> So we must have
<math|\<frak-U\>\<rightarrow\>x>
</enumerate>
<item>
<\enumerate>
<item><math|\<Rightarrow\>>Assume <math|\<frak-U\>\<succ\>x> then if
<math|U> is open such that <math|x\<in\>U> then
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have
<math|A<rsub|\<alpha\>><big|cap>U\<neq\>\<emptyset\>> and using
<reference|neighborhood filterbase> and <math|<reference|construction
of new filterbases>> we have <math|>that
<math|\<frak-W\>=\<frak-U\><big|cap>\<frak-U\>(x)={A<rsub|\<alpha\>><big|cap>U}<rsub|(\<alpha\>,U)\<in\>\<bbb-U\>\<times\>\<frak-U\>(x)>>
is a filter base <math|\<frak-W\>\<gg\>\<frak-U\>> [if
<math|\<alpha\>\<in\>\<bbb-U\>> then
<math|A<rsub|\<alpha\>><big|cap>X\<subseteq\>A<rsub|a>]>, also as a
filter base is not empty (<math|\<bbb-U\>\<neq\>\<emptyset\>)> we
have the existence of a <math|\<alpha\>\<in\>\<bbb-U\>> where
<math|A<rsub|\<alpha\>><big|cap>U\<subseteq\>U> proving that
<math|\<frak-W\>\<rightarrow\>x>\
<item><math|\<Leftarrow\>>If there exist a
<math|\<frak-W\>\<gg\>\<frak-U\>> with
<math|\<frak-W\>\<rightarrow\>x> then using <reference|converging and
accumulation> we have <math|\<frak-W\>\<succ\>x> and using
<reference|converging and accumulation> again that
<math|\<frak-U\>\<succ\>x>
</enumerate>
</enumerate>
</proof>
<\definition>
<index|filterbase on a set>Let X be a topological space and
<math|A\<subseteq\>X> then a filter base
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> is
said to be a on <math|A> if and only <math|\<forall\>\<alpha\>\<in\>\<bbb-U\>>
we have <math|A<rsub|\<alpha\>>\<subseteq\>A>\
</definition>
<\theorem>
<label|filterbase and closure>Let <math|X> be a topological space and let
<math|A\<subseteq\>X> then we have the following equivalence
<math|y\<in\><wide|A|\<bar\>>\<Leftrightarrow\>\<exists\>\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
a filter base on A with <math|\<frak-U\>\<rightarrow\>x>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Let <math|y\<in\><wide|A|\<bar\>>> then <math|\<forall\>U> open with
<math|y\<in\>U> we have <math|A<big|cap>U\<neq\>\<emptyset\>> so that
using <reference|trivial filterbase>,<reference|neighborhood filterbase>
and <reference|construction of new filterbases> we have that
<math|\<frak-W\>={A}<big|cap>\<frak-U\>(y)> is a filter base, which is on
A [<math|A<big|cap>U\<subseteq\>A]> and if <math|U> is open with
<math|x\<in\>U> then <math|A<big|cap>U\<subseteq\>U> and
<math|A<big|cap>U\<in\>\<frak-W\>> so <math|\<frak-W\>\<rightarrow\>y>\
<math|\<Leftarrow\>>
If <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
is a filter base <with|font-series|bold|on> A with
<math|\<frak-U\>\<rightarrow\>y> then if <math|U> is open and
<math|y\<in\>U> there exists <math|\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U> and as
<math|A<rsub|\<alpha\>>\<neq\>\<emptyset\> > and
<math|A<rsub|\<alpha\>>\<subseteq\>A>
<math|\<Rightarrow\>\<emptyset\>\<neq\>A<rsub|\<alpha\>>=A<rsub|\<alpha\>><big|cap>A\<subseteq\>U<big|cap>A\<Rightarrow\>y\<in\><wide|A|\<bar\>>>
</proof>
<\theorem>
<label|mapping of a filterbase>Let <math|X,Y> be topological spaces and
let <math|f:X\<rightarrow\>Y> \ be a function then if <math|\<frak-U\>>
is a filter base then <math|f(\<frak-U\>)={f(A<rsub|\<alpha\>>)}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
is also a filter base
</theorem>
<\proof>
First as <math|\<bbb-U\>\<neq\>\<emptyset\>><math|> and
<math|\<emptyset\>\<neq\>A<rsub|\<alpha\>>\<Rightarrow\>f(A<rsub|\<alpha\>>)\<neq\>\<emptyset\>>
so <math|{f(A<rsub|\<alpha\>>)}<rsub|\<alpha\>\<in\>\<bbb-U\>>> is a
nonempty family of non empty sets. Second if
<math|\<alpha\>,\<beta\>\<in\>\<bbb-U\>> then
<math|\<exists\>\<gamma\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>\<Rightarrow\>f(A<rsub|\<gamma\>>)\<subseteq\>f(A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>)\<subseteq\>f(A<rsub|\<alpha\>>)<big|cap>f(A<rsub|\<beta\>>)>
</proof>
<\theorem>
<label|filterbase and continuity 1>Let <math|X,Y> be topological space
and there be a function <math|f:X\<rightarrow\>Y> then <math|f> is
continuous at x \ if and only <math|f(\<frak-U\>(x))\<rightarrow\>f(x)>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Assume that <math|f> is continuous and let <math|V> open with
<math|f(x)\<in\>V> then <math|\<exists\>U> open with <math|x\<in\>U> and
<math|f(U)\<subseteq\>V\<Rightarrow\>f(\<frak-U\>(x))\<rightarrow\>f(x)>
<math|\<Leftarrow\>>
Let <math|V> be open and <math|f(x)\<in\>V> then as
<math|f(\<frak-U\>(x))\<rightarrow\>f(x)> there exists a
<math|A\<in\>f(\<frak-U\>(x))> such that <math|A\<subseteq\>f> and by
definition of <math|f(\<frak-U\>(x))> there exists a <math|U> open with
<math|x\<in\>U> and <math|A=f(U)> thus proving the continuity
</proof>
<\theorem>
<label|filterbase and continuity 2>Let <math|X,Y> be topological spaces
and let <math|f:X\<rightarrow\>Y> then <math|f> is continuous on <math|X>
if and only if <math|\<forall\>x\<in\>X> and <math|\<forall\>\<frak-U\>,
\ \<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> a filter
base with <math|\<frak-U\>\<rightarrow\>x> we have
<math|f(\<frak-U\>)\<rightarrow\>f(x)>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Assume <math|f> is continuous and let <math|\<frak-U\>\<rightarrow\>x>
then by <reference|properties of two filterbases> we have
<math|\<frak-U\>\<gg\>\<frak-U\>(x)> so if <math|U> is open and
<math|x\<in\>U> then <math|\<exists\>\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>U> and then
<math|f(A<rsub|\<alpha\>>)\<subseteq\>f(U)> and thus
<math|f(\<frak-U\>)\<gg\>f(\<frak-U\>(x))>. Then by using
<reference|filterbase and continuity 1> we have that
<math|f(\<frak-U\>(x))\<rightarrow\>f(x)> and then using
<reference|converging and accumulation> we have
<math|f(\<frak-U\>)\<rightarrow\>f(x)>
<math|\<Leftarrow\>>
Let <math|A\<subseteq\>X> and let <math|y\<in\>f(<wide|A|\<bar\>>)> then
<math|\<exists\>x\<in\><wide|A|\<bar\>>> with <math|y=f(x)> then by
<reference|filterbase and closure> there exists a filter base
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> on
<math|A> such that <math|\<frak-U\>\<rightarrow\>x> and then by the
hypothesis we have <math|f(\<frak-U\>)\<rightarrow\>f(x)> and because
<math|\<frak-U\>> is a filter base on <math|A> we have
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> that
<math|A<rsub|\<alpha\>>\<subseteq\>A> so that
<math|f(A<rsub|\<alpha\>>)\<subseteq\>f(A)> and thus <math|f(\<frak-U\>)>
is on <math|f(A)> and this together with
<math|f(\<frak-U\>)\<rightarrow\>f(x)> means by <reference|filterbase and
closure> that <math|f(x)\<in\><wide|f(A)|\<bar\>>> or
<math|f(<wide|A|\<bar\>>)\<subseteq\><wide|f(A)|\<bar\>>> and thus by
<reference|continuity characterization> that <math|f> is continuous
</proof>
<\theorem>
<label|filterbases and products>Let <math|{X<rsub|i>}<rsub|i\<in\>I>> be
a family of topological spaces then a filter base in
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> in
<math|<big|prod><rsub|i\<in\>I>X<rsub|i>> converges to
<math|y\<in\><big|prod><rsub|i\<in\>I>X<rsub|i>> (using the product
topology) if and only if <math|\<forall\>i\<in\>I> such that
<math|\<pi\><rsub|i>(\<frak-U\>)\<rightarrow\>\<pi\><rsub|i>(y)=y(i)=y<rsub|i>>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Let <math|\<frak-U\>\<rightarrow\>y> then as <math|\<pi\><rsub|i>> is
continuous (using the product topology on the product) we have by
<reference|filterbase and continuity 2> that
<math|\<pi\><rsub|i>(\<frak-U\>)\<rightarrow\>\<pi\><rsub|i>(y)>
<math|\<Leftarrow\>>
Let <math|U> be open with <math|y\<in\>U> then there exists a <math|B> in
the basis of the product topology such that <math|y\<in\>B\<subseteq\>U>
and by the definition of this basis there exists a finite set
<math|\<cal-A\>\<subseteq\>I> such that
<math|B=<big|cap><rsub|i\<in\>\<cal-A\>>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)>
where <math|U<rsub|i>> is a open set in <math|X<rsub|i>> then
<math|\<forall\>i\<in\>\<cal-A\>> we have
<math|\<pi\><rsub|i>(y)\<in\>U<rsub|i>> and as by the hypothesis we have
<math|\<forall\>i\<in\>I> that <math|\<pi\><rsub|i>(\<frak-U\>)\<rightarrow\>\<pi\><rsub|i>(y)>
there exists <math|\<forall\>i\<in\>\<cal-A\>> a
<math|\<alpha\><rsub|i>\<in\>\<bbb-U\>> such that
<math|\<pi\><rsub|i>(A<rsub|\<alpha\><rsub|i>>)\<subseteq\>U<rsub|i>\<Rightarrow\>A<rsub|\<alpha\><rsub|i>>\<subseteq\>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)\<Rightarrow\><big|cap><rsub|i\<in\>\<cal-A\>>A<rsub|\<alpha\><rsub|i>>\<subseteq\><big|cap><rsub|i\<in\>\<cal-A\>>\<pi\><rsub|i><rsup|-1>(U<rsub|i>)=B>
and by <reference|finite subset of filterbase><math|> and the finiteness
of <math|\<cal-A\>> there exists a <math|\<gamma\>\<in\>\<bbb-U\>> such
that <math|A<rsub|\<gamma\>>\<subseteq\><big|cap><rsub|i\<in\>\<cal-A\>>A<rsub|\<alpha\>i>\<subseteq\>B\<subseteq\>U\<Rightarrow\>\<frak-U\>\<rightarrow\>y>
</proof>
<\definition>
<index|maximal filterbase>A filter base <math|\<frak-U\>> in X is called
maximal (or a ultra fiddlers) if <math|\<forall\>\<frak-W\>,\<frak-W\>> a
filter base in <math|X> with <math|\<frak-W\>\<gg\>\<frak-U\>> we have
<math|\<frak-U\>\<gg\>\<frak-W\>> (in other words <math|\<frak-U\>> does
not has a proper subordinate filter base)
</definition>
<\theorem>
<label|maximal filterbase characterization>A filter base
<math|\<frak-U\>> in <math|X> is maximal if and only
<math|\<forall\>A\<subseteq\>Y> one and only one of the two sets
<math|A,X<mid|\\>A> contains a member of <math|\<frak-U\>>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Assume <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
is maximal and let <math|A\<subseteq\>X> then if there exists a
<math|\<alpha\>,\<beta\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>A> and
<math|A<rsub|\<beta\>>\<subseteq\>X<mid|\\>A> then
<math|\<emptyset\>\<neq\><rsub|<reference|finite subset of
filterbase>>A<rsub|\<alpha\>><big|cap>A<rsub|\<beta\>>\<subseteq\>A<big|cap>(X<mid|\\>A)=\<emptyset\>>
which is a contradiction, so we can not have a
<math|\<alpha\>,\<beta\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>A,A<rsub|\<beta\>>\<subseteq\>X<mid|\\>A>.
Assume now that <math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have
<math|A<rsub|\<alpha\>>\<nsubseteq\>A> then we have
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> that
<math|(X<mid|\\>A)<big|cap>A<rsub|\<alpha\>>\<neq\>\<emptyset\>> so
<math|\<frak-W\>={(X<mid|\\>A)<big|cap>A<rsub|a>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
forms a filter base (see <reference|trivial filterbase> and
<reference|construction of new filterbases>) and trivially
<math|\<frak-W\>\<gg\>\<frak-U\>> then we must by maximality have that
<math|\<frak-U\>\<gg\>\<frak-W\>>. And thus
<math|\<forall\>\<alpha\>\<in\>\<bbb-U\>> we have a
<math|\<beta\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<beta\>>\<subseteq\>(X<mid|\\>A)<big|cap>A<rsub|\<beta\>>\<subseteq\>(X<mid|\\>A)>
<math|\<Leftarrow\>>
Let <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<beta\>>> be a filter base
such that \ <math|\<forall\>A\<subseteq\>Y> one and only one of the two
sets <math|A,X<mid|\\>A> contains a member of <math|\<frak-U\>> and let
<math|\<frak-W\>={B<rsub|\<beta\>>}<rsub|\<beta\>\<in\>\<bbb-W\>>> such
that <math|\<frak-W\>\<gg\>\<frak-U\>> then if
<math|\<beta\>\<in\>\<bbb-W\>> then <math|\<exists\>\<alpha\>\<in\>\<bbb-U\>>
such that either <math|A<rsub|\<alpha\>>\<subseteq\>B<rsub|\<beta\>>> or
<math|A<rsub|\<alpha\>>\<subseteq\>X<mid|\\>B<rsub|\<beta\>>>. The last
case would mean that <math|A<rsub|\<alpha\>><big|cap>B<rsub|\<beta\>>=\<emptyset\>>
contradicting <reference|properties of two filterbases>, so we are left
with <math|A<rsub|\<alpha\>>\<subseteq\>B<rsub|\<beta\>>> or we have
<math|\<frak-U\>\<gg\>\<frak-W\>> and thus <math|\<frak-U\>> is maximal
</proof>
<\theorem>
<label|existance of maximal filterbase>Let
<math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>> be
any filter base in a topological space <math|X> then there exists a
maximal filter base <math|\<frak-W\>> in <math|X> with
<math|\<frak-W\>\<gg\>\<frak-U\>>
</theorem>
<\proof>
Let <math|\<cal-A\>={\<frak-W\>\|\<frak-W\>> a filter base such that
<math|\<frak-W\>\<gg\>\<frak-U\>>} then
<math|\<cal-A\>\<neq\>\<emptyset\>> because
<math|\<frak-U\>\<gg\>\<frak-U\>> and thus
<math|\<frak-U\>\<in\>\<cal-A\>>. We define now a preorder
<math|\<leqslant\>> on <math|\<cal-A\>> defined by
<math|\<frak-W\>\<leqslant\>\<frak-W\><rprime|'>\<Leftrightarrow\>\<frak-W\><rprime|'>\<gg\>\<frak-W\>>,
this is a preorder because\
<\enumerate>
<item>(reflexivity) <math|\<frak-W\>\<gg\>\<frak-W\>\<Rightarrow\>\<frak-W\>\<leqslant\>\<frak-W\>>
<item>(transitivity) Assume that <math|\<frak-W\>\<leqslant\>\<frak-W\><rprime|'>>
and <math|\<frak-W\><rprime|'>\<leqslant\>\<frak-W\><rprime|''>\<Rightarrow\>\<frak-W\><rprime|''>\<gg\>\<frak-W\><rprime|'>>
and <math|\<frak-W\><rprime|'>\<gg\>\<frak-W\>> and then using
<reference|properties of two filterbases> (4) we have
<math|\<frak-W\><rprime|''>\<gg\>\<frak-W\>\<Rightarrow\>\<frak-W\>\<leqslant\>\<frak-W\><rprime|''>>
</enumerate>
Let now <math|\<cal-C\>\<subseteq\>\<cal-A\>> be any chain in
<math|\<cal-A\>> and prove that it has a upper bound, if
<math|\<cal-C\>=\<emptyset\>> then we have vacuous that
<math|\<forall\>\<frak-C\>\<in\>\<cal-C\>> we have
<math|\<frak-C\>\<leqslant\>\<frak-U\>> so
<math|\<frak-U\>\<in\>\<cal-A\>> is a upper bound of <math|\<cal-C\>>. So
we are left with the case <math|\<cal-C\>\<neq\>\<emptyset\>>. Define
<math|\<frak-B\>={A\|\<exists\>\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>\<in\>\<cal-C\>\<vdash\>\<exists\>\<gamma\>\<in\>\<bbb-C\>A
with =C<rsub|\<gamma\>>}> then <math|\<frak-B\>={B}<rsub|B\<in\>\<frak-B\>>>
(meaning <math|\<frak-B\>> indexed by itself) is a filter base because
<\enumerate>
<item>If <math|B\<in\>\<frak-B\>> then
<math|\<exists\>\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>\<in\>\<cal-C\>>
such that <math|\<exists\>\<gamma\>\<in\>\<bbb-C\>> with
<math|B=C<rsub|\<gamma\>>\<neq\>\<emptyset\>>
<item>Now if <math|B<rsub|1>,B<rsub|2>\<in\>\<frak-B\>> then
<math|\<exists\>\<frak-C\><rsub|1>={C<rsub|\<gamma\>><rsup|1>}<rsub|\<gamma\>\<in\>\<bbb-C\><rsub|1>>\<in\>\<cal-C\>,\<exists\>\<frak-C\><rsub|2>={C<rsup|2><rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\><rsub|2>>\<in\>\<cal-C\>>
such that <math|\<exists\>\<gamma\><rsub|1>\<in\>\<bbb-C\><rsub|1>,\<exists\>\<gamma\><rsub|2>\<in\>\<bbb-C\><rsub|2>>
and <math|B<rsub|1>=C<rsub|\<gamma\><rsub|1><rsub|>><rsup|1>,B<rsub|2>=C<rsub|\<gamma\><rsub|2><rsup|>><rsup|2>>.
Now as <math|\<cal-C\>> is a chain we have either\
<\enumerate>
<item><math|\<frak-C\><rsub|1>\<leqslant\>\<frak-C\><rsub|2>\<Rightarrow\>\<frak-C\><rsub|2>\<gg\>\<frak-C\><rsub|1>\<Rightarrow\>\<exists\>\<gamma\>\<in\>\<bbb-C\><rsub|2>>
such that <math|C<rsub|\<gamma\>><rsup|2>\<subseteq\>C<rsub|\<gamma\><rsub|1>><rsup|1>=B<rsub|1><rsub|>>
also because <math|\<frak-C\><rsub|1>> is a filter base we have
<math|\<exists\>\<beta\>\<in\>\<bbb-C\><rsub|2>> such that
<math|C<rsub|\<beta\>><rsup|2>\<subseteq\>C<rsub|\<gamma\>><rsup|2><big|cap>C<rsub|\<gamma\><rsub|2>><rsup|2>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
and we have then <math|B<rsub|3>=C<rsub|\<beta\>><rsup|2>\<in\>\<frak-B\>>
such that <math|B<rsub|3>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
<item><math|\<frak-C\><rsub|2>\<leqslant\>\<frak-C\><rsub|1>\<Rightarrow\>\<frak-C\><rsub|1>\<gg\>\<frak-C\><rsub|2>\<Rightarrow\>\<exists\>\<gamma\>\<in\>\<bbb-C\><rsub|1>>
such that <math|C<rsub|\<gamma\>><rsup|1>\<subseteq\>C<rsub|\<gamma\><rsub|2>><rsup|2>=B<rsub|2>>
also because <math|\<frak-C\><rsub|2>> is a filter base we have
<math|\<exists\>\<beta\>\<in\>\<bbb-C\><rsub|1>> such that
<math|C<rsub|\<beta\>><rsup|1>\<subseteq\>C<rsub|\<gamma\>><rsup|1><big|cap>C<rsub|\<gamma\><rsub|1>><rsup|1>\<subseteq\>B<rsub|2><big|cap>B<rsub|1>=B<rsub|1><big|cap>B<rsub|2>>
and we have then <math|B<rsub|3>=C<rsub|\<beta\>><rsup|1>\<in\>\<frak-B\>>
with <math|B<rsub|3>\<subseteq\>B<rsub|1><big|cap>B<rsub|2>>
</enumerate>
</enumerate>
As <math|\<cal-C\>\<neq\>\<emptyset\>\<Rightarrow\>\<exists\>\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>\<in\>\<cal-C\>\<subseteq\>\<cal-A\>>
so <math|\<frak-C\>\<gg\>\<frak-U\>> and thus if
<math|\<alpha\>\<in\>\<bbb-U\>> we have a <math|\<gamma\>\<in\>\<bbb-C\>>
such that <math|C<rsub|\<gamma\>>\<subseteq\>A<rsub|\<alpha\>>> and as
<math|C<rsub|\<gamma\>>\<in\>\<frak-B\>> (by definition) we have that
<math|\<frak-B\>\<gg\>\<frak-U\>\<Rightarrow\>\<frak-B\>\<in\>\<cal-A\>>.
Finally <math|\<forall\>\<frak-C\>={C<rsub|\<gamma\>>}<rsub|\<gamma\>\<in\>\<bbb-C\>>\<in\>\<cal-C\>>
we have that <math|\<forall\>\<gamma\>\<in\>\<bbb-C\>> that
<math|C<rsub|\<gamma\>>\<subseteq\>C<rsub|\<gamma\>>> and because
<math|C<rsub|\<gamma\>>\<in\>\<frak-B\>> by the definition of
<math|\<frak-B\>> we have thus <math|\<frak-B\>\<gg\>\<frak-C\>> or
<math|\<frak-C\>\<leqslant\>\<frak-B\>>. So we have proved that every
chain in <math|\<cal-A\>> has a upper bound and thus by Zorn's lemma
(<reference|zorn's lemma>) there exists a maximal element in
<math|\<cal-A\>>, or said otherwise <math|\<exists\>\<frak-W\>\<in\>\<cal-A\>>
such that <math|\<forall\>\<frak-A\>\<in\>\<cal-A\>> we have
<math|\<frak-A\>\<leqslant\>\<frak-W\>> or
<math|\<frak-W\>\<gg\>\<frak-A\>> and as
<math|\<frak-A\>\<gg\>\<frak-U\>> (<math|\<frak-A\>\<in\>\<cal-A\>>) we
have <math|\<frak-W\>\<gg\>\<frak-U\>>. Also if
<math|\<frak-Q\>\<gg\>\<frak-W\>> then as
<math|\<frak-W\>\<gg\>\<frak-U\>> we have
<math|\<frak-Q\>\<gg\>\<frak-U\>> and then
<math|\<frak-Q\>\<in\>\<cal-A\>> and thus
<math|\<frak-W\>\<gg\>\<frak-Q\>> proving that <math|\<frak-W\>> is a
maximal filter base with <math|\<frak-W\>\<gg\>\<frak-U\>>
</proof>
<\theorem>
<label|maximum filterbase and convergence>Let <math|X> be a topological
space and <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
be a maximal filter base then <math|\<frak-U\>\<succ\>x\<Leftrightarrow\>\<frak-U\>\<rightarrow\>x>
</theorem>
<\proof>
Because of <reference|converging and accumulation> we must only proof
that <math|\<frak-U\>\<succ\>x\<Rightarrow\>\<frak-U\>\<rightarrow\>x>
for a maximal filter base. So let <math|x\<in\>U,U> open then by
<reference|maximal filterbase characterization> there exists a
<math|\<alpha\>\<in\>\<bbb-U\>> such that either
<math|A<rsub|\<alpha\>>\<subseteq\>U> or
<math|A<rsub|\<alpha\>>\<subseteq\>X<mid|\\>U> now as
<math|\<frak-U\>\<succ\>x> we must have
<math|A<rsub|a><big|cap>U\<neq\>\<emptyset\>> which contradicts
<math|A<rsub|\<alpha\>>\<subseteq\>X<mid|\\>U> and thus
<math|A<rsub|\<alpha\>>\<subseteq\>U> or <math|\<frak-U\>\<rightarrow\>x>
</proof>
<\theorem>
<label|maximal filterbase and mapping>Let <math|X,Y> be topological
spaces and let <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
be a maximal filter base in <math|X> then for any function
<math|f:X\<rightarrow\>Y> we have that <math|f(\<frak-U\>)> is a maximal
filter base in <math|Y>
</theorem>
<\proof>
By <reference|mapping of a filterbase> we have that
<math|f(\<frak-U\>)={f(A<rsub|\<alpha\>>)<rsub|\<alpha\>\<in\>\<bbb-U\>>>
is a filter base, now to prove maximality we use <reference|maximal
filterbase characterization> so let <math|A\<subseteq\>Y> and take then
<math|f<rsup|-1>(A)> and <math|X <mid|\\>f<rsub|1>(A)> then by maximality
of <math|\<frak-U\>> we the existence of a
<math|\<alpha\>\<in\>\<bbb-U\>> such that
<math|A<rsub|\<alpha\>>\<subseteq\>f<rsup|-1>(A)> or
<math|A<rsub|\<alpha\>>\<subseteq\>X<mid|\\>f<rsup|-1>(A)\<equallim\><rsub|<reference|preimage
of difference>>f<rsup|-1>(Y<mid|\\>A)\<Rightarrow\>>we have either
<math|f(A<rsub|\<alpha\>>)\<subseteq\>A> or
<math|f(A<rsub|\<alpha\>>)\<subseteq\>Y<mid|\\>A> proving that
<math|f(\<frak-U\>)> is maximal
</proof>
<\theorem>
<label|filterbases and compactness>Let <math|X> be a topological space
then the following are equivalent
<\enumerate>
<item><math|X> is compact
<item>For each family <math|{A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>>
of closed sets in <math|X> with <math|<big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>A<rsub|\<alpha\>>=\<emptyset\>>
then there exists a finite family <math|\<cal-B\>={\<alpha\><rsub|1>,\<ldots\>,\<alpha\><rsub|n>}\<subseteq\>\<cal-A\>>
such that <math|<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>=\<emptyset\><rsub|>>
(otherwise said if <math|{A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>>
is a family such that for all finite
<math|\<cal-B\>\<subseteq\>\<cal-A\>> with
<math|<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>\<neq\>\<emptyset\>>
then <math|<big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>A<rsub|\<alpha\>>\<neq\>\<emptyset\>>
(finite intersection property))
<item>Each filter base <math|{A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
in <math|X> has at least one accumulation point
<item>Each maximal filter base in <math|X> converges
</enumerate>
</theorem>
<\proof>
\;
<math|1\<Rightarrow\>2>
Let <math|{A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>> be a family
of closed sets with <math|<big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>A<rsub|\<alpha\>>=\<emptyset\>>
then <math|X=X<mid|\\>\<emptyset\>=X<mid|\\>(<big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>A<rsub|\<alpha\>>)=<big|cup><rsub|\<alpha\>\<in\>\<cal-A\>>(X<mid|\\>A<rsub|\<alpha\>>)>
and because <math|X<mid|\\>A<rsub|\<alpha\>>> is open
<math|\<forall\>\<alpha\>\<in\>\<cal-A\>> we have by compactness the
existence of a finite <math|\<cal-B\>\<subseteq\>\<cal-A\>> such that
<math|X=<big|cup><rsub|\<beta\>\<in\>\<cal-B\>>(X<mid|\\>A<rsub|\<beta\>>)\<Rightarrow\>\<emptyset\>=X<mid|\\>X=X<mid|\\>(<big|cup><rsub|\<beta\>\<in\>\<cal-B\>>(X<mid|\\>A<rsub|\<beta\>>))=<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>(X<mid|\\>(X<mid|\\>A<rsub|\<beta\>>))=<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>>
<math|2\<Rightarrow\>1>
Let <math|{U<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>> be a family
of open sets such that <math|X=<big|cup><rsub|\<alpha\>\<in\>\<cal-A\>>U<rsub|\<alpha\>>\<Rightarrow\>\<emptyset\>=X<mid|\\>X=X<mid|\\>(<big|cup><rsub|\<alpha\>\<in\>\<cal-A\>>U<rsub|\<alpha\>>)=<big|cap><rsub|\<alpha\>\<in\>A>(X<mid|\\>U<rsub|\<alpha\>>)>
and as we have that <math|\<forall\>\<alpha\>\<in\>\<cal-A\>>
<math|X<mid|\\>U<rsub|\<alpha\>>> is closed we have by the hypothesis
that <math|\<exists\>\<cal-B\>\<subseteq\>\<cal-A\>>, <math|\<cal-B\>>
finite such that <math|\<emptyset\>=<big|cap><rsub|\<beta\>>(X
<mid|\\>U<rsub|\<beta\>>)=X=X \ <mid|\\>\<emptyset\>=X<mid|\\>(<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>(X<mid|\\>U<rsub|\<beta\>>))=<big|cup><rsub|\<beta\>\<in\>\<cal-B\>>(X<mid|\\>(X<mid|\\>U<rsub|\<beta\>>))=<big|cup><rsub|\<beta\>\<in\>\<cal-B\>>U<rsub|\<beta\>>>
proving compactness
<math|2\<Rightarrow\>3>
Let <math|\<frak-U\>={A<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<bbb-U\>>>
be a filter base, then <math|\<forall\>\<bbb-B\>\<subseteq\>\<bbb-U\>\<vdash\>\<bbb-B\>>
is finite we have that <math|\<emptyset\>\<neq\><big|cap><rsub|\<beta\>\<in\>\<bbb-B\>>A<rsub|\<beta\>>\<neq\>\<emptyset\>>
(see <reference|finite subset of filterbase>) then
<math|\<emptyset\>\<neq\><big|cap><rsub|\<beta\>\<in\>\<bbb-B\>>A<rsub|\<beta\>>\<subseteq\><big|cap><rsub|\<beta\>\<in\>\<bbb-B\>><wide|A<rsub|\<beta\>>|\<bar\>>>
and then by (2) we have <math|\<emptyset\>=<big|cap><rsub|\<alpha\>\<in\>\<bbb-A\>><wide|A<rsub|\<alpha\>>|\<bar\>>\<Rightarrow\>\<exists\>x\<in\><big|cap><rsub|\<alpha\>\<in\>>>
and then by <reference|accumulation and closure> we have that <math|x> is
a accumulation point of <math|\<frak-U\>>.\
<math|3\<Rightarrow\>4>
Let <math|\<frak-U\>> be a maximal filter base then by (3) there exists a
<math|x\<in\>X> such that <math|\<frak-U\>\<succ\>x>, and then using
<reference|maximum filterbase and convergence> we have
<math|\<frak-U\>\<rightarrow\>x>
<math|4\<Rightarrow\>3>
Let <math|\<frak-U\>> be a filter base then using <reference|existance of
maximal filterbase> there exist a maximal filter base
<math|\<frak-W\>\<gg\>\<frak-U\>> and then by (4) we have then
<samp|<math|\<exists\>x\<in\>X\<succ\>\<frak-W\>\<rightarrow\>x\<Rightarrowlim\><rsub|<reference|maximum
filterbase and convergence>>\<frak-W\>\<succ\>x>> and then using
<reference|converging and accumulation> we have
<math|\<frak-U\>\<succ\>x>
<math|3\<Rightarrow\>2>
Let <math|{F<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>> be a family
of closed sets such that <math|\<forall\>\<cal-B\>\<subseteq\>\<cal-A\>,\<cal-B\>>
finite we have <math|<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>F<rsub|\<beta\>>\<neq\>\<emptyset\>>
then if <math|\<frak-U\>={<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>A<rsub|\<beta\>>\|\<cal-B\>\<subseteq\>\<cal-A\>,\<cal-B\>>
is finite} we form the family <math|\<frak-U\>={A}<rsub|A\<in\>\<frak-U\>>>.
We prove now that <math|\<frak-U\>> is a filter base, first of all we
have by the hypothesis that <math|<big|cap><rsub|\<beta\>\<in\>\<cal-B\>>F<rsub|\<beta\>>\<neq\>\<emptyset\>>,
also if <math|B<rsub|1>,B<rsub|2>\<in\>\<frak-U\>> then
<math|\<exists\>\<cal-B\><rsub|1>,\<cal-B\><rsub|2>\<subseteq\>\<cal-A\>,\<cal-B\><rsub|1>,\<cal-B\><rsub|2>>
finite and thus <math|\<cal-B\><rsub|1><big|cup>\<cal-B\><rsub|2>> is
finite such that <math|B<rsub|1><big|cap>B<rsub|2>=(<big|cap><rsub|\<beta\>\<in\>\<cal-B\><rsub|1>>F<rsub|\<beta\>>)<big|cap>(<big|cap><rsub|\<beta\>\<in\>\<cal-B\><rsub|2>>F<rsub|\<beta\>>)=<big|cap><rsub|\<beta\>\<in\>\<cal-B\><rsub|1><big|cup>\<cal-B\><rsub|2>>F<rsub|\<beta\>>\<in\>\<frak-U\>>
so by (3) <math|\<exists\>x\<in\>X> such that <math|\<frak-U\>\<succ\>x>
or using <reference|accumulation and closure> we have
<math|x\<in\><big|cap><rsub|B\<in\>\<frak-U\>><wide|B|\<bar\>>> so
<math|\<forall\>B\<in\>\<frak-U\>> we have <math|x\<in\><wide|B|\<bar\>>>
now as <math|\<forall\>\<alpha\>\<in\>\<cal-A\>> we have
<math|{\<alpha\>}> is finite and thus
<math|F<rsub|\<alpha\>>=<big|cap><rsub|\<beta\>\<in\>{\<alpha\>}>F<rsub|\<alpha\>>\<in\>\<frak-U\>\<Rightarrow\>x\<in\><wide|F<rsub|\<alpha\>>|\<bar\>>\<equallim\><rsub|A<rsub|\<alpha\>>
is closed>F<rsub|\<alpha\>>\<Rightarrow\>x\<in\><big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>F<rsub|\<alpha\>>\<Rightarrow\><big|cap><rsub|\<alpha\>\<in\>\<cal-A\>>F<rsub|\<alpha\>>\<neq\>\<emptyset\>>
</proof>
<subsection|Product of compacts subsets>
All the work concerning filter bases was there to prove finally the
following theorem
<\theorem>
<label|Tychonoff><dueto|Tychonoff's><index|Tychonoff's theorem>Let
<math|{X<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>> be a family of
non empty topological spaces then <math|\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>>
is compact in the product topology if and only if
<math|\<forall\>\<alpha\>\<in\>\<cal-A\>\<succ\>X<rsub|\<alpha\>>> is
compact
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
As the projection map <math|\<pi\><rsub|\<beta\>>:<big|prod><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>\<rightarrow\>X<rsub|\<beta\>>>
is continuous (see <reference|projection map is continuous>) we have
using <reference|continuous image of compact sets> that
<math|X<rsub|\<beta\>>\<equallim\><rsub|<reference|image of
projection>>\<pi\><rsub|\<beta\>>(<big|prod><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>)>
is compact
<math|\<Leftarrow\>>
Assume that <math|\<forall\>\<alpha\>\<in\>\<cal-A\>\<succ\>X<rsub|\<alpha\>>>
is compact, and let <math|\<frak-U\>> be a maximum filter base in
<math|\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|a>> then using
<math|<reference|mapping of a filterbase> > we have that
<math|\<forall\>\<beta\>\<in\>\<cal-A\>\<succ\>\<pi\><rsub|\<beta\>>(\<frak-U\>)>
is a maximum filter base in <math|X<rsub|\<beta\>>> and then by
compactness of <math|X<rsub|\<beta\>>> we have using
<reference|filterbases and compactness> that
<math|\<exists\>x<rsub|\<beta\>>\<in\>X<rsub|\<beta\>>> such that
<math|\<pi\><rsub|\<beta\>>(\<frak-U\>)\<rightarrow\>x<rsub|\<beta\>>>.
Define then <math|x\<in\>\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>>
by <math|x(\<alpha\>)=x<rsub|\<alpha\>>> so that
<math|\<pi\><rsub|\<alpha\>>(\<frak-U\>)\<rightarrow\>\<pi\><rsub|\<alpha\>>(x)=x<rsub|\<alpha\>>>
and then by <reference|filterbases and products>
<math|\<frak-U\>\<rightarrow\>x>. finally using <reference|filterbases
and compactness> again we have that <math|\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>>
is compact.
</proof>
<\corollary>
<label|product of compact subspaces>Let
<math|{X<rsub|\<alpha\>>}<rsub|\<alpha\>\<in\>\<cal-A\>>> be a family of
non empty topological spaces then if <math|A<rsub|\<alpha\>>\<subseteq\>X<rsub|\<alpha\>>>
\ we have that <math|\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>>
is compact if and only if <math|\<forall\>\<alpha\>\<in\>\<cal-A\>\<succ\>A<rsub|\<alpha\>>>
is compact
</corollary>
<\proof>
\;
First note that the subspace topology of
<math|\<Pi\><rsub|\<alpha\>\<in\>\<cal-A\>>X<rsub|\<alpha\>>> is equal to
the product topology of the subspace topologies (see <reference|product
topology of subspace topolgies>) and as by definition a subset if it is
compact in the subspace topology we can use the previous theorem to prove
this theorem.
</proof>
<\theorem>
<label|compact subsets of the reals><dueto|Heine-Borel><index|Heine-Borel
theorem>Let <math|\<bbb-R\><rsup|n>> be the set of real tuples together
with the product topology (based on the trivial normed topology on
<math|\<bbb-R\>,\| \|>) then a subset of
<math|C\<subseteq\>\<bbb-R\><rsup|n>> is compact if and only if <math|C>
is bounded and closed
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
Using <reference|compact space in a metric space is bounded> we have that
<math|C> is bounded. As using <reference|metric space is
hausdorf>,<reference|product of metric spaces> we have that
<math|\<bbb-R\><rsup|n>> is Hausdorff and then using
<reference|compactness and closed sets> we have that <math|C> is closed.
<math|\<Leftarrow\>>
Assume now that <math|C> is closed and bounded. If <math|C> is empty then
it is trivially compact, so for the rest of the proof we assume that
<math|C> is not empty. So there exists a <math|c\<in\>C> and then by
boundedness we have for the product norm
<math|\<shortparallel\>\<shortparallel\>> defined by
<math|\<shortparallel\>x\<shortparallel\>=max(\|\<pi\><rsub|i>(x)\|\|i\<in\>{1,\<ldots\>,n})>
and thus the metric <math|d(x,y)=\<shortparallel\>x-y\<shortparallel\>=max(\|\<pi\><rsub|i>(x)-\<pi\><rsub|i>(y)\|
\| i\<in\>{1,\<ldots\>,n})> that <math|\<exists\>M\<gtr\>0> such that
<math|\<forall\>y\<in\>C\<succ\>d(x,y)\<leqslant\>M> or
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|\|\<pi\><rsub|i>(x)-\<pi\><rsub|i>(y)\|\<leqslant\>M>, or
<math|\<pi\><rsub|i>(y)\<in\>[\<pi\><rsub|i>(x)-M,\<pi\><rsub|i>(x)+M]>
and thus <math|C\<subseteq\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>[\<pi\><rsub|i>(x)-M,\<pi\><rsub|i>(x)+M]>
which is compact by <math|<reference|[a,b] is compact>> and
<reference|product of compact subspaces> (Tychonoff's theorem) and as
<math|C=C<big|cap>(\<Pi\><rsub|i\<in\>{1,\<ldots\>,n}>[\<pi\><rsub|i>(x)-M,\<pi\><rsub|i>(x)+M])>
is closed in the subspace topology of the compact set
<math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>[\<pi\><rsub|i>(x)-M,\<pi\><rsub|i>(x)+M]>
we use <reference|closed subsets are compact> to prove that <math|C> is
compact.
</proof>
<\theorem>
<label|real continuous mappings on a compact subset>Let
<math|f:X\<rightarrow\>\<bbb-R\>> be a continuous real function on a
topological space <math|X>. Then <math|\<forall\>C\<subseteq\>X,C>
compact in <math|X> we have <math|\<exists\>M<rsub|C>\<gtr\>0> such that
<math|\<forall\>x\<in\>C> we have <math|\|f(x)\|\<leqslant\>M<rsub|C>>
</theorem>
<\proof>
\;
Using <reference|continuous image of compact sets> and <reference|compact
space in a metric space is bounded> we have that <math|f(C)> is bounded
so <math|\<exists\>k\<gtr\>0> such that
<math|\<forall\>x,y\<in\>C\<succ\>\|x-y\|\<leqslant\>k>. Now if
<math|C=\<emptyset\>> then the theorem is vacuously satisfied so assume
that <math|\<exists\>c<rsub|0>\<in\>C> take then
<math|M<rsub|C>=k+\|f(c<rsub|0>)\|>. Then we have
<math|\<forall\>c\<in\>C\<succ\>\|f(c)\|\<leqslant\>\|f(c)-f(c<rsub|0>)\|+\|f(c<rsub|0>)\|\<leqslant\>k+\|f(c<rsub|0>)\|=M<rsub|C>>
</proof>
<\theorem>
<label|existance of maximum and minimum in the image of a continuous
function of a compect set>Let <math|f:X\<rightarrow\>\<bbb-R\>> be a
continuous function on a topological space <math|X>. Then for every non
empty compact subset <math|C\<subseteq\>X> we have
<math|\<exists\>x<rsub|1>,x<rsub|2>\<in\>C> such that
<math|\<forall\>c\<in\>C\<succ\>f(x<rsub|1>)\<leqslant\>f(c)\<leqslant\>f(x<rsub|2>)>
</theorem>
<\proof>
\;
According to <reference|continuous image of compact sets> we have that
<math|f(C)> is compact and then using <reference|compact subsets of the
reals> we have that <math|f(C)> is bounded and closed. Also using
<reference|real continuous mappings on a compact subset> we have that
<math|\<forall\>c\<in\>C> that <math|-M<rsub|C>\<leqslant\>f(c)\<leqslant\>M<rsub|C>>
and thus by the lower (upper) bound property of the real's we have that
<math|M<rsub|1>=inf(f(c)\|c\<in\>C},M<rsub|2>=sup(f(c)\|c\<in\>C}>
exists. Now if <math|]a,b[> is a open interval containing
<math|M<rsub|1>> (or <math|M<rsub|2>>) then there exists <math|c\<in\>C>
such that <math|a\<less\>M<rsub|1>\<leqslant\>f(c)\<less\>b> (or
<math|a\<less\>f(c)\<leqslant\>M<rsub|2>\<less\>b>) this means that
<math|M<rsub|1>,M<rsub|2>\<in\><wide|f(C)|\<bar\>>\<equallim\><rsub|f(C)
is closed>f(C)>. Hence <math|\<exists\>x<rsub|1>,x<rsub|2>> such that
<math|f(x<rsub|1>)=M<rsub|1>,f(x<rsub|2>)=M<rsub|2>> and
<math|\<forall\>c\<in\>C> we have <math|f(x<rsub|1>)=M<rsub|1>\<leqslant\>f(c)\<leqslant\>M<rsub|2>=f(x<rsub|2>)>
</proof>
<\theorem>
<label|equivalence of norms on product of real numbers>All norms on
<math|\<bbb-R\><rsup|n>> are equivalent
</theorem>
<\proof>
Given a norm <math|\<shortparallel\>\<shortparallel\><rsup|\<ast\>>> on
<math|\<bbb-R\><rsup|n>> then using <reference|norms mappings are
continuous in the product of reals> we have that
<math|\<shortparallel\>.\<shortparallel\><rsup|\<ast\>>> is a continuous
function in the maximum norm topology
<math|\<shortparallel\>\<shortparallel\><rsub|n>> on
<math|\<bbb-R\><rsup|n>>. Take then <math|S<rsup|n-1>={x\<in\>\<bbb-R\><rsup|n>\|
\<shortparallel\>x\<shortparallel\><rsub|n>=1}> which is closed because
<math|S<rsup|n-1>=\<shortparallel\>\<shortparallel\><rsub|n><rsup|-1>({1})>,<math|\<shortparallel\>.\<shortparallel\>:\<bbb-R\><rsup|n>\<rightarrow\>\<bbb-R\>>
is continuous by <reference|continuity of vectorspace operations>,
<math|{x}> is closed in <math|\<bbb-R\>> (see <reference|metric space is
hausdorf> and <reference|every finite subset of a hausdorf space is
closed>) and by <reference|continuity characterization>. It is also
bounded because <math|\<forall\>x,y\<in\>S<rsup|n-1>> we have
<math|\<shortparallel\>x-y\<shortparallel\><rsub|m>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|m>+\<shortparallel\>y\<shortparallel\><rsub|m>\<leqslant\>2>
so according to <reference|compact subsets of the reals>
<math|S<rsup|n-1>> is compact and thus from <reference|existance of
maximum and minimum in the image of a continuous function of a compect
set> <math|\<exists\>x<rsub|1>,x<rsub|2>> with
<math|\<alpha\>=\<shortparallel\>x<rsub|1>\<shortparallel\><rsup|\<ast\>>,\<beta\>=\<shortparallel\>x<rsub|2>\<shortparallel\><rsup|\<ast\>>>
and <math|\<alpha\>\<leqslant\>\<shortparallel\>x\<shortparallel\>\<leqslant\>\<beta\>>
for all <math|x\<in\>S<rsup|n-1>> (meaning
<math|\<shortparallel\>x\<shortparallel\>=1>) we have then for
<math|x\<in\>\<bbb-R\><rsup|n>> either\
<\enumerate>
<item><math|x=0\<Rightarrow\>\<shortparallel\>x\<shortparallel\>=0=\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<Rightarrow\>\<alpha\>.\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<leqslant\>\<beta\>\<shortparallel\>x\<shortparallel\>>
<item><math|x\<neq\>0\<Rightarrow\>\<shortparallel\>x\<shortparallel\>\<neq\>0\<Rightarrow\>\<shortparallel\><frac|1|\<shortparallel\>x\<shortparallel\>>x\<shortparallel\>=1\<Rightarrow\>\<alpha\>\<leqslant\>\<shortparallel\><frac|1|\<shortparallel\>x\<shortparallel\>>x\<shortparallel\><rsup|\<ast\>>\<leqslant\>\<beta\>\<Rightarrow\>\<alpha\>\<leqslant\><frac|1|\<shortparallel\>x\<shortparallel\>>\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<leqslant\>\<beta\>\<Rightarrow\>\<alpha\>.\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsup|\<ast\>>\<leqslant\>\<beta\>.\<shortparallel\>x\<shortparallel\>>
</enumerate>
Using <reference|equivalence of norms> we find that
<math|\<shortparallel\>\<shortparallel\>> and
<math|\<shortparallel\>\<shortparallel\><rsup|\<ast\>>> are equivalent,
so every norm generates the same topology as
<math|\<shortparallel\>\<shortparallel\>> and thus all norms are
equivalent
</proof>
<\corollary>
<label|linear maps between finite dimensional spaces are continuous>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be finite dimensional
spaces then every linear function <math|L:X\<rightarrow\>Y> is
continuous.
</corollary>
<\proof>
Using <reference|there exists a isometry between a finite dimensional
normed space and the product of reals with a induced norm> there exists
<math|\<varphi\><rsub|1>:X\<rightarrow\>\<bbb-R\><rsup|n>,
\<varphi\><rsub|2>:Y\<rightarrow\>\<bbb-R\><rsup|m>> where
<math|\<varphi\><rsub|1>,\<varphi\><rsub|2>> are isometries,
homeomorphism and isomorphisms (using norms
<math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
on <math|\<bbb-R\><rsup|n>,\<bbb-R\><rsup|m>)>. So given a linear map
<math|L:X\<rightarrow\>Y> we have that
<math|L<rprime|'>=\<varphi\><rsub|2>\<circ\>L\<circ\>\<varphi\><rsub|1><rsup|-1>:\<bbb-R\><rsup|n>\<rightarrow\>\<bbb-R\><rsup|m>>
is linear (isomorphisms are linear). And we use <reference|linear maps
between products of the real spaces are continuous> to prove that
<math|L<rprime|'>> is continuous using the maximum norm on
<math|\<bbb-R\><rsup|n>,\<bbb-R\><rsup|m>>. The equivalence of norms on
<math|\<bbb-R\><rsup|n>,\<bbb-R\><rsup|m>> (see <reference|equivalence of
norms on product of real numbers>) means then that <math|L<rprime|'>> is
also continuous in the norms <math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
on <math|\<bbb-R\><rsup|n>,\<bbb-R\><rsup|m>>. Then as
<math|L=\<varphi\><rsup|-1><rsub|2>\<circ\>L<rprime|'>\<circ\>\<varphi\><rsub|1>>
we have by the fact that <math|\<varphi\><rsub|1>,\<varphi\><rsub|2>> is
a homeomorphism (and thus <math|\<varphi\><rsub|1>,\<varphi\><rsub|2><rsup|-1>>
are continuous) that <math|L> is indeed continuous by
<reference|continuitity of composition>.
</proof>
<\corollary>
Every finite n-dimensional normed space
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> is isomorphic with
every other n-dimension normed space <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
</corollary>
<\proof>
Using <reference|there exists a isometry between a finite dimensional
normed space and the product of reals with a induced norm> there exists
norms <math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
on <math|\<bbb-R\><rsup|n>> and functions
<math|\<varphi\><rsub|1>:X\<rightarrow\>\<bbb-R\><rsup|n>,\<varphi\><rsub|2>:Y\<rightarrow\>\<bbb-R\><rsup|n>>
such that <math|\<varphi\><rsub|1>,\<varphi\><rsub|2>> are isomorphisms
and homeomorphisms. Then using <reference|equivalence of norms on product
of real numbers> we have that the topologies generated by
<math|\<shortparallel\>\<shortparallel\><rsub|1>,
\<shortparallel\>\<shortparallel\><rsub|2>> are equivalent and thus
<math|\<varphi\><rsub|1>,\<varphi\><rsub|2>,\<varphi\><rsub|2><rsup|-1>>
are also homeomorphism using the topology generated by
<math|\<shortparallel\>\<shortparallel\><rsub|1>> so that
<math|\<varphi\><rsub|2><rsup|-1>\<circ\>\<varphi\><rsub|1>:X\<rightarrow\>Y>
\ is a homeomorphism (by <reference|composition of homeomorphisms is a
homemorphism>) and a isomorphism (using <reference|inverse of
isomorphisme> and <reference|composition of isomorphisme>)
</proof>
<section|Complete Spaces>
<\notation>
Given a <math|m\<in\>\<bbb-N\>=\<bbb-N\><big|cup>{0}> we note by
<math|{m,\<ldots\>,\<infty\>}> the set
<math|{n\<in\>\<bbb-N\>\|n\<geqslant\>m}>.\
For example <math|{0,\<ldots\>,\<infty\>}=\<bbb-N\>> and
<math|{1\<ldots\>,\<infty\>}=\<bbb-N\><rsub|0>>
</notation>
<\definition>
<dueto|Cauchy property><index|Cauchy property>Let <math|X,d> be a metric
space then a sequence <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
of points in <math|X> is Cauchy iff <math|\<forall\>\<varepsilon\>\<gtr\>0>
there exists <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n,m\<geqslant\>N> we have
<math|d(x<rsub|n>,x<rsub|m>)\<less\>\<varepsilon\>>
</definition>
<\note>
<label|othere definition of Cauchy>A equivalent definition of the Cauchy
property is the following <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy if <math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n\<geqslant\>N> and <math|m\<in\>\<bbb-N\><rsub|0>> we
have <math|d(x<rsub|n>,x<rsub|n+m>)\<less\>\<varepsilon\>>
</note>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>If <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy then given <math|\<varepsilon\>\<gtr\>0> there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|n,m\<geqslant\>N> we have <math|d(x<rsub|n>x<rsub|m>)\<less\>\<varepsilon\>>.
So if <math|n\<geqslant\>N> and <math|k\<in\>\<bbb-N\><rsub|0>\<Rightarrow\>n+k\<geqslant\>N>
and thus <math|d(x<rsub|n>,x<rsub|n+k>)\<less\>\<varepsilon\>>
<item><math|\<Leftarrow\>> Let <math|\<varepsilon\>\<gtr\>0> then there
exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|n\<geqslant\>N> and <math|k\<in\>\<bbb-N\><rsub|0>> then
<math|d(x<rsub|n>,x<rsub|n+k>)\<less\>\<varepsilon\>> then if also
<math|m\<geqslant\>N> we have the following cases
<\enumerate>
<item><math|m=n\<Rightarrow\>d(x<rsub|n>,x<rsub|m>)=d(x<rsub|n>,x<rsub|n>)=0\<less\>\<varepsilon\>>
<item><math|m\<less\>n\<Rightarrow\>n=m+k> and as
<math|m\<geqslant\>N\<Rightarrow\>d(x<rsub|n>,x<rsub|m>)=d(x<rsub|m>,x<rsub|m+k>)\<less\>\<varepsilon\>>
<item><math|n\<less\>m\<Rightarrow\>m=n+k and as
><math|n\<geqslant\>N\<Rightarrow\>d(x<rsub|n>,x<rsub|m>)=d(x<rsub|n>,x<rsub|n+k>)\<less\>\<varepsilon\>>
</enumerate>
</enumerate>
</proof>
<\definition>
<dueto|Convergence property><index|Convergence property>Let <math|X,d> be
a metric space then a sequence <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is convergent to a point <math|x\<in\>X\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0>
there exists <math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>i\<geqslant\>N> we have
<math|d(x,x<rsub|i>)\<less\>\<varepsilon\>>
</definition>
<\lemma>
Let <math|X,d> be a metric space then given a sequence
<math|{x<rsub|i>}<rsub|i\<in\>{n,\<ldots\>,\<infty\>}>> if for
\ <math|m\<geqslant\>n> we have that <math|{x<rsub|i>}<rsub|i\<in\>{m,\<ldots\>,\<infty\>}>>
is convergent to <math|x\<in\>X> then we have that
<math|{x<rsub|i>}<rsub|i\<in\>{n,\<ldots\>,\<infty\>}>> is convergent
</lemma>
<\proof>
This is trivial because if <math|\<varepsilon\>\<gtr\>0> we can find a
<math|N\<in\>{m,\<ldots\>,\<infty\>}> with if
<math|\<forall\>n\<geqslant\>N> we have
<math|d(x<rsub|n>,x)\<less\>\<varepsilon\>> and as
<math|{m,\<ldots\>,\<infty\>}\<subseteq\>{n,\<ldots\>,\<infty\>}> we have
also <math|N\<in\>{m,\<ldots\>,\<infty\>}>
</proof>
<\example>
<label|example of convergent sequence>In <math|\<bbb-R\>, \| \|> the
sequence <math|{<frac|1|n>}<rsub|n\<in\>\<bbb-N\><rsub|0>>> converges to
<math|0> (or <math|lim<rsub|n\<rightarrow\>\<infty\>><frac|1|n>>)
</example>
<\proof>
Given <math|\<varepsilon\>\<gtr\>0> then using <reference|make inverse of
natural lower then e> there exists a <math|N\<in\>\<bbb-N\><rsub|0>> such
that <math|0\<less\><frac|1|N>\<less\>\<varepsilon\>> and if
<math|n\<geqslant\>N\<gtr\>0\<Rightarrow\><frac|1|n>\<leqslant\><frac|1|N>\<less\>\<varepsilon\>>
and thus <math|\|<frac|1|n>-0\|=\|<frac|1|n>\|=<frac|1|n>\<less\>\<varepsilon\>>
<math|>
</proof>
<\theorem>
<label|uniqueness of limit>Let <math|X,d> be a metric space then if a
sequence <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> is
convergent to <math|x,y\<in\>X> then <math|x=y>. If
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> is convergent
then we note the unique element it converges to the limit of the sequence
and note it as <math|lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>>
</theorem>
<\proof>
Assume <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> converges
to <math|x,y> then if <math|x\<neq\>y> we have
<math|\<varepsilon\>=d(x,y)\<gtr\>0> and there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|d(x,x<rsub|N>)\<less\><frac|\<varepsilon\>|2>> and then
<math|d(x,y)\<leqslant\>d(x,x<rsub|N>)+d(x<rsub|N>,y)\<less\>\<varepsilon\>=d(x,y)>
a contradiction
</proof>
<\theorem>
<label|limit and continuity>Let <math|X,d<rsub|X>,Y,d<rsub|Y>> be metric
spaces <math|f:X\<rightarrow\>Y> a continuous function and
<math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> a convergent
sequence then <math|{f(x<rsub|i>)}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is convergent and <math|lim<rsub|i\<rightarrow\>\<infty\>>f(x<rsub|i>)=f(lim<rsub|i\<rightarrow\>\<infty\>>x<rsub|i>)>
</theorem>
<\proof>
Let <math|x=lim<rsub|i\<rightarrow\>\<infty\>>x<rsub|i> >and let
<math|\<varepsilon\>\<gtr\>0> then by continuity there exists a
<math|\<delta\>\<gtr\>0> such that if
<math|d<rsub|X>(x,y)\<less\>\<delta\>\<Rightarrow\>d(f(x),f(y))\<less\>\<varepsilon\>>.
So lets take <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|d<rsub|X>(x,x<rsub|i>)\<less\>\<delta\>> for all
<math|i\<geqslant\>N><math|> then if <math|i\<geqslant\>N> we have
<math|d<rsub|Y>(f(x),f(x<rsub|i>))\<less\>\<varepsilon\>> proving our
theorem
</proof>
<\theorem>
<label|sequence of points in a set of a metric space>Let <math|X,d> be a
metric space and <math|A> a set and <math|{a<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
be a sequence of points in <math|A> (<math|a<rsub|i>\<in\>A>) with
<math|lim<rsub|i\<rightarrow\>\<infty\>>a<rsub|i>=a> then
<math|a\<in\><wide|A|\<bar\>>>
</theorem>
<\proof>
If <math|x\<in\>U>, <math|U> open then there exists a
<math|\<delta\>\<gtr\>0> such that <math|x\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>U>
then there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|n\<geqslant\>N> we have <math|d(x,x<rsub|n>)\<less\>\<delta\>\<Rightarrow\>x<rsub|n>\<in\>B<rsub|d>(x,\<delta\>)\<subseteq\>U\<Rightarrow\>x\<in\><wide|A|\<bar\>>>
</proof>
<\definition>
<index|complete metric space>A metric space <math|X,d> is complete if
every Cauchy sequence is convergent
</definition>
<\definition>
<index|Banach space>Let <math|X,\<shortparallel\>\<shortparallel\>> be a
normed space then a sequence <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is convergent (Cauchy) if it is convergent (Cauchy) in the associated
metric space. A normed space is Banach if it is complete. Using the fact
that the associated metric is defined by
<math|d(x,y)=\<shortparallel\>x-y\<shortparallel\>> we have that
<math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is\
<\enumerate>
<item>Convergent to <math|x\<in\>X\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0>
there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>]> such that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>},n\<geqslant\>N> we have
<math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\>\<varepsilon\>>
<item><math|Cauchy\<Leftrightarrow\>\<forall\>\<varepsilon\>\<gtr\>0>
there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n,m\<geqslant\>N> we have
<math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\>\<less\>\<varepsilon\>>
</enumerate>
</definition>
<\theorem>
<label|limit preserves inequality (1)>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed space and if
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> is a convergent
sequence with <math|lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>\<rightarrow\>x>
then if <math|\<forall\>i\<in\>{k,\<ldots\>,\<infty\>}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\>\<leqslant\>A>
we have <math|\<shortparallel\>x\<shortparallel\>\<leqslant\>A>
\;
</theorem>
<\proof>
Assume that <math|\<shortparallel\>x\<shortparallel\>\<gtr\>A> take then
<math|\<varepsilon\>=\<shortparallel\>x\<shortparallel\>-A\<gtr\>0> then
there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>x-x<rsub|N>\<shortparallel\>+\<shortparallel\>x<rsub|N>\<shortparallel\>\<less\>\<varepsilon\>+\<shortparallel\>x<rsub|N>\<shortparallel\>\<leqslant\>\<varepsilon\>+A=\<shortparallel\>x\<shortparallel\>-A+A=\<shortparallel\>x\<shortparallel\>>
or <math|\<shortparallel\>x\<shortparallel\>\<less\>\<shortparallel\>x\<shortparallel\>>
a contradiction, so we have <math|\<shortparallel\>x\<shortparallel\>\<leqslant\>A>
</proof>
<\theorem>
<label|limit preserves inequality (2)>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed space and let
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>,
{y<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> be convergent sequences
with <math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\>\<shortparallel\>y<rsub|n>\<shortparallel\>>
then <math|\<shortparallel\>lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>\<shortparallel\>\<leqslant\>\<shortparallel\>lim<rsub|n\<rightarrow\>\<infty\>>y<rsub|n>\<shortparallel\>>
</theorem>
<\proof>
Let <math|x=lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>,y=lim<rsub|n\<rightarrow\>\<infty\>>y<rsub|n>>
and assume that <math|\<shortparallel\>x\<shortparallel\>\<gtr\>\<shortparallel\>y\<shortparallel\>>
so <math|0\<less\>\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>=\<varepsilon\>>
and find <math|N<rsub|1>,N<rsub|2>\<in\>{k,\<ldots\>,\<infty\>}> such
that <math|\<forall\>n,m\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>n\<geqslant\>N<rsub|1>,m\<geqslant\>N<rsub|2>>
we have <math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>,
<math|\<shortparallel\>y<rsub|n>-y\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>.
Then <math|\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>x<rsub|n>-x\<shortparallel\>+\<shortparallel\>x<rsub|n>\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\><frac|\<varepsilon\>|2>+\<shortparallel\>y<rsub|n>\<shortparallel\>\<leqslant\><frac|\<varepsilon\>|2>+\<shortparallel\>y\<shortparallel\>+\<shortparallel\>y-y<rsub|n>\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+\<shortparallel\>y\<shortparallel\>+<frac|\<varepsilon\>|2>=\<varepsilon\>+\<shortparallel\>y\<shortparallel\>=\<shortparallel\>x\<shortparallel\>-\<shortparallel\>y\<shortparallel\>+\<shortparallel\>y\<shortparallel\>=\<shortparallel\>x\<shortparallel\>>
so we reach the contradiction <math|\<shortparallel\>x\<shortparallel\>\<less\>\<shortparallel\>x\<shortparallel\>>
</proof>
<\theorem>
<label|limit of a sum>Let <math|X,\<shortparallel\>\<shortparallel\>> be
a normed space and given two sequences
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>,
<math|{y<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> with
<math|lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>=x>,
<math|lim<rsub|n\<rightarrow\>\<infty\>>y<rsub|n>=y> then
<math|{x<rsub|n>+y<rsub|n>}<rsub|n\<in\>\<bbb-N\><rsub|0>>> is convergent
to <math|x+y> (or <math|lim<rsub|n\<rightarrow\>\<infty\>>(x<rsub|n>+y<rsub|n>)=x+y)>.
Second <math|\<forall\>\<alpha\>\<in\>\<bbb-K\>> we have
<math|lim<rsub|n\<rightarrow\>\<infty\>>(\<alpha\>.x<rsub|n>)=\<alpha\>.lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>>
and third if <math|z\<in\>X> then <math|lim<rsub|n\<rightarrow\>\<infty\>>(z+x<rsub|n>)=z+lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>>
</theorem>
<\proof>
\;
<\enumerate>
<item>Given <math|\<varepsilon\>\<gtr\>0> find a
<math|N<rsub|1>,N<rsub|2>\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>n\<geqslant\>N<rsub|1>>
we have <math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
and <math|\<forall\>m\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>m\<geqslant\>N<rsub|2>>
we have <math|\<shortparallel\>y<rsub|n>-y\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
then if <math|i\<geqslant\>N=max(N<rsub|1>,N<rsub|2>)> we have
<math|\<shortparallel\>(x<rsub|i>+y<rsub|i>)-(x+y)\<shortparallel\>\<leqslant\>\<shortparallel\>x<rsub|i>-x\<shortparallel\>+\<shortparallel\>y<rsub|i>-y\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
<item>If <math|\<alpha\>=0> then <math|\<alpha\>.x<rsub|i>=0> and it is
trivial to prove that <math|lim<rsub|n\<rightarrow\>\<infty\>>(\<alpha\>.x<rsub|n>)=0=\<alpha\>.lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>>.
So assume <math|\<alpha\>\<neq\>0> then <math|\|\<alpha\>\<nvDash\>0>
and given <math|\<varepsilon\>\<gtr\>0> find a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> so that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>n\<geqslant\>N>
we have <math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\><frac|\<varepsilon\>|\|\<alpha\>\|>>
then if <math|n\<geqslant\>N,n\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>\<alpha\>.x<rsub|n>-\<alpha\>.x\<shortparallel\>=\<shortparallel\>\<alpha\>.(x<rsub|n>-x)\<shortparallel\>=\|\<alpha\>\|.\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\>\|\<alpha\>\|.<frac|\<varepsilon\>|\|\<alpha\>\|>=\<varepsilon\>>
<item>Given <math|\<varepsilon\>\<gtr\>1> then there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|n\<geqslant\>N,n\<in\>{k,\<ldots\>,\<infty\>}> that
<math|\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<Rightarrow\>\<shortparallel\>(z+x<rsub|n>)-(z+x)\<shortparallel\>=\<shortparallel\>x<rsub|n>-x\<shortparallel\>\<less\>\<varepsilon\>>
</enumerate>
\;
</proof>
<\definition>
<label|definition of a serie><dueto|Definition of a
Series><index|series>Let <math|X,\<shortparallel\>\<shortparallel\>> be a
normed space then if <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
is a sequence then if <math|lim<rsub|n\<rightarrow\>\<infty\>>(<big|sum><rsub|i=k><rsup|n>x<rsub|i>)>
exists we say that <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> is
a convergent series and we note <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<equallim\><rsub|notation>lim<rsub|n\<rightarrow\>\<infty\>>(<big|sum><rsub|i=k><rsup|n>x<rsub|i>)>
</definition>
<\note>
As a shorthand if say that for the sequence
<math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> we have that
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>=s> then we mean that
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> is a convergent
series and that <math|s=lim<rsub|n\<rightarrow\>\<infty\>>(<big|sum><rsub|i=k><rsup|n>x<rsub|i>)>
</note>
<\theorem>
<label|terms of a convergent serie goes to 0>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed space and
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>\<infty\>}>> be such that the
series <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> converges then
<math|lim<rsub|n\<rightarrow\>\<infty\>>x<rsub|n>=0>
</theorem>
<\proof>
Given <math|\<varepsilon\>\<gtr\>0> then because
<math|{<big|sum><rsub|i=k><rsup|n>x<rsub|i>}<rsub|n\<in\>{k,\<ldots\>\<infty\>}>>
has a limit it is Cauchy so we can find a
<math|N\<in\>{k,\<ldots\>\<infty\>}> such that if
<math|n\<geqslant\>N,m\<in\>\<bbb-N\><rsub|0>> we have
<math|\<shortparallel\><big|sum><rsub|i=k><rsup|n+m>x<rsub|i>-<big|sum><rsub|i=k><rsup|n>x<rsub|i>\<shortparallel\>\<less\>\<varepsilon\>>
so if <math|n\<geqslant\>N+1> then <math|n,n-1\<geqslant\>N> and
<math|\<shortparallel\>x<rsub|n>\<shortparallel\>=\<shortparallel\><big|sum><rsub|i=k><rsup|n>x<rsub|i>-<big|sum><rsub|i=k><rsup|n-1>x<rsub|i>\<shortparallel\>\<less\>\<varepsilon\>>
</proof>
<\theorem>
<label|subserie property>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a normed space and <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
a sequence and <math|m\<in\>{k,\<ldots\>,\<infty\>}> then
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> is convergent if and
only if we have <math|<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>> is
a convergent series. Also if any of the two sequences is convergent we
have <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>=(<big|sum><rsub|i=k><rsup|m>x<rsub|i>)+<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>Given <math|\<varepsilon\>\<gtr\>0> then
<math|\<exists\>N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|n\<geqslant\>N> we have <math|\<shortparallel\><big|sum><rsub|i=k><rsup|n>x<rsub|i>-<big|sum><rsub|i=k><rsup|\<infty\>>\<shortparallel\>\<less\>\<varepsilon\>>.
Now take <math|N<rprime|'>=N+(m-k)+1\<in\>{m+1,\<ldots\>.,\<infty\>}>
and <math|n\<geqslant\>N<rprime|'>\<geqslant\>N> then
<math|\<shortparallel\><big|sum><rsub|i=m+1><rsup|n>x<rsub|i>-(<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>-<big|sum><rsub|i=k><rsup|m>x<rsub|i>)\<shortparallel\>=\<shortparallel\><big|sum><rsub|i=m+1><rsup|n>x<rsub|i>+<big|sum><rsub|i=k><rsup|m>x<rsub|i>-<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<equallim\><rsub|<reference|split
of a sum>>\<shortparallel\><big|sum><rsub|i=k><rsup|n>x<rsub|i>-<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<less\>\<varepsilon\>>
proving that <math|<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>>
exists and is equal to <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>-<big|sum><rsub|i=k><rsup|m>x<rsub|i>>
<item><math|\<Leftarrow\>> Given <math|\<varepsilon\>\<gtr\>0> then
<math|\<exists\>N\<in\>{m+1,\<ldots\>,\<infty\>}> such that if
<math|n\<geqslant\>N> we have <math|\<shortparallel\><big|sum><rsub|i=m+1><rsup|n>x<rsub|i>-<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<less\>\<varepsilon\>>.
Then also <math|N\<in\>{k,\<ldots\>,\<infty\>}> and
<math|\<shortparallel\><big|sum><rsub|i=k><rsup|n>x<rsub|i>-(<big|sum><rsub|i=k><rsup|m>x<rsub|i>+<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>)\<shortparallel\>\<equallim\><rsub|<reference|split
of a sum>>\<shortparallel\><big|sum><rsub|i=m+1><rsup|n>x<rsub|i>-<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<less\>\<varepsilon\>>
proving that <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> exist
and is equal to <with|mode|math|<big|sum><rsub|i=k><rsup|m>x<rsub|i>+<big|sum><rsub|i=m+1><rsup|\<infty\>>x<rsub|i>>
</enumerate>
</proof>
<\theorem>
<label|linear combination of convergent series>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed space over
<math|\<bbb-K\>>, <math|\<alpha\>,\<beta\>\<in\>\<bbb-K\>> and
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>,
{y<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> such that
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> and
<math|<big|sum><rsub|i=k><rsup|\<infty\>>y<rsub|i>> are convergent then
<math|<big|sum><rsub|i=k><rsup|\<infty\>>(\<alpha\>x<rsub|i>+\<beta\>y<rsub|i>)>
converges and is equal to <math|\<alpha\><big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>+\<beta\><big|sum><rsub|i=k><rsup|\<infty\>>y<rsub|i>>
</theorem>
<\proof>
Given <math|n\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|<big|sum><rsub|i=k><rsup|n>(\<alpha\>x<rsub|i>+\<beta\>y<rsub|i>)\<equallim\><rsub|<reference|sum
of a linear combination>>\<alpha\><big|sum><rsub|i=k><rsup|n>x<rsub|i>+\<beta\><big|sum><rsub|i=k><rsup|n>y<rsub|i>>
and using <reference|limit of a sum> we have that
<math|lim<rsub|n\<rightarrow\>\<infty\>><big|sum><rsub|i=k><rsup|n>*(\<alpha\>x<rsub|i>+\<beta\>y<rsub|i>)>
exists and is equal to <math|\<alpha\>.lim<rsub|n\<rightarrow\>\<infty\>><big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>+\<beta\>.lim<rsub|n\<rightarrow\>\<infty\>><big|sum><rsub|i=k><rsup|n>y<rsub|i>>
proving our theorem
</proof>
<\theorem>
<label|convergent criteria of a serie in a Banach space>If
<math|X,\<shortparallel\>\<shortparallel\>> is a normed Banach space and
if <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is a sequence,
and if <math|{s<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is a
sequence in <math|\<bbb-R\>> such that
<math|\<forall\>i\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>x<rsub|i>\<shortparallel\>\<less\>s<rsub|i>> and
<math|<big|sum><rsub|i=k><rsup|\<infty\>>s<rsub|i>> converges then
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> converges and
<math|\<shortparallel\><big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=k><rsup|\<infty\>>s<rsub|i>>.
</theorem>
<\proof>
First we prove that <math|{\<sigma\><rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
where <math|\<sigma\><rsub|i>=<big|sum><rsub|j=k><rsup|i>x<rsub|i>> is
Cauchy. So let <math|\<varepsilon\>\<gtr\>0> then as
<math|<big|sum><rsub|i=k><rsup|\<infty\>>s<rsub|i>> converges we have
that <math|lim<rsub|i\<rightarrow\>\<infty\>>(<big|sum><rsub|j=k><rsup|i>s<rsub|j>)>
exists and thus that <math|{<big|sum><rsub|j=k><rsup|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy so there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such
that if <math|n\<geqslant\>N> and <math|m\<in\>\<bbb-N\><rsub|0>> then
<math|><math|\|<big|sum><rsub|i=n+1><rsup|n+m>\|\<equallim\><rsub|<reference|split
of a sum>>\|<big|sum><rsub|i=k><rsup|n+m>s<rsub|i>-<big|sum><rsub|i=k><rsup|n>s<rsub|i>\|\<less\>\<varepsilon\>>.
Now <math|\<shortparallel\>\<sigma\><rsub|n+m>-\<sigma\><rsub|n>\<shortparallel\>=\<shortparallel\><big|sum><rsub|i=k><rsup|n+m>x<rsub|i>-<big|sum><rsub|i=k><rsup|n>x<rsub|i>\<shortparallel\>\<equallim\><rsub|<reference|split
of a sum>>\<shortparallel\><big|sum><rsub|i=n+1><rsup|n+m>x<rsub|i>\<shortparallel\>\<leqslant\><rsub|<reference|norm
of a finite sum>><big|sum><rsub|i=n+1><rsup|n+m>\<shortparallel\>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=n+1><rsup|n+m>s<rsub|i>\<less\>\<varepsilon\>>
proving the Cauchy condition. Then because <math|X> is complete
<math|{\<sigma\><rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is
convergent and its limit exists and thus
<math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>> exists.
Also for <math|n\<in\>{k,\<ldots\>,\<infty\>}>
<math|\<shortparallel\><big|sum><rsub|i=k><rsup|n>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=k><rsup|n>\<shortparallel\>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=k><rsup|n>s<rsub|i>>
we have by <reference|limit preserves inequality> that
<math|\<shortparallel\><big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<shortparallel\>=\<shortparallel\>lim<rsub|n\<rightarrow\>\<infty\>>(<big|sum><rsub|i=k><rsup|n>x<rsub|i>)\<shortparallel\>\<leqslant\>\|lim<rsub|n\<rightarrow\>\<infty\>>(<big|sum><rsub|i=k><rsup|n>s<rsub|i>)\|=\|<big|sum><rsub|i=k><rsup|\<infty\>>s<rsub|i>\|=<big|sum><rsub|i=k><rsup|\<infty\>>s<rsub|i>>
</proof>
<\corollary>
If <math|X,\<shortparallel\>\<shortparallel\>> is a normed Banach space
and if <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is a
sequence, and if <math|<big|sum><rsub|i=k><rsup|\<infty\>>\<shortparallel\>x<rsub|i>\<shortparallel\>>
converge then <math|<big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>>
converges and <math|\<shortparallel\><big|sum><rsub|i=k><rsup|\<infty\>>x<rsub|i>\<shortparallel\>\<leqslant\><big|sum><rsub|i=k><rsup|\<infty\>>\<shortparallel\>x<rsub|i>\<shortparallel\>>
</corollary>
<\proof>
This follows from the previous theorem as
<math|\<shortparallel\>x<rsub|i>\<shortparallel\>\<leqslant\>\<shortparallel\>x<rsub|i>\<shortparallel\>>
</proof>
<\theorem>
Let <math|X> be a space with two equivalent compatibly norms
<math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
then <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is
convergent to <math|x\<in\>X> (<math|Cauchy)> in
<math|\<shortparallel\>\<shortparallel\><rsub|1>> if it is convergent to
<math|x> (Cauchy) in <math|\<shortparallel\>\<shortparallel\><rsub|2>>
</theorem>
<\proof>
Using (<reference|equivalence of norms>) we know that there exists a
<math|M<rsub|1>,M<rsub|2>\<gtr\>0> such that
<math|M<rsub|1>\<shortparallel\>x\<shortparallel\><rsub|1>\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|2>\<leqslant\>M<rsub|2>\<shortparallel\>x\<shortparallel\><rsub|1>>
so we have\
<\enumerate>
<item>Convergence, let <math|\<varepsilon\>\<gtr\>0> then by
convergence in <math|\<shortparallel\>\<shortparallel\><rsub|1>> there
exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}.i\<geqslant\>N> we have
<math|\<shortparallel\>x<rsub|i>-x\<shortparallel\><rsub|1>\<less\><frac|\<varepsilon\>|M<rsub|2>>\<Rightarrow\>\<shortparallel\>x<rsub|i>-x\<shortparallel\><rsub|2>\<leqslant\>M<rsub|2>\<shortparallel\>x<rsub|i>-x\<shortparallel\><rsub|1>\<less\><frac|\<varepsilon\>|M<rsub|2>>M<rsub|2>=\<varepsilon\>>
<item>Cauchy, let <math|\<varepsilon\>\<gtr\>0> then by Cauchy in
<math|\<shortparallel\>\<shortparallel\><rsub|1>> we have the existence
of a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n,m\<in\>{k,\<ldots\>,\<infty\>},m,n\<geqslant\>N> we
have <math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\>\<less\><frac|\<varepsilon\>|M<rsub|2>>>
and thus <math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\><rsub|2>\<leqslant\>M<rsub|2>\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\><rsub|1>\<less\><frac|\<varepsilon\>|M<rsub|2>>M<rsub|2>=\<varepsilon\>>
</enumerate>
</proof>
<\theorem>
<label|limit point of sequences>Let <math|X,d> be a metric space and let
<math|x> be a limit point of a sequence
<math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> then
<math|\<forall\>N\<in\>{k,\<ldots\>,\<infty\>}> there exists a
<math|n\<in\>{k,\<ldots\>,\<infty\>},n\<gtr\>N> such that
<math|d(x,x<rsub|n>)\<less\>\<varepsilon\>>
</theorem>
<\proof>
Let <math|A<rsub|\<varepsilon\>>={i\<in\>{k,\<ldots\>,\<infty\>}\|0\<less\>d(x,x<rsub|i>)\<less\>\<varepsilon\>}\<subseteq\>{k,\<ldots\>,\<infty\>}>
now if <math|A<rsub|\<varepsilon\>>> is infinite then we have proved our
theorem [for if there exists a <math|N<rsub|0>\<in\>{k,\<ldots\>,\<infty\>}>
we have <math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>},n\<gtr\>N<rsub|0>>
that <math|d(x,x<rsub|n>)\<geqslant\>\<varepsilon\>\<Rightarrow\>n\<nin\>A<rsub|\<varepsilon\>>>
and thus <math|A<rsub|\<varepsilon\>>\<subseteq\>[k,\<ldots\>,N<rsub|0>]>
so <math|A<rsub|\<varepsilon\>>> would be finite a contradiction].\
We prove that <math|A<rsub|\<varepsilon\>>> is infinite by contradiction,
so assume that <math|A<rsub|\<varepsilon\>>> is finite. As <math|x> is a
limit point there exist <math|x<rsub|i>\<neq\>x> with
<math|x<rsub|i>\<in\>B<rsub|d>(x,\<varepsilon\>)> or
<math|0\<less\>d(x,x<rsub|i>)\<less\>\<varepsilon\>> so
<math|i\<in\>A<rsub|\<varepsilon\>>> and
<math|A<rsub|\<varepsilon\>>\<neq\>\<emptyset\>> so
<math|\<delta\>=min{d(x,x<rsub|i>)\|i\<in\>A<rsub|\<varepsilon\>>}>
exists (<math|A<rsub|\<varepsilon\>>> is finite and non empty) and is
<math|\<gtr\>0>. then there exists a <math|x<rsub|j>\<neq\>x> such that
<math|x<rsub|j>\<in\>B<rsub|d>(x,\<delta\>)\<Rightarrow\>0\<less\>d(x,x<rsub|j>)\<less\>\<delta\>\<Rightarrow\>j\<in\>A<rsub|\<varepsilon\>>\<Rightarrow\>d(x,x<rsub|j>)\<in\>{d(x,x<rsub|i>)\|i\<in\>A<rsub|\<varepsilon\>>}\<Rightarrow\>\<delta\>\<leqslant\>d(x,x<rsub|j>)\<less\>\<delta\>>
a contradiction so <math|A<rsub|\<varepsilon\>>> must be finite.
</proof>
<\theorem>
<label|cauchy and uniqueness>Let <math|X,d> be a metric space then every
Cauchy sequence has a maximum of one limit point
</theorem>
<\proof>
Assume that <math|x,y> are different limit points of the Cauchy sequence
<math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> then as
<math|x\<neq\>y> then we have <math|\<varepsilon\>=d(x,y)\<gtr\>0> now by
the Cauchy condition we can find a <math|N\<in\>{k,\<ldots\>,\<infty\>}>
such that <math|\<forall\>i,j\<in\>{k,\<ldots\>,\<infty\>},i,j\<geqslant\>N>
we have <math|d(x<rsub|i>,x<rsub|j>)\<less\><frac|\<varepsilon\>|3>>. Now
by <reference|limit point of sequences> there exists
<math|n,m\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|n\<gtr\>N,m\<gtr\>N> and <math|d(x,x<rsub|n>)\<less\><frac|\<varepsilon\>|3>,d(y,x<rsub|m>)\<less\><frac|\<varepsilon\>|3>\<Rightarrow\>\<varepsilon\>\<less\>d(x,y)\<less\>d(x,x<rsub|n>)+d(x<rsub|n>,x<rsub|m>)+d(y,x<rsub|m>)\<less\><frac|\<varepsilon\>|3>+<frac|\<varepsilon\>|3>+<frac|\<varepsilon\>|3>=\<varepsilon\>>
a contradiction so we must have that <math|x=y>
</proof>
<\theorem>
<label|every compact metric space is complete>Every compact metric space
<math|X,d> is complete
</theorem>
<\proof>
Let <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> be a
arbitrary Cauchy sequence and let <math|S={x<rsub|i>\|i\<in\>{k,\<ldots\>,\<infty\>}}>
we can consider now two exclusive cases
<\enumerate>
<item><math|S> is finite, then there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>i,j\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>i,j\<geqslant\>N>
we have <math|x<rsub|i>=x<rsub|j>.> [We prove this by contradiction, so
assume that <math|\<forall\>N\<in\>{k,\<ldots\>,\<infty\>}> there
exists <math|i,j\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>i,j\<geqslant\>N>
such that <math|x<rsub|i>\<neq\>x<rsub|j>> define then
<math|A={d(x,y)\|(x,y)\<in\>S\<times\>S, x\<neq\>y}> which is finite
because <math|S\<times\>S> is finite (see <reference|product of finite
sets is finite> and <reference|map of finite set is finite>) . Also if
<math|N=1> then <math|\<exists\>i\<geqslant\>1> such that
<math|x<rsub|1>\<neq\>x<rsub|i>> so <math|S> must have at least 2
elements and thus <math|A> is not empty, so
<math|\<varepsilon\>=min(A)> exists (minimum of finite sets exists, see
<reference|maximum (minumu,) of finite sets>) and is <math|\<gtr\>0>.
By the Cauchy condition there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>i,j\<in\>{k,\<ldots\>,\<infty\>}\<vdash\>i,j\<geqslant\>N>
we have <math|d(x<rsub|i>,x<rsub|j>)\<less\>\<varepsilon\>> and by our
assumption we have <math|\<exists\>i<rprime|'>,j<rprime|'>\<in\>{k,\<ldots\>,\<infty\>},i<rprime|'>,j<rprime|'>\<geqslant\>N>
and <math|x<rsub|i<rprime|'>\<neq\>>x<rsub|j<rprime|'>>> \ and thus
<math|d(x<rsub|i<rprime|'>>,x<rsub|j<rprime|'>>)\<in\>A> which together
with <math|d(x<rsub|i<rprime|'>>,x<rsub|j<rprime|'>>)\<less\>\<varepsilon\>>
means that <math|\<varepsilon\>\<leqslant\>d(x<rsub|i<rprime|'>>,x<rsub|j<rprime|'>>)\<less\>\<varepsilon\>>
which is a contradiction]. Now given a <math|\<varepsilon\>\<gtr\>0> if
<math|i\<in\>{k,\<ldots\>,\<infty\>},i\<geqslant\>N\<Rightarrow\>x<rsub|i>=x<rsub|N>\<Rightarrow\>0=d(x<rsub|i>,x<rsub|N>)\<less\>\<varepsilon\>>
proving that the Cauchy sequence is convergent to <math|x<rsub|N>>.
<math|>
<item><math|S> is infinite. Then using <reference|compactness and
sequenctial compactness> we have the existence of a limit points for
<math|S> and then by <reference|cauchy and uniqueness> it has a unique
limit point <math|x>. Let now <math|\<varepsilon\>\<gtr\>0> then by the
Cauchy property there <math|\<exists\>N\<in\>{k,\<ldots\>,\<infty\>}>
such that <math|\<forall\>i,j\<in\>{k,\<ldots\>,\<infty\>},i,j\<geqslant\>N>
we have <math|d(x<rsub|i>,x<rsub|j>)\<less\><frac|\<varepsilon\>|2>>.
Now by <reference|limit point of sequences> there exists a
<math|n\<in\>{k,\<ldots\>,\<infty\>}> such that <math|n\<geqslant\>N>
and <math|d(x<rsub|>,x<rsub|n>)\<less\><frac|\<varepsilon\>|2>> and
then <math|\<forall\>m\<in\>{k,\<ldots\>\<infty\>},m\<geqslant\>N> we
have <math|d(x,x<rsub|m>)\<leqslant\>d(x,x<rsub|n>)+d(x<rsub|n>,x<rsub|m>)\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
proving convergence
</enumerate>
</proof>
<\theorem>
<label|real space is complete>The topological space <math|\<bbb-R\>,\|
\|> of the real numbers together with the default metric is complete.
</theorem>
<\proof>
Let <math|{x<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> be a Cauchy
sequence in <math|\<bbb-R\>> then there exists a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\|x<rsub|i>-x<rsub|j>\|\<less\>1>
<math|\<forall\>i,j\<in\>{k,\<ldots\>,\<infty\>],i,j\<geqslant\>N> Taken
then <math|A=max(\|x<rsub|1>\|,\<ldots\>,\|x<rsub|N-1>\|,\|x<rsub|N>\|+1}>
and <math|I=[\<um\>A,A]>. Then if <math|i\<geqslant\>N> we have
<math|\|x<rsub|i>\|\<less\>\|x<rsub|i>-x<rsub|N>\|+\|x<rsub|N>\|\<less\>1+\|x<rsub|N>\|\<leqslant\>A\<Rightarrow\>x<rsub|i>\<in\>[-A,A]>
and if <math|i\<less\>N\<Rightarrow\>\|x<rsub|i>\|\<leqslant\>A\<Rightarrow\>x<rsub|i>\<in\>[\<um\>A,A]\<Rightarrow\>\<forall\>i\<in\>{k,\<ldots\>,\<infty\>}>
we have <math|x<rsub|i>\<in\>[\<um\>A,A]>. Now by <reference|[a,b] is
compact> <math|[-A,A]> is compact in <math|\<bbb-R\>> and thus also
compact in the subspace topology and then by <reference|every compact
metric space is complete> <math|[\<um\>A,A]> is complete and thus by
Cauchy we have the existence of a <math|x\<in\>[\<um\>M,M]> such
<math|\<exists\>M\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>m\<in\>\<bbb-N\><rsub|0>,m\<geqslant\>M> that
<math|\|x-x<rsub|m>\|=\|x-x<rsub|m>\|<rsub|A>\<less\>\<varepsilon\>>\
</proof>
<\definition>
<index|Banach space>A normed vector space is a Banach space if it is a
complete space using the metric associated with the norm
</definition>
<\lemma>
Let <math|E,\<shortparallel\>\<shortparallel\>> be a normed space and
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> be a Cauchy
sequence then there exists a <math|k\<gtr\>0> such that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}> we have that
<math|\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\>k>
</lemma>
<\proof>
Take <math|\<varepsilon\>=1\<gtr\>0> then because of the Cauchy condition
there exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n,m\<geqslant\>N> we have
<math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\>\<leqslant\>1>
and thus \ if <math|n\<geqslant\>N> that
<math|\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\>\<shortparallel\>x<rsub|n>-x<rsub|N>\<shortparallel\>+\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\>1+\<shortparallel\>x<rsub|N>\<shortparallel\>>
so if we take <math|k=max{\<shortparallel\>x<rsub|1>\<shortparallel\>,\<ldots\>,\<shortparallel\>x<rsub|N-1>\<shortparallel\>,\<shortparallel\>x<rsub|N>\<shortparallel\>+1}>
then we have <math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>}> that
<math|\<shortparallel\>x<rsub|n>\<shortparallel\>\<leqslant\>k>
</proof>
<\theorem>
<math|\<bbb-R\><rsup|n>, \<shortparallel\>\<shortparallel\>> be the
maximum (or product) norm in <math|\<shortparallel\>x\<shortparallel\>=max{\|x<rsub|i>\|
\|i\<in\>{1,\<ldots\>,n}}> is a Banach space (and as all norms in
<math|\<bbb-R\>> are equivalent) it is a Banach space for all norms\
</theorem>
<\proof>
If <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> is a Cauchy
sequence then by the previous lemma there exists a <math|k\<gtr\>0> such
that <math|\<forall\>i\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>x<rsub|i>\<shortparallel\>\<leqslant\>k> or
<math|\<forall\>i\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|x<rsub|i>\<in\>K={x\<in\>\<bbb-R\><rsup|n>\|\<shortparallel\>x<rsub|>\<shortparallel\>\<leqslant\>k}>
which is a closed and bounded, hence compact by Hein Borel
(<reference|compact subsets of the reals>) and thus by <reference|every
compact metric space is complete> <math|K> together with the induced norm
is complete and hence has a limit <math|x\<in\>K\<subseteq\>\<bbb-R\><rsup|n>>
which is of course also a limit in <math|\<bbb-R\><rsup|n>> of the
sequence (<math|><math|\<shortparallel\>a\<shortparallel\><rsub|\|K>=\<shortparallel\>a\<shortparallel\>)>
</proof>
<\theorem>
Every finite dimensional normed vector space is a Banach space
</theorem>
<\proof>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be a
n-dimensional finite normed space then using <reference|there exists a
isometry between a finite dimensional normed space and the product of
reals with a induced norm> there exists a norm
<math|\<shortparallel\>\<shortparallel\>> on <math|\<bbb-R\><rsup|n>> and
a isometry <math|\<varphi\>> (which is also a isomorphism) between
<math|X> and <math|\<bbb-R\><rsup|n>>. So if
<math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>> is a Cauchy
sequence in <math|X> then given <math|\<varepsilon\>\<gtr\>0> there
exists a <math|N> such that if <math|n,m\<geqslant\>N> we have
<math|\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\><rsub|X>\<less\>\<varepsilon\>>
and thus <math|\<shortparallel\>\<varphi\>(x<rsub|n>)-\<varphi\>(x<rsub|m>)\<shortparallel\>=\<shortparallel\>\<varphi\>(x<rsub|n>-x<rsub|m>)\<shortparallel\>=\<shortparallel\>x<rsub|n>-x<rsub|m>\<shortparallel\><rsub|X>\<less\>\<varepsilon\>>
and thus <math|{\<varphi\>(x<rsub|n>)}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
is a Cauchy sequence in <math|\<bbb-R\><rsup|n>> so that by the previous
theorem we have <math|{\<varphi\>(x<rsub|n>)}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
has a limit <math|y> or <math|\<forall\>\<varepsilon\>\<gtr\>0> there
exists a <math|N\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>n\<in\>{k,\<ldots\>,\<infty\>},n\<geqslant\>N> we have
<math|\<shortparallel\>\<varphi\>(x<rsub|n>)-y\<shortparallel\>\<less\>\<varepsilon\>>
then for <math|x=\<varphi\><rsup|-1>(y)> we have
<math|\<forall\>\<varepsilon\>\<gtr\>0> and
<math|n\<in\>{k,\<ldots\>,\<infty\>},n\<geqslant\>N> that
<math|\<shortparallel\>x<rsub|n>-x\<shortparallel\><rsub|X>=\<shortparallel\>\<varphi\>(x<rsub|n>-x)\<shortparallel\>=\<shortparallel\>\<varphi\>(x<rsub|n>)-\<varphi\>(x)\<shortparallel\>=\<shortparallel\>\<varphi\>(x<rsub|n>)-y\<shortparallel\>\<less\>\<varepsilon\><rsub|>>
proving that <math|{x<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
converges to <math|x> and thus proving that <math|X> is Cauchy and thus a
Banach space
</proof>
<\theorem>
<label|space of linear continuous maps to a Banach space is Banach>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
be a set of normed vector spaces and let
<math|Y,\<shortparallel\>\<shortparallel\>> be a Banach space then
<math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> is a Banach space using the
operator norm
</theorem>
<\proof>
Let <math|{L<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> be a Cauchy
sequence in <math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> and
<math|(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
then <math|\<shortparallel\>L<rsub|k>(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|l>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=\<shortparallel\>(L<rsub|k>-L<rsub|l>)(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|k>-L<rsub|l>\<shortparallel\>.<big|prod><rsub|i\<in\>{1,.,,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>.
It is now easy to prove that <math|{L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy. We have two cases\
<\enumerate>
<item>If there is a <math|i\<in\>{1,\<ldots\>,n}> such that
<math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=0> then
<math|\<forall\>l,m\<in\>{k,\<ldots\>,\<infty\>}\<succ\>\<shortparallel\>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|n>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=0>
and thus <math|{L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)}<rsub|i\<in\>{k,.\<ldots\>,\<infty\>}>>
is Cauchy.
<item>If <math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<neq\>0> then
<math|0\<less\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>
then because <math|{L<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>> is
Cauchy we can given a <math|\<varepsilon\>\<gtr\>0> find a
<math|N\<in\>{k,\<ldots\>,\<infty\>}> such that if
<math|m,l\<geqslant\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>L<rsub|m>-L<rsub|l>\<shortparallel\>\<less\><frac|\<varepsilon\>|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>>>
and thus if <math|m,l\<geqslant\>N> we have
<math|\<shortparallel\>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|l>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<less\>\<varepsilon\>>
proving again that <math|{L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy.
</enumerate>
\ We use now the fact that <math|Y> is a Banach space to find a limit
depending on <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)> for
<with|mode|math|{L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
and define <math|L(x<rsub|1>,\<ldots\>,x<rsub|n>)> to be
<math|lim<rsub|i\<rightarrow\>\<infty\>>L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)=L(x<rsub|1>,\<ldots\>,x<rsub|m>)>
we prove now that the function <math|L:<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>\<rightarrow\>Y>
defined by <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)\<rightarrow\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)>
is linear and continuous.
<\enumerate>
<item>(Multilinearity) Given <math|m\<in\>{k,\<ldots\>,\<infty\>}> and
<math|i\<in\>{1,\<ldots\>,n}> we have because of multilinearity of
<math|L<rsub|m>> that <math|L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<alpha\>.x+\<beta\>.y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=\<alpha\>.L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
and thus using <reference|limit of a sum> we have
<math|L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<alpha\>.x+\<beta\>.y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=lim<rsub|m\<rightarrow\>\<infty\>>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<alpha\>.x+\<beta\>.y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=lim<rsub|m\<rightarrow\>\<infty\>>(\<alpha\>.L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,y,x<rsub|i+1>,\<ldots\>,x<rsub|n>))\<equallim\><rsub|<reference|limit
of a sum>>\<alpha\>.lim<rsub|m\<rightarrow\>\<infty\>>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.lim<rsub|m\<rightarrow\>\<infty\>>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=\<alpha\>.L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)+\<beta\>.L(x<rsub|1>,\<ldots\>,x<rsub|i-1>,y,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
proving multilinearity
<item>(Continuity) As <math|\|\<shortparallel\>L<rsub|m>\<shortparallel\>-\<shortparallel\>L<rsub|l>\<shortparallel\>\|\<leqslant\>\<shortparallel\>L<rsub|m>-L<rsub|l>\<shortparallel\>>
we have because <math|{L<rsub|n>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy that <math|{\<shortparallel\>L<rsub|n>\<shortparallel\>}<rsub|n\<in\>{k,\<ldots\>,\<infty\>}>>
is Cauchy in <math|\<bbb-R\>,\<\|\|\>> and thus there exists a
<math|lim<rsub|m\<rightarrow\>\<infty\>>\<shortparallel\>L<rsub|m>\<shortparallel\>=A>,
so using <reference|limit of a sum> we have that
<math|{\<shortparallel\>L<rsub|m>\<shortparallel\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)}<rsub|m\<in\>\<bbb-N\><rsub|0>>>
is convergent with limit <math|A.(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)>.
Now using <reference|limit preserves inequality (2)> and
<math|\<shortparallel\>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsub|m>\<shortparallel\>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)>
we have <math|\<shortparallel\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>=\<shortparallel\>lim<rsub|m\<rightarrow\>\<infty\>>L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>lim<rsub|m\<rightarrow\>\<infty\>>L<rsub|m>(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>\<shortparallel\>)=A.(<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>)>
proving that <math|L> is indeed continuous and that
<math|\<shortparallel\>L\<shortparallel\>\<leqslant\>A>\
</enumerate>
We prove now that <math|{L<rsub|i>}<rsub|i\<in\>{k,\<ldots\>,\<infty\>}>>
converges to <math|L>. Let <math|(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
with <math|\<forall\>i\<in\>{1,\<ldots\>,n}\<succ\>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=1>
then given <math|m,l\<in\>{k,\<ldots\>,\<infty\>}> we have
<math|\<shortparallel\>(L-L<rsub|m>)(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|l+m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>+\<shortparallel\>L<rsub|l+m>(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<leqslant\>\<shortparallel\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|l+m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>+\<shortparallel\>L<rsub|l+m>-L<rsub|m>\<shortparallel\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|i>=\<shortparallel\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|l+m>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>+\<shortparallel\>L<rsub|l+m>-L<rsub|m>\<shortparallel\>>,
now given <math|\<varepsilon\>\<gtr\>0> then use the convergence of
<math|{L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
to find a <math|N<rsub|1>\<in\>{k,\<ldots\>,\<infty\>}> such that
<math|\<forall\>i\<in\>{k,\<ldots\>,\<infty\>},i\<geqslant\>N<rsub|1>> we
have <math|\<shortparallel\>L(x<rsub|1>,\<ldots\>,x<rsub|n>)-L<rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
and by Cauchy a <math|N<rsub|2>\<in\>{k,\<ldots\>,\<infty\>}> with
<math|\<forall\>j,m\<in\>{k,\<ldots\>,\<infty\>},j\<geqslant\>N<rsub|2>>
we have <math|\<shortparallel\>L<rsub|j>-L<rsub|m>\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
then if we take <math|N=max(N<rsub|1>,N<rsub|2>)> then if
<math|l,k\<geqslant\>N\<Rightarrow\>l+k\<geqslant\>N<rsub|1>,N<rsub|2>>
and <math|k\<geqslant\>N<rsub|2>> and thus
<math|\<shortparallel\>(L-L<rsub|k>)(x<rsub|1>,\<ldots\>,x<rsub|n>)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
and by the definition of the operator norm we we have that
<math|\<shortparallel\>(L-L<rsub|k>)\<shortparallel\>\<less\>\<varepsilon\>>
proving convergence.
</proof>
<\definition>
<index|contraction>Let <math|(S,d)> be a metric space, a function
<math|f:S\<rightarrow\>S> is a contraction if
<math|\<exists\>\<lambda\>\<in\>[0,1]>
(<math|0\<leqslant\>\<lambda\>\<leqslant\>1>) such that
<math|\<forall\>x,y\<in\>S> we have <math|d(f(x),f(y))\<leqslant\>\<lambda\>d(x,y)>
</definition>
<\lemma>
Let <math|\<lambda\>\<in\>\<bbb-R\>> such that <math|\<lambda\>\<neq\>1>
then if <math|k\<in\>\<bbb-N\><rsub|0>> we have
<math|<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>=\<lambda\><rsup|n><frac|1-\<lambda\><rsup|k>|1-\<lambda\>>>
</lemma>
<\proof>
We prove this by induction
<\enumerate>
<item><math|k=1> then <math|<big|sum><rsub|i=1><rsup|1>\<lambda\><rsup|n+(i-1)>=\<lambda\><rsup|n>=\<lambda\><rsup|n><frac|1-\<lambda\><rsup|>|1-\<lambda\>>=\<lambda\><rsup|n><frac|1-\<lambda\><rsup|1>|1-\<lambda\>>>
<item>Assume the theorem true for <math|k> and prove it for <math|k+1>\
<\eqnarray*>
<tformat|<table|<row|<cell|<big|sum><rsub|i=1><rsup|k+1>\<lambda\><rsup|n+(i-1)>>|<cell|=>|<cell|(<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>)+\<lambda\><rsup|n+k>>>|<row|<cell|>|<cell|=>|<cell|\<lambda\><rsup|n><frac|1-\<lambda\><rsup|k>|1-\<lambda\>>+\<lambda\><rsup|n+k>>>|<row|<cell|>|<cell|=>|<cell|\<lambda\><rsup|n>(<frac|1-\<lambda\><rsup|k>|1-\<lambda\>>+\<lambda\><rsup|k>)>>|<row|<cell|>|<cell|=>|<cell|\<lambda\><rsup|n><frac|1-\<lambda\><rsup|k>+(1-\<lambda\>)\<lambda\><rsup|k>|1-y>>>|<row|<cell|>|<cell|=>|<cell|\<lambda\><rsup|n><frac|1-\<lambda\><rsup|k+1>|1-\<lambda\>>>>>>
</eqnarray*>
<math|>
</enumerate>
</proof>
<\corollary>
<label|example of a convergent serie>Let
<math|\<lambda\>\<in\>\<bbb-R\>,0\<less\>\<lambda\>\<less\>1> then
<math|<big|sum><rsub|i=1><rsup|\<infty\>>\<lambda\><rsup|i>> converges
and <math|<big|sum><rsub|i=1><rsup|\<infty\>>\<lambda\><rsup|i>=<frac|1|1-\<lambda\>>-1=<frac|\<lambda\>|1-\<lambda\>>>
</corollary>
<\proof>
Given <math|n> we find using the previous lemma with <math|n=1> that
<math|\|<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|i>-(<frac|\<lambda\>|1-\<lambda\>>)\|=\|<frac|\<lambda\>(1-\<lambda\><rsup|k>)|1-\<lambda\>>-<frac|\<lambda\>|1-\<lambda\>>\|=<frac|\<lambda\><rsup|k>|1-\<lambda\>>>
. Given <math|\<varepsilon\>\<gtr\>0> we find using <reference|power of
x\<less\>1> the existence of a <math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>k\<geqslant\>N> we have
<math|\<lambda\><rsup|k>\<less\>\<varepsilon\>(1-\<lambda\>)\<Rightarrow\><frac|\<lambda\><rsup|k>|1-\<lambda\>>\<less\>\<varepsilon\><rsup|>>
proving convergence
</proof>
<\theorem>
<label|the Banach Fixed Point Theorem><dueto|The Banach Fixed Point
Theorem><index|Banach fixed point theorem>In a complete metric space
<math|S,d> then any contraction <math|f:S\<rightarrow\>S> then
<math|\<exists\>!x<rsub|0>\<in\>S> such that
<math|f(x<rsub|0>)=x<rsub|0>> (<math|f> has exactly one fixed point). If
<math|x\<in\>S> then <math|lim<rsub|n\<rightarrow\>\<infty\>>f<rsup|n>x\<rightarrow\>x<rsub|0>>.
Where <math|f<rsup|n>> is defined recursively by\
<\eqnarray*>
<tformat|<table|<row|<cell|f<rsup|1>>|<cell|=>|<cell|f>>|<row|<cell|f<rsup|n+1>>|<cell|=>|<cell|f\<circ\>f<rsup|n>>>|<row|<cell|>|<cell|>|<cell|>>>>
</eqnarray*>
</theorem>
<\proof>
We prove this in steps
<\enumerate>
<item>First we prove that <math|\<forall\>n\<in\>\<bbb-N\><rsub|0>> we
have <math|d(f<rsup|n>(x),f<rsup|n+1>(x))\<leqslant\>\<lambda\><rsup|n>d(x,f(x))>
<\proof>
We do this by induction on n
<\enumerate>
<item><math|n=1> then by the definition of contraction we have
<math|d(f(x),f<rsup|2>(x))=d(f(x),f(f(x)))\<leqslant\>\<lambda\>d(x,f(x))>
<item>Assume that the theorem is <math|n> lets prove it for
<math|n+1>. Now by the definition of a contraction
<math|d(f<rsup|n+1>(x),f<rsup|(n+1)+1>(x))=d(f(f<rsup|n>(x)),f(f<rsup|n+1>(x)))\<leqslant\>\<lambda\>d(f<rsup|n>(x),f<rsup|n+1>(x))\<leqslant\><rsub|induction
hypothese>\<lambda\>.\<lambda\><rsup|n>d(x,f(x))=\<lambda\><rsup|n+1>d(x,f(x))>
</enumerate>
</proof>
<item>Next given a <math|n,k\<in\>\<bbb-N\>> we prove if
<math|\<lambda\>\<in\>[0,1]> that <math|d(f<rsup|n>(x),f<rsup|n+k>(x))\<leqslant\>(<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>)d(x,f(x))>\
<\proof>
Again we use induction on <math|k> to prove this
<\enumerate>
<item><math|k=1> In this case we have by (1)
<math|d(f<rsup|n>(x),f<rsup|n+1>(x))\<leqslant\>\<lambda\><rsup|n>d(x,f(x))=(<big|sum><rsub|i=1><rsup|1>\<lambda\><rsup|n+(i-1)>)d(x,f(x))>
<item>Assume that the theorem is true for <math|k> and prove it for
<math|n+1> now\
<\eqnarray*>
<tformat|<table|<row|<cell|d(f<rsup|n>(x),f<rsup|n+(k+1)>)>|<cell|\<leqslant\>>|<cell|d(f<rsup|n>(x),f<rsup|n+k>(x))+d(f<rsup|n+k>(x),f<rsup|(n+k)+1>(x))>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>)d(x,f(x))+d(f<rsup|n+k>(x),f<rsup|(n+k)+1>(x))
(induction hypothese)>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>)d(x,f(x))+\<lambda\><rsup|n+k>d(x,f(x))
(1)>>|<row|<cell|>|<cell|=>|<cell|(<big|sum><rsub|i=1><rsup|k+1>\<lambda\><rsup|n+(i-1)>)d(x,f(x))>>>>
</eqnarray*>
</enumerate>
</proof>
<item>Next we prove that if <math|\<lambda\>\<in\>[0,1[> that
<math|d(f<rsup|n>(x),f<rsup|n+k>(x))\<leqslant\><frac|\<lambda\><rsup|n>|1-\<lambda\>>d(x,f(x))>
<\proof>
Using (2) we have\
<\eqnarray*>
<tformat|<table|<row|<cell|d(f<rsup|n>(x),f<rsup|n+k>(x))>|<cell|\<leqslant\>>|<cell|(<big|sum><rsub|i=1><rsup|k>\<lambda\><rsup|n+(i-1)>)d(x,f(x))>>|<row|<cell|>|<cell|\<equallim\><rsub|previous
lemma>>|<cell|\<lambda\><rsup|n><frac|1-\<lambda\><rsup|k>|1-\<lambda\>>d(x,f(x))
>>|<row|<cell|>|<cell|\<less\><rsub|\<lambda\><rsup|k>\<gtr\>0>>|<cell|<frac|\<lambda\><rsup|n>|1-\<lambda\>>d(x,f(x))>>>>
</eqnarray*>
\;
</proof>
<item>Next we prove that <math|{f<rsup|n>(x)}<rsub|n\<in\>\<bbb-N\><rsub|0>>>
is Cauchy , so let <math|\<varepsilon\>\<gtr\>0> then by
(<reference|power of number bigger then one>) there exists a <math|N>
such that if <math|n\<geqslant\>N> then
<math|\<lambda\><rsup|n>\<less\><frac|1-\<lambda\>|d(x,f(x))>\<varepsilon\>\<Rightarrow\>d(f<rsup|n>(x),f<rsup|n+k>(x))\<less\>\<varepsilon\>>
proving Cauchy by <reference|othere definition of Cauchy>
<item>Because <math|S> is Cauchy we have that there exists a
<math|x<rsub|0>\<in\>S> such that <math|lim<rsub|n\<rightarrow\>\<infty\>>f<rsup|n>(x)=x<rsub|0>>
now by <reference|limit and continuity> we have
<math|lim<rsub|n\<rightarrow\>\<infty\>>f<rsup|<rsup|n+1>>(x)\<rightarrow\>f(x<rsub|0>)>.
Now if the second equation is true then if
<math|\<varepsilon\>\<gtr\>0> there exists a
<math|N\<in\>\<bbb-N\><rsub|0>> and <math|n\<geqslant\>N >such that
<math|d(f<rsup|n+1>,f(x<rsub|0>))\<less\>\<varepsilon\>\<Rightarrow\>>
if <math|m\<geqslant\>N+1\<Rightarrow\>d(f<rsup|m>(x),f(x<rsub|0>))> so
<math|{f<rsup|n>(x)}<rsub|n\<in\>\<bbb-N\><rsub|0>>> converges to
<math|x<rsub|0>> and <math|f(x<rsub|0>)> and because of
<reference|uniqueness of imit> we have <math|x<rsub|0>=f(x<rsub|0>)>
and found the fixed point
<item>The only thing left to prove is uniqueness assume that there is
also a <math|x<rsub|1>\<in\>S> such that <math|f(x<rsub|1>)=x<rsub|1>>
then if <math|x<rsub|1>\<neq\>x<rsub|0>> then
<math|0\<less\>d(x<rsub|1>,x<rsub|0>)=d(f(x<rsub|0>),f(x<rsub|1>))\<less\>\<lambda\>d(x<rsub|0>,x<rsub|1>)\<leqslant\>d(x<rsub|0>,x<rsub|1>)>
where <math|\<lambda\>\<in\>[0,1]> because of the definition of a
contraction <math|\<Rightarrow\> d(x<rsub|1>,x<rsub|0>)\<less\>d(x<rsub|1>,x<rsub|0>)>
a contradiction so <math|x<rsub|0>=x<rsub|1>>
</enumerate>
</proof>
<\lemma>
Let <math|X,d> be complete metric spaces and let
<math|{A<rsub|i>}<rsub|i\<in\>n>> a sequence of non empty closed sets
with <math|A<rsub|i>\<subseteq\>A<rsub|i-1>,i\<in\>\<bbb-N\><rsub|0><mid|\\>{1}>
and <math|lim<rsub|n\<rightarrow\>\<infty\>>diam(A<rsub|i>)=0> then
<math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>>
</lemma>
<\proof>
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>> choose a
<math|y<rsub|i>\<in\>A<rsub|i>> (axiom of choice). So if
<math|\<varepsilon\>\<gtr\>0> we can find a <math|N> if
<math|n\<geqslant\>N> such that <math|diam(A<rsub|n>)\<less\>\<varepsilon\>>
and thus if <math|n,m\<geqslant\>N> we have
<math|y<rsub|n>,y<rsub|m>\<in\>A<rsub|N>\<Rightarrow\>d(y<rsub|n>,y<rsub|m>)\<less\>\<varepsilon\>>
proving that <math|{y<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> is a
Cauchy sequence so because of Cauchy we have the existence of a <math|x>
such that <math|lim<rsub|i\<rightarrow\>\<infty\>>x<rsub|i>=x>. Then if
<math|k\<in\>\<bbb-N\><rsub|0>> we define
<math|y<rsub|i>=x<rsub|k+(i-1)>> and we have for
<math|{y<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> that
<math|lim<rsub|i\<rightarrow\>\<infty\>>y<rsub|i>=x> (for if
<math|\<varepsilon\>\<gtr\>0> there exists a <math|N> such that if
<math|n\<geqslant\>N> we have <math|d(x<rsub|n>,x)\<less\>\<varepsilon\>>
and of course <math|n+(k-1)\<geqslant\>N> so
<math|d(y<rsub|n>,x)=d(x<rsub|n+k-1>,x)\<less\>\<varepsilon\>>. As
<math|y<rsub|i>=x<rsub|k+(i-1)>\<in\>A<rsub|k>> we use
<reference|sequence of points in a set of a metric space> to find that
<math|x\<in\><wide|A<rsub|i>|\<bar\>>=A<rsub|i>> and thus
<math|x\<in\><big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>>
</proof>
<\theorem>
<label|Baire category theorem><dueto|Baire Category Theorem><index|Baire
Category Theorem>If <math|X> is a compact Hausdorff topological space or
<math|X,d> is a complete metric space then <math|X> is a Baire space
</theorem>
<\proof>
First note that <math|X> is always regular (see <reference|a metric space
is regular> and <reference|compact Hausdorf space is regular and
normal>). Let <math|{A<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> be a
sequence of closed sets of <math|X> with
<math|\<forall\>i\<in\>\<bbb-N\><rsub|0>
A<rsup|\<circ\>><rsub|i>=\<emptyset\>> we must then show that
<math|(<big|cup><rsub|i\<in\>\<bbb-N\><rsub|0>>A<rsub|i>)<rsup|\<circ\>>=\<emptyset\>>,
to do this , given a non empty open set <math|<rsub|>U<rsub|0>> we must
show there exists a <math|x\<in\>U<rsub|0>> with
<math|x\<nin\>A<rsub|i>,i\<in\>\<bbb-N\><rsub|0>>.
Consider the set <math|A<rsub|1>> then because
<math|A<rsub|1><rsup|\<circ\>>=\<emptyset\>> then
<math|U<rsub|0>\<nsubseteq\>A<rsub|1>\<Rightarrow\>\<exists\>y\<in\>U<rsub|0>
such that y\<nin\>A<rsub|1>> and then because of regularity there exists
<math|V<rsub|1>,U<rsub|1><rprime|'>> open spaces with
<math|y\<in\>U<rsub|1><rprime|'>,A<rsub|1>\<subseteq\>V<rsub|1>> and
<math|U<rsub|1><rprime|'><big|cap>V<rsub|1>=\<emptyset\>\<Rightarrow\>U<rsub|1><rprime|'>\<subseteq\>X<mid|\\>V<rsub|1>>
closed so also <math|<wide|U<rsub|1><rprime|'>|\<bar\>>\<subseteq\>X<mid|\\>V<rsub|1>\<Rightarrow\><wide|U<rsub|1><rprime|'>|\<bar\>><big|cap>A\<subseteq\><wide|U<rsub|1><rprime|'>|\<bar\>><big|cap>V=\<emptyset\>>
so we have found a non empty open neighborhood
\ \ <math|U<rsub|1>=U<rsub|1><rprime|'><big|cap>U<rsub|0>\<ni\>y> (in
case of a metric space we use <math|U<rsub|1>=U<rprime|'><rsub|1><big|cap>U<rsub|0><big|cap>B<rsub|d>(y,<frac|1|2>)\<ni\>y>
with\
<\eqnarray*>
<tformat|<table|<row|<cell|<wide|U<rsub|1>|\<bar\>><big|cap>A<rsub|1>=\<emptyset\>,>|<cell|<wide|U<rsub|1>|\<bar\>>\<subseteq\>U<rsub|0>>|<cell|and
in case of a metric space, \ diam(<wide|U<rsub|1>|\<bar\>>)\<leqslant\>1>>>>
</eqnarray*>
We proceed now by induction, so assume that there exists a set a set of
non empty open sets <math|U<rsub|i>> <math|i={1,\<ldots\>,n}> such that\
<\eqnarray*>
<tformat|<table|<row|<cell|<wide|U<rsub|i>|\<bar\>><big|cap>A<rsub|i>=\<emptyset\>,
<wide|U<rsub|i>|\<bar\>>\<subseteq\>U<rsub|i-1> and in case of a metric
space, diam(U<rsub|i>)\<leqslant\><frac|1|i>>|<cell|>|<cell|>>>>
</eqnarray*>
Then as <math|U<rsub|n>> is non empty and as
<math|A<rsup|\<circ\>><rsub|n+1>=\<emptyset\>> we have
<math|U<rsub|n>\<nsubseteq\>A<rsub|n+1>> so there exists a
<math|y\<in\>U<rsub|n>> such that <math|y\<in\>A<rsub|n+1>> and because
of regularity we find disjoint open sets
<math|V<rsub|n+1>,U<rprime|'><rsub|n+1>> with
<math|y\<in\>U<rprime|'><rsub|n+1>,A<rsub|n+1>\<subseteq\>V<rsub|n+1>>
and <math|U<rprime|'><rsub|n+1><big|cap>V<rsub|n+1>=\<emptyset\>\<Rightarrow\>U<rprime|'><rsub|n+1>\<subseteq\>X<mid|\\>V<rsub|n+1>>
closed so <math|<wide|U<rprime|'><rsub|n+1>|\<bar\>>\<subseteq\>X<mid|\\>V<rsub|n+1>\<Rightarrow\><wide|U<rprime|'><rsub|n+1>|\<bar\>><big|cap>A<rsub|n+1>\<subseteq\><wide|U<rprime|'><rsub|n+1>|\<bar\>><big|cap>V<rsub|n+1>=\<emptyset\>>
and if we take then <math|U<rsub|n+1>=U<rprime|'><rsub|n+1><big|cap>U<rsub|n>\<ni\>y>
(in case of a metric space we use <math|U<rsub|n+1>=U<rprime|'><rsub|n+1><big|cap>U<rsub|n><big|cap>B<rsub|d>(y,<frac|1|2(n+1)>)\<ni\>y>)
giving\
<\eqnarray*>
<tformat|<table|<row|<cell|<wide|U<rsub|n+1>|\<bar\>><big|cap>A<rsub|n+1>=\<emptyset\>,<wide|U<rsub|n+1>|\<bar\>>\<subseteq\>U<rsub|n>>|<cell|and
in case of a metric space, diam(<wide|U<rsub|n+1>|\<bar\>>)\<leqslant\><frac|1|n>>|<cell|>>>>
</eqnarray*>
So we have created a sequence <math|{U<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|i>>>
such that <with|mode|math|<wide|U<rsub|i>|\<bar\>><big|cap>A<rsub|i>=\<emptyset\>,
<wide|U<rsub|i>|\<bar\>>\<subseteq\>U<rsub|i-1> and in case of a metric
space, diam(<wide|U<rsub|i>|\<bar\>>)\<less\><frac|1|i>>. If we can prove
that <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>>U<rsub|i>\<neq\>\<emptyset\>>
we have proved our theorem, for if <math|x\<in\><big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>><wide|U|\<bar\>><rsub|i>>
then <math|x\<in\>U<rsub|0>> and <math|\<forall\>i\<in\>\<bbb-N\><rsub|0>>
we have <math|x\<nin\>A<rsub|i>>.\
Now to prove that <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>><wide|U<rsub|i>|\<bar\>>\<neq\>\<emptyset\>>
we consider the two cases
<\enumerate>
<item><math|X> is compact Hausdorff. Then we have
<math|<wide|U<rsub|n>|\<bar\>>\<subseteq\><big|cap><rsub|i=1,\<ldots\>,n><wide|U<rsub|i>|\<bar\>>>
(as <math|<wide|U<rsub|n>|\<bar\>>\<subseteq\>U<rsub|n-1>\<subseteq\><wide|U<rsub|n-1>|\<bar\>>\<subseteq\>\<ldots\>\<subseteq\><wide|U<rsub|1>|\<bar\>>>
so <math|{<wide|U<rsub|i>|\<bar\>>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> has
the finite intersection property and using <reference|filterbases and
compactness> we have <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>><wide|U<rsub|i>|\<bar\>>\<neq\>\<emptyset\>>
<item><math|X,d> is a metric space, then as
<math|diam(<wide|U<rsub|i>|\<bar\>>)\<less\><frac|1|i>> so we can use
the previous lemma to prove that <math|<big|cap><rsub|i\<in\>\<bbb-N\><rsub|0>><wide|U<rsub|i>|\<bar\>>\<neq\>\<emptyset\>>
</enumerate>
\;
</proof>
<\theorem>
<label|Open Mapping Theorem><dueto|Open Mapping Theorem><index|Open
Mapping Theorem>Let <math|X,Y> be Banach spaces over <math|\<bbb-K\>> and
<math|L:X\<rightarrow\>Y> a surjective continuous linear function then
<math|L> is also a open function
</theorem>
<\proof>
The proof is done in different steps
<\enumerate>
<item>We show that given <math|\<delta\>\<gtr\>0> and
\ <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>)<rsub|>>
\ be a open ball around <math|0> then there exists a
<math|\<lambda\>\<gtr\>0> with <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>)|\<bar\>>>
<\proof>
First of all we have that <math|X=<big|cup><rsub|n\<in\>\<bbb-N\><rsub|0>>nB<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)>
[If <math|x\<in\>X> then using the Archimedean ordering property
there exists a <math|n\<in\>\<bbb-N\><rsub|0>> with
<math|\<shortparallel\>x\<shortparallel\><frac|2|\<delta\>>\<less\>n>
and thus <math|\<shortparallel\>x\<shortparallel\>\<less\><frac|n\<delta\>|2>\<Rightarrow\>if
y=<frac|x|n>\<Rightarrow\>x=y.n\<Rightarrow\>\<shortparallel\>n.y\<shortparallel\>\<less\><frac|n\<delta\>|2>\<Rightarrow\>\<shortparallel\>y\<shortparallel\>\<less\><frac|\<delta\>|2>\<Rightarrow\>x\<in\>nB<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)\<Rightarrow\>x\<in\><big|cup><rsub|n\<in\>\<bbb-N\><rsub|0>>nB<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)>].
Then as <math|L> is surjective we have that
<math|Y=L(X)=<big|cup><rsub|n\<in\>\<bbb-N\><rsub|0>>L(nB<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))\<equallim\><rsub|L
is linear>nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)\<Rightarrow\>Y=<big|cup><rsub|n\<in\>\<bbb-N\><rsub|0>><wide|nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>|\<bar\>>)>
, now following the Baire Category Theorem as a Banach space is a
complete metric space we have that <math|Y> is a Baire space and as
<math|0\<in\>Y=Y<rsup|0>> there must be a
<math|n<rsub|>\<in\>\<bbb-N\><rsub|0>> such that
<math|(<wide|nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>)<rsup|\<circ\>>\<neq\>\<emptyset\>>
so there exists a <math|z\<in\>(<wide|nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>)<rsup|\<circ\>>>
which is open so there exists a <math|\<beta\>\<gtr\>0> such that
<math|z\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(z,\<beta\>)\<subseteq\>(<wide|nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(z,\<beta\>))|\<bar\>>)<rsup|\<circ\>>\<subseteq\><below|<wide|nL(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>\<equallim\><rsub|<reference|normed
space properties>>n<wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>\<Rightarrow\><frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Z>>(z,\<beta\>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>|>>.
Now as <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)-B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>)>
[if <math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|x>>(0,<frac|\<delta\>|2>)-B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)\<Rightarrow\>x=y<rsub|1>-y<rsub|2>>
with <math|\<shortparallel\>y<rsub|1>\<shortparallel\><rsub|X>\<less\><frac|\<delta\>|2>>
and <math|\<shortparallel\>y<rsub|2>\<shortparallel\><rsub|X>\<less\><frac|\<delta\>|2>\<Rightarrow\>\<shortparallel\>x\<shortparallel\>\<leqslant\>\<shortparallel\>y<rsub|1>\<shortparallel\>+\<shortparallel\>y<rsub|2>\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>)>]
and thus <math|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))-L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))=L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>)-B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>))>.
Further <math|<big|cup><rsub|x\<in\><frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(z,\<beta\>)>(x-<frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Z>>(z,\<beta\>))\<subseteq\><frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Z>>(z,\<beta\>)-<frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Z>>(z,\<beta\>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>-<wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>\<subseteq\><rsub|<reference|normed
space properties>><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))-L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<delta\>|2>))|\<bar\>>\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>))|\<bar\>>>
and thus we have proved that for <math|U=<big|cup><rsub|x\<in\><frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(z,\<beta\>)>(x-<frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(z,\<beta\>))>
we have that <math|U\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>))|\<bar\>>>.
Now using <reference|normed space properties> again we find that
<math|x-<frac|1|n>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(z,\<beta\>)>
is open and thus that <math|U> is a open set, also
<math|0=z-z\<in\>U> and thus we find a <math|\<lambda\>\<gtr\>0> such
that <math|0\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\>)\<subseteq\>U\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\>))|\<bar\>>>
as we set out to prove
</proof>
<item>We refine (1) to show that given a
<math|2\<sigma\><rsub|0>\<gtr\>0> and
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2\<sigma\><rsub|0>)>
a open ball around <math|0> then there exists a
<math|\<lambda\>\<gtr\>0> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\>)\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2\<sigma\><rsub|0>)>
<\proof>
Using <reference|make inverse of natural lower then e> we find a
<math|N\<in\>\<bbb-N\><rsub|0>> such that
<math|<frac|1|N>\<less\>\<sigma\><rsub|0>> then if we take
<math|\<sigma\><rsub|n>=(<frac|1|N+1>)<rsup|n>> then we have using
<reference|example of a convergent serie> and the fact that
<math|<big|sum><rsub|i=1><rsup|\<infty\>>\<sigma\><rsub|i>=<big|sum><rsub|i=1><rsup|\<infty\>>(<frac|1|N+1>)<rsup|i>=<frac|<frac|1|N+1>|1-<frac|1|N+1>>=<frac|<frac|1|N+1>|<frac|N+1-1|N+1>>=<frac|<frac|1|N+1>|<frac|N|N+1>>=<frac|1|N>\<less\>\<sigma\><rsub|0>>.
Now using (1) we have for every <math|n\<in\>\<bbb-N\><rsub|>=\<bbb-N\><big|cup>{0}>
the existence of a <math|\<lambda\><rprime|'><rsub|n>> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\><rprime|'><rsub|n>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|n>))|\<bar\>>>
and if we take <math|\<lambda\><rsub|n>=min(\<lambda\><rprime|'><rsub|n>,<frac|1|n+1>)\<less\><frac|1|n+1>>
\ we have also <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\><rsub|n>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|n>))|\<bar\>>>
and using <reference|example of convergent sequence> and
<reference|limit preserves inequality (2)> we have that
<math|lim<rsub|n\<rightarrow\>\<infty\>>\<lambda\><rsub|n>=0>.
\ Now if <math|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\><rsub|0>)\<Rightarrow\>y\<in\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|0>))|\<bar\>>>
and thus there exists a <math|y<rsub|0>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(y,\<lambda\><rsub|1>)<big|cap>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|0>))>
or there exists a <math|x<rsub|0>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|0>)>
such that <math|y<rsub|0>=L(x<rsub|0>)> and
<math|\<shortparallel\>y-L(x<rsub|0>)\<shortparallel\><rsub|Y>=\<shortparallel\>y-y<rsub|0>\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|1>>.
To summarise\
<\eqnarray*>
<tformat|<table|<row|<cell|x<rsub|0>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|0>)>|<cell|,>|<cell|\<shortparallel\>y-L(x<rsub|0>)\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|1><rsub|>>>>>
</eqnarray*>
if we now assume the existence of a
<math|{x<rsub|0>,\<ldots\>,x<rsub|n>}> such that\
<\eqnarray*>
<tformat|<table|<row|<cell|x<rsub|i>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<sigma\><rsub|i>|2>)>|<cell|,>|<cell|\<shortparallel\>y-L(<big|sum><rsub|j=0><rsup|i>x<rsub|j>)\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|i+1>>>>>
</eqnarray*>
then we can as <math|\<shortparallel\>y-L(<big|sum><rsub|j=0><rsup|n>x<rsub|i>)\<shortparallel\>\<less\>\<lambda\><rsub|n+1>>
we have that <math|y-L(<big|sum><rsub|j=0><rsup|n>x<rsub|j>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\><rsub|i+1>)\<subseteq\><wide|L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|n+1>))|\<bar\>>>
and thus there exists a <math|y<rsub|n+1>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(y-L(<big|sum><rsub|j=0><rsup|n>x<rsub|j>),\<lambda\><rsub|n+2>)<big|cap>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|n+1>))\<Rightarrow\>\<exists\>x<rsub|n+1>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|n+1>)\<vdash\>y<rsub|n+1>=L(x<rsub|n+1>)>
with <math|\<shortparallel\>y-L(<big|sum><rsub|j=0><rsup|n+1>x<rsub|i>)\<shortparallel\><rsub|Y>=\<shortparallel\>(y-L(<big|sum><rsub|j=0><rsup|n>x<rsub|j>))-L(x<rsub|n+1>)\<shortparallel\><rsub|Y>=\<shortparallel\>(y-L(<big|sum><rsub|j=0><rsup|n>x<rsub|j>))-y<rsub|n+1>\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|n+2>=\<lambda\><rsub|(n+1)+1>>
so by recursion we have constructed a sequence
<math|{x<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> with the property
that <math|\<forall\>i\<in\>\<bbb-N\>>
<\eqnarray*>
<tformat|<table|<row|<cell|x<rsub|i>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|i>)>|<cell|,>|<cell|\<shortparallel\>y-L(<big|sum><rsub|j=0><rsup|i>x<rsub|i>)\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|i+1>>>>>
</eqnarray*>
Using <reference|subserie property> we have that
<math|<big|sum><rsub|i=0><rsup|\<infty\>>\<sigma\><rsub|i>=\<sigma\><rsub|0>+<big|sum><rsub|i=1><rsup|\<infty\>>\<sigma\><rsub|i>>.
Now as <math|x<rsub|i>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\><rsub|i>)>
then <math|\<shortparallel\>x<rsub|i>\<shortparallel\><rsub|X>\<less\>\<sigma\><rsub|i>>
and thus using <reference|convergent criteria of a serie in a Banach
space> we have that <math|<big|sum><rsub|i=0><rsup|\<infty\>>x<rsub|i>>
converges and <math|\<shortparallel\><big|sum><rsub|i=0><rsup|\<infty\>>x<rsub|i>\<shortparallel\><rsub|X>\<leqslant\>\<sigma\><rsub|0>+<big|sum><rsub|i=1><rsup|\<infty\>>\<sigma\><rsub|i>\<less\>\<sigma\><rsub|0>+\<sigma\><rsub|0>=2\<sigma\><rsub|0>>
and thus if <math|x=<big|sum><rsub|i=0><rsup|\<infty\>>x<rsub|i>> we
have <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<less\>2\<sigma\>\<Rightarrow\>x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2\<sigma\><rsub|0>)>.
Now assume that <math|><math|\<shortparallel\>y-L(x)\<shortparallel\>\<gtr\>0>
then take <math|\<varepsilon\>=\<shortparallel\>y-L(x)\<shortparallel\><rsub|Y>>
then as <math|lim<rsub|i\<rightarrow\>\<infty\>>\<lambda\><rsub|i>>
there exists a <math|N<rsub|1>\<in\>\<bbb-N\><rsub|>> such that if
<math|n\<gtr\>N<rsub|1>> we have <math|\<lambda\><rsub|n+1>\<less\><frac|\<varepsilon\>|2>>,
also there exists a <math|N<rsub|2>\<in\>\<bbb-N\>> such that
<math|\<shortparallel\>x-<big|sum><rsub|i=0><rsup|n>x<rsub|i>\<shortparallel\><rsub|X>\<less\><frac|\<varepsilon\>|2\<shortparallel\>L\<shortparallel\>>>
if <math|n\<geqslant\>N<rsub|2>> . If now
<math|n\<geqslant\>max(N<rsub|1>,N<rsub|2>)> then
<math|\<varepsilon\>=\<shortparallel\>y-L(x)\<shortparallel\><rsub|Y>=\<shortparallel\>y-L(<big|sum><rsub|i=0><rsup|n>x<rsub|i>)+L(<big|sum><rsub|i=0><rsup|n>x<rsub|i>)-L(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>y-L(<big|sum><rsub|i=0><rsup|n>x<rsub|i>)\<shortparallel\><rsub|Y>+\<shortparallel\>L(<big|sum><rsub|i=0><rsup|n>x<rsub|i>-x)\<shortparallel\><rsub|Y>\<less\>\<lambda\><rsub|n+1>+\<shortparallel\>L\<shortparallel\>\<shortparallel\><big|sum><rsub|i=0><rsup|n>x<rsub|i>-x\<shortparallel\><rsub|X>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
a contradiction, so we must have that
<math|\<shortparallel\>y-L(x)\<shortparallel\>=0\<Rightarrow\>y=L(x)\<Rightarrow\>y\<in\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2\<sigma\><rsub|0>))\<Rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\><rsub|0>)\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2\<sigma\><rsub|0>))>
proving our assertion if we use <math|\<lambda\><rsub|0>=\<lambda\>>
<item>Now take <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,1)=B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,2<frac|1|2>)>
and we find using (2) the existence of a <math|\<lambda\>\<gtr\>0>
such that <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|Y>>(0,\<lambda\>0\<subseteq\>L(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,1))>
and using <reference|linear open mappings> we find that <math|L> is
open which we set out to prove.
\;
</proof>
</enumerate>
</proof>
<\corollary>
<label|inverse of continuous linear mappings between Banach spaces is
continuous>Let <math|X,Y> be Banach spaces over <math|\<bbb-K\>> and
<math|L:X\<rightarrow\>Y> a continuous linear isomorphism then
<math|L<rsup|-1>> is continuous\
</corollary>
<\proof>
This is trivial using the open function theorem (<reference|Open Mapping
Theorem>) and <reference|inverse of open map is continuous>
</proof>
<section|Integration in Banach space>
<\note>
In the following we assume that with <math|[a,b]> we mean a closed and
bounded interval in <math|\<bbb-R\>> that is not empty
<math|(a\<less\>b)>
</note>
<\definition>
<index|partition <math|\<cal-P\>> of <math|[a,b]>>A partition
<math|\<cal-P\>> of <math|[a,b]> is a function
<math|t:{1,\<ldots\>,n}\<rightarrow\>[a,b]> (where <math|n\<gtr\>1)> such
that <math|t<rsub|1>=a,t<rsub|n>=b> and
<math|\<forall\>i\<in\>{1,\<ldots\>,n-1}> we have
<math|t<rsub|i>\<less\>t<rsub|i+1>> \ (here we use the usual notation of
<math|t<rsub|i>=t(i)> and <math|\<cal-P\>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>).
</definition>
<\lemma>
Let <math|{t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> be such that
<math|\<forall\>i\<in\>{1,\<ldots\>,n-1}> we have
<math|t<rsub|i>\<less\>t<rsub|i+1>> then\
<\enumerate>
<item><math|\<forall\>i,j\<in\>{1,\<ldots\>,n}\<vdash\>i\<less\>j\<Rightarrow\>t<rsub|i>\<less\>t<rsub|j>>
<item>The function <math|t> is injective (ie if
<math|t<rsub|i>=t<rsub|j>\<Rightarrow\>i=j>
<item>If <math|i\<in\>{1,\<ldots\>,n-1},k\<in\>{1,\<ldots\>,n}> and
<math|t<rsub|k>\<in\>[t<rsub|i>,t<rsub|i+1>]> then <math|k=i> or
<math|k=i+1>
</enumerate>
</lemma>
<\proof>
\;
<\enumerate>
<item>If <math|i\<less\>j> as we have in this case that
<math|j=i+k,1\<leqslant\>k> we must prove that if <math|k\<geqslant\>1>
such that <math|i,i+k\<in\>{1,\<ldots\>,n}> then we have
<math|t<rsub|i>\<less\>t<rsub|i+k>>. We prove this by induction
<\enumerate>
<item><math|k=1>, this is trivial as we have
<math|t<rsub|i>\<less\>t<rsub|i+1>=t<rsub|i+k>>
<item>Assume that it is true for <math|k> and assume that
<math|i,i+(k+1)\<in\>{1,\<ldots\>,n}> then as <math|k\<geqslant\>1>
we have <math|1\<leqslant\>i\<less\>i+1\<leqslant\>i+k\<less\>i+(k+1)\<leqslant\>n>
so we can apply the induction hypothesis to prove that
<math|t<rsub|i>\<less\>t<rsub|i+k>>. Then we have because
<math|t<rsub|i+k>\<less\>t<rsub|(i+k)+1>=t<rsub|i+(k+1<rsub|>)>> that
<math|t<rsub|i>\<less\>t<rsub|i+(k+1>)>\
</enumerate>
<item>This follows easily from (1) for if <math|t<rsub|i>=t<rsub|k>>
and <math|i\<neq\>k> then either <math|i\<less\>k\<Rightarrow\>t<rsub|i>\<less\>t<rsub|k>>
or <math|k\<less\>i\<Rightarrow\>t<rsub|k>\<less\>t<rsub|i>> which both
yields <math|t<rsub|i>\<neq\>t<rsub|k>> a contradiction
</enumerate>
</proof>
<\definition>
<index|<math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>>>Let
<math|\<cal-P\><rsub|1>={t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,\<cal-P\><rsub|2>={t<rsup|2><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>>
be partitions of <math|[a,b]> then we define the combined partition
<math|\<cal-P\>=\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> as follows. Let
<math|P={t<rsup|1><rsub|i>\|i\<in\>{1,\<ldots\>,n<rsub|1>}<big|cup>{t<rsub|i><rsup|2>\|i\<in\>{1,\<ldots\>,n<rsub|2>}\<subseteq\>[a,b]>
which is finite so by <reference|ordering of a finite set> there exists a
bijection <math|t:{1,\<ldots\>,n}\<rightarrow\>P\<subseteq\>[a,b]> such
that <math|\<forall\>i\<in\>{1,\<ldots\>,n-1}> we have
<math|t<rsub|i>\<less\>t<rsub|i+1>>, also as <math|a,b\<in\>P> there
exists a <math|i,j\<in\>{1,\<ldots\>,n}> such that
<math|t<rsub|i>=a,t<rsub|j>=b>. We prove now that <math|i=1,j=n>. If
<math|i\<neq\>1\<Rightarrow\>i\<gtr\>1\<Rightarrowlim\><rsub|previous
lemma>a=t<rsub|i>\<gtr\>t<rsub|1>\<geqslant\>a> a contradiction, likewise
if <math|j\<neq\>n\<Rightarrow\>j\<less\>n\<Rightarrow\>b=t<rsub|j>\<less\>t<rsub|n>\<leqslant\>b>
a contradiction. So we have proved that
<math|{t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> is a partition of
<math|[a,b]> and we define then <math|\<cal-P\>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
</definition>
<\lemma>
Let <math|\<cal-P\><rsub|1>={t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,\<cal-P\><rsub|2>={t<rsup|2><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>>
be partitions of <math|[a,b]> \ and <math|\<cal-P\>=\<cal-P\><rsub|1>#\<cal-P\><rsub|2>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
then for every <math|i\<in\>{1,\<ldots\>,n-1}> there exists unique
<math|i<rsub|1>\<in\>{1,\<ldots\>,n<rsub|1>-1},i<rsub|2>\<in\>{1,\<ldots\>,n<rsub|2>-1}>
such that <math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|i<rsub|1>>,t<rsup|1><rsub|i<rsub|1>+1>],[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|2><rsub|i<rsub|2>>,t<rsup|2><rsub|i<rsub|2>+1>]>
</lemma>
<\proof>
We prove this for <math|i<rsub|1>> (the proof for <math|i<rsub|2>)> is
similar.\
First we prove the existence. Let <math|i\<in\>{1,\<ldots\>,n-1}> then by
the definition of <math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> and the fact
that <math|i\<neq\>n\<Rightarrow\>t<rsub|i>\<neq\>b> we have the
following two cases\
<\enumerate>
<item><math|\<exists\>k\<in\>{1,\<ldots\>,n<rsub|1>-1}> such that
<math|t<rsup|1><rsub|k>=t<rsub|i>>. Now again by the definition of
<math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> there exists a
<math|j\<in\>{1,\<ldots\>,n}> such that
<math|t<rsub|j>=t<rsup|1><rsub|k+1>>. Suppose now that
<math|t<rsub|j>=t<rsup|1><rsub|k+1>\<less\>t<rsub|i+1>> then
<math|t<rsub|i>=t<rsup|1><rsub|k>\<less\>t<rsup|1><rsub|k+1>=t<rsub|j>\<less\>t<rsub|i+1>>
and thus <math|t<rsub|j>\<in\>[t<rsub|i>,t<rsub|i+1>]> so that by the
previous lemma and the fact that <math|t<rsub|j>\<less\>t<rsub|j>> we
have <math|j=i+1> or <math|t<rsub|i+1>=t<rsub|j>=t<rsup|1><rsub|k+1>\<less\>t<rsub|i+1>>
a contradiction so we must have <math|t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|k+1>>
or we have found that <math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|k>,t<rsup|1><rsub|k+1>]>
so we can take <math|i<rsub|1>=k>
<item><math|\<exists\>k\<in\>{1,\<ldots\>,n<rsub|2>-1}> such that
<math|t<rsup|2><rsub|k>=t<rsub|i>>. We define then
<math|B={i\<in\>{1,\<ldots\>,n<rsub|1>}\|t<rsup|1><rsub|i>\<leqslant\>t<rsub|i>}>
then as <math|a=t<rsub|1><rsup|1>=t<rsub|1>\<leqslant\>t<rsub|i>> we
have <math|1\<in\>B\<Rightarrow\>B\<neq\>\<emptyset\>> also as
<math|t<rsub|i>\<less\>t<rsub|n>=b=t<rsup|1><rsub|n<rsub|1>>> we have
<math|i\<in\>{1,\<ldots\>,n<rsub|1>-1}> and as <math|B> is finite as a
subset of a finite set there exists a
<math|m=max(B)\<in\>{1,\<ldots\>,n<rsub|1>-1}>. Then we have
<math|t<rsup|1><rsub|m>\<leqslant\>t<rsub|i>\<less\>t<rsup|1><rsub|m+1>>.
Assume now that <math|t<rsup|1><rsub|m+1>\<less\>t<rsub|i+1>> then by
definition of <math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> there exists a
<math|j\<in\>{1,\<ldots\>,n}> such that
<math|t<rsup|1><rsub|m+1>=t<rsub|j>> and then as
<math|t<rsub|i>\<less\>t<rsup|1><rsub|m+1>=t<rsub|j>\<less\>t<rsub|i+1>>
we have that <math|t<rsub|j>\<in\>[t<rsub|i>,t<rsub|i+1>]> so using the
previous lemma and <math|t<rsub|i>\<less\>t<rsub|j>> we have that
<math|j=i+1> but then <math|t<rsub|i+1>=t<rsub|j>=t<rsup|1><rsub|m+1>\<less\>t<rsub|i+1>>
a contradiction. So must have <math|t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|m+1>>
and thus <math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|m>,t<rsup|1><rsub|m+1>]>
so we can take <math|i<rsub|1>=m>
</enumerate>
Next we prove uniqueness so assume that there exists a
<math|i<rsub|1>,i<rsub|1><rprime|'>> such that
<math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|i<rsub|1>>,t<rsup|1><rsub|i<rsub|1>+1>],[t<rsup|1><rsub|i<rprime|'><rsub|1>>,t<rsup|1><rsub|i<rprime|'><rsub|1>+1>]>
and assume that <math|i<rsub|1>\<neq\>i<rsub|1><rprime|'>>, we have then
two cases
<\enumerate>
<item><math|i<rsub|1>\<less\>i<rsub|1><rprime|'>> then by the previous
lemma we have <math|t<rsup|1><rsub|i<rsub|1>>\<less\>t<rsup|1><rsub|i<rprime|'><rsub|1>>><math|>
and so <math|t<rsup|1><rsub|i<rsub|1>>\<less\>t<rsup|1><rsub|i<rsub|1><rprime|'>>\<leqslant\>t<rsub|i>\<less\>t<rsub|i+1>\<less\>t<rsup|1><rsub|i<rsub|1>+1>\<Rightarrow\>t<rsub|i<rprime|'><rsub|1>><rsup|1>\<in\>[t<rsup|1><rsub|i<rsub|1>>,t<rsup|1><rsub|i<rsub|1>+1>]>
which gives using the previous lemma and
<math|t<rsup|1><rsub|i<rsub|1>>\<less\>t<rsup|1><rsub|i<rprime|'><rsub|1>>>
that <math|i<rprime|'><rsub|1>=i<rsub|1>+1> but then
<math|t<rsup|1><rsub|i<rsub|1>+1>=t<rprime|'><rsub|i<rsub|1>>\<less\>t<rsub|i>\<less\>t<rsup|1><rsub|i<rsub|1>+1>>
a contradiction<math|>
<item><math|i<rprime|'><rsub|1>\<less\>i<rsub|1>> then by the previous
lemma we have <math|t<rsup|1><rsub|i<rprime|'><rsub|1>>\<less\>t<rsup|1><rsub|i<rsub|1>>>
and so <math|t<rsup|1><rsub|i<rprime|'><rsub|1>>\<less\>t<rsup|1><rsub|i<rsub|1>>\<leqslant\>t<rsub|i>\<less\>t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|i<rprime|'><rsub|1>+1>\<Rightarrow\>t<rsup|1><rsub|i<rsub|1>>\<in\>[t<rsup|1><rsub|i<rprime|'><rsub|1>>,t<rsup|1><rsub|i<rprime|'><rsub|1>+1>]>
which gives using the previous lemma and
<math|t<rsup|1><rsub|i<rprime|'><rsub|1>>\<less\>t<rsup|1><rsub|i<rsub|1>>>
that <math|i<rsub|1>=i<rprime|'><rsub|1>+1> but then
<math|t<rsup|1><rsub|i<rprime|'><rsub|1>+1>=t<rsup|1><rsub|i<rsub|1>>\<leqslant\>t<rsub|i>\<less\>t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|i<rprime|'><rsub|1>+1>>
a contradiction
</enumerate>
So as we have in both cases a contradiction we must conclude that
<math|i<rsub|1>=i<rprime|'><rsub|1>> what we set out to prove
</proof>
<\lemma>
Let <math|\<cal-P\><rsub|1>={t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,\<cal-P\><rsub|2>={t<rsup|2><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>>
be partitions of <math|[a,b]> \ and <math|\<cal-P\>=\<cal-P\><rsub|1>#\<cal-P\><rsub|2>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
then using the previous lemma we have for every
<math|i\<in\>{1,\<ldots\>,n-1}> \ unique
<math|i<rsub|1>\<in\>{1,\<ldots\>,n<rsub|1>-1},
i<rsub|2>\<in\>{1,\<ldots\>,n<rsub|2>-1}> such that
<math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|i<rsub|1>>,t<rsup|1><rsub|i<rsub|1>+1>],[t<rsup|2><rsub|i<rsub|2>>,t<rsup|2><rsub|i<rsub|2>+1>]>.
This defines functions <math|i<rsub|1>:{1,\<ldots\>,n-1}\<rightarrow\>{1,\<ldots\>,n<rsub|1>-1},i<rsub|2>:{1,\<ldots\>,n-1}\<rightarrow\>{1,\<ldots\>,n<rsub|2>-1}>
such that <math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|i<rsub|1>(i)>,t<rsup|1><rsub|i<rsub|1>(i)+1>],[t<rsup|2><rsub|i<rsub|2>(i)>,t<rsup|2><rsub|i<rsub|2>(i)+1>]>.
We have then that\
<\enumerate>
<item><math|i<rsub|1>,i<rsub|2>> are surjective
<item><math|\<forall\>i\<in\>{1,\<ldots\>,n<rsub|1>-1} we have
i<rsub|1><rsup|-1>(i)={m<rsub|1>(i),\<ldots\>,M<rsub|1>(i)},\<forall\>i\<in\>{1,\<ldots\>,n<rsub|2>-1}
we have i<rsub|2><rsup|-1>(i)={m<rsub|2>(i),\<ldots\>,M<rsub|2>(i)}>
where <math|t<rsub|m<rsub|1>(i)>=t<rsup|1><rsub|i>,t<rsub|M<rsub|1>(i)+1>=t<rsup|1><rsub|i+1>,t<rsub|m<rsub|2>(i)>=t<rsup|2><rsub|i>,t<rsub|M<rsub|2>(i)+1>=t<rsup|2><rsub|i+1>>
<item><math|i<rsup|-1><rsub|1>(i)<big|cap>i<rsup|-1><rsub|1>(j)=\<emptyset\>>
if <math|i\<neq\>j>, <math|i<rsup|-1><rsub|2>(i)<big|cap>i<rsup|-1><rsub|2>(j)=\<emptyset\>>
if <math|i\<neq\>j>
</enumerate>
<\proof>
We prove this for <math|i<rsub|1>> the prove for <math|i<rsub|2>> is
similar
<\enumerate>
<item>Let <math|k\<in\>{1,\<ldots\>,n<rsub|1>-1}> then for
<math|t<rsup|1><rsub|k>> we have by the definition of
<math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> that there exists a
<math|i\<in\>{1,\<ldots\>,n}> such that
<math|t<rsup|1><rsub|k>=t<rsub|i>> and as
<math|k\<less\>n\<Rightarrow\>t<rsup|1><rsub|k-1>\<less\>t<rsup|1><rsub|n>=b=t<rsub|n>>
we have that <math|i\<in\>{1,\<ldots\>,n-1}>. Now assume that
<math|t<rsub|k+1><rsup|1>\<less\>t<rsub|i+1>> then again by the
definition of <math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>> there exists
a <math|j\<in\>{1,\<ldots\>,n}> such that
<math|t<rsup|1><rsub|k+1>=t<rsub|j>>. we have then
<math|t<rsub|i>\<less\>t<rsup|1><rsub|k+1>=t<rsub|j>\<less\>t<rsub|i+1>>
or <math|t<rsub|j>\<in\>[t<rsub|i>t<rsub|i+1>]\<Rightarrow\>j=i+1>
but then <math|t<rsub|i+1>=t<rsub|j>=t<rsup|1><rsub|k+1>\<less\>t<rsub|i+1>>
a contradiction. So we must have <math|t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|k+1>>
or <math|t<rsup|1><rsub|k>=t<rsub|i>\<less\>t<rsub|i+1>\<leqslant\>t<rsup|1><rsub|k+1>\<Rightarrow\>[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|k>,t<rsup|1><rsub|k+1>]>
our <math|i<rsub|1>(i)=k> proving surjectivity<math|>
<item>Let <math|i\<in\>{1,\<ldots\>,n<rsub|1>-1}>. As
<math|i<rsub|1>> is surjective we have
<math|i<rsub|1><rsup|-1>(i)\<neq\>\<emptyset\> > and
<math|i<rsub|1><rsup|-1>(i)\<subseteq\>{1,\<ldots\>,n-1}>means that
<math|i<rsub|1><rsup|-1>(i)> is finite so there exists a minimum and
maximum of <math|i<rsub|1><rsup|-1>(i)> let
<math|m<rsub|1>(i)=min(i<rsup|-1><rsub|1>(i)),M<rsub|1>=max(i<rsub|1><rsup|-1>(i))>.
Lets now prove that <math|i<rsub|1><rsup|-1>(i)={m<rsub|1>(i),\<ldots\>,M<rsub|1>(i)}>
<\enumerate>
<item><math|i<rsub|1><rsup|-1>(i)\<subseteq\>{m<rsub|1>(i),M<rsub|1>(i)}>.
This is trivial by the definition of minimum and maximum
<item><math|{m<rsub|1>(i),M<rsub|1>(i)}\<subseteq\>i<rsub|1><rsup|-1>(i)>.
So let <math|k\<in\>{m<rsub|1>(i),M<rsub|1>(i)}> then
<math|m<rsub|1>(i)\<leqslant\>k\<leqslant\>M<rsub|1>(i)\<Rightarrow\>t<rsub|m<rsub|1>(i)>\<leqslant\>t<rsub|k>>
and <math|k+1\<leqslant\>M<rsub|1>(i)\<Rightarrow\>t<rsub|k+1>\<leqslant\>t<rsub|M<rsub|1>(i)+1>>.
As we have <math|[t<rsub|m<rsub|1>(i)>,t<rsub|m<rsub|1(i)+1>>],[t<rsub|M<rsub|1>(i)>,t<rsub|M<rsub|1>(i)+1>]\<subseteq\>[t<rsup|1><rsub|i>,t<rsup|1><rsub|i+1>]>
we have <math|t<rsup|1><rsub|i>\<leqslant\>t<rsub|m<rsub|1>(i)>\<leqslant\>t<rsub|k>\<less\>t<rsub|k+1>\<leqslant\>t<rsup|1><rsub|M<rsub|1>(i)+1>\<leqslant\>t<rsup|1><rsub|i+1>\<Rightarrow\>[t<rsub|k>,t<rsub|k+1>]\<subseteq\>[t<rsup|1><rsub|i>,t<rsup|1><rsub|i+1>]\<Rightarrow\>i<rsub|1>(k)=i\<Rightarrow\>k\<in\>i<rsub|1><rsup|-1>(i)>
</enumerate>
\ Next using the definition of <math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>>
we can given a <math|i\<in\>{1,\<ldots\>,n<rsub|1>-1}> find a
<math|k\<in\>{1,\<ldots\>,n-1}> such that
<math|t<rsub|k>=t<rsup|1><rsub|i>>, then using the proof of (1) we
find that <math|k\<in\>i<rsub|1><rsup|-1>(i)={m<rsub|1>(i),\<ldots\>,M<rsub|1>(i))\<Rightarrow\>m<rsub|1>(i)\<leqslant\>k\<leqslant\>M<rsub|1>(i)>
now assume that <math|m<rsub|1>(i)\<less\>k> then
<math|t<rsub|m<rsub|1>(i)>\<less\>t<rsub|k>=t<rsup|1><rsub|i>\<leqslant\><rsub|[t<rsub|m<rsub|1>(i)>,t<rsub|m<rsub|1>(i)>]\<subseteq\>[t<rsub|i><rsup|1>,t<rsup|1><rsub|i+1>]>\<leqslant\>t<rsub|m<rsub|1>(k)>>
a contradiction, so we must have <math|m<rsub|1>(i)=k\<Rightarrow\>t<rsub|m<rsub|1>(i)>=t<rsub|k>=t<rsup|1><rsub|i>>.
Next as <math|[t<rsub|M<rsub|1>(i)>,t<rsub|M<rsub|1>(i)+1>]\<subseteq\>[t<rsup|1><rsub|i>,t<rsup|1><rsub|i+1>]\<Rightarrow\>t<rsup|1><rsub|i>\<leqslant\>t<rsub|M<rsub|1>(i)+1>\<leqslant\>t<rsup|1><rsub|i+1>>.
Assume now that <math|t<rsub|M<rsub|1>(i)+1>\<less\>t<rsup|1><rsub|i+1>\<leqslant\>b=t<rsub|n>\<Rightarrow\>M<rsub|1>(i)+1\<less\>n>.
Now find a <math|l\<in\>{1,\<ldots\>,n<rsub|1>-1}> such that
<math|t<rsup|1><rsub|l>=t<rsub|M<rsub|1>(i)+1>\<Rightarrow\>t<rsup|1><rsub|l>=t<rsub|M<rsub|1>(i)+1>\<less\>t<rsub|i+1><rsup|1>\<Rightarrow\>l\<less\>i+1\<Rightarrow\>l\<leqslant\>i>
and from <math|t<rsub|i><rsup|1>\<leqslant\>t<rsub|M<rsub|1>(i)+1>=t<rsub|l>\<Rightarrow\>i\<leqslant\>l\<Rightarrow\>i=l\<Rightarrow\>t<rsub|M<rsub|1>(i)+1>=t<rsup|1><rsub|l>=t<rsub|i><rsup|1>=t<rsub|m(i)>\<less\>t<rsub|M(i)>\<leqslant\>t<rsub|M(i)+1>>
a contradiction so we must have \ <math|t<rsup|1><rsub|i+1>=t<rsup|><rsub|M<rsub|1>(i)+1>><math|>
<item>Assume that <math|k\<in\>i<rsub|1><rsup|-1>(i)<big|cap>i<rsub|2><rsup|-1>(j),i\<neq\>j>
then <math|[t<rsub|k>,t<rsub|k+1>]\<subseteq\>[t<rsup|1><rsub|i>,t<rsup|1><rsub|i+1>],[t<rsup|1><rsub|j>,t<rsup|1><rsub|j+1>]>
which because of the previous lemma means <math|i=j> a contradiction
(actually from the definition of a function this would also be
trivial)
</enumerate>
</proof>
</lemma>
<\definition>
<index|norm of partition><index|<math|\<mu\>(\<cal-P\>)>>Let <math|[a,b]>
be a interval in <math|\<bbb-R\>> with <math|a\<less\>b>,
<math|\<cal-P\>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> a partition of
<math|[a,b]> then the norm <math|\<mu\>(\<cal-P\>)> of the partition is
defined by <math|\<mu\>(\<cal-P\>)=max{\|t<rsub|i+1>-t<rsub|i>\<\|\|\>i\<in\>{1,\<ldots\>,n-1}\<equallim\><rsub|t<rsub|i>\<less\>t<rsub|i+1>>max{t<rsub|i+1>-t<rsub|i>\|i\<in\>{1,\<ldots\>,n-1}>
</definition>
<\definition>
<index|tag><index|tagged partition>Let
<math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b> and
<math|\<cal-P\>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> be a partition
of <math|[a,b]> then a tag is a function
<math|s:{1,\<ldots\>,n-1}\<rightarrow\>[a,b]> such that
<math|s<rsub|i>\<equallim\><rsub|notation>s(i)\<in\>[t<rsub|i>,t<rsub|i+1>]>.
A tagged partition <math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,s)=(\<cal-P\>,s)>
of <math|[a,b]> is a partition together with a tag. We define again the
norm of <math|\<bbb-P\>=(\<cal-P\>,s)> as
<math|\<mu\>(\<bbb-P\>)=\<mu\>(\<cal-P\>)>
</definition>
<\definition>
<label|Riemann Sum><dueto|Riemann Sum><index|Riemann sum>Let <math|[a,b]>
be a interval in <math|R> with <math|a\<less\>b>,
<math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,s)> a tagged
partition of <math|[a,b]>, <math|X,\<shortparallel\>\<shortparallel\>> a
normed real vector and <math|f:[a,b]\<rightarrow\>X> a function then
<math|\<cal-S\>(f,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n-1>f(s<rsub|i>)(t<rsub|i+1>-t<rsub|i>)>
is called a Riemann sum of <math|f>
</definition>
<\lemma>
<label|difference of Rieman sums>Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b>
, <math|X,\<shortparallel\>\<shortparallel\>> a real normed vector space
and <math|f:[a,b]\<rightarrow\>X> a function and two Riemann sums
<math|\<cal-S\>(f,\<bbb-P\><rsub|1>),\<cal-S\>(f,\<bbb-P\><rsub|2>)>,
where <math|\<bbb-P\><rsub|1>=({t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,s<rsub|1>),\<bbb-P\><rsub|2>=({t<rsup|2><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|2>}>,s<rsub|2>)>
then if we take <math|\<cal-P\>=\<cal-P\><rsub|1>#\<cal-P\><rsub|2>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
\ we have <math|\<cal-S\>(f,\<bbb-P\><rsub|1>)-\<cal-S\>(f,\<bbb-P\><rsub|2>)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(f(s<rsub|1>(i<rsub|1>(i)))-f(s<rsub|2>(i<rsub|2>(i))))(t<rsub|i+1>-t<rsub|i>)>
</lemma>
<\proof>
If we proof that <math|\<cal-S\>(f,\<bbb-P\><rsub|1>)=<big|sum><rsub|i=1><rsup|n>f(s<rsub|1>(i<rsub|1>(i)))(t<rsub|i+1>-t<rsub|i>),
\<cal-S\>(f,\<bbb-P\><rsub|2>)=<big|sum><rsub|i=1><rsup|n>f(s<rsub|2>(i<rsub|2>(i)))(t<rsub|i+1>-t<rsub|i>)>
then we can use <reference|sum of a linear combination> to prove that
<math|\<cal-S\>(f,\<bbb-P\><rsub|1>)-\<cal-S\>(f,\<bbb-P\><rsub|2>)=<big|sum><rsub|i=1><rsup|n>[f(s<rsub|1>(i<rsub|1>(i)))(t<rsub|i+1>-t<rsub|i>)-f(s<rsub|2>(i<rsub|2>(i)))(t<rsub|i+1>-t<rsub|i>)]=<big|sum><rsub|i=1><rsup|n>(f(s<rsub|1>(i<rsub|1>(i)))-f(s<rsub|2>(i<rsub|2>(i))))(t<rsub|i+1>-t<rsub|i>)>.
So lets prove that <math|\<cal-S\>(f,\<bbb-P\><rsub|1>)=<big|sum><rsub|i=1><rsup|n>f(s<rsub|1>(i<rsub|1>(i)))(t<rsub|i+1>-t<rsub|i>)>
(the proof for <math|\<cal-S\>(f,\<bbb-P\><rsub|2>)=<big|sum><rsub|i=1><rsup|n>f(s<rsub|2>(i<rsub|2>(i)))(t<rsub|i+1>-t<rsub|i>)>
is similar).
Using the previous lemma, we find that
<math|{1,\<ldots\>,n-1<rsub|>}=<big|cup><rsub|k\<in\>{1,\<ldots\>,n<rsub|1>-1}>i<rsub|1><rsup|-1>(k)>
where <math|i<rsup|-1><rsub|1>(k)={m<rsub|1>(k),\<ldots\>,M<rsub|1>(k)}>
where each <math|i<rsup|-1><rsub|1>(k)> are pairwise disjoint. Using
<reference|partioned sum> we have then
<math|<big|sum><rsub|i=1><rsup|n>f(s<rsub|1>(i<rsub|1>(i)))(t<rsub|i+1>-t<rsub|i>)=<big|sum><rsub|k=1><rsup|n<rsub|1>-1>(<big|sum><rsub|i=m<rsub|1>(k)><rsup|M<rsub|1>(k)>f(s<rsub|1>(i<rsub|1>(i)))(t<rsub|i+1>-t<rsub|i>))=<big|sum><rsub|k=1><rsup|n<rsub|1>-1>(<big|sum><rsub|i=m<rsub|1>(k)><rsup|M<rsub|1>(k)>f(s<rsub|1>(k))(t<rsub|i+1>-t<rsub|i>))\<equallim\><rsub|<reference|sum
of a linear combination>><big|sum><rsub|k=1><rsup|n<rsub|1>-1>(f(s<rsub|1>(k)(<big|sum><rsub|i=m<rsub|1>(k)><rsup|M<rsub|1>(k)>(t<rsub|i+1>-t<rsub|i>))\<equallim\><rsub|<reference|sum
of differences>><big|sum><rsub|k=1><rsup|n-1>f(s<rsub|1>(k))(t<rsub|M<rsub|1>(k)+1>-t<rsub|m<rsub|1>(k)>)=<big|sum><rsub|k=1><rsup|n-1>f(s<rsub|1>(k))(t<rsup|1><rsub|k+1>-t<rsup|1><rsub|k>)=\<cal-S\>(f,\<bbb-P\><rsub|1>)>
what we set out to prove.
</proof>
<\lemma>
Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b>,
<math|X,\<shortparallel\>\<shortparallel\>> a real normed vector space
and <math|f:[a,b]\<rightarrow\>X> a continuous function then for every
<math|\<varepsilon\>\<gtr\>0> there exists a <math|\<delta\>\<less\>0>
such that <math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)-\<cal-S\>(f,\<bbb-P\><rsub|2>)\<shortparallel\>\<less\>\<varepsilon\>>
for all <math|\<bbb-P\><rsub|1>,\<bbb-P\><rsub|2> > (tagged partitions of
<math|[a,b]>) such that <math|\<mu\>(\<bbb-P\><rsub|1>)\<less\>\<delta\>>,
<math|\<mu\>(\<bbb-P\><rsub|2>)\<less\>\<delta\>>
</lemma>
<\proof>
Using the fact that <math|[a,b]> is compact (see <reference|[a,b] is
compact>) and <math|f> is continuous we have that <math|f> is uniform
continuous on <math|[a,b]>(see <reference|continuous functions on a
compact set are uniform continuous>). There exists a
<math|\<delta\>\<gtr\>0> such that <math|\<forall\>x,x<rprime|'>\<in\>[a,b]\<vdash\>\|x-x<rprime|'>\|\<less\>2\<delta\>\<Rightarrow\>\<shortparallel\>f(x)-f(x<rprime|'>)\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>>.
Let now <math|\<mu\>(\<bbb-P\><rsub|1>)\<less\>\<delta\>,\<mu\>(\<bbb-P\><rsub|2>)\<less\>\<delta\>>
then using the previous lemma we have if
<math|\<bbb-P\><rsub|1>=({t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,s<rsub|1>),\<bbb-P\><rsub|2>=({t<rsup|2><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|2>}>,s<rsub|2>)>
that <math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)-\<cal-S\>(f,\<bbb-P\><rsub|2>)\<shortparallel\>=\<shortparallel\><big|sum><rsub|i=1><rsup|n>(f(s<rsub|1>(i<rsub|1>(i)))-f(s<rsub|2>(i<rsub|2>(i)))(t<rsub|i+1>-t<rsub|i>)\<shortparallel\>>.
Now as <math|[t<rsub|i>,t<rsub|i+1>]\<subseteq\>[t<rsup|1><rsub|i<rsub|1>(i)>,t<rsup|1><rsub|i<rsub|1>(i)+1>],[t<rsup|2><rsub|i<rsub|2>(i)>,t<rsup|2><rsub|i<rsub|2>(i)+1>]>
we have that <math|\<exists\>t\<in\>[t<rsup|1><rsub|i<rsub|1>(i)>,t<rsup|1><rsub|i<rsub|1>(i)+1>]<big|cap>[t<rsup|2><rsub|i<rsub|2>(i)>,t<rsup|2><rsub|i<rsub|2>(i)>+1]>
so as <math|s<rsub|1>(i<rsub|1>(i))\<in\>[t<rsup|1><rsub|i<rsub|1>(i)>,t<rsup|1><rsub|i<rsub|1>(i)+1>],s<rsub|2>(i<rsub|2>(i))\<in\>[t<rsup|2><rsub|i<rsub|2>(i)>,t<rsup|2><rsub|i<rsub|2>(i)+1>]\<Rightarrow\>\|s<rsub|1>(i<rsub|1>(i))-s<rsub|2>(i<rsub|2>(i))\|\<leqslant\>\|s<rsub|1>(i<rsub|1>(i))-t\|+\|t-s<rsub|2>(i<rsub|2>(i))\|\<less\><frac|\<delta\>|2>+<frac|\<delta\>|2>=\<delta\>\<Rightarrow\>\<shortparallel\>f(s<rsub|1>(i<rsub|1>(i)))-f(s<rsub|2>(i<rsub|2>(i)))\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>\<Rightarrow\>\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)-\<cal-S\>(f,\<bbb-P\><rsub|2>)\<shortparallel\>\<less\><big|sum><rsub|i=1><rsup|n>(<frac|\<varepsilon\>|b-a>)(t<rsub|i+1>-t<rsub|i>)=<frac|\<varepsilon\>|b-a><big|sum><rsub|i=1><rsup|n>(t<rsub|i+1>-t<rsub|i>)=<frac|\<varepsilon\>|b-a>(b-a)=\<varepsilon\>>
</proof>
<\theorem>
<\dueto>
Existence of Riemann Integral
</dueto>
<index|Riemann integral>
Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b,
X,\<shortparallel\>\<shortparallel\>> a normed real Banach space and
<math|f:[a,b]\<rightarrow\>X> then there exists a unique
<math|I(f)\<in\>X> which satisfies the following condition
<\enumerate>
<item><math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\>\<gtr\>0> such that for all Riemann sums
<math|\<cal-S\>(f,\<bbb-P\>)> with <math|\<mu\>(\<bbb-P\>)\<less\>\<delta\>>
(where <math|\<bbb-P\>> is a tagged partition of <math|[a,b]>) we have
<math|\<shortparallel\>I(f)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>\<less\>\<varepsilon\>>
<item>There exists a <math|{\<bbb-P\><rsub|i>}<rsub|i\<in\>\<bbb-N\>>>
familly of tagged partition of <math|[a,b]> such that
<math|lim<rsub|i\<rightarrow\>\<infty\>>\<mu\>(\<bbb-P\><rsub|i>)=0>
<item>For every <math|{\<bbb-P\><rsub|i>}<rsub|i\<in\>\<bbb-N\>>> a
tagged partition of <math|[a,b]> such that
<math|lim<rsub|i\<rightarrow\>\<infty\>>\<mu\>(\<bbb-P\><rsub|i>)> we
have <math|lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(f,\<bbb-P\><rsub|i>)=I(f)>
</enumerate>
</theorem>
We note <math|I(f)=<big|int><rsub|a><rsup|b>f=<big|int><rsub|a><rsup|b>f(t)dt>
[note that in the last notation we can interchange <math|t> with any symbol
we like, note that <math|<big|int><rsub|a><rsup|b>E(t)dt> directly
indicates that we are actually dealing with
<math|<big|int><rsub|a><rsup|b>f> where <math|f:t\<rightarrow\>E(t)>], this
is called the integral (Riemann) of f.
<\proof>
The proof happens in different stages
<\enumerate>
<item>First we prove that there is a family
<math|{(\<cal-P\><rsub|i>,s<rsub|i>)}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
such that <math|lim<rsub|i\<rightarrow\>\<infty\>>\<mu\>(\<cal-P\><rsub|i>)=0>.
To prove this define for each <math|k\<in\>\<bbb-N\><rsub|0>>
<math|{t<rsup|k><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,k}>> by
<math|t<rsub|i><rsup|k>=(i-1)<frac|(b-a)|k>+a> if
<math|i\<in\>{1,\<ldots\>,k+1=n<rsub|k>}> (note that then
<math|t<rsup|k><rsub|k+1>=b> and <math|><math|t<rsup|k><rsub|1>=a>)
then <math|t<rsup|k><rsub|i+1>-t<rsup|k><rsub|i>=((i+1-1)<frac|(b-a)|k>+a)-(i-1)<frac|b-a|k>+a)=(i-(i-1))<frac|b-a|k>=<frac|b-a|k>\<gtr\>0>
proving that <math|t<rsup|k><rsub|i>\<less\>t<rsup|k><rsub|i+1>> and as
for every <math|\<varepsilon\>\<gtr\>0> we can take
<math|N\<gtr\><frac|b-a|\<varepsilon\>>\<Rightarrow\>if
k\<geqslant\>N\<Rightarrow\>\|t<rsup|k><rsub|i+1>-t<rsup|k><rsub|i>=<frac|b-a|k>\<less\><frac|b-a|(<frac|b-a|\<varepsilon\>>)>=\<varepsilon\>>.
Finally we define <math|s<rsub|k>:{1,\<ldots\>,n<rsub|k>-1}\<rightarrow\>[a,b]>
by <math|s<rsub|k>(i)=t<rsup|k><rsub|i>>
<item>For every family<math|{\<bbb-P\><rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
such that <math|lim<rsub|i\<rightarrow\>\<infty\>>\<mu\>(\<bbb-P\><rsub|i>)=0>
we have that <math|{\<cal-S\>(f,\<bbb-P\><rsub|i>)}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
forms a Cauchy sequence. To prove this let
<math|\<varepsilon\>\<gtr\>0> then using the previous statement there
exists a <math|\<delta\>\<gtr\>0> such that for all
<math|\<bbb-P\>,\<bbb-P\><rprime|'>> be so that
<math|\<mu\>(\<bbb-P\><rsub|>),\<mu\>(\<bbb-P\><rprime|'>)\<less\>\<delta\>
we have \<shortparallel\>\<cal-S\>(f,\<bbb-P\>)-\<cal-S\>(f,\<bbb-P\><rprime|'>)\<shortparallel\>\<less\>\<varepsilon\>>.
Now we can find a <math|N\<in\>\<bbb-N\><rsub|0>> such that if
<math|k\<geqslant\>N> then <math|\<mu\>(\<bbb-P\><rsub|k>)\<less\>\<delta\>\<Rightarrow\>\<forall\>k,l\<geqslant\>N>
we have <math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|k>)-\<cal-S\>(f,\<bbb-P\><rsub|l>)\<shortparallel\>>.
<item>As by (1) and (2) there exists a family
<math|{\<bbb-P\><rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>> such that
<math|{\<cal-S\>(f,\<bbb-P\><rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|0>>>
is a Cauchy sequence in <math|X> which is complete. So there exists a
<math|I(f)=lim<rsub|k\<rightarrow\>\<infty\>>\<cal-S\>(f,\<bbb-P\><rsub|i>)>
<item>Let <math|S(f,\<bbb-P\>)> be a Riemann sum and
<math|\<varepsilon\>\<gtr\>0> then there exists a
<math|N<rsub|1>\<in\>\<bbb-N\><rsub|0>> such that
<math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|i>)-I(f)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
if <math|i\<geqslant\>N<rsub|1>>. Also we can find a <math|\<delta\>>
such that for all <math|\<bbb-P\>,\<bbb-P\><rprime|'>> with
<math|\<mu\>(\<bbb-P\>),\<mu\>(\<bbb-P\><rprime|'>)\<less\><frac|\<varepsilon\>|2>>.
As <math|lim<rsub|i\<rightarrow\>\<infty\>>\<mu\>(\<bbb-P\><rsub|i>)=0>
we can find a <math|N<rsub|2>\<in\>\<bbb-N\><rsub|0>> such that
<math|\<forall\>k\<geqslant\>N<rsub|2>> we have
<math|\<mu\>(\<bbb-P\><rsub|k>)\<less\>\<delta\>> and then if we take
<math|N=max(N<rsub|1>,N<rsub|2>)> \ <math|\<shortparallel\>I(f)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>\<leqslant\>\<shortparallel\>I(f)-\<cal-S\>(f,\<bbb-P\><rsub|N>)\<shortparallel\>+\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|n>)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
<item>Uniqueness let <math|I(f),I<rprime|'>(f)> be fulfilling (1) in
the theorem and <math|I(f)\<neq\>I<rprime|'>(f)>. Take then
<math|\<varepsilon\>=\<shortparallel\>I(f)-I<rprime|'>(f)\<shortparallel\>\<gtr\>0>,
as we have the existence of a <math|\<delta\>\<gtr\>0> such that
<math|\<shortparallel\>I(f)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
if <math|\<shortparallel\>\<bbb-P\>\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<varepsilon\>=\<shortparallel\>I(f)-I<rprime|'>(f)\<shortparallel\>\<leqslant\>\<shortparallel\>I(f)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>+\<shortparallel\>I<rprime|'>(f)-\<cal-S\>(f,\<bbb-P\>)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
a contradiction, so <math|I(f)=I<rprime|'>(f)>
</enumerate>
</proof>
<\note>
Up to now the Riemann integral is defined on a
<math|f\<in\>\<cal-C\>([a,b],X)> where
<math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b> and <math|X> is a Banach
space. If <math|a=b> then we define <math|<big|int><rsub|a><rsup|a>f=0>
(<math|0\<in\>X> not in <math|\<bbb-R\>>) for every
<math|f\<in\>\<cal-C\>([a,b],X)>
</note>
<\note>
If <math|\<cal-C\>([a,b],X)> is the set of all continuous functions from
<math|[x,y]> to the Banach space <math|X> then
<math|<big|int><rsub|a><rsup|b>>is a function from
<math|\<cal-C\>([a,b],X)\<rightarrow\>X> defined by
<math|f\<rightarrow\><big|int><rsub|a><rsup|b>f>
</note>
<\theorem>
<label|linearity of integral>Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<leqslant\>b>
, <math|X,\<shortparallel\>\<shortparallel\>> a Banach space then
<math|<big|int><rsub|a><rsup|b>:\<cal-C\>([a,b],X)\<rightarrow\>X> is
multilinear. (or <math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,><math|\<forall\>f,g\<in\>\<cal-C\>([a,b],X)>
we have <math|<big|int><rsub|a><rsup|b>(\<alpha\>.f+\<beta\>.g)=\<alpha\><big|int><rsub|a><rsup|b>f+\<beta\><big|int><rsub|a><rsup|b>>
</theorem>
<\proof>
First in the case <math|a=b> we have <math|<big|int><rsub|a><rsup|a>(\<alpha\>.f+\<beta\>.g)=0=\<alpha\>.0+\<beta\>.0=\<alpha\>.<big|int><rsub|a><rsup|a>f+\<beta\>.<big|int><rsub|a><rsup|a>g>
We are thus left with the case <math|a\<less\>b>. If
<math|\<bbb-P\>=(\<cal-P\>,s)> where <math|\<cal-P\>={t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
is a partition of <math|[a,b]> then <math|\<cal-S\>(\<alpha\>.f+\<beta\>.g,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n>(\<alpha\>.f+\<beta\>.g)(s(i))(t<rsub|i+1>-t<rsub|i>)=\<alpha\>(<big|sum><rsub|i=1><rsup|n>f(s(i))(t<rsub|i+1>-t<rsub|i>))+\<beta\>(<big|sum><rsub|i=1><rsup|n>g(s(i))(t<rsub|i+1>-t<rsub|i>)=\<alpha\>,\<cal-S\>(f,\<bbb-P\>)+\<beta\>.\<cal-S\>(g,\<bbb-P\>)>.
Now <math|I(\<alpha\>.f+\<beta\>.g)=lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(\<alpha\>.f+\<beta\>.g,\<bbb-P\><rsub|i>)\<equallim\><rsub|<reference|limit
of a sum>>\<alpha\>.lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(f,\<bbb-P\><rsub|i>)+\<beta\>.lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(g,\<bbb-P\><rsub|i>)=\<alpha\><big|int><rsub|a><rsup|b>f+\<beta\><big|int><rsub|a><rsup|b>g>
</proof>
<\theorem>
<label|property of integral>Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<leqslant\>b>
, <math|X,\<shortparallel\>\<shortparallel\>> a Banach space then we have
<\enumerate>
<item><math|\<forall\>\<varphi\>:X\<rightarrow\>\<bbb-R\>> be a
continuous linear function and <math|\<forall\>f\<in\>\<cal-C\>([a,b],X)>
then <math|<big|int><rsub|a><rsup|b>(\<varphi\>\<circ\>f)=\<varphi\>(<big|int><rsub|a><rsup|b>f)>,
note here we must have <math|a\<less\>b>
<item><math|><math|\<forall\>f\<in\>\<cal-C\>([a,b],X)> we have
<math|\<shortparallel\><big|int><rsub|a><rsup|b>f\<shortparallel\>\<leqslant\><big|int><rsub|a><rsup|b>\<shortparallel\>f\<shortparallel\>>
<item>If <math|m\<in\>\<bbb-R\>> and <math|f\<in\>\<cal-C\>([a,b],X)>
such that <math|\<shortparallel\>f\<shortparallel\>\<leqslant\>m> then
<math|\<shortparallel\><big|int><rsub|a><rsup|b>f\<shortparallel\>\<leqslant\>m(b-a)>
</enumerate>
Not that from (3) it follows that <math|<big|int><rsub|a><rsup|b>f-<big|int><rsub|a><rsup|c>f=(<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f)-<big|int><rsub|a><rsup|c>f=<big|int><rsub|c><rsup|b>f>
</theorem>
<\proof>
\;
We prove first the case for <math|a=b>
<\enumerate>
<item>As we require <math|a\<less\>b> we don't have to prove this
<item><math|\<shortparallel\><big|int><rsub|a><rsup|a>f\<shortparallel\>=\<shortparallel\>0\<shortparallel\>\<leqslant\>0=<big|int><rsub|a><rsup|a>\<shortparallel\>f\<shortparallel\>>
<item><math|\<shortparallel\><big|int><rsub|a><rsup|a>f\<shortparallel\>=0\<leqslant\>m(b-a)>
as <math|m\<geqslant\>\<shortparallel\>f\<shortparallel\>\<geqslant\>0>
</enumerate>
If <math|a\<less\>b> then we have
<\enumerate>
<item>First if <math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,s)>
then by linearity of <math|\<varphi\>> we have
<math|\<cal-S\>(\<varphi\>\<circ\>f,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n>\<varphi\>(f(s(i))(t<rsub|i+1>-t<rsub|i>)=\<varphi\>(<big|sum><rsub|i=1><rsup|n>f(s(i))(t<rsub|i+1>-t<rsub|i>))=\<varphi\>(\<cal-S\>(f,\<bbb-P\>))>.
Now as <math|\<varphi\>,f> are continuous
<math|\<varphi\>\<circ\>f:[a,b]\<rightarrow\>\<bbb-R\>> is continuous
so the integral exists. Then <math|<big|int><rsub|a><rsup|b>(\<varphi\>\<circ\>f)=lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(\<varphi\>\<circ\>f,\<bbb-P\><rsub|i>)\<equallim\>lim<rsub|i\<rightarrow\>\<infty\>>\<varphi\>(\<cal-S\>(f,\<bbb-P\><rsub|i>))\<equallim\><rsub|<reference|limit
and continuity>>\<varphi\>(lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(f,\<bbb-P\><rsub|i>)=\<varphi\>(<big|int><rsub|a><rsup|b>f)>
<item>First of all given <math|x,y\<in\>X>
<math|\|\<shortparallel\>f(x)\<shortparallel\>-\<shortparallel\>f(y)\<shortparallel\>\|\<leqslant\>\<shortparallel\>f(x)-f(y)\<shortparallel\>>
from which it follows that if <math|f> is continuous then
<math|\<shortparallel\>f\<shortparallel\>:[a,b]\<rightarrow\>\<bbb-R\>>
is also a continuous function, so its integral exists. Second we have
given <math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>\<bbb-N\><rsub|o>>,s)>
that <math|\<shortparallel\><big|sum><rsub|i=1><rsup|n>f(s(i))(t<rsub|i+1>-t<rsub|i>)\<shortparallel\>\<leqslant\><big|sum><rsub|i=1><rsup|n>\<shortparallel\>f(s(i))\<shortparallel\>(t<rsub|i+1>-t<rsub|i>)=\<cal-S\>(\<shortparallel\>f\<shortparallel\>,\<bbb-P\>)>
. So <math|\<shortparallel\><big|int><rsub|a><rsup|b>f\<shortparallel\>=\<shortparallel\>lim<rsub|i\<rightarrow\>\<infty\>>S(f,\<bbb-P\><rsub|i>)\<shortparallel\>\<equallim\><rsub|continuity
of \<shortparallel\>\<shortparallel\> and <reference|limit and
continuity>>lim<rsub|i\<rightarrow\>\<infty\>>\<shortparallel\>S(f,\<bbb-P\><rsub|i>)\<shortparallel\>\<leqslant\><rsub|<reference|limit
preserves inequality>>lim<rsub|i\<rightarrow\>\<infty\>>S(\<shortparallel\>f\<shortparallel\>,\<bbb-P\><rsub|i>)=<big|int><rsub|a><rsup|b>\<shortparallel\>f\<shortparallel\>>
<item>First using (2) we have <math|\<shortparallel\><big|int><rsub|a><rsup|b>f\<shortparallel\>\<leqslant\><big|int><rsub|a><rsup|b>\<shortparallel\>f\<shortparallel\>>.
Now if <math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,s)> is
a partition of <math|[a,b]> then <math|S(\<shortparallel\>f\<shortparallel\>,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n-1>\<shortparallel\>f(s<rsub|i>)\<shortparallel\>(t<rsub|i+1>-t<rsub|i>)\<leqslant\>m(<big|sum><rsub|i=1><rsup|n>(t<rsub|i+1>-t<rsub|i>))=m(b-a)\<Rightarrow\><big|int><rsub|a><rsup|b>\<shortparallel\>f\<shortparallel\>=lim<rsub|i\<rightarrow\>\<infty\>>S(\<shortparallel\>f\<shortparallel\>,\<bbb-P\><rsub|i>)\<leqslant\>m(b-a)>
</enumerate>
</proof>
<\theorem>
<label|integral of function to linear operators>If
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> is a normed space and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> a Banach space (then
by <reference|space of linear continuous maps to a Banach space is
Banach> we have that <math|L(X,Y)> is a Banach space). If
<math|f\<in\>\<cal-C\>[a,b],L(X,Y))> then <math|\<forall\>x\<in\>X> we
have<math|(<big|int><rsub|a><rsup|b>f)(x)=<big|int><rsub|a><rsup|b>f(.)(x)>
(where <math|\<forall\>x\<in\>X \ f(.):[a,b]\<rightarrow\>Y> is defined
by <math|t\<rightarrow\>f(.)(x)(t)=f(t)(x)> [Or using our alternative
syntax we have <math|(<big|int><rsub|a><rsup|b>f(t)dt)(x)=<big|int><rsub|a><rsup|b>f(t)(x)dt>]
</theorem>
<\proof>
\;
Note that given <math|x\<in\>X> the mapping
<math|\<varphi\><rsub|x>:L(X,Y)\<rightarrow\>Y> defined by
<math|f\<rightarrow\>\<varphi\><rsub|x>(f)=f(x)> is continuous as
<math|\<shortparallel\>\<varphi\><rsub|x>(f)-\<varphi\><rsub|x>(g)\<shortparallel\><rsub|Y>=\<shortparallel\>f(x)-g(x)\<shortparallel\><rsub|Y>=\<shortparallel\>(f-g)(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>f-g\<shortparallel\>\<shortparallel\>x\<shortparallel\><rsub|X>>
and also linear [<math|\<varphi\><rsub|x>(\<alpha\>.f+\<beta\>.g)=(\<alpha\>.f+\<beta\>.g)(x)=\<alpha\>.f(x)+\<beta\>.g(x)=\<alpha\>.\<varphi\><rsub|x>(f)+\<beta\>.\<varphi\><rsub|x>(g)>.
\ Then as <math|f(.)(x)=\<varphi\><rsub|x>\<circ\>f:[a,b]\<rightarrow\>Y>
is continuous its integral exists and
\ <math|\<cal-S\>(\<varphi\><rsub|x>\<circ\>f,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n>\<varphi\><rsub|x>(f(s(i))(t<rsub|i+1>-t<rsub|i>)=\<varphi\><rsub|x>(<big|sum><rsub|i=1><rsup|n>f(s(i))(t<rsub|i+1>-t<rsub|i>))=\<varphi\><rsub|x>(S(f,\<bbb-P\>))>.
Then from <math|<big|int><rsub|a><rsup|b>(\<varphi\><rsub|x>\<circ\>f)=lim<rsub|i\<rightarrow\>\<infty\>>S(\<varphi\><rsub|x>\<circ\>f,\<bbb-P\><rsub|i>)=lim<rsub|i\<rightarrow\>\<infty\>>\<varphi\><rsub|x>(S(f,\<bbb-P\><rsub|i>))\<equallim\><rsub|\<varphi\><rsub|x>
is continuous>\<varphi\><rsub|x>(lim<rsub|i\<rightarrow\>\<infty\>>S(f,\<bbb-P\><rsub|i>))=\<varphi\><rsub|x>(<big|int><rsub|a><rsup|b>f)=(<big|int><rsub|a><rsup|b>f)(x)=(<big|int><rsub|a><rsup|b>f(t)dt)(x)>
</proof>
<\theorem>
<label|integration of constant>Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<leqslant\>b,
X,\<shortparallel\>\<shortparallel\>> a Banach space and
<math|C:[a,b]\<rightarrow\>X> a constant function
(<math|\<forall\>x\<in\>[a,b]\<Rightarrow\>C(x)=C>) then
<math|<big|int><rsub|a><rsup|b>C=C(b-a)>\
</theorem>
<\proof>
\;
First if <math|a=b> then <math|<big|int><rsub|a><rsup|a>C=0=C.0=0>
Second, constant functions are trivial continuous so the integral is well
defined, thirdly if <math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>,s)>
is a tagged partition then <math|\<cal-S\>(C,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n>C(t<rsub|i+1>-t<rsub|i>)=C<big|sum><rsub|i=1><rsup|n>(t<rsub|i+1>-t<rsub|i>)=C(b-a)\<Rightarrow\><big|int><rsub|a><rsup|b>C=lim<rsub|i\<rightarrow\>\<infty\>>\<cal-S\>(C,\<bbb-P\><rsub|i>)=lim<rsub|i\<rightarrow\>\<infty\>>(C(b-a))=C(b-a)>
</proof>
\;
<\theorem>
<label|splitting of a integral>Let <math|[a,b]\<subseteq\>\<bbb-R\>,a\<leqslant\>b,
X,\<shortparallel\>\<shortparallel\>> a Banach space \ and
<math|c\<in\>[a,b]> and <math|f:[a,b]\<rightarrow\>X> is continuous then
<math|<big|int><rsub|a><rsup|b>f=<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f>
(here we have noted <math|f<rsub|\|[a,c]>> and <math|f<rsub|\|[c,d]>>
also as <math|f>
</theorem>
<\proof>
\;
First lets look at the following cases
<\enumerate>
<item><math|a=b>, in this case <math|a=b=c\<Rightarrow\><big|int><rsub|a><rsup|b>f=<big|int><rsub|a><rsup|a>f=0=0+0=<big|int><rsub|a><rsup|a>f+<big|int><rsub|a><rsup|a>f=<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f>
<item><math|a\<less\>b,><math|c=a> then
<math|<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f=<big|int><rsub|a><rsup|a>f+<big|int><rsub|a><rsup|b>=0+<big|int><rsub|a><rsup|b>f=<big|int><rsub|a><rsup|b>f>
<item><math|a\<less\>b,c=b> then <math|<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f=<big|int><rsub|a><rsup|b>f+<big|int><rsub|b><rsup|b>f=<big|int><rsub|a><rsup|b>f+0=<big|int><rsub|a><rsup|b>f>
</enumerate>
Let's now look at the case where <math|a\<less\>b> and
<math|c\<in\>]a,b[>
First we prove the following:
If <math|\<bbb-P\><rsub|1>=({t<rsup|1><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>}>,s<rsup|1>)>
is a tagged partition of <math|[a,c]> and
<math|\<bbb-P\><rsub|2>=({t<rsub|i><rsup|2>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|2>}>,s<rsup|2>}>
is a tagged partition of <math|[c,b]> then
<math|\<bbb-P\>=({t<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n<rsub|1>+n<rsub|2>-1}>,s)\<equallim\><rsub|notation>\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>>
\ where\
<\eqnarray*>
<tformat|<table|<row|<cell|t<rsub|i>=>|<cell|t<rsub|i><rsup|1>>|<cell|i\<in\>{1,\<ldots\>,n<rsub|1>}>>|<row|<cell|>|<cell|t<rsup|2><rsub|i-n<rsub|1>+1>>|<cell|i\<in\>{n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>-1}>>>>
</eqnarray*>
and\
<\eqnarray*>
<tformat|<table|<row|<cell|s(i)=>|<cell|s<rsup|1>(i)>|<cell|i\<in\>{1,\<ldots\>,n<rsub|1>-1}>>|<row|<cell|>|<cell|s<rsup|2>(i-n<rsub|1>+1)>|<cell|i\<in\>{n<rsub|1>,\<ldots\>,n<rsub|1>+n<rsub|2>-2}>>>>
</eqnarray*>
is a partition of <math|[a,b]> and <math|\<mu\>(\<bbb-P\>)=max(\<mu\>(\<bbb-P\><rsub|1>),\<mu\>(\<bbb-P\><rsub|2>))>.
To prove this note first that <math|t<rsub|1>=t<rsup|1><rsub|1>=a> and
<math|t<rsub|n<rsub|1>+n<rsub|2>-1>=t<rsup|2><rsub|n<rsub|1>+n<rsub|2>-1-n<rsub|1>+1>=t<rsup|2><rsub|n<rsub|2>>=b>,
also if <math|i\<in\>{1,\<ldots\>,n<rsub|1>-1}> then
<math|t<rsub|i>=t<rsup|1><rsub|i>\<less\>t<rsup|1><rsub|i+1>=t<rsub|i+1>>,
if <math|i=n<rsub|1>> then <math|t<rsub|i>=t<rsub|n<rsub|1>>=t<rsup|1><rsub|n<rsub|1>>=c\<less\>t<rsup|2><rsub|2>=t<rsub|n<rsub|1>+1>=t<rsub|i+1>>,
if <math|i\<in\>{n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>-1}> we have
<math|t<rsub|i>=t<rsup|2><rsub|i-n<rsub|1>+1>\<less\>t<rsup|2><rsub|(i+1)-n<rsub|1>+1>=t<rsub|i+1>>.
Also if <math|i\<in\>{1,\<ldots\>,n<rsub|1>-1}> then
<math|s(i)=s<rsup|1>(i)\<in\>[t<rsup|1><rsub|i>,t<rsup|1><rsub|i+1>]=[t<rsub|i>,t<rsub|i+1>]>,
if <math|i=n<rsub|1>> then <math|s<rsup|>(i)=s(n<rsub|1>)=s<rsup|2><rsub|>(1)\<in\>[t<rsup|2><rsub|1>,t<rsup|2><rsub|2>]=[c,t<rsub|n<rsub|1>+1>]=[t<rsub|n<rsub|1>>,t<rsub|n<rsub|1>+1>]=[t<rsub|i>,t<rsub|i+1>]>,
if <math|i\<in\>{n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>-2}={n<rsub|1>+1,\<ldots\>,n<rsub|1>+n<rsub|2>-1}>
then <math|s(i)=s<rsup|2>(i-n<rsub|1>+1)\<in\>[t<rsup|2><rsub|i-n<rsub|1>+1>,t<rsup|2><rsub|i-n<rsub|1>+1+1>]=[t<rsub|i>,t<rsub|i+1>]>
proving that <math|\<bbb-P\>> is indeed a tagged partition. Now
<math|{\|t<rsub|i+1>-t<rsub|i>\| \| i\<in\>{1,\<ldots\>,n<rsub|1>+n<rsub|2>-1-1}}={\|t<rsub|i+1>-t<rsub|i>\|
\|i\<in\>{1,\<ldots\>,n<rsub|1>-1}}<big|cup>{\|t<rsub|i+1>-t<rsub|i>\| \|
i\<in\>{n<rsub|1>,\<ldots\>,n<rsub|1>+n<rsub|2>-1-1}}\<equallim\><rsub|t<rsub|n<rsub|1>>=c=t<rsup|2><rsub|1>>{\|t<rsup|1><rsub|i+1>-t<rsup|i1><rsub|i>\|\|i\<in\>{1,\<ldots\>,n<rsub|1>-1}}<big|cup>{\|t<rsup|2><rsub|i+1>,t<rsup|2><rsub|i>\|\|<rsup|>i\<in\>{1,\<ldots\>,n<rsub|2>-1}}\<Rightarrow\>\<mu\>(\<bbb-P\>)=max(\<mu\>(\<bbb-P\><rsub|1>),\<mu\>(\<bbb-P\><rsub|2>))>
Note also that <math|\<cal-S\>(f,\<bbb-P\>)=\<cal-S\>(f<rsub|1>,\<bbb-P\><rsub|1>)+\<cal-S\>(f<rsub|2>,\<bbb-P\>)>
as <math|\<cal-S\>(f,\<bbb-P\>)=<big|sum><rsub|i=1><rsup|n<rsub|1>+n<rsub|2>-2>f(s(i))(t<rsub|i+1>-t<rsub|i>)=(<big|sum><rsub|i=1><rsup|n<rsub|1>-1>f(s(i))(t<rsub|i+1>-t<rsub|i>))+<big|sum><rsub|i=n<rsub|1>><rsup|n<rsub|1>+n<rsub|2>-2>f(s(i))(t<rsub|i+1>-t<rsub|i>)=(<big|sum><rsub|i=1><rsup|n<rsub|1>-1>f(s<rsup|1>(i))(t<rsup|1><rsub|i+1>-t<rsup|1><rsub|i>))+<big|sum><rsub|i=1><rsup|n<rsub|2>-1>f(s<rsup|2>(i))(t<rsup|2><rsub|i+1>-t<rsub|i><rsup|2>)=\<cal-S\>(f,\<bbb-P\><rsub|1>)+\<cal-S\>(f,\<bbb-P\><rsub|2>)>
Secondly if <math|\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\><rsub|1>\<gtr\>0,\<delta\><rsub|2>\<gtr\>0> such that if
<math|\<bbb-P\><rsub|1>> is a tagged partition of <math|[a,c]> with
<math|\<mu\>(\<bbb-P\><rsub|1>)\<less\>\<delta\><rsub|1>> and
<math|\<mu\>(\<bbb-P\><rsub|2>)\<less\>\<delta\><rsub|2>> so that
<math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)-<big|int><rsub|a><rsup|c>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
and <math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|2>)-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>\<Rightarrow\>\<\|\|\>\<cal-S\>(f,\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)-(<big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f)\<shortparallel\>=\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)+\<cal-S\>(f,\<bbb-P\><rsub|2>)-<big|int><rsub|a><rsup|c>f-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<leqslant\>\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1>)-<big|int><rsub|a><rsup|c>\<shortparallel\>+\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|2>)-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>.\
Assume now that <math|<big|int><rsub|a><rsup|b>f\<neq\><big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f>
then <math|\<varepsilon\>=\<shortparallel\><big|int><rsub|a><rsup|b>f-<big|int><rsub|a><rsup|c>f-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<gtr\>0>
and take <math|\<delta\><rprime|'>\<gtr\>0> such that if <math|\<bbb-P\>>
is a tagged partition of <math|[a,b]> with
<math|\<mu\>(\<bbb-P\>)\<less\>\<delta\><rprime|'>> then
<math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\>)-<big|int><rsub|a><rsup|b>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>,
also we can find a <math|\<delta\><rsub|1>,\<delta\><rsub|2>\<gtr\>0>
such that <math|\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)-<big|int><rsub|a><rsup|c>f-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
if <math|\<mu\>(\<bbb-P\><rsub|1>)\<less\>\<delta\><rsub|1>,\<mu\>(\<bbb-P\><rsub|2>)\<less\>\<delta\><rsub|2>>
then if we take <math|\<delta\>=min(\<delta\><rprime|'>,\<delta\><rsub|1>,\<delta\><rsub|2>)>
and we have if <math|\<mu\>(\<bbb-P\><rsub|1>)\<less\>\<delta\>,\<mu\>(\<bbb-P\><rsub|2>)\<less\>\<delta\>>
so that <math|\<mu\>(\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)\<less\>\<delta\>\<leqslant\>\<delta\><rprime|'>>
and <math|\<varepsilon\>=\<shortparallel\><big|int><rsub|a><rsup|b>f-<big|int><rsub|a><rsup|c>f-<big|int><rsub|c><rsup|b>\<shortparallel\>\<leqslant\>\<shortparallel\><big|int><rsub|a><rsup|b>f-\<cal-S\>(f,\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)\<shortparallel\>+\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)-<big|int><rsub|a><rsup|b>f\<shortparallel\>+\<shortparallel\>\<cal-S\>(f,\<bbb-P\><rsub|1><big|cup>\<bbb-P\><rsub|2>)-<big|int><rsub|a><rsup|c>f-<big|int><rsub|c><rsup|b>f\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>
> a contradictions so we must have <math|<big|int><rsub|a><rsup|b>f\<neq\><big|int><rsub|a><rsup|c>f+<big|int><rsub|c><rsup|b>f>
</proof>
<section|Connected Sets>
<\definition>
<index|connected topological space>A topological space <math|X> is
connected if <math|\<forall\>U<rsub|1>,U<rsub|2>> open such that
<math|U<rsub|1><big|cap>U<rsub|2>=\<emptyset\>> and
<math|X=U<rsub|1><big|cup>U<rsub|2>> then either
<math|U<rsub|1>=\<emptyset\>> or <math|U<rsub|2>=\<emptyset\>>. If
<math|X> is not connected then it is disconnected (meaning there exists
<math|U<rsub|1>,U<rsub|2>> open and non empty with
<math|U<rsub|1><big|cap>U<rsub|2>=\<emptyset\>> and
<math|X=U<rsub|1><big|cup>U<rsub|2>>
</definition>
<\theorem>
Let <math|X> be a topological space then the following are equivalent
<\enumerate>
<item><math|\<emptyset\>> and <math|X> are the only subsets of <math|X>
which are open and closed
<item><math|X> is connected
<item><math|\<forall\>A<rsub|1>,A<rsub|2>> closed sets \ with
<math|A<rsub|1><big|cap>A<rsub|2>=\<emptyset\>> and
<math|X=A<rsub|1><big|cup>A<rsub|2>> we have either
<math|A<rsub|1>=\<emptyset\>> or <math|A<rsub|2>=\<emptyset\>>
</enumerate>
</theorem>
<\proof>
We make in this proof heavy use of <reference|union of disjoint sets>
<math|1\<Rightarrow\>2>
Assume that we have disjoint open sets <math|U<rsub|1>,U<rsub|2>> with
<math|X=U<rsub|1><big|cup>U<rsub|2>>. Then
<math|U<rsub|2>=X<mid|\\>U<rsub|1>> is closed and open so either it is
<math|\<emptyset\>> or it is <math|X> and then
<math|U<rsub|2>=X<mid|\\>U<rsub|1>=X<mid|\\>X=\<emptyset\>> and thus
<math|X> is connected
<math|2\<Rightarrow\>3>
Assume that we have disjoint closed sets <math|A<rsub|1>,A<rsub|2>> with
<math|X=A<rsub|1><big|cup>A<rsub|2>> then
<math|A<rsub|1>=X<mid|\\>A<rsub|2>> is open and also
<math|A<rsub|2>=X<mid|\\>A<rsub|1>> is open and then by connectedness we
have either <math|A<rsub|1>=\<emptyset\>> or
<math|A<rsub|2>=\<emptyset\>>
<math|3\<Rightarrow\>1>
Let <math|A\<subseteq\>X> and <math|A> is open and closed then
<math|X<mid|\\>A> is closed and <math|X=A<big|cup>(X<mid|\\>A)> with
<math|A<big|cap>(X<mid|\\>A)=\<emptyset\>> then we have either
<math|A=\<emptyset\>> or <math|X<mid|\\>A=\<emptyset\>\<Rightarrow\>A=X>
</proof>
<\definition>
<index|connected set>Let <math|X> be a topological space and let
<math|A\<subseteq\>X> then <math|A> is connected if it is connected in
the subspace topology
</definition>
<\theorem>
<label|[a,b] is connected>Let <math|\<bbb-R\>,\<\|\|\>> be the real space
equipped with the topology generated by the norm <math|\| \|> then
<math|\<forall\>a,b\<in\>\<bbb-R\>> with <math|a\<leqslant\>b> we have
that <math|[a,b]> is a connected subset
</theorem>
<\proof>
First as <math|[a,b]=[<frac|a+b|2>-<frac|b-a|2>,<frac|a+b|2>+<frac|b-a|2>]=<wide|B<rsub|\<\|\|\>>(<frac|a+b|2>,<frac|b-a|2>)|\<bar\>>>
which is closed by <reference|closed balls are closed> we have that
<math|[a,b]> is closed, so if <math|A\<subseteq\>[a,b]> is closed in the
subspace topology then <math|A=A<rprime|'><big|cap>[a,b]> with
<math|A<rprime|'>> closed in <math|\<bbb-R\>> and thus <math|A> is also
closed in <math|\<bbb-R\>>. Assume now that <math|[a,b]> is not connected
then we can find a disjoint closed sets <math|A<rsub|1>,A<rsub|2>> (in
<math|[a,b]> thus also closed in <math|\<bbb-R\>>) such that
<math|[a,b]=A<rsub|1><big|cup>A<rsub|2>> and with
<math|A<rsub|1>,A<rsub|2>\<neq\>\<emptyset\>>. Now we may assume that
<math|b\<in\>A<rsub|2>> [otherwise interchange the meaning of
<math|A<rsub|1>,A<rsub|2>>]. As <math|A<rsub|1>> is bounded above by
<math|b> and is not empty we have the existence of
<math|c=sup(A<rsub|1>)>. Now if <math|c\<in\>]x,y[> then
<math|\<exists\>a<rprime|'>\<in\>A<rsub|1>> such that
<math|x\<less\>a<rprime|'>\<leqslant\>c\<Rightarrow\>a<rprime|'>\<in\>]x,y[\<Rightarrow\>A<rsub|1><big|cap>]x,y[\<neq\>\<emptyset\>\<Rightarrowlim\><rsub|A<rsub|1>is
closed>c\<in\>A<rsub|1>>. Since <math|b> is a upper bound of
<math|A<rsub|1>> and <math|b\<nin\>A<rsub|1>> we have <math|c\<less\>b>.
Now if <math|x\<gtr\>c> then <math|x\<nin\>A<rsub|1>> (otherwise <math|c>
is no upper bound)<math|\<Rightarrow\>]c,b]\<subseteq\>[a,b]<mid|\\>A<rsub|1>=A<rsub|2>>.
Hence for a an open interval <math|c\<in\>]x,y[> we have
<math|y-c,b-c\<gtr\>0\<Rightarrow\>\<varepsilon\>=min(y-c,b-c)\<gtr\>\<varepsilon\>>.
and <math|x\<less\>c\<less\>c+<frac|\<varepsilon\>|2>\<less\>c+\<varepsilon\>\<leqslant\>c+(y-c)=y>
(and <math|\<leqslant\>c+(b-c)=b>, and thus
<math|x\<less\>c+<frac|\<varepsilon\>|2>\<less\>y> (and
<math|x\<less\>c+<frac|\<varepsilon\>|2>\<less\>b> so
<math|c+<frac|\<varepsilon\>|2>\<in\>]x,y[<big|cap>]x,b[\<subseteq\>]x,y[<big|cap>]x,b]\<subseteq\>]x,y[<big|cap>A<rsub|2>\<Rightarrow\>]x,y[<big|cap>A<rsub|2>\<neq\>\<emptyset\>>
and then as <math|c\<in\><wide|A<rsub|2>|\<bar\>>\<equallim\><rsub|A<rsub|2>
is closed>A<rsub|2>> but then <math|c\<in\>A<rsub|1><big|cap>A<rsub|2>> a
contradicting the assumption that <math|A<rsub|1>,A<rsub|2>> are
disjoint.\
</proof>
<\theorem>
<label|generalized intervals and connectness>Let
<math|\<bbb-R\>,\<\|\|\>> be the real space equipped with the topology
generated by the norm <math|\| \|> then the following are equivalent for
a set <math|I\<subseteq\>\<bbb-R\>>
<\enumerate>
<item><math|I> is connected
<item><math|I> is a generalized interval
</enumerate>
</theorem>
<\proof>
\;
<math|1\<Rightarrow\>2>
Assume that <math|I> is connected and that <math|I> is not a generalized
interval, then using <reference|generalized intervals>
<math|\<exists\>x,y\<in\>I,z\<in\>\<bbb-R\><mid|\\>I> such that
<math|x\<less\>z\<less\>y> then <math|U<rsub|1>=]\<um\>\<infty\>,z[<big|cap>I>
contains <math|x> and is open in <math|I>, also
<math|U<rsub|2>=]z,\<infty\>[<big|cap>I> contains <math|y> and is open in
<math|I>. Then as <math|z\<nin\>I> we have
\ <math|I\<subseteq\>]\<um\>\<infty\>,\<infty\>[<mid|\\>{z}=]\<um\>\<infty\>,z[<big|cup>]z,\<infty\>[\<Rightarrow\>I=I<big|cap>(]\<um\>\<infty\>,z[<big|cup>]z,\<infty\>[)=U<rsub|1><big|cup>U<rsub|2>>
and from the fact that <math|U<rsub|1><big|cap>U<rsub|2>=]\<um\>\<infty\>,z[<big|cap>]z,\<infty\>[=\<emptyset\>>
and <math|U<rsub|1>,U<rsub|2> are not empty> we get that <math|I> is not
connected a contradiction.
<math|2\<Rightarrow\>1>
Assume that <math|I> is a generalized interval and that <math|I> is not
connected then we have (using the definition of subspace topology and
connectedness) open sets <math|U<rsub|1>,U<rsub|2>\<subseteq\>\<bbb-R\>>
such that <math|(U<rsub|1><big|cap>I)<big|cap>(U<rsub|2><big|cap>I)=\<emptyset\>>
and <math|I=(U<rsub|1><big|cap>I)<big|cup>(U<rsub|2><big|cap>I)> then and
<math|U<rsub|1><big|cap>I,U<rsub|2><big|cap>I> are not empty so
<math|\<exists\>x\<in\>U<rsub|1><big|cap>I\<subseteq\>I,\<exists\>y\<in\>U<rsub|2><big|cap>I\<subseteq\>I>
now we must have <math|x\<neq\>y> (otherwise
<math|(U<rsub|1><big|cap>I)<big|cap>(U<rsub|2><big|cap>)> is not empty)
so assume that <math|x\<less\>y> [for the other case interchange the
roles of <math|U<rsub|1>> and <math|U<rsub|2> >] then using
<reference|generalized intervals> we have <math|[x,y]\<subseteq\>I> and
thus <math|[x,y]=[x,y]<big|cap>I=([x,y]<big|cap>I<big|cap>U<rsub|1>)<big|cup>([x,y]<big|cap>I<big|cap>U<rsub|2>)=([x,y]<big|cap>U<rsub|1>)<big|cup>([x,y]<big|cap>U<rsub|2>)>
as <math|x\<in\>[x,y]<big|cap>U<rsub|1>,y\<in\>[x,y]<big|cap>U<rsub|2>>
and <math|([x,y]<big|cap>U<rsub|1>)<big|cap>([x,y]<big|cap>U<rsub|2>)\<subseteq\>(I<big|cap>U<rsub|1>)<big|cap>(I<big|cap>U<rsub|2>)=\<emptyset\>>
we have that <math|[x,y]> is covered by two non empty disjoint open sets,
so it is not connected in contradiction with <reference|[a,b] is
connected>. So we conclude that <math|I> must be connected.
</proof>
<\theorem>
<label|continuity and connectness>Let <math|f:X\<rightarrow\>Y> be a
continuous function of a connected topological space <math|X> to a
topological space <math|Y> then <math|f(X)> is connected.
</theorem>
<\proof>
Assume that <math|f(X)> is not connected then we have
<math|V<rsub|1>,V<rsub|2>> open in <math|Y> such that
<math|f(X)=(f(X)<big|cap>V<rsub|1>)<big|cap>(f(X)<big|cap>V<rsub|2>)\<subseteq\>V<rsub|1><big|cup>V<rsub|2>>,
<math|\<exists\>y<rsub|1>\<in\>f(X)<big|cap>V<rsub|1>,y<rsub|2>\<in\>f(X)<big|cap>V<rsub|2>>
and <math|(f(X)<big|cap>V<rsub|1>)<big|cap>(f(X)<big|cap>V<rsub|2>)=\<emptyset\>>
then <math|\<exists\>x<rsub|1>,x<rsub|2>\<in\>X> such that
<math|f(x<rsub|1>)=y<rsub|1>\<in\>V<rsub|1>,f(x<rsub|2>)=y<rsub|2>\<subseteq\>V<rsub|2>>
thus <math|x<rsub|1>\<in\>f<rsup|-1>(V<rsub|1>)=U<rsub|2>,x<rsub|2>\<in\>f<rsup|-1>(V<rsub|2>)=U<rsub|2>>
where <math|U<rsub|1>,U<rsub|2>> are open because of continuity of
<math|f>. Further if <math|x\<in\>X\<Rightarrow\>f(x)\<in\>f(X)\<subseteq\>V<rsub|1><big|cup>V<rsub|2>\<Rightarrow\>x\<in\>f<rsup|-1>(V<rsub|1>)<big|cup>f<rsup|-1>(V<rsub|2>)=U<rsub|1><big|cup>U<rsub|2>\<Rightarrow\>X=U<rsub|1><big|cup>U<rsub|2>>.
Finally assume that <math|y\<in\>U<rsub|1><big|cap>U<rsub|2>\<Rightarrow\>f(y)\<in\>V<rsub|1><big|cap>V<rsub|2><big|cap>f(X)=\<emptyset\>>
a contradiction so <math|U<rsub|1><big|cap>U<rsub|2>=\<emptyset\>>. So
finally we have proved that <math|X> is covered by tow non empty disjoint
open sets contradicting the fact that <math|X> is connected
</proof>
<\theorem>
<label|continuous mapping in [a,b]>Let
<math|f:[a,b]\<rightarrow\>\<bbb-R\>> a continuous function then
<math|f([a,b])=[c,d]> (so <math|f> retains its minimum and maximum
settings)
</theorem>
<\proof>
As <math|f> is continuous and <math|[a,b]> compact we have by
<reference|compact space in a metric space is bounded> that
<math|f([a,b])> is bounded. As by <reference|generalized intervals and
connectness> we have that <math|[a,b]> is connected, we can apply
<reference|continuity and connectness> to find that <math|f([a,b])> is
connected and then by applying <reference|generalized intervals and
connectness> we have that <math|f([a,b])> is a generalized interval and
because it is bounded we must have the existence of a
<math|c,d\<in\>\<bbb-R\>> such that <math|f([a,b])=[c,d]>
</proof>
\;
<chapter|Differentiability in normed vector spaces>
<section|Differentiability>
<\definition>
<index|<math|U<rsub|x>>>Let <math|X, \<shortparallel\>\<shortparallel\><rsub|X>>
be a normed vector space and a set <math|U\<subseteq\>X,x\<in\>U> then
<math|U<rsub|x>={h\<in\>X\|x+h\<in\>U}>.
</definition>
<\remark>
<math|U<rsub|x>=U-x={y-x\|y\<in\>U}>
</remark>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|h\<in\>U-x>|<cell|\<Rightarrow\>>|<cell|\<exists\>y\<in\>U\<succ\>h=y-x>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x+h=y\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|h\<in\>U<rsub|x>>>|<row|<cell|h\<in\>U<rsub|x>>|<cell|\<Rightarrow\>>|<cell|y=x+h\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|h=y-x,y\<in\>U>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|h\<in\>U-x>>>>
</eqnarray*>
</proof>
<\definition>
<index|<math|\<varepsilon\>-mapping>>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>,U\<subseteq\>X,x\<in\>U>
be normed spaces then a <math|\<varepsilon\>\<um\>mapping> at <math|x> is
a function <math|\<varepsilon\>:U<rsub|x>\<rightarrow\>Y> which is
continuous at <math|0\<in\>U<rsub|x>> and for which
<math|\<varepsilon\>(0)=0><math|> (using the subspace topology on
<math|U<rsub|x>>)
</definition>
<\definition>
<index|differentiable mapping at x>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be normed vector spaces and <math|U\<subseteq\>X> a open subset then a
function <math|f:U\<rightarrow\>Y> is differentiable at <math|x\<in\>U>
if and only if there exists a continuous linear function
<math|L:X\<rightarrow\>Y> and a <math|\<varepsilon\>\<um\>mapping> at
<math|x> such that <math|\<forall\>h\<in\>U<rsub|x> we have
f(x+h)-f(x)-L(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>
</definition>
<\remark>
If <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
are finite dimensional then using <reference|linear maps between finite
dimensional spaces are continuous> we have that every linear map between
<math|X> and <math|Y> is continuous, so in the final case we can drop the
continuity requirement in the definition of differentiability.
</remark>
<\theorem>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,
><math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces,
<math|U\<subseteq\>X> a open subset and <math|f:U\<rightarrow\>Y> is
differentiable at <math|x\<in\>U> then <math|f> is continuous at
<math|x>.
</theorem>
<\proof>
Given <math|\<varepsilon\>\<gtr\>0> then by continuity of
<math|\<varepsilon\>(h)> at <math|h=0> there exists a
<math|\<delta\><rsub|1>> such that <math|\<shortparallel\>\<varepsilon\>(h)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\>(h)-\<varepsilon\>(0)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>
if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>>,
also by the continuity of <math|L> we have the existence of a
<math|\<delta\><rsub|2>> such that <math|\<shortparallel\>L(x)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>
if <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>>.
Also as <math|x\<in\>U> (open) then <math|\<exists\>\<delta\><rsub|3>\<gtr\>0>
such that <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(x,\<delta\><rsub|3>)\<subseteq\>U.>
Let now <math|\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>,\<delta\><rsub|3>,1)>
then as <math|\<shortparallel\>x-y\<shortparallel\><rsub|Y>\<less\>\<delta\>\<leqslant\>\<delta\><rsub|3>>
we have that <math|y\<in\>U> and thus
<math|y=x+(y-x)\<in\>U\<Rightarrow\>h=(y-x)\<in\>U<rsub|x>> so
<math|f(y)-f(x)=f(x+h)-f(x)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|Y>+L(h)\<Rightarrow\>\<shortparallel\>f(y)-f(x)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>\<varepsilon\>(h)\<shortparallel\><rsub|Y>\<shortparallel\>h\<shortparallel\><rsub|Y>+\<shortparallel\>L(h)\<shortparallel\>><math|\<leqslant\>\<shortparallel\>\<varepsilon\>(h)\<shortparallel\><rsub|Y>.1+\<shortparallel\>L(h)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
proving the continuity of <math|f>
</proof>
<\theorem>
<index|differential>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be normed vector spaces and <math|U\<subseteq\>X> a open subset and a
function <math|f:U\<rightarrow\>Y> that is differentiable at
<math|x\<in\>U> then there exists only one <math|L> such that
<math|><math|f(x+h)-f(x)-L(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>.
This unique linear function is called the differential of
<math|f:U\<rightarrow\>Y> at <math|x\<in\>U> and noted as <math|Df(x)>
</theorem>
<\proof>
Assume there exists <math|L<rsub|1>,L<rsub|2>> and
<math|\<varepsilon\>-mappings ><math|\<varepsilon\><rsub|1>,\<varepsilon\><rsub|2>>
such that <math|\<forall\>h\<in\>U<rsub|x>> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|>|<cell|>|<cell|f(x+h)-f(x)-L<rsub|1>(h)=\<varepsilon\><rsub|1>(h).\<shortparallel\>h\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|>|<cell|f(x+h)-f(x)-L<rsub|2>(h)=\<varepsilon\><rsub|2>(h).\<shortparallel\>h\<shortparallel\><rsub|X>>>>>
</eqnarray*>
So that <math|\<forall\>h\<in\>U<rsub|x>> we have
<math|L<rsub|1>(h)-L<rsub|2>(h)=\<varepsilon\><rsub|2>(h).\<shortparallel\>h\<shortparallel\><rsub|X>-\<varepsilon\><rsub|1>(h).\<shortparallel\>h\<shortparallel\><rsub|X>=(\<varepsilon\><rsub|2>(h)-\<varepsilon\><rsub|1>(h)).\<shortparallel\>h\<shortparallel\><rsub|X>>,
now if <math|h=0> then because of linearity we have
<math|L<rsub|1>(0)=L<rsub|2>(0)=0 >(<reference|linear map applied to
neutral element yields the neutral element>) so lets assume that
<math|h\<neq\>0,h\<in\>X\<Rightarrow\>\<shortparallel\>h\<shortparallel\><rsub|X>\<neq\>0>.
Now as <math|x\<in\>U> there exists a <math|\<delta\>\<gtr\>0> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(x,\<delta\>)\<subseteq\>U>
so if we take <math|\<alpha\>\<leqslant\><frac|1|\<shortparallel\>h\<shortparallel\><rsub|X>><frac|\<delta\>|2>\<gtr\>0>
then for <math|\<shortparallel\>x+\<alpha\>.h-x\<shortparallel\><rsub|X>=\<shortparallel\>\<alpha\>.h\<shortparallel\><rsub|X>=\<alpha\>.\<shortparallel\>h\<shortparallel\><rsub|X>\<leqslant\><frac|1|\<shortparallel\>h\<shortparallel\><rsub|X>>\<shortparallel\>h\<shortparallel\><rsub|X><frac|\<delta\>|2>=<frac|\<delta\>|2>\<less\>\<delta\>\<Rightarrow\>x+\<alpha\>.h\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(x,\<delta\>)\<Rightarrow\>\<alpha\>.h\<in\>U<rsub|x>\<Rightarrow\>L<rsub|1>(\<alpha\>.h)-L<rsub|2>(\<alpha\>.h)=(\<varepsilon\><rsub|1>(\<alpha\>.h<rsub|1>)-\<varepsilon\><rsub|2>(\<alpha\>.h<rsub|2>)).\<shortparallel\>h\<shortparallel\><rsub|X>.\<alpha\>\<Rightarrowlim\><rsub|linearity>\<alpha\>(L<rsub|1>(h)-L<rsub|2>(h))=(\<varepsilon\><rsub|1>(\<alpha\>.h)-\<varepsilon\><rsub|2>(\<alpha\>.h<rsub|>)).\<shortparallel\>h\<shortparallel\><rsub|X>.\<alpha\>\<Rightarrowlim\><rsub|\<alpha\>\<gtr\>0>L<rsub|1>(h)-L<rsub|2>(h<rsub|2>)=(\<varepsilon\><rsub|1>(\<alpha\>.h)-\<varepsilon\><rsub|2>(\<alpha\>.h)\<shortparallel\><rsub|Y>.\<shortparallel\>h\<shortparallel\><rsub|X>>.
Now assume that <math|\<varepsilon\>=\<shortparallel\>L<rsub|1>(h)-L<rsub|2>(h)\<shortparallel\>\<gtr\>0>
then as <math|\<varepsilon\><rsub|1>(h),\<varepsilon\><rsub|2>(h)> is
continuous at <math|0> and <math|\<varepsilon\><rsub|1>(0)=\<varepsilon\><rsub|2>(0)=0>
we can find a <math|\<delta\><rprime|'>> such that for <math|i=1,2
\ \ \<shortparallel\>\<varepsilon\><rsub|i>(h)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\><rsub|i>(h)-\<varepsilon\><rsub|i>(0)\<shortparallel\>\<less\><frac|\<varepsilon\>|2\<shortparallel\>h\<shortparallel\><rsub|Y>>>
if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rprime|'>>
so if we choose <math|\<alpha\>=min(<frac|1|\<shortparallel\>h\<shortparallel\><rsub|X>><frac|\<delta\>|2>,<frac|\<delta\><rprime|'>|2.\<shortparallel\>h\<shortparallel\><rsub|X>>)\<Rightarrow\>\<shortparallel\>\<alpha\>.h\<shortparallel\>\<leqslant\><frac|\<delta\><rprime|'>|2.\<shortparallel\>h\<shortparallel\><rsub|X>>\<shortparallel\>h\<shortparallel\><rsub|X>=<frac|\<delta\><rprime|'>|2>\<less\>\<delta\><rprime|'>\<Rightarrow\>\<varepsilon\>=\<shortparallel\>L<rsub|1>(h)-L<rsub|2>(h)\<shortparallel\><rsub|Y>=\<shortparallel\>h\<shortparallel\><rsub|X>.\<shortparallel\>\<varepsilon\><rsub|1>(a.h)-\<varepsilon\><rsub|2>(\<alpha\>.h)\<shortparallel\>\<leqslant\>\<shortparallel\>h\<shortparallel\><rsub|X>(\<shortparallel\>\<varepsilon\><rsub|1>(\<alpha\>.h)\<shortparallel\><rsub|Y>+\<shortparallel\>\<varepsilon\><rsub|2>(\<alpha\>.h)\<shortparallel\><rsub|Y>)\<less\>\<shortparallel\>h\<shortparallel\><rsub|X>(<frac|\<varepsilon\>|2\<shortparallel\>h\<shortparallel\><rsub|X>>+<frac|\<varepsilon\>|2\<shortparallel\>h\<shortparallel\><rsub|X>>)=\<varepsilon\>\<Rightarrow\>\<varepsilon\>\<less\>\<varepsilon\>>
a contradiction so <math|\<shortparallel\>L<rsub|1>(h)-L<rsub|2>(h)\<shortparallel\>=0\<Rightarrow\>L<rsub|1>(h)-L<rsub|2>(h)=0\<Rightarrow\>L<rsub|1>(h)=L<rsub|2>(h)\<Rightarrowlim\><rsub|as
we already proved equality for h=0>L<rsub|1>=L<rsub|2>>
</proof>
<\theorem>
<label|differentiability and equivalent norms>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|1>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|2>> be normed vector
spaces with <math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>be
equivalent norms on <math|Y>, <math|x\<in\>U\<subseteq\>X>, <math|U> open
and <math|f:U\<rightarrow\>Y> be a function with a differential Df(x)
using <math|\<shortparallel\>\<shortparallel\><rsub|1>> then <math|Df(x)>
is also a differential of <math|f> using
<math|\<shortparallel\>\<shortparallel\><rsub|2>>
</theorem>
<\proof>
This is trivial as equivalent norms means that a function continue in one
norm is also continue in the other norm and the definition of a
differential (where <math|L> and <math|\<varepsilon\>> are continuous
using <math|\<shortparallel\>\<shortparallel\><rsub|1>> or
<math|\<shortparallel\>\<shortparallel\><rsub|2>>
</proof>
<\theorem>
<label|differentiability and restricted mappings>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be normed vector spaces and <math|U\<subseteq\>X> a open subset
<math|x\<in\>U> and <math|x\<in\>V\<subseteq\>U,V > open then a function
<math|f:U\<rightarrow\>Y> is differentiable at <math|x\<in\>U> if and
only if <math|f<rsub|\|V>:V\<rightarrow\>Y> is also differentiable at
<math|x> and <math|Df(x)=Df<rsub|\|V>(x)>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|\<Rightarrow\>>As <math|f> is differentiable at
<math|x\<in\>U> we have the existence of a
<math|\<varepsilon\>-mapping> <math|\<varepsilon\>:U<rsub|x>\<rightarrow\>Y>
such that <math|\<forall\>h\<in\>U<rsub|x>\<succ\>f(x+h)-f(x)-Df(x)(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>.
Now as <math|h\<in\>V<rsub|x>\<Rightarrow\>x+h\<in\>V\<subseteq\>U\<Rightarrow\>x\<in\>U<rsub|x>\<Rightarrow\>V<rsub|x>\<subseteq\>U<rsub|x>>
and using (<reference|subspace topology of subspace topology> and
<reference|continuity of restricted maps>) we have that
<math|\<varepsilon\><rprime|'>=\<varepsilon\><rsub|\|V<rsub|x>>> is
continuous at <math|0> and of course
<math|\<varepsilon\><rprime|'>(0)=\<varepsilon\><rsub|\|V<rsub|x>>(0)=\<varepsilon\>(0)=0>
so <math|\<varepsilon\><rprime|'>> is a <math|\<varepsilon\>-mapping>
on <math|V>. Finally we have <math|\<forall\>h\<in\>V<rsub|x>\<succ\>f<rsub|\|V>(x+h)-f<rsub|\|V>(x)-Df(x)(h)=f(x+h)-f(x)-Df(x)(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>=e<rsub|\|V<rsub|x>>(h)\<shortparallel\>h\<shortparallel\><rsub|X>=\<varepsilon\><rprime|'>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>
<item><math|\<Leftarrow\>> If <math|f<rsub|\|V>> is differentiable at
<math|x\<in\>V\<subseteq\>U> then there exists a
<math|\<varepsilon\>->function <math|\<varepsilon\><rsub|V>:V\<rightarrow\>Y>
such that <math|\<forall\>h\<in\>V<rsub|x>\<succ\>f(x+h)-f(x)-Df<rsub|V>(x)(h)=f<rsub|\|V>(x+h)-f<rsub|\|V>(x)-Df<rsub|V>(x)(h)=\<varepsilon\><rsub|V>(h)\<shortparallel\>h\<shortparallel\><rsub|X>,
\<varepsilon\><rsub|V>(0)=0 and \<varepsilon\><rsub|V> \ is continue at
0. We define then the mapping > <math|><math|\<varepsilon\>:U<rsub|x>\<rightarrow\>Y>
by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(h)>|<cell|=>|<cell|\<varepsilon\><rsub|V>(h),
h\<in\>V<rsub|x> >>|<row|<cell|>|<cell|=>|<cell|<frac|f(x+h)-f(x)-Df<rsub|V>(x)(h)|\<shortparallel\>h\<shortparallel\><rsub|X>>,
h\<in\>U<rsub|x><mid|\\>V<rsub|x>\<nni\>0>>>>
</eqnarray*>
Then clearly <math|f(x+h)-f(x)-Df<rsub|V>(x)(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>.
And given a <math|\<varepsilon\>\<gtr\>0> we can find a
<math|\<delta\><rsub|1>> such that <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<delta\><rsub|1>)\<subseteq\>V<rsub|x>>
and by continuity of <math|\<varepsilon\><rsub|v>> a
<math|\<delta\><rsub|2>> such that <math|\<shortparallel\>\<varepsilon\><rsub|V>(h)\<shortparallel\><rsub|Y>\<less\>\<varepsilon\>>
if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|2>>,
so if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>)>
then <math|\<shortparallel\>\<varepsilon\>(h)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\><rsub|V>(h)\<shortparallel\><rsub|Y>\<less\>\<varepsilon\>>
proving continuity of <math|\<varepsilon\>> at <math|0> and also that
<math|f> is differentiable at <math|x> and that
<math|Df(x)=Df<rsub|\|V>(x)>
</enumerate>
</proof>
Basically what this theorem says that if you want to prove
differentiability of <math|f> at <math|x\<in\>U, U> open it is sufficient
to prove it at a open <math|V> containing <math|x> or that
differentiability is a local property
<\theorem>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|1>>,
<math|X,\<shortparallel\>\<shortparallel\><rsub|2>> and
<math|Y,\<shortparallel\>\<shortparallel\>> be normed vector spaces with
<math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>be
equivalent norms on <math|Y>, <math|x\<in\>U\<subseteq\>X>, <math|U> open
and <math|f:U\<rightarrow\>Y> be a function with a differential Df(x)
using <math|\<shortparallel\>\<shortparallel\><rsub|1>> then <math|Df(x)>
is also a differential of <math|f> using
<math|\<shortparallel\>\<shortparallel\><rsub|2>>
</theorem>
<\proof>
This is trivial as equivalent norms means that a function continue in one
norm is also continue in the other norm and the definition of a
differential (where <math|L> and <math|\<varepsilon\>> are continuous
using <math|\<shortparallel\>\<shortparallel\><rsub|1>> or
<math|\<shortparallel\>\<shortparallel\><rsub|2>>
</proof>
<\example>
<label|Differential of constant function><dueto|Differential of constant
function><index|differential of constant>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be normed vector spaces and <math|U> open in <math|X><math| > define then
<math|\<forall\>y\<in\>Y> <math|C<rsub|y,U>:U\<rightarrow\>Y> by
<math|\<forall\>x\<in\>U\<succ\>C<rsub|y>(x)=y> then
<math|\<forall\>x\<in\>U> we have <math|DC<rsub|y>(x)=C<rsub|0,X>>
</example>
<\proof>
As <math|C<rsub|0,X>:X\<rightarrow\>Y> is linear
<math|(C<rsub|0>(\<alpha\>.x+\<beta\>.y)=0=\<alpha\>.0+\<beta\>.0=\<alpha\>.C<rsub|0>(x)+\<beta\>.C<rsub|0>(y))>
and <math|\<shortparallel\>C<rsub|0>(x)\<shortparallel\><rsub|Y>=0=0.\<shortparallel\>x\<shortparallel\><rsub|X>>
implies <math|C<rsub|0,X>\<in\>L(X,Y)>, also as
<math|\<shortparallel\>C<rsub|0,U<rsub|X>>(h)-C<rsub|0,U<rsub|X>>(0)\<shortparallel\><rsub|Y>=\<shortparallel\>0-0\<shortparallel\><rsub|Y>=0\<Rightarrow\>C<rsub|0,U<rsub|X>>>
is continuous at <math|C<rsub|0,U<rsub|X>>> and of course
<math|C<rsub|0,U<rsub|X>>(0)=0> so <math|C<rsub|0,U<rsub|X>>:U<rsub|X>\<rightarrow\>Y>
is a <math|\<varepsilon\>-mapping> at <math|x>. Then
<math|\<forall\>h\<in\>U<rsub|X>> we have
<math|C<rsub|y,U>(x+h)-C<rsub|y,U>(x)-C<rsub|0,X>(h)=y-y-0=0=0.\<shortparallel\>h\<shortparallel\><rsub|X>=C<rsub|0,U<rsub|X>>\<shortparallel\>h\<shortparallel\><rsub|X>>
proving that <math|DC<rsub|y>(x)=C<rsub|0,X>>
</proof>
<\example>
<label|differential of linear mapping><dueto|Differential of a linear
function><index|differential of linear function>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,
Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector spaces and
<math|U> open in <math|X> then for all <math|L\<in\>L(X,Y)> we have
<math|\<forall\>x\<in\>U\<succ\>DL<rsub|\|U><rsub|>(x)=L>
</example>
<\proof>
<math|\<forall\>h\<in\>U<rsub|h>> we have
<math|L<rsub|\|U>(x+h)-L<rsub|\|U>(x)-L(h)=L(x+h)-L(x)-L(h)=L(x)+L(h)-L(h)-L(x)=0=0.\<shortparallel\>h\<shortparallel\><rsub|X>=C<rsub|0,U<rsub|X>>(0).\<shortparallel\>h\<shortparallel\><rsub|X>>
proving our little theorem (together with the fact that <math|L> is
continuous).<math|>
</proof>
<\note>
From now one we note <math|\<bbb-K\>,\<\|\|\>> to be either
<math|\<bbb-C\>,\<\|\|\>> or <math|\<bbb-R\>,\<\|\|\>>
</note>
<\definition>
<index|derivate><index|<math|<frac|df(x)|dx>(y)>>Given
<math|\<bbb-K\>,\<\|\|\>> \ <math|U\<subseteq\>\<bbb-K\>> open,
<math|x\<in\>U> and <math|Y,\<shortparallel\>\<shortparallel\>> be a
normed space and <math|f:U\<rightarrow\>Y> then the derivative of
<math|f> at <math|x> (if it exists) noted as <math|f<rprime|'>(x)> is
defined by <math|lim<rsub|h\<rightarrow\>0><frac|f(x+h)-f(x)|h>> . Say it
otherwise <math|f> has a derivative <math|f<rprime|'>(x)> if and only if
<math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\>\<gtr\>0> such that <math|\<forall\>h\<in\>U<rsub|x>\<vdash\>0\<less\>\|h\|\<less\>\<delta\>>
we have that <math|\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rprime|'>(x)\<shortparallel\>=\<less\>\<varepsilon\>>
</definition>
<\note>
In most cases we note <math|f<rprime|'>(y)> as <math|<frac|df(x)|dx>(y)>.
Actually if <math|f(x)> is a expression then <math|<frac|df(x)|dx>(y)> is
to interpreted as <math|f<rprime|'>(y)> where
<math|f:x\<rightarrow\>f(x)> is the function for which the derivate is
defined.
</note>
<\theorem>
<label|differentiability of real or complex functions>Given
<math|\<bbb-K\>,\<\|\|\>> <math|U\<subseteq\>\<bbb-K\>> open and
<math|x\<in\>U>, <math|Y,\<shortparallel\>\<shortparallel\>> be a normed
space (real if <math|\<bbb-K\>=\<bbb-R\>,> complex if
<math|\<bbb-K\>=\<bbb-C\>>) then <math|Df(x)> exists
<math|\<Leftrightarrow\>f<rprime|'>(x)> exists and
<math|Df(x)(1)=f<rprime|'>(x)>.
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>
By the existence of <math|Df(x)> we have the existence of a
<math|\<varepsilon\>-function> <math|\<varepsilon\>(.)> and a
<math|Df(x)\<in\>L(\<bbb-K\>,Y)> such that
<math|\<forall\>h\<in\>U<rsub|x>\<succ\>
f(x+h)-f(x)-Df(x)(h)=\<varepsilon\>(h).\|h\|>. Now
<math|Df(x)(h)=Df(x)(h.1)=h.Df(x)(1)\<Rightarrow\>f(x+h)-f(x)-h.Df(x)(1)=\<varepsilon\>(h).\|h\|>,
also <math|\<varepsilon\>(.)> is continuous at <math|0> with
<math|\<varepsilon\>(0)=0><math|\<Rightarrow\>\<forall\>\<varepsilon\>\<gtr\>0\<Rightarrow\>\<exists\>\<delta\>\<gtr\>0\<succ\>\<forall\>h\<in\>U<rsub|x>\<vdash\>\|h\|\<succ\>\<shortparallel\>\<varepsilon\>(h)\<shortparallel\>=\<shortparallel\>\<varepsilon\>(h)-\<varepsilon\>(0)\<shortparallel\>\<less\>\<varepsilon\>>.
So if <math|0\<less\>\|h\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\><frac|f(x+h)-f(x)|h>-Df(x)(1)\<shortparallel\>=\<shortparallel\><frac|f(x+h)-f(x)-h.Df(x)(1)|h>\<shortparallel\>=<frac|\<shortparallel\>f(x+h)-f(x)-h.Df(x)(1)\<shortparallel\>|\|h\|>=\<shortparallel\>\<varepsilon\>(h)\<shortparallel\>\<less\>\<varepsilon\>>
proving that <math|f<rprime|'>(x)> exists and is equal to <math|Df(x)(1)>
<math|\<Leftarrow\>>
Define <math|Df(x):\<bbb-K\>\<rightarrow\>Y> by
<math|h\<rightarrow\>f<rprime|'>(x).h> which is trivially linear we
define now <math|\<varepsilon\>(.):U<rsub|x>\<rightarrow\>Y> by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(h)>|<cell|=0>|<cell|h=0\<in\>U<rsub|x>>>|<row|<cell|\<varepsilon\>(h)>|<cell|=<frac|f(x+h)-f(x)-h.f<rprime|'>(x)|\|h\|>>|<cell|h\<in\>U<rsub|x><mid|\\>{0}>>>>
</eqnarray*>
then by the definition of the derivative we have
<math|\<forall\>\<varepsilon\>\<gtr\>0\<succ\>\<exists\>\<delta\>\<gtr\>0>
such that <math|\<forall\>h\<in\>U<rsub|x>> with
<math|0\<less\>\|h\|\<less\>\<delta\>> we have
<math|\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rprime|'>(x)\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\><frac|\<shortparallel\>f(x+h)-f(x)-h.f<rprime|'>(x)\<shortparallel\>|\|h\|>=\<shortparallel\>\<varepsilon\>(h)\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\>\<varepsilon\>(.)>
is continuous at <math|0> with <math|\<varepsilon\>(0)=0> and
<math|\<varepsilon\>(.)> is thus a <math|\<varepsilon\>-function>. But by
the definition of <math|\<varepsilon\>(.)> we have for
<math|h\<in\>U<rsub|x><mid|\\>{0}> that
<math|f(x+h)-f(x)-h.f<rprime|'>(x)=\<varepsilon\>(h).\|h\|> and if
<math|h=0> then <math|\<varepsilon\>(0).\|0\|=0=f(x+0)-f(x)-0.f<rprime|'>(x)>
so we have <math|\<forall\>h\<in\>U<rsub|x>> that
<math|f(x+h)-f(x)-h.f<rprime|'>(x)=\<varepsilon\>(h).\|h\|> proving that
<math|f> is differentiable and that <math|Df(x)(h)=h.f<rprime|'>(x)>
\;
</proof>
<\lemma>
<label|sum of partial function>Given a family of sets
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> then
<math|x=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n+1>)(x<rsub|i>)>,
where <math|(0<rsub|1>,\<ldots\>,0<rsub|i-1>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)\<in\>{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>>
is the null vector (see for the notation <reference|function of ith
variable> )
</lemma>
<\proof>
\;
<\eqnarray*>
<tformat|<table|<row|<cell|(0<rsub|1>,\<ldots\>,0<rsub|i-1>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(x<rsub|i>))<rsub|j>>|<cell|=x<rsub|i>>|<cell|if
j=i>>|<row|<cell|>|<cell|=0>|<cell|if j\<neq\>i>>>>
</eqnarray*>
so
<\eqnarray*>
<tformat|<table|<row|<cell|(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(x<rsub|i>))<rsub|j>=>|<cell|<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(x<rsub|i>))<rsub|j>=>|<cell|x<rsub|j>>>>>
</eqnarray*>
proving our theorem.
</proof>
<\theorem>
<label|partial coordinates are differentiable>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}><rsub|>>
be a finite family of normed vector spaces and let <math|Y> be a normed
space and <math|X=<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> be
the product space together with the product topology, a norm
<math|\<shortparallel\>\<shortparallel\>> that generates this product
topology then <math|\<forall\>i\<in\>{1,\<ldots\>,n},\<forall\>x\<in\>\<Pi\><rsub|i\<in\>{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>X<rsub|i>,\<forall\>t\<in\>X<rsub|i>>
we have that <math|(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>):X<rsub|i>\<rightarrow\>\<Pi\><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>
is differentiable at <math|t> and <math|D(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t)=(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>
</theorem>
<\proof>
As\
<\eqnarray*>
<tformat|<table|<row|<cell|((0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(\<alpha\>.x+\<beta\>.y))<rsub|j>=>|<cell|\<alpha\>.x+\<beta\>.y>|<cell|if
j=i>>|<row|<cell|>|<cell|0>|<cell|if
j\<neq\>i>>|<row|<cell|(\<alpha\>.0(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(x)<rsub|j>+\<beta\>.(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(y))<rsub|j>=>|<cell|\<alpha\>.x+\<beta\>.y>|<cell|if
j=i>>|<row|<cell|>|<cell|0>|<cell|if j\<neq\>i>>>>
</eqnarray*>
or <math|(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)\<in\>L(X,\<Pi\><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>)>
and we have
<\enumerate>
<item><math|j=i\<Rightarrow\>><with|mode|math|((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t+h)-(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(h)-(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h))<rsub|j>=((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t+h))<rsub|j>+((x<rsub|1>\<ldots\>,x<rsub|i-1>x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t))<rsub|j>-((0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h))<rsub|j>=t+h-t-h=0>
<item><math|i\<neq\>j><with|mode|math|\<Rightarrow\>((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t+h)-(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(h)-(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h))<rsub|j>=((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t+h))<rsub|j>+((x<rsub|1>\<ldots\>,x<rsub|i-1>x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t))<rsub|j>-((0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h))<rsub|j>=x<rsub|i>-x<rsub|i>-0=0>
</enumerate>
thus we have that <math|\<forall\>t\<in\>X\<succ\>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t+h)-(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(h)-(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h)=0.\<shortparallel\>x\<shortparallel\>>
and the constant function <math|0> is obviously a
<math|\<varepsilon\>-mapping>. \
</proof>
<\theorem>
<label|sum of differentiable functions is differentiable>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-K\>>, <math|U\<subseteq\>X> a open set and
<math|f:U\<rightarrow\>Y,g\|U\<rightarrow\>Y> be functions differentiable
at <math|x\<in\>U> then <math|f+g> is differentiable and
<math|D(f+g)(x)=Df(x)+Dg(x)>. Also if <math|\<alpha\>\<in\>\<bbb-K\>>
then <math|\<alpha\>.f> is differentiable and
<math|D(\<alpha\>.f)(x)=\<alpha\>.Df(x)>
</theorem>
<\proof>
\;
By assumption we have that if <math|h\<in\>U<rsub|x>> that there exists
<math|\<varepsilon\><rsub|1>,\<varepsilon\><rsub|2>>
<math|\<varepsilon\>-mappings> with\
<\eqnarray*>
<tformat|<table|<row|<cell|f(x+h)-f(x)-Df(x)(h)>|<cell|=>|<cell|\<varepsilon\><rsub|1>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>>|<row|<cell|g(x+h)-g(x)-Dg(x)(h)>|<cell|=>|<cell|\<varepsilon\><rsub|2>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>>>>
</eqnarray*>
then using <reference|sum of continuous functions is continuous> we have
that <math|\<varepsilon\>=\<varepsilon\><rsub|1>+\<varepsilon\><rsub|2>>
is continuous at <math|0> and <math|\<varepsilon\>=\<varepsilon\><rsub|1>(0)+\<varepsilon\><rsub|2>(0)=0>
so <math|\<varepsilon\>> is a <math|\<varepsilon\>-mapping> then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|(f+g)(x+h)-(f+g)(x)-(Df(x)+Dg(x))(h)>|<cell|=>|<cell|f(x+h)-f(x)-Df(x)(h)+g(x+h)-g(x)-Dg(x)(h)>>|<row|<cell|>|<cell|=>|<cell|\<varepsilon\><rsub|1>(h)\<shortparallel\>h\<shortparallel\><rsub|X>+\<varepsilon\><rsub|2>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|=>|<cell|\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>>>>
</eqnarray*>
\;
For the second part define the <math|\<varepsilon\>-mapping>
<math|\<alpha\>.\<varepsilon\><rsub|1>> then
<math|(\<alpha\>.f)(x+h)-(\<alpha\>.f)
(x)-(\<alpha\>.D(f))(h)=\<alpha\>(f(x+h)-f(x)-Df(x)(h))=\<alpha\>.(\<varepsilon\><rsub|1>(h)\<shortparallel\>h\<shortparallel\><rsub|X>)=(\<alpha\>.h)(h)\<shortparallel\>h\<shortparallel\><rsub|X>>
\;
</proof>
<\corollary>
<label|sum of derivatives>Given <math|\<bbb-K\>,\<\|\|\>>
<math|U\<subseteq\>\<bbb-K\>> open and <math|x\<in\>U>,
<math|Y,\<shortparallel\>\<shortparallel\>> be a normed space (real if
<math|\<bbb-K\>=\<bbb-R\>,> complex if <math|\<bbb-K\>=\<bbb-C\>>) and
<math|f:U\<rightarrow\>Y,g:U\<rightarrow\>Y,\<alpha\>,\<beta\>\<in\>\<bbb-K\>>
then <math|<frac|d(\<alpha\>.f(t)+\<beta\>.g(t))|dt>(x)=\<alpha\><frac|df(t)|dx>(x)+\<beta\><frac|dg(t)|dt>(x)>
(or in another notation <math|(\<alpha\>.f+\<beta\>.g)<rprime|'>=\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>>)
are defined.
</corollary>
<\proof>
This is trivial using <reference|sum of differentiable functions is
differentiable> <math|<frac|d(\<alpha\>f(t)+\<beta\>g(t))|dt>(x)=D(\<alpha\>f+\<beta\>g)(x)(1)=(\<alpha\>.Df(x)+\<beta\>Dg(x))(1)=\<alpha\>.Df(x)(1)+\<beta\>.Dg(x)(1)=\<alpha\><frac|df(t)|dt>(x)+\<beta\><frac|dg(t)|dt>(x)>
</proof>
<\theorem>
<label|differentiability of composition of differentiable
mappings><verbatim|><dueto|The Chain Rule><index|chain rule>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> and
<math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be normed spaces,
<math|U\<subseteq\>X>, <math|V\<subseteq\>Y> be open spaces and
<math|f:U\<rightarrow\>Y,g:V\<rightarrow\>Z> with
<math|f(U)\<subseteq\>V>, <math|f> differentiable at <math|x\<in\>U> with
differential <math|Df(x)> and <math|g> differentiable at <math|f(x)> with
differential <math|Dg(f(x))> then <math|f\<circ\>g> is differentiable at
<math|x> and <math|D(f\<circ\>g)(x)=Df(x)\<circ\>Dg(f(x))>
</theorem>
<\proof>
By differentiability of <math|f> at <math|x> and <math|g> at <math|f(x)>
then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<forall\>h\<in\>U<rsub|x>\<succ\>\<exists\>\<varepsilon\>-mapping
\<varepsilon\> such that f(x+h)-f(x)-Df(x)(h)=\<varepsilon\><rsub|f>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>|<cell|>|<cell|>>|<row|<cell|\<forall\>k\<in\>V<rsub|f(x)>\<succ\>\<exists\>\<varepsilon\>-mapping
\<gamma\> such that g(f(x)+k)-g(f(x))-Dg(f(x))(k)=\<varepsilon\><rsub|g>(k)\<shortparallel\>k\<shortparallel\><rsub|Y>>|<cell|>|<cell|>>>>
</eqnarray*>
now if <math|h\<in\>U<rsub|x>\<Rightarrow\>x+h\<in\>U\<Rightarrow\>f(x+h)\<in\>F(U)\<subseteq\>V>
take <math|k=f(x+h)-f(x)\<Rightarrow\>k+f(x)=f(x+h)\<in\>V> \ and thus
<math|k\<in\>V<rsub|f(x)>> and thus <math|g(f(x)+k)-g(f(x))-Dg(f(x))(k)=\<varepsilon\><rsub|g>(k)\<shortparallel\>k\<shortparallel\><rsub|Y>\<Rightarrowlim\><rsub|replace
k with f(x+h)-f(x)>g(f(x)+f(x+h)-f(x))-g(f(x))-Dg(f(x))(f(x+h)-f(x))=\<varepsilon\><rsub|g>(f(x+h)-f(x))\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>\<Rightarrow\>g(f(x+h))-g(f(x))-Dg(f(x))(\<varepsilon\><rsub|f>(h)\<shortparallel\>h\<shortparallel\><rsub|X>+Df(x)(h))=\<varepsilon\><rsub|g>(f(x+h)-f(x))\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>\<Rightarrow\>(g\<circ\>f)(x+h)-(g\<circ\>f)(x)-Dg(f(x))(Df(x)(h))=\<varepsilon\><rsub|g>(f(x+h)-f(x))\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>+\<shortparallel\>h\<shortparallel\><rsub|X>.Dg(f(x))(\<varepsilon\><rsub|f>(h))>
if we define then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<zeta\>(h)=\<varepsilon\><rsub|g>(f(x+h)-f(x))<frac|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>|\<shortparallel\>h\<shortparallel\><rsub|X>>+(Dg(f(x))(\<varepsilon\><rsub|f>(h))>|<cell|if
h\<neq\>0>|<cell|>>|<row|<cell|\<zeta\>(h)=0>|<cell|if h=0>|<cell|>>>>
</eqnarray*>
then we have <math|(g\<circ\>f)(x+h)-(g\<circ\>f)(x)-(Dg(f(x))\<circ\>Df(x))(h)=\<zeta\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>
so if we prove that <math|\<zeta\>> is continuous at <math|h=0> we have
proved our theorem. So let given <math|\<varepsilon\>\<gtr\>0>.\
First we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>>|<cell|=>|<cell|\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\>h\<shortparallel\><rsub|X>+Df(x)(h)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\><rsub|X>\<shortparallel\>h\<shortparallel\><rsub|X>+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>\<shortparallel\>h\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|=>|<cell|(\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\><rsub|X>+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>)\<shortparallel\>h\<shortparallel\><rsub|X>>>>>
</eqnarray*>
now as <math|\<varepsilon\><rsub|f>> is a continuous at <math|0> we have
\ given <math|1> the existences a <math|\<delta\><rsub|1>\<gtr\>0> such
that <math|\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\><rsub|Y>\<less\>1>
if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>>
then if <math|0\<less\>\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|1>>
we have <math|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>\<less\>(1+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>)\<shortparallel\>h\<shortparallel\><rsub|X>>
and\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varepsilon\><rsub|g>(f(x+h)-f(x))<frac|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>|\<shortparallel\>h\<shortparallel\><rsub|X>>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>\<varepsilon\><rsub|g>(f(x+h)-f(x))\<shortparallel\><rsub|Y>(1+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>)>>>>
</eqnarray*>
now we can again by continuity of <math|\<varepsilon\><rsub|g>> the
existence of a <math|\<delta\><rsub|2>\<gtr\>0> such that
<math|\<shortparallel\>\<varepsilon\><rsub|g>(k)\<shortparallel\><rsub|Z>\<less\><frac|\<varepsilon\>|2><frac|1|(1+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>>>
if <math|\<shortparallel\>k\<shortparallel\><rsub|Y>\<less\>\<delta\><rsub|2>>
and by continuity of <math|f> there exists a <math|\<delta\><rsub|3>>
such that if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\><rsub|3>>
such that <math|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>\<less\>\<delta\><rsub|2>>
and thus <math|><math|\<shortparallel\>\<varepsilon\><rsub|g>(f(x+h)-f(x))\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2><frac|1|(1+\<shortparallel\>Df(x)\<shortparallel\><rsub|Y>>>
or if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>min(\<delta\><rsub|1>,\<delta\><rsub|3>)\<Rightarrow\>\<shortparallel\>\<varepsilon\><rsub|g>(f(x+h)-f(x))<frac|\<shortparallel\>f(x+h)-f(x)\<shortparallel\><rsub|Y>|\<shortparallel\>h\<shortparallel\><rsub|X>>\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>.
Secondly we have <math|\<shortparallel\>Dg(f(x))(\<varepsilon\><rsub|f>(h))\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>Dg(f(x))\<shortparallel\><rsub|Z>\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\><rsub|Y>>
and by continuity of <math|\<varepsilon\><rsub|f>> at <math|0> we have
the existence of a <math|\<delta\><rsub|4>\<gtr\>0> such that
<math|\<shortparallel\>\<varepsilon\><rsub|f>(h)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2\<shortparallel\>Dg(f(x))\<shortparallel\><rsub|Z>>>
and thus if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>min(\<delta\><rsub|1>,\<delta\><rsub|3>,\<delta\><rsub|4>)>
then <math|\<shortparallel\>Dg(f(x))(\<varepsilon\><rsub|f>(h))\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>|2>>
and thus <math|\<shortparallel\>\<zeta\>(h)\<shortparallel\><rsub|Z>\<less\><frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>
proving our theorem.
</proof>
<\theorem>
<label|differentiability of products of metric spaces>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
be a finite family of normed spaces, <math|Y,\<shortparallel\>\<shortparallel\>>
a normed space and <math|U\<subseteq\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>>,
<math|U> open then if <math|f:U\<rightarrow\>Y> is differentiable at
<math|x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>x<rsub|i>>
with <math|Df(x)> as its differential then we have that
<math|\<forall\>i\<in\>{1,\<ldots\>,n}
\ f<rsup|(i)>=f(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)<rsub|\|\<pi\><rsub|i>(U)>:\<pi\><rsub|i>(U)\<rightarrow\>Y>
is differentiable at <math|x<rsub|i>> and
<math|Df<rsup|(i)>(x<rsub|i>)=Df(x)\<circ\>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>
(or <math|Df<rsup|(i)>(x<rsub|i>)(t)=Df(x)(0<rsub|1>,\<ldots\>,0<rsub|i-1>,t,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>.<math|>
Further using <reference|sum of partial function> we have
<math|Df(x)(h)=Df(x)(<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(h<rsub|i>))=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>Df(x)((0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>\<ldots\>,0<rsub|n>)(h<rsub|i>))=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>Df<rsup|(i)>(x)(h<rsub|i>)>
</theorem>
<\notation>
<index|partial differential>We note <math|Df<rsup|(i)>(x<rsub|i>)\<equallim\><rsub|notation>D<rsub|i>f(x)>
and call it the partial differential of <math|f> so that
<math|Df(x)(h)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>D<rsub|i>f(x)(h<rsub|i>)>
or <math|Df(x)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>Df<rsub|i>(f(x))\<circ\>\<pi\><rsub|i>>
</notation>
<\proof>
Using <reference|partial coordinates are differentiable>,
<reference|differentiability and restricted mappings>,
(<math|x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,.,x<rsub|n>)>
is differentiable and the restriction of a differentiable map to a open
set is differentiable and has the same differential), the fact that
<math|\<pi\><rsub|i>> is a open function (<reference|projection map is
continuous>) and thus <math|\<pi\><rsub|i>(U)> is open we have that
<math|(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)<rsub|\|\<pi\><rsub|i>(U)>>
is differentiable and <math|><math|D(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)<rsub|\|\<pi\><rsub|i>(U)>(x<rsub|i>)=(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>,
and as <math|f> is differentiable with differential <math|Df(x)> then
using (<reference|differentiability of composition of differentiable
mappings> differential of composed functions) we have that
<math|f<rsup|(i)>> is differentiable and
<math|Df<rsup|(i)>(x<rsub|i>)=D(f\<circ\>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)<rsub|\|\<pi\><rsub|i>(U)>)(x<rsub|i>)=Df((x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)<rsub|\|\<pi\><rsub|i>(U)><rsub|>(x<rsub|i>))\<circ\>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)=Df(x)\<circ\>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>
</proof>
If we apply this theorem with all <math|X<rsub|i>=\<bbb-K\>> and the
relation between the differential on <math|\<bbb-K\>> and the derivative
then we can prove the following corollary.
<\corollary>
<label|differential of multiparameter function>If
<math|f:U\<subseteq\>\<bbb-K\><rsup|n>\<rightarrow\>Y> where we have the
product norm on <math|\<bbb-K\><rsup|n>> and <math|Y> is a normed space
with norm <math|\<shortparallel\>\<shortparallel\><rsub|Y>> and <math|f>
is differentiable at <math|x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>U>.
Then <math|D<rsub|i>f(x)(k)=k.<frac|df<rsup|(i)>(t)|dt>(x<rsub|i>)=k.f<rsup|(i)><rprime|'>(x<rsub|i>)=k.Df(x)(e<rsub|i>)>
(where <math|<frac|df<rsup|(i)>(t)|dt>(x)=lim<rsub|h\<rightarrow\>0><frac|f<rsup|(i)>(x<rsub|i>+h)-f<rsup|(i)>(x<rsub|i>)|h>=lim<rsub|h\<rightarrow\>\<infty\>><frac|f(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i>+h,x<rsub|i+1>,\<ldots\>x<rsub|n>)-f(x<rsub|1>,\<ldots\>,x<rsub|n>)|h>>)
and <math|{e<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> is the canonical
basis on <math|\<bbb-R\><rsup|n>> defined by
<math|(e<rsub|i>)<rsub|j>=\<delta\><rsub|i,jj>>) (we note
<math|(<frac|df<rsup|(i)>(t)|dx>> as <math|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x)>)
and thus <math|Df<rsup|>(x)(h)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>h<rsub|i>.<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dx<rsub|i>>(x)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>h<rsub|i>.Df(x)(e<rsub|i>)>.
So if we note <math|h=<matrix|<tformat|<table|<row|<cell|h<rsub|1>>>|<row|<cell|\<ldots\>>>|<row|<cell|h<rsub|n>>>>>>>
and <math|<frac|df|dx>(x)=<matrix|<tformat|<table|<row|<cell|<frac|df(t<rsub|1>,\<ldots\>t<rsub|n>)|dt<rsub|i>>(x),\<ldots\>,<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n)>|dt<rsub|i>>(x)>>>>>>
then we can write <math|Df(x)(h)> as the matrix product
<math|<frac|df|dx>(x).h>).
</corollary>
<\proof>
This is trivial as <math|f<rsup|(i)>:\<pi\><rsub|i>(U)\<rightarrow\>Y>
where <math|\<pi\><rsub|i>(U)\<subseteq\>\<bbb-K\>> and thus
<math|><math|Df<rsup|(i)>(x<rsub|i>)(k)\<equallim\><rsub|<reference|differentiability
of real or complex functions>><frac|df|dx<rsub|i>>(x<rsub|i>)\<equallim\><rsub|<reference|differentiability
of products of metric spaces>>(Df(x)\<circ\>(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>))(k)=Df(x)((0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(k))=Df(x)(k.(0<rsub|1>,\<ldots\>,0<rsub|i-1>,\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)(1))=k.Df(x)(0<rsub|1>,\<ldots\>,0<rsub|i-1>,1,0<rsub|i+1>,\<ldots\>,0<rsub|n>)=k.Df(x)(e<rsub|i>)>.
Also using <math|Df(x)(h)\<equallim\><rsub|<reference|differentiability
of products of metric spaces>><big|sum><rsub|i\<in\>{1,\<ldots\>,n}>D<rsub|i>f(x)(h<rsub|i>)=<big|sum><rsub|i\<in\>{1,\<ldots\>,n}>h<rsub|i>.<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x)<rsub|>>
</proof>
<\note>
Again if we see <math|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x)>
then we know that we are dealing with a function
<math|f:t\<rightarrow\>f(t<rsub|1>,\<ldots\>,t<rsub|n>),x=(x<rsub|1>,\<ldots\>,x<rsub|n>)>
and <math|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n)>|dt<rsub|i>>(x)=lim<rsub|h\<rightarrow\>0><frac|f(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x<rsub|i>+h,x<rsub|i+1>,\<ldots\>,x<rsub|n>)-f(x<rsub|1>,\<ldots\>,x<rsub|n>)|h>>
</note>
<\theorem>
<label|differentiability of map to product>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be a normed vector
space, <math|{Y<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
a family of normed spaces and <math|\<shortparallel\>\<shortparallel\>>
the maximum norm defined on <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>>
(making it a normed space). Then given <math|x\<in\>U\<subset\>X, U> open
we have that <math|f:U\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>>
is differentiable at <math|x> iff <math|\<forall\>i\<in\>{1,\<ldots\>,n}
f<rsub|i>=\<pi\><rsub|i>\<circ\>f> is differentiable at <math|x> and if
<math|Df(x)> exists we have <math|Df<rsub|i>(x)=\<pi\><rsub|i>\<circ\>Df(x)>
(In other words if <math|f=(f<rsub|1>,\<ldots\>,f<rsub|n>)> then we have
that <math|Df(x)=(Df<rsub|1>(x),\<ldots\>,Df<rsub|n>(x))>
</theorem>
<\proof>
\;
Assume that <math|f> is differentiable at <math|x> then there exists a
<math|\<varepsilon\>-function> <math|\<varepsilon\>:U<rsub|x>\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>>such
that <math|f(x+h)-f(x)-Df(x)(h)=\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>then
<math|\<pi\><rsub|i>(f(x+h)-f(x)-Df(x)(h))=\<pi\><rsub|i>(f(x+h))-\<pi\><rsub|i>(f(x))-\<pi\><rsub|i>(Df(x)(h))=f<rsub|i>(x+h)-f<rsub|i>(x)-(\<pi\><rsub|i>\<circ\>Df(x))(h)=\<pi\><rsub|i>(\<varepsilon\>(h).\<shortparallel\>h\<shortparallel\>)=\<pi\><rsub|i>(\<varepsilon\>(h)).\<shortparallel\>h\<shortparallel\>>
which given the fact that <math|\<pi\><rsub|i> > is continuous and the
fact that the composition of continuous functions are continue that
<math|f<rsub|i>> is differentiable at <math|x> and
<math|Df<rsub|i>(x)=\<pi\>\<circ\>Df(x)>
Assume now that <math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have that
<math|f<rsub|i>> is differentiable. Then
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
<math|(\<pi\><rsub|i>\<circ\>f)(x+h)-(\<pi\><rsub|i>\<circ\>f)(x)-D(\<pi\><rsub|i>\<circ\>f)(x)(h)=\<varepsilon\><rsub|i>(h)\<shortparallel\>h\<shortparallel\>>
now if for <math|\<forall\>h\<in\>U<rsub|x>> we define
<math|\<varepsilon\>(h)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>>
as <math|\<varepsilon\>(h)(i)=\<varepsilon\><rsub|i>(h)> then
<math|\<pi\><rsub|i>\<circ\>\<varepsilon\>=\<varepsilon\>> and using
<reference|continuous mappings to a product space> we have that
<math|\<varepsilon\>> is continuous at <math|0>, also define
<math|Df(x)(h)\<in\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>> by
<math|Df(x)(h)(i)=D(\<pi\><rsub|i>\<circ\>f)(x)(h)> which is linear,
because <math|Df(x)(\<alpha\>.h+\<beta\>.g)(i)=D(\<pi\><rsub|i>\<circ\>f)(x)(\<alpha\>.h+\<beta\>.g)=\<alpha\>.D(\<pi\><rsub|i>\<circ\>f)(x)(h)+\<beta\>.D(\<pi\><rsub|i>\<circ\>f)(x)(g)=\<alpha\>.Df(x)(h)+\<beta\>.Df(x)(g)>
and thus we have proved that <math|f(x+h)-f(x)-Df(x)(h)=\<varepsilon\>(h).\<shortparallel\>h\<shortparallel\>>
proving that <math|f> is differentiable at <math|x> and that
<math|Df(x)=\<pi\><rsub|i>\<circ\>Df(x)>
</proof>
<\theorem>
<label|Differential of a billinear mapping>Let
<math|L\<in\>L(X<rsub|1>,X<rsub|2>;Y)> (<math|L> is a continuous bilinear
function) and <math|X<rsub|1>,\<shortparallel\>\<shortparallel\><rsub|1>,X<rsub|2>,\<shortparallel\>\<shortparallel\><rsub|2>>
and <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces
and <math|(x<rsub|1>,x<rsub|2>)\<in\>U\<subseteq\>X<rsub|1>\<times\>X<rsub|2>>
then <math|L> is differentiable at <math|x> and
<math|DL(x<rsub|1>,x<rsub|2>)=L(x<rsub|1>,.)+L(.,x<rsub|2>)> where
<math|(L(x<rsub|1>,.)+L(x<rsub|2>,.))(h<rsub|1>,h<rsub|2>)=L(x<rsub|1,>h<rsub|2>)+L(h<rsub|1>,x<rsub|2>)>
if we consider the maximum norm <math|\<shortparallel\>
\<shortparallel\>> on <math|X<rsub|1>\<times\>X<rsub|2>>
</theorem>
<\proof>
First we prove that <math|L(x<rsub|1>,.)+L(.,x<rsub|2>)> is linear. Now
<math|(L(x<rsub|1>,.)+L(.,x<rsub|2>))((h<rsub|1>,h<rsub|2>)+(g<rsub|1>+g<rsub|2>))=L(x<rsub|1>,h<rsub|2>+g<rsub|2>)+L(h<rsub|1>+g<rsub|1>,x<rsub|2>)=L(x<rsub|1>,h<rsub|2>)+L(h<rsub|1>,x<rsub|2>)+L(x<rsub|1>,g<rsub|2>)+L(g<rsub|1>,x<rsub|2>)=(L(x<rsub|1>,.)+L(.,x<rsub|2>))((h<rsub|1>,h<rsub|2>))+(L(x<rsub|1>,.)+L(.,x<rsub|2>))((g<rsub|1>,g<rsub|2>))>
and <math|(L(x<rsub|1>,.)+L(x<rsub|2>,.))(\<alpha\>.(h<rsub|1>,h<rsub|2>))=L(x<rsub|1>,\<alpha\>.h<rsub|2>)+L(\<alpha\>.h<rsub|1>,x<rsub|2>)=\<alpha\>.L(x<rsub|1>,h<rsub|2>)+\<alpha\>.L(x<rsub|1>,h)=\<alpha\>(L(x<rsub|1>,.)+L(.,x<rsub|2>))((h<rsub|1>,h<rsub|2>))>
proving linearity.
Second <math|L(x<rsub|1>,.)+L(.,x<rsub|2>)> is continuous for take
<math|(r,s)\<in\>X<rsub|1>\<times\>X<rsub|2>> such that
<math|max(\<shortparallel\>r\<shortparallel\>,\<shortparallel\>s\<shortparallel\>)=\<shortparallel\>(r,s)\<shortparallel\>=1>
\ or <math|\<shortparallel\>r\<shortparallel\>\<leqslant\>1,\<shortparallel\>s\<shortparallel\>\<leqslant\>1>
then <math|\<shortparallel\>L(x<rsub|1>,.)+L(.,x<rsub|2>)(r,s)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x<rsub|1>,s)+L(r,x<rsub|2>)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L(x<rsub|1>,s)\<shortparallel\>+\<shortparallel\>L(r,x<rsub|2>)\<shortparallel\>\<leqslant\>\<shortparallel\>L\<shortparallel\>\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|1>\<shortparallel\>s\<shortparallel\><rsub|2>+\<shortparallel\>L\<shortparallel\>\<shortparallel\>r\<shortparallel\><rsub|1>\<shortparallel\>s\<shortparallel\><rsub|2>\<leqslant\>\<shortparallel\>L\<shortparallel\>\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|1>+\<shortparallel\>L\<shortparallel\>\<shortparallel\>x<rsub|2>\<shortparallel\><rsub|2>>
proving continuity
Third <math|L(x<rsub|1>+h<rsub|1>,x<rsub|2>+h<rsub|2>)-L(x<rsub|1>,x<rsub|2>)=L(x<rsub|1>,x<rsub|2>)+L(h<rsub|1>,x<rsub|2>)+L(x<rsub|1>,h<rsub|2>)+L(h<rsub|1>,h<rsub|2>)-L(x<rsub|1>,x<rsub|2>)=(L(.,x<rsub|2>)+L(x<rsub|1>,.))(h<rsub|1>.h<rsub|2>)+L(h<rsub|1>,h<rsub|2>)\<Rightarrow\>L(x<rsub|1>+h<rsub|1>,x<rsub|2>+h<rsub|2>)-L(x<rsub|1>,x<rsub|2>)-(L(x<rsub|1>,.)+L(.,x<rsub|2>))(h<rsub|1>,h<rsub|2>)=L(h<rsub|1>,h<rsub|2>)>
now define <math|\<varepsilon\>(h<rsub|1>,h<rsub|2>)> by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(h<rsub|1>,h<rsub|2>)>|<cell|=>|<cell|<frac|L(h<rsub|1>,h<rsub|2>)|\<shortparallel\>(h<rsub|1>,h<rsub|2>)\<shortparallel\>>,
(h<rsub|1>,h<rsub|2>)\<neq\>0>>|<row|<cell|>|<cell|>|<cell|0,
(h<rsub|1>,h<rsub|2>)=0>>>>
</eqnarray*>
then if <math|(h<rsub|1>,h<rsub|2>)\<neq\>0> we
have<math|\<shortparallel\>\<varepsilon\>(h<rsub|1>,h<rsub|2>)\<shortparallel\><rsub|Y>=<frac|\<shortparallel\>L(h<rsub|1>,h<rsub|2>)\<shortparallel\><rsub|Y>|\<shortparallel\>(h<rsub|1>,h<rsub|2>)\<shortparallel\>>\<leqslant\><frac|\<shortparallel\>L\<shortparallel\>\<shortparallel\>h<rsub|1>\<shortparallel\>.\<shortparallel\>h<rsub|2>\<shortparallel\>|max(\<shortparallel\>h<rsub|1>\<shortparallel\><rsub|1>,\<shortparallel\>h<rsub|2>\<shortparallel\><rsub|2>)>\<leqslant\><frac|\<shortparallel\>L\<shortparallel\>max(\<shortparallel\>h<rsub|1>\<shortparallel\><rsub|1>,\<shortparallel\>h<rsub|2>\<shortparallel\><rsub|2>)<rsup|2>|max(\<shortparallel\>h<rsub|1>\<shortparallel\><rsub|1>,\<shortparallel\>h<rsub|2>\<shortparallel\><rsub|2>)>=\<shortparallel\>L\<shortparallel\>
\<shortparallel\>(h<rsub|1>,h<rsub|2>)\<shortparallel\>>proving
continuity of <math|\<varepsilon\>> at <math|(0,0)> so we have proved
that <math|\<shortparallel\>L(x<rsub|1>+h<rsub|1>,x<rsub|2>+h<rsub|2>)-L(x<rsub|1>,x<rsub|2>)-(L(x<rsub|1>,.)+L(.,x<rsub|2>))(h<rsub|1>,h<rsub|2>)=\<varepsilon\>(h<rsub|1>,h<rsub|2>).\<shortparallel\>(h<rsub|1>,h<rsub|2>)\<shortparallel\>>
proving our theorem
</proof>
<\corollary>
<label|differentiability of product>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be a normed space,
<math|x\<in\>U\<subseteq\>X, U open> and
<math|f,g:U\<rightarrow\>\<bbb-K\>> (<math|\<bbb-K\>> a field with a norm
like <math|\<bbb-R\>> or <math|\<bbb-C\>>) be two maps differentiable at
<math|x> then <math|f.g> is differentiable at <math|x> and
<math|D(f.g)(x)=g(x).Df(x)+f(x).Dg(x)>.\
</corollary>
<\proof>
As <math|f,g> are differentiable at <math|x> we have
<math|h:X\<rightarrow\>(Y,Y)> defined by <math|h(x)=(f(x),g(x))> is
differentiable because <math|\<pi\><rsub|1>\<circ\>h=f,\<pi\><rsub|2>\<circ\>h=g>
are differentiable at <math|x>, also the bilinear function
<math|p:Y\<times\>Y\<rightarrow\>Y:(y,z)\<rightarrow\>y.z> (the product
in the field) is differentiable at every <math|(y,z)\<in\>Y\<times\>Y>.
So as <math|f.g=p\<circ\>h> we have <math|f.g> being the composition of
maps differentiable at <math|x> and <math|(f(x),g(x))> respectively that
<math|g.g> is differentiable at <math|x> and
<math|D(f.g)(x)(s)=D(p\<circ\>h)(x)(s)=Dp(h(x))\<circ\>Dh(x)(s)=Dp(h(x))(Dh(x)(s))=p(\<pi\><rsub|1>(Dh(x)(s)),\<pi\><rsub|2>(h(x)))+p(\<pi\><rsub|1>(h(x),\<pi\><rsub|2>(Dh(x)(s)))=p(Df(x)(s),g(x))+p(f(x),Dg(x)(s))=g(x).Df(x)(s)+f(x).Dg(x)(s)=(g(x).Df(x)+f(x).Dg(x))(s)\<Rightarrow\>D(f.g)(x)=g(x).Df(x)+f(x).Dg(x)>
</proof>
<\corollary>
<label|derivative of product>If <math|f,g:\<bbb-K\>\<rightarrow\>\<bbb-K\>>
a normed space have a derivative then <math|f.g> has a derivative and
<math|(f.g)<rprime|'>(x)=f(x).g<rprime|'>(x)+g(x)f<rprime|'>(x) \ (in
another notation <frac|d(f.g)(t)|dt>(x)=f(x).<frac|dg(t)|dt>(x)+g(x).<frac|df(t)|dt>(x)><math|><math|>
</corollary>
<\proof>
Because of the previous corollary we have that <math|f.g> is
differentiable (as <math|f,g> are differentiable because they have a
derivative) and <math|<frac|d(f.g)(t)|dt>(x)=D(f.g)(x)(1)=g(x).Df(x)(1)+f(x).Dg(x)(1)=g(x).<frac|df(t)|dt>(x)+f(x).<frac|dg(t)|dt>(x)>
</proof>
<\corollary>
Let <math|\<bbb-K\>,\<\|\|\>> be the complex or real field then for
<math|n\<in\>\<bbb-Z\><rsub|0>> we have that
<math|\<forall\>x\<in\>\<bbb-K\>> <math|t<rsup|n>:\<bbb-K\>\<rightarrow\>\<bbb-K\>:t\<rightarrow\>t<rsup|n>>
has a derivative at x (and is thus differentiable) and
<math|<frac|d(t<rsup|n>)|dt>(x)=n.x<rsup|n-1>.h> (here in case of
<math|n\<less\>0> we assume that <math|x\<neq\>0>)
</corollary>
<\proof>
First consider <math|n\<in\>\<bbb-N\><rsub|0>> then we can proceed by
induction
<\enumerate-numeric>
<item><math|n=1> then as <math|x<rsup|1>:\<bbb-K\>\<rightarrow\>\<bbb-K\>:x\<rightarrow\>x>
is linear we have <math|Dt<rsup|1>(x)=t<rsup|1>> and thus
<math|<frac|dt<rsup|1>|dt>(x)=Dt<rsup|1>(x)(1)=t<rsup|1>(1)=1=x<rsup|0>=1.t<rsup|0>(x)>
<item>Assume the theorem is true for <math|n> and proof it for
<math|n+1> then as <math|t<rsup|1>> is differentiable and
<math|t<rsup|n>> is differentiable we have that
<math|t<rsup|n+1>=t<rsup|1>.t<rsup|n>> is differentiable (see
<reference|differentiability of product>) and
<math|<frac|dt<rsup|n+1>|dt>(x)=t<rsup|1>(x).<frac|dt<rsup|n>|dt>(x)+t<rsup|n>(x)<frac|dt<rsup|1>|dt>(x)=x.n.x<rsup|n-1>+x<rsup|n>=(n+1).x<rsup|n>>
concluding our proof by induction
</enumerate-numeric>
Next assume that <math|n=-m,m\<in\>\<bbb-N\><rsub|0>> and proceed by
induction on <math|m>
<\enumerate>
<item><math|m=1> then if take <math|0\<neq\>x\<in\>\<bbb-K\>> we have
for <math|h\<in\>U<rsub|x><mid|\\>{0}> that
<math|<frac|<frac|1|x+h>-<frac|1|x>|h>+<frac|1|x<rsup|2>>=<frac|1|h>(<frac|x-x-h)|(x+h).x>)+<frac|1|x<rsup|2>>=<frac|-1|x.(x+h)>+<frac|1|x<rsup|2>>=<frac|1|x>(<frac|-1|x+h>+<frac|1|x>)=<frac|1|x>(<frac|-x+x+h<rsup|>|x(x+h)>)=<frac|h|x<rsup|2>(x+h)>>
now given <math|\<varepsilon\>\<gtr\>0> take then
<math|0\<less\>\<delta\>=min(<frac|\<varepsilon\>\|x\|<rsup|3>|2>,<frac|\|x\||2>)>
then if <math|0\<less\>\|h\|\<less\>\<delta\>> we have
<math|\|h\|\<less\><frac|\|x\||2>><math|> and thus
<math|\<um\><frac|\|x\||2>\<less\>\<um\>\|h\|\<Rightarrow\><frac|\|x\||2>=\|x\|-<frac|\|x\||2>\<less\>\|x\|-\|h\|\<less\>\|x+h\|>
(because <math|\|x\|=\|x+h-h\|\<leqslant\>\|x+h\|+\|h\|\<Rightarrow\>\|x\|-\|h\|\<less\>\|x+h\|)>
and thus <math|<frac|\|x\||2>\<less\>\|x+h\|> or
<math|<frac|1|\|x+h\|>\<less\><frac|2|\|x\|>> and thus
<math|\|<frac|h|x<rsup|2>(x+h)>\|=<frac|\|h\||\|x\|<rsup|2>\|x+h\|>\<less\><frac|2\|h\||\|x\|<rsup|2>\|x\|>=2<frac|\|h\||\|x\|<rsup|3>>=2<frac|\<varepsilon\>\|x\|<rsup|3>|2\|x\|<rsup|3>>=\<varepsilon\>>
so if <math|0\<less\>\|h\|\<less\>\<delta\>\<Rightarrow\>\|<frac|<frac|1|x+h>-<frac|1|x>|h>-(\<um\><frac|1|x<rsup|2>>)\|=\|<frac|h|x<rsup|2>(x+h)>\|\<less\>\<varepsilon\>>
proving that <math|lim<rsub|h\<rightarrow\>0><frac|<frac|1|x+h>-<frac|1|x>|h>=<frac|\<um\>1|x<rsup|2>>>
and thus that <math|<frac|d<frac|1|t>|dt>(x)=<frac|\<um\>1|x<rsup|2>>>
proving the theorem for <math|m=1>
<item>Assume now that the theorem is true for <math|m> and prove it for
<math|m+1> then <math|<frac|d<frac|1|t<rsup|m+1>>|dt>(x)=<frac|d(<frac|1|t>.<frac|1|t<rsup|m>>)|dt>(x)=<frac|1|x>.<frac|d(<frac|1|t<rsup|m>>)|dt>(x)+<frac|1|x<rsup|m>>.<frac|d(<frac|1|t>)|dt>(x)\<equallim\><rsub|induction
hypothese><frac|1|x>.(\<um\>m).<frac|1|x<rsup|m+1>>+<frac|1|x<rsup|m>><frac|d(<frac|1|t>)|dt>(x)\<equallim\><rsub|case
m=1>(\<um\>m)<frac|1|x<rsup|m+2>>-<frac|1|x<rsup|m>>.<frac|1|x<rsup|2>>=(-m-2)<frac|1|x<rsup|(m+1)+1>>=(-((m+1)+1)).<frac|1|x<rsup|(m+1)+1>>>
proving the theorem for <math|m+1>
</enumerate>
</proof>
<\theorem>
<label|Jacobian><dueto|Jacobian Matrix><index|Jacobian matrix>Let
<math|x\<in\>U\<subseteq\>\<bbb-R\><rsup|n>> be a open set and
<math|f:U\<rightarrow\>\<bbb-R\><rsup|m>> differentiable at <math|x> then
<math|\<forall\>j\<in\>{1,\<ldots\>,m}
\<pi\><rsub|j>(Df(x)(h))\<equallim\><rsub|note>(Df(x)(h))<rsub|j>=<big|sum><rsub|i=1><rsup|n><frac|d(\<pi\><rsub|j>\<circ\>f)(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x).h<rsub|i>\<equallim\><rsub|note><big|sum><rsub|i=1><rsup|n><frac|df<rsub|j>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x).h<rsub|i>>
or if we take <math|J(f(x))=<matrix|<tformat|<table|<row|<cell|<frac|df<rsub|1>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|1>>(x)>|<cell|\<ldots\>>|<cell|<frac|df<rsub|1>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|n>>(x)>>|<row|<cell|\<ldots\>>|<cell|\<ldots\>.>|<cell|\<ldots\>>>|<row|<cell|<frac|df<rsub|m>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|1>>(x)>|<cell|\<ldots\>.>|<cell|<frac|df<rsub|m>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|n>>(x)>>>>>>
then <math|<matrix|<tformat|<table|<row|<cell|(Df(x)(h))<rsub|1>>>|<row|<cell|\<ldots\>>>|<row|<cell|(Df(x)(h))<rsub|m>>>>>>=Df(x)(h)=J(f(x)).h\<equallim\><rsub|matrix
multiplication><matrix|<tformat|<table|<row|<cell|<frac|df<rsub|1>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|1>>(x)>|<cell|\<ldots\>>|<cell|<frac|df<rsub|1>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|n>>(x)>>|<row|<cell|\<ldots\>>|<cell|\<ldots\>.>|<cell|\<ldots\>>>|<row|<cell|<frac|df<rsub|m>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|1>>(x)>|<cell|\<ldots\>.>|<cell|<frac|df<rsub|m>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|n>>(x)>>>>><matrix|<tformat|<table|<row|<cell|h<rsub|1>>>|<row|<cell|\<ldots\>>>|<row|<cell|h<rsub|n>>>>>>>
</theorem>
<\proof>
Using <reference|differentiability of map to product> we have that
<math|Df(x)=(Df<rsub|1>(x),\<ldots\>,Df<rsub|m>(x))> where
<math|Df<rsub|i>(x)> is the differential of
<math|\<pi\><rsub|i>\<circ\>f=f<rsub|i>:U\<subseteq\>\<bbb-R\><rsup|n>\<rightarrow\>\<bbb-R\><rsup|>>
at <math|x\<in\>U> and then using <reference|differential of
multiparameter function> we have that
<math|Df<rsub|i>(x)(h)=<big|sum><rsub|j\<in\>{1,\<ldots\>,n}><frac|df<rsub|i>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|j>>(x).h<rsub|j>>
</proof>
<section|Higher order differentiability>
<\notation>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed space
<math|{X<rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>> defined by
<math|X<rsub|i>=X> <math|i\<in\>{1,\<ldots\>,n}> then we note\
<\enumerate>
<item><math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y),\<ldots\>)\<equallim\><rsub|notation>L<rsup|n>(X,\<ldots\>,L(X,Y))>
where as usual if <math|n=1> we have
<math|L<rsup|1>(X,Y)=L(X<rsub|1>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L(X<rsub|1>,Y)=L(X,Y)>\
<\note>
If <math|n\<gtr\>1> then <math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)=L*(X<rsub|n>,L<rsub|(X<rsub|n-1>>,\<ldots\>,L(X<rsub|1>,Y)
\<ldots\>)\<Rightarrow\>L<rsup|n>(X,\<ldots\>,L(X,Y))=L(X,L<rsup|n-1>(X,\<ldots\>,L(X,Y)))>
</note>
<item><math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)\<equallim\><rsub|notation>L<rsup|n>(X,Y)>
</enumerate>
</notation>
<\definition>
<index|first derivative><index|<math|C<rsup|1>>>Let
<math|X,\<shortparallel\>\<shortparallel\>> and
<math|Y,\<shortparallel\>\<shortparallel\>> be normed sets,
<math|U\<subseteq\>X> a open set and <math|f:U\<rightarrow\>Y> then
<math|f> is differentiable on <math|U> if <math|f> is differentiable at
every <math|x\<in\>U>. The function <math|Df:U\<rightarrow\>L(X,Y)>
defined by <math|x\<rightarrow\>Df(x)> is called the first derivative
<math|f>. If in addition <math|Df> is continuous then <math|f> is
differentiable of class <math|C<rsup|1>> (or <math|f> is of class
<math|C<rsup|1>> or <math|f> is <math|C<rsup|1>>).\
</definition>
<\lemma>
Let <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be a normed space
over <math|\<bbb-K\>> then given <math|f\<in\>U\<subseteq\>\<bbb-K\>\<rightarrow\>L(\<bbb-K\>,Y)>
we can define <math|f<rsub|(1)>:U\<rightarrow\>Y> by
<math|f<rsub|(1)>(x)=f(x)(1)> we have then that <math|f> is continuous
<math|\<Leftrightarrow\>> <math|f<rsub|(1)>> is continuous
</lemma>
<\proof>
To prove this we have to prove that <math|\<forall\>x\<in\>U> we have
that <math|f> is continuous at <math|x><math|\<Leftrightarrow\>>
<math|f<rsub|(1)>> is continuous at <math|x>
<\enumerate>
<item><math|\<Rightarrow\>> By continuity of <math|f> at <math|x> we
have <math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\>\<gtr\>0> such that if
<math|\|x-y\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(x)-f(y)\<shortparallel\>\<less\>\<varepsilon\>\<Rightarrow\>\<shortparallel\>f<rsub|(1)>(x)-f<rsub|(1)>(y)\<shortparallel\><rsub|Y>=\<shortparallel\>f(x)(1)-f(y)(1)\<shortparallel\><rsub|Y>=\<shortparallel\>(f(x)-f(y))(1)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>f(x)-f(y)\<shortparallel\>\|1\|=\<shortparallel\>f(x)-f(y)\<shortparallel\>\<less\>\<varepsilon\>>
proving continuity of <math|f<rsub|(1)>>
<item><math|\<Leftarrow\>> Let <math|\<varepsilon\>\<gtr\>0> then by
continuity of <math|f<rsub|(1)>> at <math|x> there exists a
<math|\<delta\>\<gtr\>0> such that if <math|\|x-y\|\<less\>\<delta\>>
we have <math|\<shortparallel\>f<rsub|(1)>(x)-f<rsub|(1)>(y)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>.
Then as <math|\<shortparallel\>(f(x)-f(y))(r)\<shortparallel\><rsub|Y>=\<shortparallel\>(f(x)-f(y))(r.1)\<shortparallel\><rsub|Y>=\<shortparallel\>r.(f(x)-f(y))(1)\<shortparallel\><rsub|Y>=\|r\|.\<shortparallel\>f<rsub|(1)>(x)-f<rsub|(1)>(y)\<shortparallel\><rsub|Y>\<less\>\|r\|.<frac|\<varepsilon\>|2>\<Rightarrow\>\<shortparallel\>f(x)-f(y)\<shortparallel\>\<leqslant\><frac|\<varepsilon\>|2>\<less\>\<varepsilon\>>
(see the definition of <math|\<shortparallel\>\<shortparallel\>>,
<reference|definition of the norm of a linear mapping>) proving
continuity
</enumerate>
</proof>
<\theorem>
<label|C1 of real or complex functions>Let
<math|f:U\<subseteq\>\<bbb-K\>\<rightarrow\>Y>,
<math|Y,\<shortparallel\>\<shortparallel\>> be a normed vector space over
<math|\<bbb-K\> >be a function then the following are equivalent
<\enumerate>
<item><math|f> is <math|C<rsup|1>>
<item><math|\<forall\>t\<in\>U> <math|f<rprime|'>(t)> exists and
<math|f<rprime|'>:U\<rightarrow\>Y> defined by
<math|t\<rightarrow\>f<rprime|'>(t)=<frac|df(x)|dx>(t)> is continuous
on <math|U>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|1\<Rightarrow\>2> As <math|f> is <math|C<rsup|1>> we have
that <math|f> is differentiable on <math|U> and
<math|Df:x\<rightarrow\>Df(x)\<in\>L(\<bbb-K\>,Y)> is continuous. Then
using <reference|differentiability of real or complex functions>
<math|\<forall\>x\<in\>U> <math|f<rprime|'>(x)> exists and
<math|f<rprime|'>(x)=Df(x)(1)>. We can use now the previous lemma to
find that <math|f<rprime|'>> is continuous on <math|U> because
continuity of <math|Df>
<item><math|2\<Rightarrow\>1> As <math|\<forall\>x\<in\>U>
<math|f<rprime|'>(x)> exists we use <reference|differentiability of
real or complex functions> again to find the existence of <math|Df(x)>
where <math|f<rprime|'>(x)=Df(x)(1)> and using the previous lemma and
the continuity of <math|f<rprime|'> >we find that
<math|Df:x\<rightarrow\>Df(x)> is continuous.\
</enumerate>
</proof>
We define now higher order derivatives as follows
<\definition>
<index|n-th derivative><math|><index|<math|C<rsup|n>>><index|<math|C<rsup|\<infty\>>>>Let
<math|X,\<shortparallel\>\<shortparallel\>> and
<math|Y,\<shortparallel\>\<shortparallel\>> be normed spaces over
<math|\<bbb-K\>>, <math|U\<subseteq\>X> a open set and
<math|f:U\<rightarrow\>Y> then <math|f> is <math|n> times differentiable
on U and have a n-th derivative <math|D<rsup|n>f:U\<rightarrow\>L<rsup|n>(X,\<ldots\>,L(X,Y))>
if
<\enumerate>
<item><math|n=1> <math|f> is differentiable at every <math|x\<in\>U>
and <math|D<rsup|1>f=Df> and <math|Df:U\<rightarrow\>L(X,Y)=L<rsup|1>(X,\<ldots\>,L(X,Y))>
<item><math|n\<gtr\>1> <math|f> is <math|n-1> times differentiable and
<math|D<rsup|n-1>f:U\<rightarrow\>L<rsup|n-1>(X,\<ldots\>,L(X,Y))> is
differentiable on <math|U>, then <math|D(D<rsup|n-1>f):U\<rightarrow\>L(X,L<rsup|n-1>(X,\<ldots\>,L(X,Y)))=L<rsup|n>(X,\<ldots\>,L(X,Y))>
defined by <math|x\<rightarrow\>D(D<rsup|n-1>f)(x)> is defined and is
set to be <math|D<rsup|n>f> or the n-th derivative from <math|f>
</enumerate>
If also the function <math|D<rsup|n>f> is continuous then <math|f> is of
class <math|C<rsup|n>>. If <math|f> is of class <math|C<rsup|n>> for
every <math|n\<in\>\<bbb-N\><rsub|0>> then <math|f> is called
<math|C<rsup|\<infty\>>>
</definition>
<\remark>
We have seen (see <reference|equivalence of multilinear and linear
continuous mappings>) that there exists a norm preserving isomorphism
<math|\<cal-P\>> between <math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y)\<ldots\>)>
and <math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)> (and thus also between
<math|L<rsup|n>(X,\<ldots\>,L(X,Y))> and <math|L<rsup|n>(X,Y)>) defined
by <math|\<cal-P\>(f)(x<rsub|1>,\<ldots\>,x<rsub|n>)=f(x<rsub|1>:\<ldots\>:x<rsub|n>)=f(x<rsub|n>)(x<rsub|n-1>)\<ldots\>(x<rsub|1>)>
in this way we can identify <math|D<rsup|n>f(x)> also with a multilinear
function where <math|\<cal-P\>(D<rsup|n>f(x))(x<rsub|1>,\<ldots\>,x<rsub|n>)=D<rsup|n>f(x)(x<rsub|1>:\<ldots\>:x<rsub|n>)=D<rsup|n>f(x)(x<rsub|n>)(x<rsub|n-1>)\<ldots\>(x<rsub|1>)>.
In fact from now on, when we write <math|D<rsup|n>f(x)(x<rsub|1>,\<ldots\>,x<rsub|n>)>
or say <math|D<rsup|n>f(x)> is multilinear or
<math|D<rsup|n>f(x)\<in\>L<rsup|n>(X,Y)> we actually mean
<math|\<cal-P\>(D<rsup|n>f(x))(x<rsub|1>,\<ldots\>,x<rsub|n>)> or say
\ that <math|D<rsup|n>f(x)> is multilinear or
<math|\<cal-P\>(D<rsup|n>f(x))\<in\>L<rsup|n>(X,Y)>\
</remark>
<\note>
<label|restriction of C-r mapping>If <math|f> is differentiable of class
<math|C<rsup|r>> on <math|U> and we have a open set <math|V\<subseteq\>U>
then <math|f<rsub|\|V>> is differentiable of class <math|C<rsup|r>>. This
follows at once out of the definition of <math|C<rsup|r>> and
<reference|differentiability and restricted mappings>.\
</note>
<\lemma>
<label|proving f is Cn based on Cn-1 of Df>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-K\>> then for <math|n\<in\>\<bbb-N\><rsup|0>,n\<gtr\>1> we
have <math|f> is <math|C<rsup|n>> if and only <math|Df> is
<math|C<rsup|n-1>> and <math|D<rsup|n-1>(Df)=D<rsup|n>f>
</lemma>
<\proof>
First prove that <math|f> is <math|C<rsup|n>\<Rightarrow\>Df> is
<math|C<rsup|n-1>> and <math|D<rsup|n-1>(Df)=D<rsup|n>f> we prove this by
induction on <math|n>\
<\enumerate>
<item><math|n=2> then <math|D<rsup|2>f=D(D<rsup|2-1>f)=D(Df)> exists
and is continuous meaning that <math|Df> is <math|C<rsup|1>>
<item>Assume trueness for <math|n\<gtr\>1> and let <math|f> be
<math|C<rsup|n+1>> then <math|D<rsup|n>f> is differentiable and thus
continue, or <math|f> is <math|C<rsup|n>> and thus by the induction
hypothesis we have that <math|Df> is <math|C<rsup|n-1>> and
<math|D<rsup|n>f=D<rsup|n-1>(Df)> which because <math|f> is
<math|C<rsup|n+1>> has a derivative that is continuous meaning that
<math|Df> is <math|C<rsup|n>> and we have
<math|D<rsup|n+1>f=D(D<rsup|n>f)=D(D<rsup|n-1>(Df))=D<rsup|n>*(Df)>
</enumerate>
Second proof the opposite <math|Df> is <math|C<rsup|n-1>\<Rightarrow\>f>
is <math|C<rsup|n>> (because if <math|f> is <math|C<rsup|n>> we have via
(1) that <math|D<rsup|n-1>(Df)=D<rsup|n>f> we only have to prove that
<math|f> is <math|C<rsup|n>>). Again we use induction in this proof
<\enumerate>
<item><math|n=2> then if <math|Df> is <math|C<rsup|1>> the derivative
of <math|Df> exists and is continuous, meaning that <math|f> is
<math|C<rsup|2>>
<item>Assume trueness for <math|n\<gtr\>1> then if <math|Df> is
<math|C<rsup|n>> we have that <math|Df> is <math|C<rsup|n-1>> and thus
<math|f> is <math|C<rsup|n>> and by (1) we have
<math|D<rsup|n>f=D<rsup|n-1>(Df)> which because <math|Df> is
<math|C<rsup|n>> means that <math|D(D<rsup|n>f)> exists and is
continuous so that <math|f> is <math|C<rsup|n+1>>
</enumerate>
</proof>
<\theorem>
<label|C-r is a local property>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y<rsub|,>\<shortparallel\>\<shortparallel\><rsub|Y>>
be normed spaces over <math|\<bbb-K\>>, <math|U> a open set in <math|X>
and <math|f:U\<rightarrow\>Y> a function such that
<math|\<forall\>x\<in\>U> there exists a <math|V> open with
<math|x\<in\>V\<subseteq\>U> for which
<math|f<rsub|\|V>:V\<rightarrow\>Y> is differentiable on <math|V> of
class <math|C<rsup|r>> then <math|f> is of class <math|C<rsup|r>> on
<math|U> and <math|(D<rsup|r>f)<rsub|\|V>=D<rsup|r>(f<rsub|\|V>)>
</theorem>
<\proof>
We prove this by induction on <math|r>
<\enumerate>
<item>Case <math|r=1> this follows from <reference|differentiability
and restricted mappings> proving that <math|f> is differentiable at
every <math|x\<in\>U> and that <math|\<forall\>x\<in\>V\<succ\>D(f<rsub|\|V>)(x)=Df(x)>
and thus <math|(Df)<rsub|\|V>=D(f<rsub|\|V>)>, we use then
<reference|continuity is local> to prove that
<math|Df:U\<rightarrow\>L(X,Y)> is continue, proving that <math|f> is
<math|C<rsup|1>>
<item>Case assume that the theorem is proved for <math|r>, lets prove
it for <math|r+1> . So we have that if <math|x\<in\>U> there exists a
open <math|V> such that <math|x\<in\>V> and <math|f<rsub|\|V>> is of
class <math|C<rsup|r+1>> on <math|V> then <math|f<rsub|\|V>> is by
definition of class <math|C<rsup|r>> and by the induction hypothesis we
have that <math|f> is of class <math|C<rsup|r>> on <math|U> and
<math|D<rsup|r>(f<rsub|\|V>)=(D<rsup|r>f)<rsub|\|V>>. Take now
<math|g=D<rsup|r>f> then we have <math|g<rsub|\|V>=D<rsup|r>(f<rsub|\|V>)>
and as <math|f<rsub|\|V>> is <math|C<rsup|r+1>> we have that <math|g>
fulfil the requirements for the case <math|r=1> and thus <math|g> is
<math|C<rsup|1>> proving that <math|f> is <math|C<rsup|r+1>> and
<math|(Dg)<rsub|\|V>=D(g<rsub|\|V>)=D<rsup|>(D<rsup|r>(f<rsub|\|V>))=D<rsup|r+1>(f<rsub|\|V>)>
</enumerate>
</proof>
<\lemma>
<label|sum of C-r functions is C-r>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-K\>>, <math|U\<subseteq\>X> a open set and
<math|f:U\<rightarrow\>Y,g:U\<rightarrow\>Y> be functions differentiable
on <math|U> of class <math|C<rsup|r>> then <math|f+g> is differentiable
of class <math|C<rsup|r>> and <math|D<rsup|r>(f+g)=D<rsup|r>f+D<rsup|r>g>
</lemma>
<\proof>
The theorem is already proved for <math|n=1> because of <reference|sum of
differentiable functions is differentiable> and <reference|sum of
continuous functions is continuous>, so proceeding by induction, assume
the theorem is true for <math|n> and prove it for <math|n+1> then
<math|D<rsup|n+1>(f+g)=D(D<rsup|n>(f+g))=D(D<rsup|n>f+D<rsup|n>g)\<equallim\><rsub|case
of n=1>D(D<rsup|n>f)+D(D<rsup|n>g)=D<rsup|n>f+D<rsup|n>g>\
</proof>
<\lemma>
<label|linear continuous mappings are C infinity>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-K\>> then every linear continuous map <math|L\<in\>L(X,Y)> is
<math|C<rsup|\<infty\>>> where <math|D<rsup|1>L(x)=L> and
<math|D<rsup|n>L=0>(the 0 constant function) (if <math|n\<gtr\>1>)
</lemma>
<\proof>
If <math|n=1> then use (<reference|differential of linear mapping>) so
that <math|\<forall\>x\<in\>X> we have
<math|DL(x)=L\<Rightarrow\>D<rsup|1>L:x<rsub|>\<rightarrow\>L> which is a
constant function and thus continuous. Further we use
<reference|Differential of constant function> to prove that
<math|\<forall\>n\<gtr\>1> that <math|D<rsup|n>L=0> (which is continuous
and again a constant function so that it is differentiable again and that
<math|D<rsup|n+1>L=0>, so we essentially use induction here)
</proof>
<\lemma>
<label|translation is C-inifinity>Let
<math|X,\<shortparallel\>\<shortparallel\>,
Y,\<shortparallel\>\<shortparallel\><rsub|Y>,x<rsub|0>\<in\>X> be normed
spaces over <math|\<bbb-K\>> then every translation
<math|\<tau\><rsub|x<rsub|0>>:X\<rightarrow\>Y\<vdash\>x\<rightarrow\>\<tau\><rsub|x<rsub|0>>(x)=x+x<rsub|0>>
is <math|C<rsup|\<infty\>>> with <math|\<forall\>x\<in\>X we have
D\<tau\><rsub|x<rsub|0>>(x)=i<rsub|X>> and
<math|D<rsup|r>\<tau\><rsub|x<rsub|0>>(x)=0> if <math|r\<gtr\>1>
</lemma>
<\proof>
\;
For <math|r=1> we have <math|\<tau\><rsub|x<rsub|0>>(x+h)-\<tau\><rsub|x<rsub|0>>(x)-i<rsub|X>(h)=x+h+x<rsub|0>-x-x<rsub|0>-h=0>
proving as \ the constant <math|0->function is a
<math|\<varepsilon\>-mapping> that <math|D\<tau\><rsub|x<rsub|0>>(x)=i<rsub|X>>\
As <math|i<rsub|X>> is linear the rest of the proof follows from the
previous theorem (see <reference|linear continuous mappings are C
infinity>)
</proof>
<\lemma>
<label|billinear continuous mappings are C infinity>Let
<math|L\<in\>L(X<rsub|1>,X<rsub|2>;Y)> (<math|L> is a continuous bilinear
function) and <math|X<rsub|1>,\<shortparallel\>\<shortparallel\><rsub|1>,X<rsub|2>,\<shortparallel\>\<shortparallel\><rsub|2>>
and <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces
and <math|(x<rsub|1>,x<rsub|2>)\<in\>U\<subseteq\>X<rsub|1>\<times\>X<rsub|2>>
then <math|L> is <math|C<rsup|\<infty\>>>
</lemma>
<\proof>
We prove this as follows
<\enumerate>
<item><math|n=1> then we use <reference|Differential of a billinear
mapping> to prove that <math|\<forall\>(x,y)> DL(x,y) exists and is
equal to the linear continuous function <math|L(x,.)+L(y,.)> exists. So
<math|DL=(x,y)\<rightarrow\>L*(x,.)+L(y,.)> and prove that it is linear
and continuous (so that <math|L> is <math|C<rsup|1>>.\
<\enumerate>
<item>Linearity, choose <math|(r,s)> arbitrary
<math|DL(\<alpha\>.(x<rsub|1>,y<rsub|1>)+\<beta\>.(x<rsub|2>,y<rsub|2>))(r,s)=DL(\<alpha\>.x<rsub|1>+\<beta\>.x<rsub|2>,\<alpha\>.y<rsub|1>+\<beta\>.y<rsub|2>)(r,s)=(L(\<alpha\>.x<rsub|1>+\<beta\>.x<rsub|2>,.)+L(.,\<alpha\>.y<rsub|1>+\<beta\>.y<rsub|2>))(r,s)=L(\<alpha\>.x<rsub|1>+\<beta\>.x<rsub|2>,s)+L(r,\<alpha\>.y<rsub|1>+\<beta\>.y<rsub|2>)\<equallim\><rsub|billinearity>\<alpha\>L(x<rsub|1>,s)+\<beta\>L(x<rsub|2>,s)+\<alpha\>L(r,y<rsub|1>)+\<beta\>L(r,y<rsub|2>)=\<alpha\>*(L(x<rsub|1>,.)+L(.,y<rsub|1>))(r,s)+\<beta\>(L(x<rsub|2>,.)+L(.,y<rsub|2>))(r,s)=(\<alpha\>DL(x<rsub|1>,y<rsub|1>)+\<beta\>DL(x<rsub|2>,y<rsub|2>))(r,s)\<Rightarrow\>DL(\<alpha\>.(x<rsub|1>,y<rsub|1>)+\<beta\>.(x<rsub|2>,y<rsub|2>))=\<alpha\>.DL(x<rsub|1>,y<rsub|1>)+\<beta\>.DL(x<rsub|2>,y<rsub|2>)>
<item>Continuity, given <math|(x,y),(r,s)\<in\>X<rsub|1>\<times\>X<rsub|2>>
then <math|\<shortparallel\>DL(x,y)(r,s)\<shortparallel\><rsub|Y>=\<shortparallel\>L(x,s)+L(r,y)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>\<shortparallel\>x\<shortparallel\><rsub|1>\<shortparallel\>s\<shortparallel\><rsub|2>+\<shortparallel\>L\<shortparallel\>\<shortparallel\>y\<shortparallel\><rsub|2>\<shortparallel\>\<shortparallel\>r\<shortparallel\><rsub|1>\<leqslant\>\<shortparallel\>L\<shortparallel\>(\<shortparallel\>x\<shortparallel\><rsub|1>+\<shortparallel\>y\<shortparallel\><rsub|2>)max(\<shortparallel\>s\<shortparallel\><rsub|2>,\<shortparallel\>r\<shortparallel\><rsub|1>)=\<shortparallel\>L\<shortparallel\>(\<shortparallel\>x\<shortparallel\><rsub|1>+\<shortparallel\>y\<shortparallel\><rsub|2>)\<shortparallel\>(r,s)\<shortparallel\>\<Rightarrow\>\<shortparallel\>DL(x,y)\<shortparallel\>\<leqslant\>\<shortparallel\>L\<shortparallel\>(\<shortparallel\>x\<shortparallel\><rsub|1>+\<shortparallel\>y\<shortparallel\><rsub|2>)\<leqslant\>\<shortparallel\>L\<shortparallel\>max(\<shortparallel\>x\<shortparallel\><rsub|1>,\<shortparallel\>y\<shortparallel\><rsub|2>)=\<shortparallel\>L\<shortparallel\>\<shortparallel\>(x,y)\<shortparallel\>>
proving that <math|DL> is indeed continuous.
</enumerate>
<item><math|n=2> then as DL is proved in (1.a,b) to be linear and
continuous we have that <math|D<rsup|2>L:(r,s)\<rightarrow\>((x,y)\<rightarrow\>L(x,.)+L(.,y))>
a constant function and thus continuous
<item><math|n=3> then as <math|D<rsup|2>L> is a constant function we
have <math|D<rsup|3>L=0> , again a constant function
<item><math|n\<gtr\>3> then we have always the differential of a
constant function so <math|D<rsup|n>L=0>
</enumerate>
</proof>
<\theorem>
<label|projection map is C-infinite>Let
<math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
is a finite family of normed space over <math|\<bbb-K\>> then the
projection maps \ <math|\<pi\><rsub|i>:<big|prod><rsub|j\<in\>{1,\<ldots\>,n}>X<rsub|j>\<rightarrow\>X<rsub|i>>
are <math|C<rsup|\<infty\>>>
</theorem>
<\proof>
This is trivially as the projection map is linear and continuous (see
<reference|projection map is continuous>) and we can use then
<reference|linear continuous mappings are C infinity>.
</proof>
<\theorem>
<label|c-r of mapping to product >Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be a normed vector
space, <math|{Y<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>
a family of normed spaces and <math|\<shortparallel\>\<shortparallel\>>
the maximum norm defined on <math|<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>>
(making it a normed space). Then given <math|U\<subset\>X, U> open we
have that <math|f:U\<rightarrow\><big|prod><rsub|i\<in\>{1,\<ldots\>,n}>Y<rsub|i>>
is <math|C<rsup|n>> if and only if <math|\<forall\>i\<in\>1,\<ldots\>,n}\<succ\>\<pi\><rsub|i>\<circ\>f>
is <math|C<rsup|n>> and then <math|\<pi\><rsub|i>\<circ\>D<rsup|n>f=D<rsup|n>(\<pi\><rsub|i>\<circ\>f)>
</theorem>
<\proof>
We proof this by induction on <math|n>
<\enumerate>
<item><math|n=1> Using <reference|differentiability of map to product>
we find <math|\<forall\>x\<in\>U> that <math|f> is differentiable at
<math|x\<in\>U\<Leftrightarrow\>\<forall\>i\<in\>{1,\<ldots\>,n}
\<pi\><rsub|i>\<circ\>f> is differentiable at <math|x> and that
<math|\<pi\><rsub|i>\<circ\>Df(x)=D(\<pi\><rsub|i>\<circ\>f)(x)> this
together with the fact that <math|Df:x\<rightarrow\>Df(x)> is continue
if only if <math|\<pi\><rsub|i>\<circ\>Df> is continue
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> and
<math|\<pi\><rsub|i>\<circ\>Df=D(\<pi\><rsub|i>\<circ\>f)>
<item>Assume the theorem to be true for <math|n> and prove it for
<math|1\<less\>n+1>
<\enumerate>
<item>Assume that <math|f> is <math|C<rsup|n+1>> then <math|Df> is
<math|C<rsup|n>> and by the induction hypothesis we have that
<math|\<pi\><rsub|i>\<circ\>Df> is <math|C<rsup|n>> or using (1)
<math|D(\<pi\><rsub|i>\<circ\>f)> is <math|C<rsup|n>> or
<math|\<pi\><rsub|i>\<circ\>f> is <math|C<rsup|n+1>> and
<math|\<pi\><rsub|i>\<circ\>D<rsup|n+1>f=\<pi\><rsub|i>\<circ\>D<rsup|n>(Df)=D<rsup|n>(\<pi\><rsub|i>\<circ\>Df)\<equallim\><rsub|(1)>D<rsup|n>(D(\<pi\><rsub|i>\<circ\>f))=D<rsup|n+1>(\<pi\><rsub|i>\<circ\>f)>
<item>Assume that <math|\<forall\>i\<in\>{1,\<ldots\>,n}> we have
that <math|\<pi\><rsub|i>\<circ\>f> is <math|C<rsup|n+1>> then
<math|\<pi\><rsub|i>\<circ\>Df\<equallim\><rsub|(1)>D(\<pi\><rsub|i>\<circ\>f)>
is <math|C<rsup|n>> and thus by the induction hypothesis we have that
<math|Df> is <math|C<rsup|n>> and thus <math|f>is <math|C<rsup|n>>
and also that <math|D<rsup|n>(\<pi\><rsub|i>\<circ\>Df)=\<pi\><rsub|i>\<circ\>D<rsup|n>(Df)\<Rightarrow\>D<rsup|n>(D(\<pi\><rsub|i>\<circ\>f))=\<pi\><rsub|i>\<circ\>D<rsup|n+1>f\<Rightarrow\>\<pi\><rsub|i>\<circ\>D<rsup|n+1>f=\<pi\><rsub|i>\<circ\>D<rsup|n+1>f>
</enumerate>
\
</enumerate>
</proof>
<\theorem>
<label|generalized chain rule>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>,
<math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be normed spaces and
let <math|U\<subseteq\>X> and <math|V\<subseteq\>Y> be open sets in
<math|X> and <math|Y> respectively and let <math|f:U\<rightarrow\>Y> and
<math|g:V\<rightarrow\>Z> be functions such that <math|f(U)\<subseteq\>V>
so that <math|f\<circ\>g> is defined. Then if <math|f,g> are of class
<math|C<rsup|n>,n\<in\>\<bbb-N\><rsub|0>> so is <math|g\<circ\>f> of
class <math|C<rsup|n>>\
</theorem>
<\proof>
We prove this by induction on <math|n>
<\enumerate>
<item><math|n=1> then <math|\<forall\>x\<in\>U> and
<math|\<forall\>y\<in\>V> we have that the differential <math|Df(x)>
and <math|Dg(y)> exists so also as <math|f(x)\<in\>V> the differential
<math|Dg(f(x))> exists and then we use the chain rule
(<reference|differentiability of composition of differentiable
mappings>) to proof that <math|f\<circ\>g> is also differentiable at
<math|x> and the differential <math|D(g\<circ\>f)(x)=Dg(f(x))\<circ\>Df(x)>.
Now the function <math|Dg<rsub|f>:U\<rightarrow\>L(Y,Z)\|x\<rightarrow\>Dg<rsub|f>(x)=Dg(f(x))=(Dg\<circ\>f)(x)>
is continuous because <math|f> is continuous by differentiability and
<math|Dg> is continuous because <math|g> is <math|C<rsup|1>>. As also
<math|Df> is continuous because <math|f> is <math|C<rsup|1>> we find
that <math|h:U\<rightarrow\>L(Y,Z)\<times\>L(X,Y)\|x\<rightarrow\>h(x)=(Dg(f(x)),Df(x))>
is continue and as <math|D(g\<circ\>f)(x)=Dg(f(x))\<circ\>Df(x)\<equallim\><rsub|<reference|composition
of continuous linear maps is continuous>>C<rsub|\<circ\>>(Dg(f(x)),Df(x))=(C<rsub|\<circ\>>\<circ\>h)(x)>
or <math|D(g\<circ\>f)=C<rsub|\<circ\>>\<circ\>h> which is the
composition of two continuous functions and thus continuous (see
<reference|composition of continuous linear maps is continuous>)
<item>Assume the theorem now true for <math|n> and prove it for
<math|n+1> so if <math|f,g> are of class <math|C<rsup|n+1>> then
<math|Df<rsup|n+1>:x\<rightarrow\>Df<rsup|n+1>(x),Dg<rsup|n+1>:x\<rightarrow\>Dg<rsup|n+1>(x)>
exists and is continue, now take <math|f\<circ\>g> and if we prove that
<math|D(f\<circ\>g)> is <math|C<rsup|n>> we have proved that
<math|(f\<circ\>g)> is <math|C<rsup|n+1>>. Now using part of the proof
of (1) we have <math|D(g\<circ\>f)=C<rsub|\<circ\>>\<circ\>h> where
<math|h(x)=(Dg(f(x)),Df(x))> now <math|C<rsub|\<circ\>>> being a
continuous bilinear map (<reference|composition of continuous linear
maps is continuous>) it is <math|C<rsup|\<infty\>>> and thus
<math|C<rsup|n>> so if we prove that <math|h \ is C<rsup|n>> we can use
the induction hypothesis so that <math|D(g\<circ\>f)> is
<math|C<rsup|n>>. Now <math|\<pi\><rsub|1>\<circ\>h=Dg\<circ\>f> is the
composition of two <math|C<rsup|n>> functions (<math|f> is
<math|C<rsup|n+1>> and thus <math|C<rsup|n>>, <math|g> is
<math|C<rsup|n+1>\<Rightarrow\>Dg\<in\>C<rsup|n>>) which is by the
induction hypothesis again <math|C<rsup|n>>. Further
<math|\<pi\><rsub|2>\<circ\>h=Df> is <math|C<rsup|n>> because <math|f>
is <math|C<rsup|n+1>>. Finally we use the previous theorem to prove
that <math|h> is <math|C<rsup|n>><math|>
</enumerate>
</proof>
<\lemma>
<label|fv is differentiable>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>.
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces,
<math|U\<subseteq\>X> open and <math|f:U\<rightarrow\>Y> be a
differentiable function of class <math|C<rsup|r>,r\<geqslant\>1> then
<math|\<forall\>v\<in\>X> <math|f<rsub|v>:X\<rightarrow\>Y> defined by
<math|x\<rightarrow\>f<rsub|v>(x)=Df(x)(v)> is itself differentiable of
class <math|C<rsup|r-1>> on <math|U>
</lemma>
<\proof>
\;
<\enumerate>
<item><math|\<forall\>v\<in\>X> define
<math|O<rsub|v>:U\<rightarrow\>U\<times\>X> by
<math|x\<rightarrow\>O<rsub|v>(x)=(x,v)>, we prove then that
<math|O<rsub|v>> is linear and continuous\
<\enumerate>
<item>Linearity, <math|O<rsub|v>(\<alpha\>.x+\<beta\>.y)=(\<alpha\>.x+\<beta\>.y,v)=\<alpha\>.(x,v)+\<beta\>.(y,v)=\<alpha\>.O<rsub|v>(x)+\<beta\>.O<rsub|v>(y)>
<item>Continuity, <math|\<shortparallel\>O<rsub|v>(x)\<shortparallel\><rsub|U\<times\>X>=\<shortparallel\>(x,v)\<shortparallel\><rsub|U\<times\>X>=max(\<shortparallel\>v\<shortparallel\><rsub|X>,\<shortparallel\>x\<shortparallel\><rsub|X>)\<leqslant\>\<shortparallel\>x\<shortparallel\><rsub|X>>
</enumerate>
From this it follows that <math|O<rsub|v>> is n-differentiable for
<math|n\<in\>\<bbb-N\><rsub|0>> (thus of class
<math|C<rsup|\<infty\>>>)
<item>Define <math|\<varphi\>:U\<times\>X\<rightarrow\>L(X,Y)\<times\>X>
by <math|(x,v)\<rightarrow\>(Df(x),v)> then we have that
<math|\<varphi\>> is differentiable on <math|U>. For
<math|\<pi\><rsub|1>\<circ\>\<varphi\>=Df> which is differentiable of
class <math|C<rsup|r-1>> on <math|U> because <math|f> is differentiable
of class <math|C<rsup|r>>, <math|\<pi\><rsub|2>\<circ\>\<varphi\>=i<rsub|X>>
which is differentiable because it is linear and continuous (trivial).
<item>The evaluation function <math|ev:L(X,Y)\<times\>X\<rightarrow\>Y>
defined by <math|(L,v)=L(v)> is bilinear and continuous hence
n-differentiable for <math|n\<in\>\<bbb-N\><rsub|0>> (thus
<math|C<rsup|\<infty\>>>)
<\enumerate>
<item><math|ev(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>,v)=(\<alpha\>.L<rsub|1>+\<beta\>.L<rsub|2>)(v)=\<alpha\>.L)<rsub|1>(v)+\<beta\>.L<rsub|2>(v)=\<alpha\>.ev(L<rsub|1>,v)+\<beta\>.ev(L<rsub|2>,v)>
<item><math|ev(L,\<alpha\>.v<rsub|1>+\<beta\>.v<rsub|2>)=L(\<alpha\>.v<rsub|1>+\<beta\>.v<rsub|2>)\<equallim\><rsub|linearity
of L>\<alpha\>.L(v<rsub|1>)+\<beta\>.L(v<rsub|2>)=\<alpha\>.ev(L,v<rsub|1>)+\<beta\>.ev(L,v<rsub|2>)>
<item><math|\<shortparallel\>ev(L,v)\<shortparallel\><rsub|Y>=\<shortparallel\>L(v)\<shortparallel\><rsub|Y>\<leqslant\>\<shortparallel\>L\<shortparallel\>\<shortparallel\>v\<shortparallel\><rsub|X>,=max(\<shortparallel\>L\<shortparallel\>,\<shortparallel\>v\<shortparallel\><rsub|X>)=\<shortparallel\>(L,v)\<shortparallel\><rsub|L(X,Y)\<times\>X>>
</enumerate>
</enumerate>
Now <math|(ev\<circ\>\<varphi\>\<circ\>O<rsub|v>)(x)=ev(\<varphi\>(O<rsub|v>(x)))=ev(\<varphi\>(x,v))=ev(Df(x),v)=Df(x)(v)=f<rsub|v>(x)>
and thus <math|f<rsub|v>=ev\<circ\>\<varphi\>\<circ\>O<rsub|v>> which
being the composition of differentiable functions of class
<math|C<rsup|r-1>> is differentiable by the Generalized Chain rule
(<reference|generalized chain rule>)
</proof>
<\lemma>
<label|(x,h)-\<gtr\>Df(x)(h) differentiability>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces
<math|U\<subseteq\>X> open and <math|f:U\<rightarrow\>Y> differentiable
of class <math|C<rsup|r>> then <math|\<Delta\>f:X\<times\>X\<rightarrow\>Y>
defined by <math|\<Delta\>f(x,h)=Df(x)(h)> is differentiable of class
<math|C<rsup|r-1>><math|>
</lemma>
<\proof>
First define <math|\<varphi\>:U\<times\>X\<rightarrow\>L(X,Y> by
<math|\<varphi\>(x,h)=(Df(x),h)>. If we then use the evaluation function
<math|ev> introduced in the proof of <reference|fv is differentiable>t
hen we have <math|\<Delta\>f=ev\<circ\>\<varphi\>>
[<math|ev\<circ\>\<varphi\>(x,h)=ev(Df(x),h)=Df(x)(h)=\<Delta\>f(x,h)>.
Now as <math|f> is of class <math|C<rsup|r>> we have by
<reference|proving f is Cn based on Cn-1 of Df> that
<math|\<pi\><rsub|1>\<circ\>\<varphi\>> is of class <math|C<rsup|r-1>>
and as <math|\<pi\><rsub|2>\<circ\>\<varphi\>=id<rsub|U>> is of class
<math|C<rsup|\<infty\>>> we have by <reference|c-r of mapping to product
> that <math|\<varphi\>> is of class <math|C<rsup|r-1>> and using the
generalized chain rule (<reference|generalized chain rule>) and the fact
that <math|ev> is <math|C<rsup|r>> we have that <math|\<Delta\>f> is of
class <math|C<rsup|r-1>>
</proof>
<\lemma>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed spaces over
<math|\<bbb-K\>>, <math|U\<subseteq\>X>, open, <math|f:U\<rightarrow\>Y>
and <math|L> be a linear functional on <math|Y> (<math|L> is a continuous
linear function <math|Y\<rightarrow\>\<bbb-K\>> then if <math|f> is
<math|C<rsup|r>> then <math|L\<circ\>f> is <math|C<rsup|r>> and
<math|D<rsup|r>(L\<circ\>f)=L\<circ\>D<rsup|r>f>
</lemma>
<\proof>
As <math|L> is <math|C<rsup|\<infty\>>> (see <reference|linear continuous
mappings are C infinity>) we have using the chain rule (see
<reference|generalized chain rule>) that <math|L\<circ\>f> is
<math|C<rsup|r>, to prove ><math|D<rsup|r>(L\<circ\>f)=L\<circ\>D<rsup|r>f>
we use induction and the chain rule again <math|\<forall\>x\<in\>U> we
have
<\enumerate>
<item><math|r=1> then <math|D(L\<circ\>f)(x)=DL(f(x))\<circ\>Df(x)\<equallim\><rsub|DL(y)=L>L\<circ\>Df(x)\<Rightarrow\>D(L\<circ\>f)=L\<circ\>f>
<item>Assume the theorem is true for <math|r> and assume that <math|f>
is <math|C<rsup|r+1>> then <math|L\<circ\>f> is <math|C<rsup|r+1>> and
<math|D<rsup|r+1>(L\<circ\>f)=D(D<rsup|r>(L\<circ\>f))=D(L\<circ\>D<rsup|r>f)\<equallim\><rsub|case
1>L\<circ\>D(D<rsup|r>f))=L\<circ\>D<rsup|r+1>f>
</enumerate>
</proof>
<\theorem>
<label|Cr and partial derivates>Let <math|{X<rsub|i>,\<shortparallel\>\<shortparallel\><rsub|i>}<rsub|i\<in\>{1,\<ldots\>,n}>>,
<math|Y,\<shortparallel\>\<shortparallel\>> be normed space,
<math|U\<subseteq\>X=<big|prod><rsub|i\<in\>{1,\<ldots\>,n}>X<rsub|i>> is
open and <math|f:U\<rightarrow\>Y> is differentiable of class
<math|C<rsup|r>>, <math|r\<geqslant\>1> then
<math|f<rsup|(i)>:X<rsub|i>\<rightarrow\>Y:x\<rightarrow\>f(x<rsub|1>,\<ldots\>,x<rsub|i-1>,x,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
is differentiable of class <math|C<rsup|r>> further <math|D<rsub|i>f> is
differentiable of class <math|C<rsup|r-1>>
</theorem>
<\proof>
\;
Note that <math|f<rsup|(i)>=f\<circ\>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,*\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
and using <math|<reference|partial coordinates are differentiable>> we
have that <math|D(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)(t)=(0<rsub|1>,\<ldots\>,0<rsub|i-1>,*\<ast\>,0<rsub|i+1>,\<ldots\>,0<rsub|n>)>
a linear continuous function independend <math|t> from which it follows
that <math|D<rsup|2>(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)=0>
so <math|(x<rsub|1>,\<ldots\>,x<rsub|i-1>,\<ast\>,x<rsub|i+1>,\<ldots\>,x<rsub|n>)>
is differentiable of class <math|C<rsup|\<infty\>>>, using the
generalized chainrule we have then that <math|f<rsup|(i)>> is
differentiable of class <math|C<rsup|r>>. If <math|r=1> then
<math|Df<rsup|(i)>=D<rsub|i>f> is continuous so of class
<math|C<rsup|0>=c<rsup|1-1>>. From <reference|proving f is Cn based on
Cn-1 of Df> we have that <math|D<rsub|i>f=Df<rsup|(i)>> is differentiable
of class <math|C<rsup|r-1>>.\
</proof>
<\theorem>
<label|product of Cr functions is Cr>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>, be normed spaces
over <math|\<bbb-K\>,U\<subseteq\>X> open and
<math|f,g:U\<rightarrow\>\<bbb-K\>> are functions differentiable of class
<math|C<rsup|r>,r\<geqslant\>1> then <math|f.g> is of class
<math|C<rsup|r>>
</theorem>
<\proof>
We proof this by recursion\
<\enumerate>
<item><math|r=1> then using <reference|differentiability of product> we
have <math|\<forall\>x\<in\>U> that
<math|D(f.g)(x)=f(x).Dg(x)+g(x).Df(x)> \ and thus
<math|D(f.g)=f.Dg+g.Df> is continuous (as <math|f,Dg,g,Df>) are
continuous by hypothese and differentiability. So <math|f.g> is
differentiable of class <math|C<rsup|1>>
<item>Assume the theorem is true for <math|r> and proof it for
<math|r+1>, so if <math|f,g> are differentiable of class
<math|C<rsup|r+1>> then <math|D(f.g)=f.Dg+g.Df>. Now <math|f,Dg,g,Df>
are differentiable of class <math|C<rsup|r>> so by the induction
hypothese we have <math|f.Dg,g.Df> are differentiable of class
<math|C<rsup|r>> and because <reference|sum of differentiable functions
is differentiable> we have that <math|D(f.g)> is differentiable of
class <math|C<rsup|r>> and thus by <reference|proving f is Cn based on
Cn-1 of Df> we have that <math|f.g> is differentiable of class
<math|C<rsup|r+1>>
</enumerate>
</proof>
<\theorem>
<index|differentiation under the integral sign><label|differentiation
under the integral sign>Let <math|\<bbb-R\><rsup|n>,\<shortparallel\>\<shortparallel\>>
be the set of real tuples together with the maximum norm defined on
<math|\<shortmid\>\<shortmid\>>, let <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
be a Banach space, <math|U\<subseteq\>\<bbb-R\><rsup|n>> open and
<math|f:U\<times\>[a,b]\<rightarrow\>Y,a\<less\>b> a function fullfilling
the following conditions
<\enumerate>
<item><math|f> is a continuous function
<item><math|\<forall\>t\<in\>[a,b]> the function
<math|f<rsub|2>(t):U\<rightarrow\>Y> defined by
<math|f<rsub|2>(t)(x)=f(x,t)> is differentiable of class
<math|C<rsup|r>,r\<geqslant\>1>
<item><math|\<forall\>i\<vdash\>1\<leqslant\>i\<leqslant\>r> the
function <math|f<rsup|(i)>:U\<times\>[a,b]\<rightarrow\>L<rsup|i>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y)):(x,t)\<rightarrow\>D<rsup|i>(f<rsub|2>(t))(x)>
is continuous
</enumerate>
then if we define <math|\<forall\>x\<in\>U>
<math|f<rsub|1>(x):[a,b]\<rightarrow\>Y> by
<math|t\<rightarrow\>f<rsub|1>(x)(t)=f(x,t)> then the function
<math|F(x):U\<rightarrow\>Y> defined by
<math|x\<rightarrow\>F(x)=<big|int><rsub|a><rsup|b>f<rsub|1>(x)=<big|int><rsub|a><rsup|b>f(x,t)dt>
is defined and differentiable of class <math|C<rsup|r>>. Furthermore
<math|D<rsup|r>F(x)=<big|int><rsub|a><rsup|b>D<rsup|r>(f<rsub|2>(t))(x)dt>
</theorem>
<\proof>
\;
(1) First we prove that <math|\<forall\>x\<in\>U> <math|F(x)> is well
defined. From the continuity of <math|f> and <reference|continuity of
function on a product> we have that <math|f<rsub|1>(x)> is continuous so
the integral <math|<big|int><rsub|a><rsup|b>f<rsub|1>(x)> is defined.
(2) We prove the rest of the theorem by induction on
<math|r\<geqslant\>1>
\;
<strong|(Base case, r=1)>
First using <reference|space of linear continuous maps to a Banach space
is Banach> we have that <math|L(\<bbb-R\><rsup|n>,Y)> is a Banach space.
Further <math|\<forall\>x\<in\>U> we have that
\ <math|f<rsup|(1)><rsub|1>(x):[a,b]\<rightarrow\>L(\<bbb-R\>,Y)> defined
by <math|t\<rightarrow\>f<rsup|(1)><rsub|1>(t)=D(f<rsub|2>(t))(x)=f<rsup|(1)>(x,t)>
is continuous by the continuity of <math|f<rsup|(1)>> and
<reference|continuity of function on a product>. So the integral
<math|<big|int><rsub|a><rsup|b>f<rsub|1><rsup|(1)>=<big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x)dt>
exists and is well defined. Now we have if <math|h\<in\>U<rsub|x>> that
for <math|\<Delta\>(x,h)> defined as follows we have:
<\eqnarray*>
<tformat|<table|<row|<cell|\<Delta\>(x,h)>|<cell|\<equallim\><rsub|defined>>|<cell|F(x+h)-F(x)-(<big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x)dt)(h)>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|integral
of function to linear operators>>>|<cell|<big|int><rsub|a><rsup|b>f(x+h,t)dt-<big|int><rsub|a><rsup|b>f(x,t)dt-<big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x)(h)dt>>|<row|<cell|>|<cell|=>|<cell|<big|int><rsub|a><rsup|b>[f(x+h,t)-f(x,t)-D(f<rsub|2>(t))(x)(h)dt>>>>
</eqnarray*>
As given <math|t\<in\>[a,b]> the differential fo <math|f<rsub|2>(t)>
exists we have that there exists a <math|\<varepsilon\>->function
<math|\<varepsilon\>(t)(h)> such that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(t)(h)\<shortparallel\>h\<shortparallel\>>|<cell|=>|<cell|f<rsub|2>(t)(x+h)-f<rsub|2>(t)(x)-D(f<rsub|2>(t))(x)(h)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|>>|<row|<cell|\<Delta\>(x,h)>|<cell|=>|<cell|<big|int><rsub|a><rsup|b>\<varepsilon\>(t)(h)\<shortparallel\>h\<shortparallel\>dt>>>>
</eqnarray*>
Next we prove that <math|\<forall\>x\<in\>U> the function
\ <math|\<varphi\><rsub|x>:U<rsub|x>\<times\>[a,b]\<rightarrow\>Y>
defined by <math|(h,t)\<rightarrow\>\<varphi\><rsub|x>(h,t)=D(f<rsub|2>(t))(x)(h)>
is continous. If <math|(h<rprime|'>,t<rprime|'>),(h,t)\<in\>U<rsub|x>\<times\>[a,b]>
then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varphi\><rsub|x>(h<rprime|'>,t<rprime|'>)-\<varphi\><rsub|x>(h,t)\<shortparallel\><rsub|Y>>|<cell|=>|<cell|\<shortparallel\>D(f<rsub|2>(t<rprime|'>))(x)(h<rprime|'>)-D(f<rsub|2>(t))(x)(h)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>D(f<rsub|2>(t<rprime|'>))(x)(h<rprime|'>)-D(f<rsub|2>(t))(x)(h<rprime|'>)\<shortparallel\><rsub|Y>+\<shortparallel\>D(f<rsub|2>(t))(x)(h<rprime|'>)-D(f<rsub|2>(t))(x)(h)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>D(f<rsub|2>(t<rprime|'>))(x)-D(f<rsub|2>(t))(x)\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>+\<shortparallel\>D(f<rsub|2>(t))(x)\<shortparallel\>\<shortparallel\>h<rprime|'>-h\<shortparallel\>>>>>
</eqnarray*>
Take now <math|\<varepsilon\>\<gtr\>0> choosen arbitrarely and prove
continuity for every <math|(h,t)\<in\>U<rsub|x>\<times\>[a,b]> From the
continuity of <math|f<rsub|1><rsup|(1)>> there exists a
<math|\<delta\><rsub|1>\<gtr\>0> such that <math|>
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>D(f<rsub|2>(t<rprime|'>))(x)-D(f<rsub|2>(t)(x)\<shortparallel\>\<less\><frac|\<varepsilon\>|2(\<shortparallel\>h\<shortparallel\>+1)>>|<cell|>|<cell|>>>>
</eqnarray*>
If we take then <math|\<delta\>=min(1,<frac|\<varepsilon\>|2(\<shortparallel\>D(f<rsub|2>(t))(x)\<shortparallel\>+1)>,\<delta\><rsub|1>)>
then if we have <math|\<shortparallel\>h<rprime|'>-h\<shortparallel\>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>D(f<rsub|2>(t))(x)\<shortparallel\>\<shortparallel\>h<rprime|'>-h\<shortparallel\>\<less\>\<shortparallel\>D(f<rsub|2>(t))(x)\<shortparallel\><frac|\<varepsilon\>|2(\<shortparallel\>D(f<rsub|2>(t)(x)\<shortparallel\>+1)>\<less\><frac|\<varepsilon\>|2>>
and also <math|\<shortparallel\>h<rprime|'>\<shortparallel\>\<leqslant\>\<shortparallel\>h<rprime|'>-h\<shortparallel\>+\<shortparallel\>h\<shortparallel\>\<less\>1+\<shortparallel\>h\<shortparallel\>>.
It follows then that if <math|\<shortparallel\>(h<rprime|'>,t<rprime|'>)-(h,t)\<shortparallel\><rsub|max>\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>h-h\<shortparallel\>\<less\>\<delta\>,\<shortmid\>t<rprime|'>-t\<shortmid\>\<less\>\<delta\>>
and
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varphi\><rsub|x>(h<rprime|'>,t)-\<varphi\><rsub|x>(h,t)\<shortparallel\><rsub|Y>>|<cell|\<leqslant\>>|<cell|<frac|\<varepsilon\>|2(\<shortparallel\>h\<shortparallel\>+1)>\<shortparallel\>h<rprime|'>\<shortparallel\>+<frac|\<varepsilon\>|2>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|\<varepsilon\>|2(\<shortparallel\>h\<shortparallel\>+1)>(\<shortparallel\>h\<shortparallel\>+1)+<frac|\<varepsilon\>|2>>>|<row|<cell|>|<cell|=>|<cell|<frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>>>>
</eqnarray*>
proving that <math|\<varphi\><rsub|x>> is indeed continuous.
Take now <math|\<varepsilon\><rprime|'>:U<rsub|x>\<times\>[a,b]\<rightarrow\>Y>
defined by <math|(h,t)\<rightarrow\>\<varepsilon\><rprime|'>(h,t)=\<varepsilon\>(t)(h)\<shortparallel\>h\<shortparallel\>=f<rsub|2>(t)(x+h)-f<rsub|2>(t)(x)-D(f<rsub|2>(t))(x)(h)=f(x+h,t)-f(x,t)-D(f<rsub|2>(t))(x)(h)=f(x+h,t)-f(x,t)-\<varphi\><rsub|x>(h,t)>.
<math|\<varepsilon\><rprime|'>> is continuous because of the continuity
\ <math|of > <math|f> and <math|\<varphi\><rsub|x>>.\
We prove now that from the continuity of <math|\<varepsilon\><rprime|'>>
Next we prove that <math|\<varepsilon\>:U<rsub|x>\<times\>[a,b]\<rightarrow\>Y>
defined by <math|(h,t)\<rightarrow\>\<varepsilon\>(h,t)=\<varepsilon\>(t)(h)>
is continuous. To do it is enough that we prove that
<math|\<varepsilon\>> is continuous at every
<math|(h,t)\<in\>U<rsub|x>\<times\>[a,b]>, we have two cases to consider
<\enumerate>
<item><math|(h,t)\<in\>(U<rsub|x><mid|\\>{0})\<times\>[a,b]>. As
<math|h\<neq\>0> we have <math|\<shortparallel\>h\<shortparallel\>\<gtr\>0>
so that <math|\<exists\>\<lambda\>> such that
<math|\<shortparallel\>h\<shortparallel\>\<gtr\>\<lambda\>\<gtr\>0>.
Take <math|\<delta\><rsub|1>=<frac|\<shortparallel\>h\<shortparallel\>-\<lambda\>|2>>
then if <math|h<rprime|'>\<in\>U<rsub|x>> with
<math|\<shortparallel\>h-h<rprime|'>\<shortparallel\>\<less\>\<delta\><rsub|1>>
then we proof by contradiction that
<math|\<shortparallel\>h<rprime|'>\<shortparallel\>\<gtr\>\<lambda\>\<gtr\>0>.
If <math|\<shortparallel\>h<rprime|'>\<shortparallel\>\<leqslant\>\<lambda\>>
then <math|\<shortparallel\>h\<shortparallel\>\<leqslant\>\<shortparallel\>h-h<rprime|'>\<shortparallel\>+\<shortparallel\>h<rprime|'>\<shortparallel\>\<less\><frac|\<shortparallel\>h\<shortparallel\>-\<lambda\>|2>+\<lambda\>\<Rightarrow\><frac|\<shortparallel\>h\<shortparallel\>|2>\<less\><frac|\<lambda\>|2>\<less\>\<lambda\>\<less\>\<shortparallel\>h\<shortparallel\>>
indeed a contradiction. Given <math|(h<rprime|'>,t<rprime|'>)> with
<math|h<rprime|'>\<neq\>0> then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varepsilon\>(h<rprime|'>,t<rprime|'>)-\<varepsilon\>(h,t)\<shortparallel\><rsub|Y>>|<cell|=>|<cell|\<shortparallel\><frac|\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)|\<shortparallel\>h<rprime|'>\<shortparallel\>>-<frac|\<varepsilon\><rprime|'>(h,t)|\<shortparallel\>h\<shortparallel\>>\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|=>|<cell|<frac|1|\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>\<shortparallel\>\<shortparallel\>h\<shortparallel\>\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)-\<shortparallel\>h<rprime|'>\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|1|\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>[\<shortparallel\>\<shortparallel\>h\<shortparallel\>\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)-\<shortparallel\>h\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>+\<shortparallel\>\<shortparallel\>h\<shortparallel\>\<varepsilon\><rprime|'>(h,t)-\<shortparallel\>h<rprime|'>\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|1|\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>[\<shortparallel\>h\<shortparallel\>\<shortparallel\>\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)-\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>+\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>\|\<shortparallel\>h\<shortparallel\>-\<shortparallel\>h<rprime|'>\<shortparallel\>\|>>>>
</eqnarray*>
Take now <math|\<varepsilon\>\<gtr\>0>. By the continuity of
<math|\<varepsilon\><rprime|'>> we can choose a
<math|\<delta\><rsub|2>\<gtr\>0> such that
<math|\<shortparallel\>\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)-\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>\<less\><frac|\<varepsilon\>.\<lambda\>|2>\<Rightarrow\><frac|1|\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>\<shortparallel\>h\<shortparallel\>\<shortparallel\>\<varepsilon\><rprime|'>(h<rprime|'>,t<rprime|'>)-\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>\<leqslant\><frac|\<varepsilon\>.\<lambda\>|2\<shortparallel\>h<rprime|'>\<shortparallel\>>\<less\><frac|\<varepsilon\>|2>>.
Take now <math|\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>,<frac|\<varepsilon\>\<lambda\><rsup|2>|2(\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>+1)>>
then if <math|\<shortparallel\>(h<rprime|'>,t<rprime|'>)-(h,t)\<shortparallel\><rsub|max>\<less\>\<delta\>>
we have also <math|\<shortparallel\>h<rprime|'>-h\<shortparallel\>\<less\>min(\<delta\><rsub|1>,<frac|\<varepsilon\>\<lambda\><rsup|2>|2(\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>+1)>>
we have
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varepsilon\>(h<rprime|'>,t<rprime|'>)-\<varepsilon\>(h,t)\<shortparallel\><rsub|Y>>|<cell|\<less\>>|<cell|<frac|\<varepsilon\>|2>+<frac|\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>\|\<shortparallel\>h\<shortparallel\>-\<shortparallel\>h<rprime|'>\<shortparallel\>\||\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|\<varepsilon\>|2>+<frac|\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>\<shortparallel\>h-h<rprime|'>\<shortparallel\>|\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>\<shortparallel\>>>>|<row|<cell|>|<cell|\<less\>>|<cell|<frac|\<varepsilon\>|2>+\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><frac|\<varepsilon\>\<lambda\><rsup|2>|2\<shortparallel\>h\<shortparallel\>\<shortparallel\>h<rprime|'>(\<shortparallel\>\<varepsilon\><rprime|'>(h,t)\<shortparallel\><rsub|Y>+1)>>>|<row|<cell|>|<cell|\<less\>>|<cell|<frac|\<varepsilon\>|2>+<frac|\<varepsilon\>|2>=\<varepsilon\>>>>>
</eqnarray*>
proving continuity at <math|(h,t)>
\;
<item><math|(0,t)\<in\>{0}\<times\>[a,b]> then given a
<math|\<varepsilon\>\<gtr\>0> we have by the continuity of
<math|\<varepsilon\>(t)> at <math|0> with <math|\<varepsilon\>(t)(0)=0>
that <math|\<exists\>\<delta\>(t)\<gtr\>0> such that
<math|\<shortparallel\>\<varepsilon\>(t)(h<rprime|'>)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\>(t)(h<rprime|'>)-\<varepsilon\>(t)(0)\<shortparallel\><rsub|Y>\<less\>\<varepsilon\>>
if <math|\<shortparallel\>h<rprime|'>-h\<shortparallel\>\<less\>\<delta\>>
and thus <math|\<shortparallel\>\<varepsilon\>(h<rprime|'>,t<rprime|'>)-\<varepsilon\>(0,t)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\>(h<rprime|'>,t<rprime|'>)\<shortparallel\><rsub|Y>\<less\>\<varepsilon\>>
if <math|\<shortparallel\>(h<rprime|'>,t<rprime|'>)-(h,t)\<shortparallel\>\<less\>\<delta\>(t<rprime|'>)>
proving continuity at <math|(0,t)>
</enumerate>
Given now a <math|x\<in\>U> then we can find a <math|\<delta\>\<gtr\>0>
such that <math|0\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(0,\<delta\>)\<subseteq\><wide|B<rsub|\<shortparallel\>\<shortparallel\>>|\<bar\>>(0,\<delta\>)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\>>(0,2\<delta\>)\<subseteq\>U<rsub|x>>
then the set is <math|K=<wide|B<rsub|\<shortparallel\>\<shortparallel\>>|\<bar\>>(x,\<delta\>)\<times\>[a,b]>
is compact (because of <reference|product of compact
subspaces>,<reference|[a,b] is compact>,<reference|compact subsets of the
reals>) and thus by <reference|continuous functions on a compact set are
uniform continuous> we have that <math|\<varepsilon\>> is uniform
continuous on <math|K>.\
So given a <math|\<varepsilon\>\<gtr\>0> there exist a
<math|\<delta\><rprime|'>\<gtr\>0> such that for
<math|\<forall\>(h,t)\<in\>K> we have if
<math|(h<rprime|'>,t<rprime|'>)\<in\>K> is such that
<math|\<shortparallel\>(h<rprime|'>,t<rprime|'>)-(h,t)\<shortparallel\><rsub|max>
then ><math|\<shortparallel\>\<varepsilon\>(h<rprime|'>,t<rprime|'>)-\<varepsilon\>(h,t)\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>>.
More specifically <math|\<forall\>t\<in\>[a,b]> we have if
<math|\<shortparallel\>h\<shortparallel\>\<less\>min(\<delta\>,\<delta\><rprime|'>)>
then <math|\<shortparallel\>(h,t)-(0,t)\<shortparallel\><rsub|max>=\<shortparallel\>(h,0)\<shortparallel\><rsub|max>\<less\>\<delta\><rprime|'>>
and thus <math|\<shortparallel\>\<varepsilon\>(h,t)\<shortparallel\><rsub|Y>=\<shortparallel\>\<varepsilon\>(h,t)-\<varepsilon\>(0,t)\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>>
from which it follows that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\><big|int><rsub|a><rsup|b>\<varepsilon\>(h,t)dt\<shortparallel\><rsub|Y>>|<cell|\<leqslant\>>|<cell|<big|int><rsub|a><rsup|b>\<shortparallel\>\<varepsilon\>(h,t)\<shortparallel\><rsub|Y>dt>>|<row|<cell|>|<cell|\<less\>>|<cell|<big|int><rsub|a><rsup|b><frac|\<varepsilon\>|b-a>dt>>|<row|<cell|>|<cell|=>|<cell|(b-a)<frac|\<varepsilon\>|b-a>=\<varepsilon\>>>>>
</eqnarray*>
this together with\
<\eqnarray*>
<tformat|<table|<row|<cell|<big|int><rsub|a><rsup|b>\<varepsilon\>(0,t)>|<cell|=>|<cell|<big|int><rsub|a><rsup|b>0dt>>|<row|<cell|>|<cell|=>|<cell|<big|int><rsub|a><rsup|b>0dt>>|<row|<cell|>|<cell|=>|<cell|(b-a)0=0>>>>
</eqnarray*>
proves that <math|\<varepsilon\>:U<rsub|x>\<rightarrow\>Y> defined by
<math|h\<rightarrow\><big|int><rsub|a><rsup|b>\<varepsilon\>(h,t)dt> is a
<math|\<varepsilon\>-function> and thus that <math|F> is differentiable
on <math|U> and that <math|Df(x)=<big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x)>.
Now to prove that <math|F> is <math|C<rsup|1>> we must prove that
<math|DF:U\<rightarrow\>L(\<bbb-R\>,Y)> is continuous.
Let <math|x\<in\>U> then we can find a <math|\<delta\>\<gtr\>0> such that
<math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)\<subseteq\><wide|B<rsub|\<shortparallel\>\<shortparallel\>>|\<bar\>>(x,\<delta\>)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,2\<delta\>)\<subseteq\>U>
then the set is <math|K=<wide|B<rsub|\<shortparallel\>\<shortparallel\>>|\<bar\>>(x,\<delta\>)\<times\>[a,b]>
is compact (because of <reference|product of compact
subspaces>,<reference|[a,b] is compact>,<reference|compact subsets of the
reals>). From the continuity of <math|f<rsup|(1)>> we have then that
<math|f<rsup|(1)>> is unform continuous on <math|K>. So
<math|\<forall\>\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\><rprime|'>\<gtr\>0> such that if
<math|\<shortparallel\>(x<rprime|'>,t<rprime|'>)-(x<rprime|''>,t<rprime|''>)\<shortparallel\><rsub|max>\<less\>\<delta\><rprime|'>>
then <math|\<shortparallel\>f<rsup|(1)>(x<rprime|'>,t<rprime|'>)-f<rsup|(1)>(x<rprime|''>,t<rprime|''>)\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>>.
If we then take <math|\<shortparallel\>x-x<rprime|'>\<shortparallel\>\<less\>\<delta\><rprime|'>>
we have <math|\<shortparallel\>(x,t)-(x<rprime|'>,t)\<shortparallel\><rsub|max>=\<shortparallel\>(x-x<rprime|'>,0)\<shortparallel\><rsub|max>\<less\>\<delta\><rprime|'>>
and thus <math|\<shortparallel\>D(f<rsub|2>(t))(x<rprime|'>)-D(f<rsub|2>(t)(x)\<shortparallel\>=\<shortparallel\>f<rsup|(1)>(x<rprime|'>,t)-f<rsup|(1)>(x,t)\<shortparallel\>\<less\><frac|\<varepsilon\>|b-a>>
form which it follows that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>DF(x<rprime|'>)-DF(x)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\><big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x<rprime|'>)dt-<big|int><rsub|a><rsup|b>D(f<rsub|2>(t))(x)dt\<shortparallel\>>>|<row|<cell|>|<cell|=>|<cell|\<shortparallel\><big|int><rsub|a><rsup|b>(D(f<rsub|2>(t))(x<rprime|'>)-D(f<rsub|2>(t))(x))dt\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<big|int><rsub|a><rsup|b>\<shortparallel\>D(f<rsub|2>(t))(x<rprime|'>)-D(f<rsub|2>(t))(x)\<shortparallel\>dt>>|<row|<cell|>|<cell|\<less\>>|<cell|<big|int><rsub|a><rsup|b><frac|\<varepsilon\>|b-a>dt>>|<row|<cell|>|<cell|=>|<cell|\<varepsilon\>>>>>
</eqnarray*>
proving that <math|DF> is continuous and the end of the first induction
step.
\;
<strong|(Induction step, if the theorem is true for <math|k=r> we prove
it to be true for case <math|r+1>)>
Define <math|g:U\<times\>[a,b]\<rightarrow\>L<rsup|r>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))>
by <math|(x,t)\<rightarrow\>g(x,t)=D<rsup|r>(f<rsub|2>(t))(x)> then
<math|g=f<rsup|r>> and because <math|1\<leqslant\>r\<less\>r+1> we have
by the hypothese that <math|g> is continuous. Also if we define
<math|\<forall\>t\<in\>[a,b]>
<\eqnarray*>
<tformat|<table|<row|<cell|g<rsub|2>(t):U\<rightarrow\>L<rsup|r>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))>|<cell|by>|<cell|x\<rightarrow\>g<rsub|2>(t)=g(x,t)=D<rsup|r><rsub|>(f<rsub|2>(t))(x)>>>>
</eqnarray*>
then because of our induction hypothese we have that <math|g<rsub|2>> is
differentiable of class <math|C<rsup|1>>.
Also <math|g<rsup|(1)>> defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|g<rsup|(1)>:U\<times\>[a,b]\<rightarrow\>L<rsup|r+1>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))>|<cell|by>|<cell|
(x,t)\<rightarrow\>g<rsup|(1)>(x)=D(g<rsub|2>(t)(x))=D(D<rsup|r>(f<rsub|2>(t))(x)=D<rsup|r+1>(f<rsub|2>(t))(x)>>>>
</eqnarray*>
so that we have <math|g<rsup|(1)>=f<rsup|(r+1)>> and is continous by our
hypotheses. Finally note that <math|L<rsup|r+1>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))=L(\<bbb-R\><rsup|n>,L<rsup|r>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))>
so that all conditions are fullfilled for the case <math|r=1> applied on
<math|g>. So we have that <math|G:U\<rightarrow\>L<rsup|r>(\<bbb-R\><rsup|n>,\<ldots\>,L(\<bbb-R\><rsup|n>,Y))>
defined by <math|x\<rightarrow\>G(x)=<big|int><rsub|a><rsup|b>g<rsub|2>(t)(x)dt=<big|int><rsub|a><rsup|b>D<rsup|r>(f<rsub|2>(t))(x)dt>
is differentiable of class <math|C<rsup|1>> and that
<math|DG(x)=<big|int><rsub|a><rsup|b>D(g<rsub|2>(t))(x)=<big|int><rsub|a><rsup|b>D(D<rsup|r>(f<rsub|2>(t)))(x)dt=<big|int><rsub|a><rsup|b>D<rsup|r+1>(f<rsub|2>(t))(x)dt>.
Now by the induction step we have that
<math|<big|int><rsub|a><rsup|b>D<rsup|r>(f<rsub|2>(t))(x)dt=D<rsup|r>F(x)>
so that we have that as \ <math|DG(x)> is defined also that
<math|DG(x)=D(D<rsup|r>F(x))=D<rsup|r+1>F(x)=<big|int><rsub|a><rsup|b>D<rsup|r+1>(f<rsub|2>(t))(x)dt>.
Finally as <math|DG=D<rsup|r+1>f> is continuous we have the <math|F> is
<math|C<rsup|r+1>> which prooves out little theorem.\
\;
</proof>
<section|Differentiability on general sets>
<\definition>
<label|Cr on arbitrary sets><index|<math|C<rsup|r>> on set>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\>> be normed spaces over
<math|\<bbb-K\>> <math|S\<subseteq\>X> and <math|f:S\<rightarrow\>Y> then
<math|f> is <math|C<rsup|r>> if there exists a <math|U> open such that
<math|S\<subseteq\>U\<subseteq\>X> and a
<math|f<rsup|U>:U\<rightarrow\>Y> with <math|f<rsup|U><rsub|\|S>=f> such
that <math|f<rsup|U>> is <math|C<rsup|r>>. In other words <math|f> is
<math|C<rsup|r>> if it can be extended \ to a <math|C<rsup|r>> function
on a open set containing <math|S>
</definition>
<\theorem>
<label|composition of derivates on arbitrary sets>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>>,
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>,
Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be normed spaces over
<math|\<bbb-K\>>. <math|S\<subseteq\>X,T\<subseteq\>Y> and
<math|f:S\<rightarrow\>Y,g:T\<rightarrow\>Z> are <math|C<rsup|r>> on
<math|S,T> and <math|f(S)\<subseteq\>T> then <math|g\<circ\>f> is
<math|C<rsup|r>> on <math|S>
</theorem>
<\proof>
By definition there exists a <math|U> open in <math|X> and a <math|V>
open in <math|Y> such that <math|S\<subseteq\>U,T\<subseteq\>V> and
<math|f<rsup|U>> and <math|f<rsup|V>> are <math|C<rsup|r>> and
<math|f<rsup|U><rsub|\|S>=f,g<rsup|V><rsub|\|T>=g>. Take then
<math|W=U<big|cap>g<rsup|-1>(V)> which is open because <math|g> is
continuous on <math|V> because it is differentiable on <math|V>. As
<math|g(S)\<subseteq\>T\<subseteq\>V\<Rightarrow\>S\<subseteq\>g<rsup|-1>(V)\<Rightarrow\>S\<subseteq\>W>
and we have then <math|g(W)\<subseteq\>g(g<rsup|-1>(V))\<subseteq\>V>, if
we the define <math|f<rsup|W>:W\<rightarrow\>Y> by
<math|f<rsup|W>=f<rsup|U><rsub|\|W>> which is trivially again
<math|C<rsup|r>> and is also a extension of <math|f> (as
<math|S\<subseteq\>W>). We use then the chain rule (see
<reference|generalized chain rule>) to prove that
<math|g<rsup|V>\<circ\>f<rsup|W>> is <math|C<rsup|r>> on <math|W> and as
<math|(g<rsup|V>\<circ\>f<rsup|W>)<rsub|\|S>=g\<circ\>f> we have found
that <math|g\<circ\>f> is indeed <math|C<rsup|r>> on <math|S>
</proof>
<\theorem>
<label|differentiability on [a,b]>Let
<math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b> and
<math|Y\<comma\>\<shortparallel\>\<shortparallel\>> be a normed space
over <math|\<bbb-R\>> then the following are equivalent for
<math|f:[a,b]\<rightarrow\>Y>
<\enumerate>
<item><math|><math|f> is <math|C<rsup|1>> on <math|[a,b]>
<item><math|>We have the following
<\enumerate>
<item><math|\<forall\>t\<in\>]a,b[> we have the existence of a
derivative <math|f<rsub|\|]a,b[><rprime|'>(t)=<frac|df<rsub|\|]a,b[>(x)|dx>(t)>
<item>There exists a right, left derivative at
<math|f<rprime|'><rsub|+>(a)>, <math|f<rprime|'><rsub|->(b)> at
<math|a,b>
<item><math|f<rprime|'>:[a,b]> defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|f<rprime|'>(a)>|<cell|=>|<cell|f<rprime|'><rsub|+>(a)>>|<row|<cell|f<rprime|'>(b)>|<cell|=>|<cell|f<rprime|'><rsub|->(b)>>|<row|<cell|f<rprime|'>(t)>|<cell|=>|<cell|f<rprime|'><rsub|\|]a,b[>(t),
t\<in\>]a,b[>>>>
</eqnarray*>
is continue on <math|[a,b]>
</enumerate>
</enumerate>
Where the left and right derivative's are defined by
<\enumerate>
<item><math|f<rprime|'><rsub|+>(a)> is the right derivative of <math|f>
at <math|a> if and only if <math|\<forall\>\<varepsilon\>\<gtr\>0>
there exists a <math|\<delta\>\<gtr\>0> such that for all <math|h> with
<math|0\<less\>h\<less\>\<delta\>> and <math|a+h\<in\>[a,b]> we have
<math|\<shortparallel\><frac|f(a+h)-f(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>\<less\>\<varepsilon\>>
<item><math|f<rprime|'><rsub|->(a)> is the left derivative of <math|f>
at <math|b> if and only if <math|\<forall\>\<varepsilon\>\<gtr\>0>
there exists a <math|\<delta\>\<gtr\>0> such that for all <math|h> with
<math|0\<less\>h\<less\>\<delta\>> and <math|a-h\<in\>[a,b]> we have
<math|\<shortparallel\><frac|f(b)-f(b-h)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>\<less\>\<varepsilon\>>
</enumerate>
Further if if <math|f> is <math|C<rsup|1>> then if <math|U> open such
that <math|[a,b]\<subseteq\>U> \ and <math|f<rsup|U>> such that
<math|f<rsup|U><rsub|\|[a,b]>=f>, then
<math|f<rsup|U><rprime|'><rsub|\|[a,b]>=f<rprime|'>>
</theorem>
<\proof>
The proof is trivial but rather elaborated\
<\enumerate>
<item><math|1\<Rightarrow\>2> By definition we have a open set
<math|U\<subseteq\>\<bbb-R\>> with <math|[a,b]\<subseteq\>U> such that
<math|f<rsup|U>:U\<rightarrow\>Y> is <math|C<rsup|1>>. Then using
<reference|C1 of real or complex functions> we have that the derivative
<math|f<rsup|U<rprime|'>>:U\<rightarrow\>Y> is defined on every
<math|U> and is continuous on <math|U> and
<math|Df<rsup|U>(x)(1)=f<rsup|U<rprime|'>>(x)>. We proceed now to prove
a,b,c
<\enumerate>
<item>Now if <math|t\<in\>]a,b[\<subseteq\>U> then given
<math|\<varepsilon\>\<gtr\>0> there exists a <math|\<delta\>\<gtr\>0>
such that <math|\<forall\>h\<in\>U<rsub|x>\<vdash\>0\<less\>\|h\|\<less\>\<delta\>>
we have <math|\<shortparallel\><frac|f<rsup|U>(x+h)-f<rsup|U>(x)|h>-f<rsup|U<rprime|'>>(x)\<shortparallel\>\<less\>\<varepsilon\>>,
but then <math|\<forall\>h\<in\>]a,b[<rsub|x>\<subseteq\>U<rsub|x>>
we have <math|\<shortparallel\><frac|f<rsub|\|]a,b[>(x+h)-f<rsub|\|]a,b[>(x)|h>-f<rsup|U><rprime|'>(x)\<shortparallel\>=\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rsup|U><rprime|'>\<shortparallel\>=\<shortparallel\><frac|f<rsup|U>(x+h)-f<rsup|U>(x)|h>-f<rsup|U><rprime|'>(x)\<shortparallel\>\<less\>\<varepsilon\>>
proving that <math|f> has a derivative at <math|x\<in\>]a,b[> and
<math|f<rprime|'>(x)=f<rsup|U><rprime|'>(x)>.
<item>Let <math|\<varepsilon\>\<gtr\>0> then\
<\enumerate>
<item>As <math|a\<in\>U> there exists a <math|\<delta\>\<gtr\>0>
such that <math|\<forall\>h\<in\>U<rsub|a>\<vdash\>0\<less\>\|h\|\<less\>\<delta\>>
we have <math|\<\|\|\><frac|f<rsup|U>(a+h)-f<rsup|U>(a)|h>-f<rsup|U><rprime|'>(a)\<shortparallel\>\<less\>\<varepsilon\>>.
So if <math|h> is such that <math|0\<less\>h\<less\>\<delta\>> and
<math|a+h\<in\>[a,b]\<subseteq\>U<rsub|a>\<Rightarrow\>0\<less\>\|h\|\<less\>\<delta\>>
and <math|h\<in\>U<rsub|x>\<Rightarrow\>\<shortparallel\><frac|f(a+h)-f(a)|h>-f<rsup|U><rprime|'>(a)\<shortparallel\>=\<shortparallel\><frac|f<rsup|U>(a+h)-f<rsup|U>(a)|h>-f<rsup|U><rprime|'>(a)\<shortparallel\>\<less\>\<varepsilon\>>
and thus <math|f<rprime|'><rsub|+>(a)> exists and
<math|f<rprime|'><rsub|+>(a)=f<rsup|U<rprime|'>>(a)>
<item>As <math|b\<in\>U> there exists a <math|\<delta\>\<gtr\>0>
such that <math|\<forall\>h\<in\>U<rsub|b>\<vdash\>0\<less\>\|h\|\<less\>\<delta\>>
we have <math|\<shortparallel\><frac|f<rsup|U>(b+h)-f<rsup|U>(b)|h>-f<rsup|U><rprime|'>(b)\<shortparallel\>\<less\>\<varepsilon\>>.
So if <math|h> is such that <math|0\<less\>h\<less\>\<delta\>> and
<math|b+(-h)=b-h\<in\>[a,b]\<subseteq\>U\<Rightarrow\>0\<less\>\|-h\|=h\<less\>\<delta\>,-h\<in\>U<rsub|b>\<Rightarrow\>\<shortparallel\><frac|f(b-h)-f(b)|h>-f<rsup|U><rprime|'>(b)\<shortparallel\>=\<shortparallel\><frac|f<rsup|U>(b-h)-f<rsup|U>(b)|h>-f<rsup|U><rprime|'>(b)\<shortparallel\>\<less\>\<varepsilon\>>
and thus <math|f<rprime|'><rsub|->(b)=f<rsup|U<rprime|'>>(b)>
</enumerate>
</enumerate>
So we have proved a,b but also that
<math|f<rprime|'>:[a,b]\<rightarrow\>U> is \ equal to
<math|f<rsup|U><rprime|'><rsub|\|[a,b]>> and thus is continuous (see
<reference|continuity of restricted maps>).
<item><math|2\<Rightarrow\>1> Given <math|f> defined on <math|[a,b]>
fulfilling (2) define <math|f<rsup|]a-1,b+1[>> on
<math|]a-1,b+1[\<supseteq\>[a,b]> by\
<\eqnarray*>
<tformat|<table|<row|<cell|f<rsup|]a-1,b+1[>(x)=>|<cell|f(x)>|<cell|x\<in\>[a,b]>>|<row|<cell|>|<cell|f(a)+f<rprime|'><rsub|+>(a)(a-x)>|<cell|x\<in\>]a-1,a[>>|<row|<cell|>|<cell|f(b)+f<rprime|'><rsub|->(b)(x-b)>|<cell|x\<in\>]b,b+1[>>>>
</eqnarray*>
then we first prove that <math|f<rsup|]a-1,b+1[>> has a derivative at
every <math|x\<in\>]a-1,b+1[>. We have the following cases to consider
<\enumerate>
<item><math|x=a> Given a <math|\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\>\<gtr\>0> such that if
<math|0\<less\>h\<less\>\<delta\>> and <math|a+h\<in\>[a,b]> then
<math|\<shortparallel\><frac|f(a+h)-f(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>\<less\>\<varepsilon\>>.
Take then <math|\<delta\><rprime|'>=min(\<delta\>,b-a)\<gtr\>0> [as
<math|a\<less\>b>]. Then if <math|0\<less\>\|h\|\<less\>\<delta\><rprime|'>>
and <math|h\<in\>]a-1,b+1[<rsub|a>> we have then two sub-cases
<\enumerate>
<item><math|h\<less\>0> then <math|a+h\<in\>]a-1,a[><math|\<shortparallel\><frac|f<rsup|]a-1,b+1[>(a+h)-f<rsup|]a-1,b+1[>(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\><frac|f(a)+f<rprime|'><rsub|+>(a)(a-(a+h))-f(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\><frac|f<rprime|'><rsub|+>(a)h|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=0\<less\>\<varepsilon\>>
<item><math|h\<gtr\>0> then <math|0\<less\>h\<less\>\<delta\><rprime|'>\<leqslant\>b-a\<Rightarrow\>a\<less\>a+h\<leqslant\>b\<Rightarrow\>a+h\<in\>[a,b]\<Rightarrow\>\<shortparallel\><frac|f<rsup|]a-1,b+1[>(a+h)-f<rsup|]a-1,b+1[>(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\><frac|f(a+h)-f(a)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>\<less\>\<varepsilon\>>
</enumerate>
so we have proved that <math|f<rsup|]a-1,b+1[><rprime|'>(a)> exists
at <math|a> and is equal to <math|f<rprime|'><rsub|+>(a)>
<item><math|x=b> Given a <math|\<varepsilon\>\<gtr\>0> there exists a
<math|\<delta\>\<gtr\>0> such that if
<math|0\<less\>h\<less\>\<delta\>> and <math|b-h\<in\>[a,b]> then
<math|\<shortparallel\><frac|f(b)-f(b-h)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>\<less\>\<varepsilon\>>.
Take then <math|\<delta\><rprime|'>=min(\<delta\>,b-a)\<gtr\>0>. Then
if <math|0\<less\>\|h\|\<less\>\<delta\><rprime|'>> and
<math|h\<in\>]a-1,b+1[<rsub|b>>, we have then two sub-cases
<\enumerate>
<item><math|h\<less\>0>, then <math|0\<less\>\<um\>h=\|h\|\<less\>\<delta\><rprime|'>\<leqslant\>b-a,\<delta\>\<Rightarrow\>a-h\<leqslant\>b\<Rightarrow\>a\<leqslant\>b+h=b-\|h\|\<less\>b\<Rightarrow\>b+h=b-\|h\|\<in\>[a,b]\<Rightarrow\>\<shortparallel\><frac|f<rsup|]a-1,b+1[>(b+h)-f<rsup|]1-1,b+1[>(b)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f(b-\|h\|)-f(b)|-\|h\|>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f(b)-f(b-\|h\|)|\|h\|>-f<rprime|'><rsub|->(b)\<shortparallel\>\<less\>\<varepsilon\>>
<item><math|h\<gtr\>0> then <math|b+h\<in\>]b,b+1[> and
<math|\<shortparallel\><frac|f<rsup|]a-1,b+1[>(b+h)-f<rsup|]a-1,b+1[>(b)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f(b)+f<rprime|'><rsub|->(b)(b+h-b)-f(b)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f<rprime|'><rsub|->(b).h|h>-f<rprime|'>(b)\<shortparallel\>=0\<less\>\<varepsilon\>>
</enumerate>
so we have proved that <math|f<rsup|]a-1,b+1[><rprime|'>(b)> exists
at <math|b> and is equal to <math|f<rsub|-><rprime|'>(b)>
<item><math|x\<in\>]a-1,a[> Given <math|\<varepsilon\>\<gtr\>0> take
then <math|\<delta\>=a-x\<gtr\>0> then for a <math|h> such that
<math|0\<less\>\|h\|\<less\>\<delta\>> and
<math|x\<in\>]a-1,b+1[<rsub|x>> we have <math|x+h\<in\>]a-1,b+1[> and
<math|x+h\<leqslant\>x+\|h\|\<less\>x+\<delta\>=x+(a-x)=a\<Rightarrow\>x+h\<in\>]a-1,a[>
. Then <math|\<\|\|\><frac|f<rsup|]a-1,b+1[>(x+h)-f<rsup|]a-1,b+1[>(x)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\><frac|f(a)+f<rprime|'><rsub|+>(a)(a-x-h)-(f(a)+f<rprime|'><rsub|+>(a)(a-x)|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\><frac|f<rprime|'><rsub|+>(a)h|h>-f<rprime|'><rsub|+>(a)\<shortparallel\>=0\<less\>\<varepsilon\>>.
So we have proved that <math|f<rsup|]a-1,b+1[><rprime|'>(x)> exists
and is equal to <math|f<rprime|'><rsub|+>(a)>
<item><math|x\<in\>]b,b+1[> Given <math|\<varepsilon\>\<gtr\>0> take
then <math|\<delta\>=1\<gtr\>0> then for <math|h> such that
<math|0\<less\>\|h\|\<less\>\<delta\>> and
<math|x\<in\>]a-1,b+1[<rsub|x>> we have <math|x+h\<in\>]b,b+1[> so
that <math|\<shortparallel\><frac|f<rsup|]a-1,b+1[>(x+h)-f<rsup|]a-1,b+1[>(x)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f(b)+f<rprime|'><rsub|->(b)(x+h-b)-(f(b)+f<rprime|'><rsub|->(b)(x-b)|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\><frac|f<rprime|'><rsub|->(b)h|h>-f<rprime|'><rsub|->(b)\<shortparallel\>=0\<less\>\<varepsilon\>>
proving that <math|f<rsup|]a-1,b+1[>(x)> exists and is equal to
<math|f<rprime|'><rsub|->(b)>
<item><math|x\<in\>]a,b[>. Given a <math|\<varepsilon\>\<gtr\>0> we
can find then a <math|\<delta\>> such that if
<math|0\<less\>\|h\|\<less\>\<delta\>> and
<math|h\<in\>]a,b[<rsub|x>> then <math|\<shortparallel\><frac|f<rsub|\|]a,b[>(x+h)-f<rsub|]a,b[>(x)|h>-f<rprime|'><rsub|\|]a,b[>(x)\<shortparallel\>\<less\>\<varepsilon\>>.
Then if <math|\<delta\><rprime|'>=min(\<delta\>,x-a,b-x)> we have if
<math|0\<less\>\|h\|\<less\>\<delta\><rprime|'>> then we have
<math|0\<less\>\|h\|\<less\>x-a,b-x> so and thus
<math|a\<less\>x-\|h\|\<leqslant\>x+h> and
<math|x+h\<leqslant\>x+\|h\|\<less\>x+b-x=b> (so
<math|x+h\<in\>]a,b[>). So it follows that
<math|\<shortparallel\><frac|f<rsup|]a-1,b+1[>(x+h)-f<rsup|]a-1,b+1[>(x)|h>-f<rprime|'><rsub|\|]a,b[>(x)\<shortparallel\>=\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rprime|'><rsub|\|]a,b[>(x)\<shortparallel\>=\<shortparallel\><frac|f<rsub|\|]a,b[>(x+h)-f<rsub|]a,b[>(x)|h>-f<rprime|'><rsub|]a,b[>(x)\<shortparallel\>\<less\>\<varepsilon\>>
proving that <math|f<rsup|]a-1,b+1[>(x)> exists and is equal to
<math|f<rprime|'><rsub|\|]a,b[>(x)>
</enumerate>
So we have proved that <math|f<rsup|]a-1,b+1><rprime|'>(x)> exists for
all <math|x\<in\>]a-1,b+1[> and is equal to
<math|f<rprime|'><rsub|+>(a)> on <math|]a-1,a]>,
<math|f<rprime|'><rsub|->(b)> on <math|[b,b+1[> and to
<math|f<rprime|'><rsub|\|]a,b[>(x)> on <math|]a,b[>. Now we must prove
that it is continue at every <math|x\<in\>]a-1,b+1[>. We have the
following cases to consider
<\enumerate>
<item><math|x=a> As <math|f<rprime|'>> is continuous on <math|[a,b]>
we have <math|\<forall\>\<varepsilon\>\<gtr\>0> the existence of a
<math|\<delta\>\<gtr\>0> such that if <math|y\<in\>[a,b]> and
<math|\|y-a\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f<rprime|'>(y)-f<rprime|'>(a)\<shortparallel\>\<less\>\<varepsilon\>>.
So lets take <math|\<delta\><rprime|'>=min(\<delta\>,<frac|b-a|2>)>
and take <math|y> such that <math|\|a-y\|\<less\>\<delta\><rprime|'>>
then we have the following cases
<\enumerate>
<item><math|y\<less\>a> Then <math|\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(y)-f<rsup|]a-1,b+1[><rprime|'>(a)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|+>(a)-f<rprime|'><rsub|+>(a)\<shortparallel\>=0\<less\>\<varepsilon\>>
<item><math|a\<less\>y> Then as
<math|\|a-y\|\<less\>\<delta\><rprime|'>\<leqslant\><frac|b-a|2>\<Rightarrow\>y-a\<leqslant\><frac|b-a|2>\<Rightarrow\>y\<leqslant\><frac|b+a|2>\<less\>b>
and then <math|\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(y)-f<rsup|]a-1,b+1[><rprime|'>(a)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|\|]a-b[>(y)-f<rprime|'><rsub|+>(a)\<shortparallel\>=\<shortparallel\>f<rprime|'>(y)-f<rprime|'>(a)\<shortparallel\>\<less\>\<varepsilon\>>
</enumerate>
This proves continuity at <math|x=a>
<item><math|x=b> As <math|f<rprime|'>> is continuous on <math|]a,b[>
we have <math|\<forall\>\<varepsilon\>\<gtr\>0> the existence of a
<math|\<delta\>\<gtr\>0> such that if <math|y\<in\>[a,b]> and
<math|\|y-b\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f<rprime|'>(y)-f<rprime|'>(b)\<shortparallel\>\<less\>\<varepsilon\>>.
Take now <math|\<delta\><rprime|'>=min(\<delta\>,<frac|b-a|2>)> and
lets take <math|y> such that <math|\|b-y\|\<less\>\<delta\><rprime|'>>
then we have the following cases
<\enumerate>
<item><math|y\<less\>b> Then as
<math|\|b-y\|\<less\>\<delta\><rprime|'>=<frac|b-a|2>\<Rightarrow\>b-y\<less\><frac|b-a|2>\<Rightarrow\>b+<frac|a-b|2>-y\<less\>0\<Rightarrow\><frac|a+b|2>\<less\>y\<Rightarrow\>a\<less\>y>
so we have <math|\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(y)-f<rsup|]a-1,b+1[><rprime|'>(b)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|\|]a,b[>(y)-f<rprime|'><rsub|->(b)\<shortparallel\>=\<shortparallel\>f<rprime|'>(y)-f<rprime|'>(b)\<shortparallel\>\<less\>\<varepsilon\>>
<item><math|b\<less\>y> Then <math|\<shortparallel\>f<rsup|]a-1,b+1[>(y)-f<rsup|]a-1,b+1[>(b)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|->(b)-f<rprime|'><rsub|->(b)\<shortparallel\>=0\<less\>\<varepsilon\>>
proving continuity at <math|x=b>
</enumerate>
<item><math|x\<in\>]a-1,a[> Take then
<math|\<delta\>=min(a-x,x-(a-1))\<gtr\>0> so that if
<math|\|y-x\|\<less\>\<delta\>> we have
<math|-\<delta\>\<less\>y-x\<less\>\<delta\>\<Rightarrow\>x-\<delta\>\<less\>y\<less\>\<delta\>+x\<Rightarrow\>x-(x-(a-1))\<less\>y\<less\>a\<Rightarrow\>a-1\<less\>y\<less\>a\<Rightarrow\>\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(y)-f<rsup|]a-1,b+1[><rprime|'>(x)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|+>(a)-f<rprime|'><rsub|+>(a)\<shortparallel\>=0\<less\>\<varepsilon\>>
<math|>proving continuity at x.
<item><math|x\<in\>]b,b+1[> Take then
<math|\<delta\>=min(x-b,b+1-x)\<gtr\>0> so that if
<math|\|y-x\|\<less\>\<delta\>> we have
<math|-\<delta\>\<less\>y-x\<less\>\<delta\>\<Rightarrow\>-\<delta\>+x\<less\>y\<less\>x+\<delta\>\<Rightarrow\>-(x-b)+x\<less\>y\<less\>x+b+1-x\<Rightarrow\>b\<less\>y\<less\>b+1\<Rightarrow\>\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(y)-f<rsup|]a-1,b+1[><rprime|'>(x)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|->(b)-f<rprime|'><rsub|->(b)\<shortparallel\>=0><math|\<less\>\<varepsilon\>>
proving continuity at x
<item><math|x\<in\>]a,b[> Because of continuity of
<math|f<rprime|'><rsub|\|]a,b[>> on <math|]a,b[> we have the
existence of a <math|\<delta\>\<gtr\>0> such that if
<math|y\<in\>]a,b[> and <math|\|x-y\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f<rprime|'><rsub|\|]a,b[>(x)-f<rsub|\|]a,b[><rprime|'>(y)\<shortparallel\>\<less\>\<varepsilon\>>.
Then if we take <math|\<delta\><rprime|'>=min(\<delta\>,x-a,b-x)> and
let <math|y> such that <math|\|x-y\|\<less\>\<delta\><rprime|'>> then
on one hand <math|\|y-x\|\<less\>\<delta\>> and on the other hand we
have <math|-\<delta\><rprime|'>\<less\>y-x\<less\>\<delta\><rprime|'>\<Rightarrow\>-\<delta\><rprime|'>+x\<less\>y\<less\>\<delta\><rprime|'>+x\<Rightarrow\>-(x-a)+x\<less\>y\<less\>b-x+x\<Rightarrow\>a\<less\>y\<less\>b>
so that <math|\<shortparallel\>f<rsup|]a-1,b+1[><rprime|'>(x)-f<rsup|]a-1,b+1[><rprime|'>(y)\<shortparallel\>=\<shortparallel\>f<rprime|'><rsub|\|]a,b[>(x)-f<rsup|]a-1,b+1[><rprime|'>(y)\<shortparallel\>\<less\>\<varepsilon\>>,
proving continuity at x.
</enumerate>
</enumerate>
\;
</proof>
<\example>
<label|example of derivative>Let <math|Y,\<shortparallel\>\<shortparallel\>>
be a normed vector space over <math|\<bbb-R\>> and <math|x,y\<in\>Y>,
<math|\<sigma\>:[a,b]\<rightarrow\>Y> defined by
<math|\<sigma\>(t)=x+t.y> is <math|C<rsup|1>> then
<math|\<sigma\><rprime|'>(t)=y> <math|\<forall\>t\<in\>[a,b]>
</example>
<\proof>
We consider the following cases
<\enumerate>
<item><math|t=a> . In this case given <math|\<varepsilon\>\<gtr\>0> let
<math|\<delta\>=1> and <math|0\<less\>h\<less\>\<delta\>> such that
<math|a+h\<in\>[a,b]> then <math|\<shortparallel\><frac|\<sigma\>(t+h)-\<sigma\>(t)|h>-y\<shortparallel\>=\<shortparallel\><frac|h.y|y>-y\<shortparallel\>=0\<less\>\<varepsilon\>>
<item><math|t=b> In this case given <math|\<varepsilon\>\<gtr\>0> let
<math|\<delta\>=1> and <math|0\<less\>h\<less\>\<delta\>> such that
<math|b-h\<in\>[a,b]> then <math|\<shortparallel\><frac|\<sigma\>(t)-\<sigma\>(t-h)|h>-y\<shortparallel\>=\<shortparallel\><frac|h.y|h>-y\<shortparallel\>=0\<less\>\<varepsilon\>>
<item><math|t\<in\>]a,b[> Let then <math|\<delta\>=1> and
<math|0\<less\>\|h\|\<less\>\<delta\>> such that <math|t+h\<in\>]a,b[>
then <math|\<shortparallel\><frac|\<sigma\>(t+h)-\<sigma\>(t)|h>-y\<shortparallel\>=<frac|h.y|h>-y>\|\|=0\<less\>e
</enumerate>
Proving indeed our little theorem
</proof>
<chapter|Inverse function theorem>
<section|Intermediate value theorem>
<\definition>
<index|local weak minimum><index|local weak maximum>Let <math|f:U
open\<subseteq\>X\<rightarrow\>\<bbb-R\>> be a function then
<math|x\<in\>U> is a local weak minimum (maximum) of <math|f> if there
exists a <math|\<delta\>\<gtr\>0> such that
<math|B<rsub|\<delta\>>(x)\<subseteq\>U> and
<math|\<forall\>y\<in\>B<rsub|\<delta\>>(x)> we have
\ <math|f(y)\<leqslant\>f(x)> (or <math|f(x)\<leqslant\>f(y))>
</definition>
<\theorem>
<label|extremum and derivate>Let <math|f:U
open\<subseteq\>\<bbb-R\>\<rightarrow\>\<bbb-R\>> be a function, <math|x>
be a local weak maximum (minimum) and <math|f<rprime|'>(x)> exists then
<math|f<rprime|'>(x)=0><math|>
</theorem>
<\proof>
We have two cases
<\enumerate>
<item>Local weak maximum: Let <math|\<delta\>\<gtr\>0> be such that
<math|]x-\<delta\>,x+\<delta\>[\<subseteq\>U> we have
<math|\<forall\>y\<in\>]x-\<delta\>,x+\<delta\>[> that
<math|f(y)\<leqslant\>f(x)> \ Assume now that
<math|f<rprime|'>(x)\<neq\>0> we have then two cases
<\enumerate>
<item><math|f<rprime|'>(x)\<gtr\>0> , take then
<math|\<varepsilon\>=f<rprime|'>(x)> and choose any <math|h\<gtr\>0>
such that <math|0\<less\>h\<less\>\<delta\>> then
<math|<frac|f(x+h)-f(x)|h>\<leqslant\>0\<less\>f<rprime|'>(x)> so
<math|\|f<rprime|'>(x)-<frac|f(x+h)-f(x)|h>\|=f<rprime|'>(x)-<frac|f(x+h)-f(x)|h>><math|\<geqslant\>f<rprime|'>(x)+0=\<varepsilon\>>
\ contradicting the existence of a derivative\
<item><math|f<rprime|'>(x)\<less\>0>, take then <math|\<varepsilon\>=
\<um\>f<rprime|'>(x)\<gtr\>0> and choose any <math|h\<gtr\>0> such
that <math|0\<less\>h\<less\>\<delta\>> then
<math|<frac|f(x-h)-f(x)|-h>\<geqslant\>0\<Rightarrow\>\|<frac|f(x-h)-f(x)|-h>-f<rprime|'>(x)\|=<frac|f(x-h)-f(x)|-h>-f<rprime|'>(x)\<geqslant\>-f<rprime|'>(x)=\<varepsilon\>>
again contradicting the existence of a derivative
</enumerate>
So the only possibility left is <math|f<rprime|'>(x)=0>
<item>Local weak minimum: Let <math|\<delta\>\<gtr\>0> be such that
<math|]x-\<delta\>,x+\<delta\>[\<subseteq\>U> and
<math|\<forall\>y\<in\>]x-\<delta\>,x+\<delta\>[> we have
<math|f(x)\<leqslant\>f(y)>. Assume now that
<math|f<rprime|'>(x)\<neq\>0> we have the two case
<\enumerate>
<item><math|f<rprime|'>(x)\<gtr\>0>, take then
<math|\<varepsilon\>=f<rprime|'>(x)\<gtr\>0> and choose any
<math|0\<less\>h\<less\>\<delta\>> then
<math|<frac|f(x-h)-f(x)|-h>\<leqslant\>0\<less\>f<rprime|'>(x)\<Rightarrow\>\|f<rprime|'>(x)-<frac|f(x-h)-f(x)|-h>=f<rprime|'>(x)-<frac|f(x-h)-f(x)|-h>\<geqslant\>f<rprime|'>(x)=\<varepsilon\>>
contradicting the existence of a derivative
<item><math|f<rprime|'>(x)\<less\>0> take then
<math|\<varepsilon\>=-f<rprime|'>(x)\<gtr\>0> and choose any
<math|0\<less\>h\<less\>\<delta\>> then
<math|<frac|f(x-h)-f(x)|h>\<geqslant\>0\<Rightarrow\>\|<frac|f(x-h)-f(x)|h>-f<rprime|'>(x)\|=<frac|f(x-h)-f(x)|h>-f<rprime|'>(x)\<geqslant\>-f<rprime|'>(x)=\<varepsilon\>>
contradicting the existence of a derivative
</enumerate>
So the only possibility left is <math|f<rprime|'>(x)=0>
</enumerate>
</proof>
<\theorem>
<label|rolles theorem><dueto|Rolle's theorem><index|Rolle's theorem>Let
<math|[a,b]\<subseteq\>\<bbb-R\>,a\<less\>b> and <math|R, \| \|> the
classical Banach space and <math|f\<in\>\<cal-C\>([a,b],\<bbb-R\>)> such
that <math|f(a)=f(b)=0> and <math|f> be differentiable on <math|]a,b[>
then there exists a <math|\<zeta\>\<in\>]a,b[> such that
<math|0=f<rprime|'>(\<zeta\>)>
</theorem>
<\proof>
Since <math|f> is continuous on <math|[a,b]> we can use
<reference|continuous mapping in [a,b]> so we have <math|f([a,b])=[c,d]>
then as <math|c\<leqslant\>f(a)=f(b)=0\<leqslant\>d> we have the
following cases
<\enumerate>
<item><math|c=d=0> then <math|f> is constant and equal to <math|0> so
<math|f<rprime|'>(\<zeta\>)=0> <math|\<forall\>\<zeta\>\<in\>]a,b[>
<item><math|c\<less\>0> As <math|f(a)=f(b)=0> and <math|f([a,b])=[c,d]>
there exists a <math|\<zeta\>\<in\>]a,b[> such that
<math|f(\<zeta\>)=c> and it is trivial that <math|\<zeta\>> is then a
local minimum, so using <reference|extremum and derivate> we have
<math|f<rprime|'>(\<zeta\>)=0>
<item><math|d\<gtr\>0> As <math|f(a)=f(b)> and <math|f([a,b])=[c,d]>
there exists a <math|\<zeta\>\<in\>]a,b[> such that
<math|f(\<zeta\>)=d> and it is trivial that <math|\<zeta\>> is then a
local maximum, so using <reference|extremum and derivate> we have
<math|f<rprime|'>(\<zeta\>)=0>
</enumerate>
</proof>
<\theorem>
<label|lagranges theorem><dueto|Lagrange's theorem><index|Lagrange's
theorem>Let <math|f:[a,b]\<rightarrow\>\<bbb-R\>> be a real valued
continuous function which has a derivative everywhere in <math|]a,b[>
then there exists a <math|\<zeta\>\<in\>]a,b[> such that
<math|f(b)-f(a)=f<rprime|'>(\<zeta\>)(b-a)>
</theorem>
<\proof>
Define the function <math|g:[a,b]\<rightarrow\>\<bbb-R\>> by
<math|g(x)=f(x)-(f(a)+<frac|f(b)-f(a)|b-a>(x-a))> which is differentiable
in <math|]a,b[> and continuous on <math|[a,b]> (because constant
functions and linear functions (in a finite dimensional space) are
continuous and differentiable and the same goes for sum of differentiable
(continuous) function). Now we have <math|g(a)=f(a)-(f(a)+<frac|f(b)-f(a)|b-a>(a-a))=f(a)-f(a)=0>
also <math|g(b)=f(b)-(f(a)+<frac|f(b)-f(a)|b-a>(b-a))=f(b)-f(b)=0>, this
means that we can apply Rolle's theorem (<reference|rolles theorem>) and
find that <math|g<rprime|'>(x)=0\<Rightarrow\>(f(x)-(f(a)+<frac|f(b)-f(a)|b-a>(x-a))=f<rprime|'>(x)-<frac|f(b)-f(a)|b-a>=0\<Rightarrow\>f(b)-f(a)=f<rprime|'>(\<zeta\>)(b-a)>
</proof>
<\lemma>
Let <math|f:[a,b]\<subseteq\>\<bbb-R\>\<rightarrow\>X> , <math|a\<less\>b
and \ X,\<shortparallel\>\<shortparallel\>,> a normed real Banach space
then <math|F:[a,b]\<rightarrow\>X> where
<math|F(x)=<big|int><rsub|a><rsup|x>f> is <math|C<rsup|1>> on
<math|[a,b]> and <math|F<rprime|'>=f>
</lemma>
<\proof>
\;
Take the following cases for <math|x\<in\>[a,b]>
<\enumerate>
<item><math|x=a> Given <math|\<varepsilon\>\<gtr\>0> then because of
continuity of <math|f> at <math| a> there exists a
<math|\<delta\>\<gtr\>0> \ such that if
<math|\|y-a\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(y)-f(a)\<shortparallel\>\<less\>\<varepsilon\>>
or if <math|h> is such that <math|0\<less\>h\<less\>\<delta\>> and
<math|a+h\<in\>[a,b]> then if <math|y\<in\>[a,a+h]> we have
<math|\<shortparallel\>f(y)-f(a)\<shortparallel\>\<less\>\<varepsilon\>><math|>,
also<math|\<shortparallel\><frac|F(a+h)-F(a)|h>-f(a)\<shortparallel\>=\<shortparallel\><frac|<big|int><rsub|a><rsup|h>f-<big|int><rsub|a><rsup|a>f|h>-f(a)\<shortparallel\>\<equallim\><rsub|<reference|splitting
of a integral>>\<shortparallel\><frac|<big|int><rsub|a><rsup|h>f|h>-<frac|h.f(a)|h>\<shortparallel\>\<equallim\><rsub|<reference|integration
of constant>>\<shortparallel\><frac|<big|int><rsub|a><rsup|h>f-<big|int><rsub|a><rsup|h>f(a)|h>\<shortparallel\>=\<shortparallel\><frac|<big|int><rsub|a><rsup|h>(f-f(a))|h>\<shortparallel\>\<leqslant\><rsub|<reference|property
of integral>,h\<gtr\>0><frac|<big|int><rsub|a><rsup|h>\<shortparallel\>f-f(a)\<shortparallel\>|h>\<leqslant\><rsub|<reference|property
of integral>,h\<gtr\>0><frac|\<varepsilon\>.h|h>=\<varepsilon\>>
proving that <math|F<rprime|'><rsub|+>(a)=f(a)>
<item>Take <math|x=b>. Given <math|\<varepsilon\>\<gtr\>0> then because
of continuity of <math|f>at b there exists a <math|\<delta\>\<gtr\>0>
such that if <math|\|y-b\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(y)-f(b)\<shortparallel\>\<less\>\<varepsilon\>>
or if <math|h> is such that <math|0\<less\>h\<less\>\<delta\>> and
<math|a-h\<in\>[a,b]> then if <math|y\<in\>[b-h,b]> we have
<math|\<shortparallel\>f(y)-f(b)\<shortparallel\>\<less\>\<varepsilon\>>,
also <math|\<shortparallel\><frac|F(b)-F(b-h)|h>-f(b)\<shortparallel\>\<equallim\><rsub|h\<gtr\>0><frac|\<shortparallel\><big|int><rsub|a><rsup|b>f-<big|int><rsub|a><rsup|b-h>f-h.f(b)<rsub|>|h>\<shortparallel\>\<equallim\><rsub|<reference|splitting
of a integral>,<reference|integration of
constant>><frac|\<shortparallel\><big|int><rsub|b-h><rsup|b>f-<big|int><rsub|b-h><rsup|b>f(b)\<shortparallel\>|h>\<equallim\><rsub|<reference|linearity
of integral>><frac|\<shortparallel\><big|int><rsub|b-h><rsup|b>(f-f(b))\<shortparallel\>|h>\<leqslant\><rsub|<reference|property
of integral>><frac|<big|int><rsub|b-h><rsup|b>\<shortparallel\>f-f(b)\<shortparallel\>|h>\<leqslant\><rsub|<reference|property
of integral>><frac|\<varepsilon\>.h|h>=\<varepsilon\>> . This proves
that <math|F<rprime|'><rsub|->(b)=f(b)>
<item>Take <math|x\<in\>]a,b[> then because of continuity of <math|f>
at <math|x> we have the existence of a <math|\<delta\>\<gtr\>0> such
that if <math|\|y-x\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(y)-f(x)\<shortparallel\>\<less\>\<varepsilon\>>
. Now if <math|h> is such that <math|0\<less\>\|h\|\<less\>\<delta\>>
and <math|x+h\<in\>]a,b[> then we have the following two cases
<\enumerate>
<item><math|h\<gtr\>0> In this case if <math|y\<in\>[x,x+h]> we have
<math|\|y-x\|\<less\>h\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(y)-f(x)\<shortparallel\>\<less\>\<varepsilon\>>
and also <math|\<\|\|\><frac|F(x+h)-F(x)|h>-f(x)\<shortparallel\>\<equallim\><rsub|h\<gtr\>0><frac|\<shortparallel\><big|int><rsub|a><rsup|x+h>f-<big|int><rsub|a><rsup|x>f-hf(x)\<shortparallel\>|h>\<equallim\><rsub|<reference|splitting
of a integral>><frac|\<shortparallel\><big|int><rsub|x><rsup|x+h>f-hf(x)\<shortparallel\>|h>\<equallim\><rsub|<reference|integration
of constant>><frac|\<shortparallel\><big|int><rsub|x><rsup|x+h>(f-f(x))\<shortparallel\>|h>\<leqslant\><frac|<big|int><rsub|x><rsup|x+h>\<shortparallel\>f-f(x)\<\|\|\>|h>\<leqslant\><rsub|<reference|property
of integral>><frac|\<varepsilon\>h|h>=\<varepsilon\>>\
<item><math|h\<less\>0> In this case <math|\|h\|=-h> and if
<math|y\<in\>[x-\|h\|,x]> we have
<math|\|y-x\|\<less\>\|h\|\<less\>\<delta\>\<Rightarrow\>\<shortparallel\>f(y)-f(x)\<shortparallel\>\<less\>\<varepsilon\>>.
Also <math|\<shortparallel\><frac|F(x+h)-F(x)|h>-f(x)\<shortparallel\>\<equallim\><rsub|h=-\|h\|><frac|\<shortparallel\>F(x-\|h\|)-F(x)+\|h\|f(x)\<\|\|\>|\|h\|>=<frac|\<\|\|\><big|int><rsub|a><rsup|x-\|h\|>f-<big|int><rsub|a><rsup|x>f+\|h\|f(x)\<shortparallel\>|h>=<frac|\<shortparallel\>\|h\|f(x)-<big|int><rsub|x-\|h\|><rsup|x>f\<shortparallel\>|\|h\|>=<frac|\<shortparallel\><big|int><rsub|x-\|h\|><rsup|x>f(x)-<big|int><rsub|x-\|h\|><rsup|x>f\<shortparallel\>|\|h\|>=<frac|\<shortparallel\><big|int><rsub|x-\|h\|><rsup|x>(f(x)-f)\<shortparallel\>|\|h\|>\<leqslant\><frac|<big|int><rsub|x-\|h\|><rsup|x>\<shortparallel\>f(x)-f)\<shortparallel\>|\|h\|>\<leqslant\><frac|\<varepsilon\>\|h\||\|h}>=\<varepsilon\>>
</enumerate>
(a) and (b) proves that <math|F<rprime|'>(x)=f(x)>
</enumerate>
So by (1),(2) and (3) we have proved that <math|F<rprime|'>> exists on
<math|[a,b]> and <math|F<rprime|'>=f> from which it follows that
<math|F<rprime|'>> is also continue on <math|[a,b]> and thus that
<math|F<rprime|'>> is <math|C<rsup|1>>
</proof>
<\theorem>
<label|fundamental theorem of calculus (1)><dueto|Fundamental Theorem of
Calculus (classical)><index|Fundamental theorem of Calculus
(classical)>Suppose that <math|g:[a,b]\<rightarrow\>Y> is differentiable
of class <math|C<rsup|1>> (see <reference|Cr on arbitrary sets> and
<reference|differentiability on [a,b]>) (here
<math|Y,\<shortparallel\>\<shortparallel\>> is a normed real Banach
space). Then we have that <math|g(b)-g(a)=<big|int><rsub|a><rsup|b>g<rprime|'>>
</theorem>
<\proof>
First note that as <math|g<rprime|'>> is defined and continue on
<math|[a,b]> we have together with the fact that <math|Y> is a Banach
space that <math|f:[a,b]\<rightarrow\>Y> defined by
<math|f(x)=g(a)+<big|int><rsub|a><rsup|x>g<rprime|'>> is well defined.
Also using the previous lemma we have
<math|f<rprime|'>=(g(a))<rprime|'>+(<big|int><rsup|x><rsub|a>g<rprime|'>)=0+g<rprime|'>=g<rprime|'>>
and that <math|f<rprime|'>>. Note also that
<math|f(a)=g(a)+<big|int><rsub|a><rsup|a>g<rprime|'>=g(a)+0=g(a)>.
Consider now <math|\<varphi\>:]a,b[\<rightarrow\>\<bbb-R\> defined by
\<varphi\>(x)=\<shortparallel\>g(x)-f(x)\<shortparallel\>> and choose a
<math|0\<less\>h> such that <math|x+h\<in\>]a,b[> then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\|<frac|\<varphi\>(x+h)-\<varphi\>(x)|h>\|>|<cell|=>|<cell|\|<frac|\<shortparallel\>g(x+h)-f(x+h)\<shortparallel\>-\<shortparallel\>g(x)-f(x)\<shortparallel\>|h>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<\|\|\><frac|\<shortparallel\>g(x+h)-f(x+h)-g(x)-f(x)\<shortparallel\>|h>>>|<row|<cell|>|<cell|=>|<cell|\<shortparallel\><frac|g(x+h)-g(x)|h>-<frac|f(x+h)-f(x)|h>\<shortparallel\>>>|<row|<cell|>|<cell|\<equallim\><rsub|f<rprime|'>=g<rprime|'>>>|<cell|\<\|\|\><frac|g(x+h)-g(x)|h>-g<rprime|'>(x)+(<frac|f(x+h)-f(x)|h>-f<rprime|'>(x))\<\|\|\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\><frac|g(x+h)-g(x)|h>-g<rprime|'>(x)\<shortparallel\>+\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rprime|'>(x)\<shortparallel\>>>>>
</eqnarray*>
Now take <math|\<varepsilon\>\<gtr\>0> then there exists a
<math|\<delta\><rsub|1>,\<delta\><rsub|2>\<gtr\>0> such that
<math|\<shortparallel\><frac|g(x+h)-g(x)|h>-g<rprime|'>(x)\<shortparallel\>\<less\><frac|\<varepsilon\>|e>>
and <math|\<shortparallel\><frac|f(x+h)-f(x)|h>-f<rprime|'>(x)\<shortparallel\>\<less\><frac|\<varepsilon\>|2>>
and thus if <math|0\<less\>h\<less\>\<delta\>=min(\<delta\><rsub|1>,\<delta\><rsub|2>)>
we have <math|\|<frac|\<varphi\>(x+h)-\<varphi\>(x)|h>-0\|\<less\>\<varepsilon\>>
proving that <math|\<varphi\><rprime|'>(x)=0> for every
<math|x\<in\>]a,b[>. Now if <math|x,y\<in\>[a,b]> <math|x\<less\>y> then
we have by Lagrange Theorem that there exists a <math|t\<in\>]x,y[> such
that <math|\<varphi\>(x)-\<varphi\>(y)=\<varphi\><rprime|'>(t)(x-y)=0(x-y)=0>.
This proves that <math|\<varphi\>> is constant on <math|[a,b]>, and as
<math|\<varphi\>(a)=\<shortparallel\>f(a)-g(a)\<shortparallel\>=0> we
have that <math|\<varphi\>(x)=0> <math|\<forall\>x\<in\>[a,b]\<Rightarrow\>f(x)=g(x)
\<forall\>x\<in\>[a,b]\<Rightarrow\>g(b)=f(b)=g(a)+<big|int><rsub|a><rsup|b>g<rprime|'>\<Rightarrow\>g(b)-g(a)=<big|int><rsub|a><rsup|b>g<rprime|'>>
</proof>
<\theorem>
<label|fundamental theorem of calculus (2)><dueto|Fundamental Theorem of
Calculus (first extension)><index|Fundamental theorem of calculus>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be a normed vector
space, <math|U\<subseteq\>X> a open set and
<math|f:U\<rightarrow\>\<bbb-R\>> be a function differentiable on
<math|U>. If <math|x,y\<in\>U> such that <math|\<forall\>t\<in\>[0,1]> we
have <math|x+t(y-x)\<in\>U> then there exists a
<math|t<rsub|0>\<in\>[0,1]> such that if <math|z=x+t<rsub|0>(x-y)> we
have <math|f(y)-f(x)=Df(z)(y-x)>
</theorem>
<\proof>
Let <math|\<sigma\>:[0,1]\<rightarrow\>U> be defined by
<math|\<sigma\>(t)=x+t(y-x)> then <math|\<sigma\>(0)=x>,
<math|\<sigma\>(1)=y>. Take then <math|g=f\<circ\>\<sigma\>:[0,1]\<rightarrow\>\<bbb-R\>>
, this function is continuous on <math|[0,1]> and differentiable in
<math|]0,1[> (as <math|\<sigma\>> is differentiable on <math|]0,1[>,
continue on <math|[0,1]> and <math|f> is differentiable and continue on
<math|U> and <math|\<sigma\>([0,1])\<subseteq\>U>). By the Lagrange
theorem (<reference|lagranges theorem>) \ there exists a
<math|\<zeta\>\<in\>]0,1[> such that <math|g(1)-g(0)=g<rprime|'>(\<zeta\>)>.
Now using the chain rule we have <math|g<rprime|'>(\<zeta\>)=(D(g\<circ\>\<sigma\>)(\<sigma\>))(1)=(Df(\<sigma\>(\<zeta\>))\<circ\>D\<sigma\>(\<zeta\>))1=Df(\<sigma\>(\<zeta\>))(D\<sigma\>(\<zeta\>)(1))=Df(\<sigma\>(\<zeta\>))(\<sigma\><rprime|'>(\<zeta\>))\<equallim\><rsub|<reference|example
of derivative>>Df(\<sigma\>(\<zeta\>))(y-x), now
><math|z=\<sigma\>(\<zeta\>)=x+\<zeta\>(y-x),
\<zeta\>\<subset\>]0,1[\<subseteq\>[0,1]> and we have
<math|g(1)=f(y),g(0)=f(x)\<Rightarrow\>f(y)-f(x)=Df(z)(y-x)> as must be
proved
</proof>
<\definition>
A subset <math|C> of a vector space <math|X> over <math|\<bbb-R\>> is
convex if for every <math|t\<in\>[0,1]> we have
<math|\<forall\>x,y\<in\>C> that <math|x+t(y-x)\<in\>C>
</definition>
<\theorem>
<label|balls are convex><index|convex set>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed vector space over
<math|\<bbb-R\>>, <math|x\<in\>X,\<delta\>\<gtr\>0> then
<math|B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)<rsub|>> and
<math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)>
are convex
</theorem>
<\proof>
Let <math|t\<in\>[0,1]> then for <math|x,y,z\<in\>X> we have
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>y+t(z-y)-x\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\>y+t(z-y)-x+t
x-t x\<shortparallel\>>>|<row|<cell|>|<cell|=>|<cell|\<shortparallel\>y(1-t)-x(1-t)+t
z-t \ x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>y(1-t)-x(1-t)\<shortparallel\>+\<shortparallel\>t
z - t x\<shortparallel\>>>|<row|<cell|>|<cell|\<equallim\><rsub|t\<leqslant\>1>>|<cell|(1-t)\<shortparallel\>y-x\<shortparallel\>+t\<shortparallel\>z-x\<shortparallel\>>>>>
</eqnarray*>
then if <math|y,z\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)>
we have <math|\<shortparallel\>y+t(z-y)-x\<shortparallel\>\<less\>(1-t)\<delta\>+t\<delta\>=\<delta\>\<Rightarrow\>y+t(z-y)\<in\>B<rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)>.
Also if <math|y,z\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)>
we have <math|\<shortparallel\>y+t(z-y)-x\<shortparallel\>\<leqslant\>(1-t)\<delta\>+t\<delta\>=\<delta\>\<Rightarrow\>y+t(z-y)\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\>>(x,\<delta\>)>
\;
</proof>
<\definition>
<label|the fundamental theorem of calculus (3)><dueto|The fundamental
theorem of calculus (generalized)><index|Fundamental theorem of claculus
(generalized)>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>> be
a normed space over <math|\<bbb-R\>> and let
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be a Banach space
over <math|\<bbb-R\>>, <math|U\<subseteq\>X> a open set and let
<math|f:U\<rightarrow\>Y> be a differentiable function of class
<math|C<rsup|1>>. Assume now that there is a convex subset
<math|C\<subseteq\>U> and a constant <math|k\<in\>\<bbb-R\>> such that
<math|\<shortparallel\>Df(x)\<shortparallel\>\<leqslant\>k> for all
<math|x\<in\>C> then we have the equality
<math|\<shortparallel\>f(y)-f(x)\<shortparallel\><rsub|Y>\<leqslant\>k\<shortparallel\>y-x\<shortparallel\><rsub|X>
\<forall\>x,y\<in\>C>
</definition>
<\proof>
Let <math|x,y\<in\>C> define then <math|\<sigma\><rprime|'>:\<bbb-R\>\<rightarrow\>X>
by <math|\<alpha\>(t)=x+t(y-x)> then <math|\<alpha\>([0,1])\<subseteq\>C\<subseteq\>U>
and take then <math|V=\<alpha\><rsup|-1>(U)> which is open because
<math|\<alpha\>> is continue, so let take
<math|\<sigma\>=\<alpha\><rsub|\|V>>, then
<math|\<sigma\>(V)\<subseteq\>U> and also because
<math|\<alpha\>([0,1])\<subseteq\>C\<subseteq\>U> we have
<math|[0,1]\<subseteq\>V>. If we define now
<math|g<rsup|V>=f\<circ\>\<sigma\>> then this is of class
<math|C<rsup|1>> because of the chain rule and also
<math|Dg<rsup|V>(t)=Df(\<sigma\>(t))\<circ\>D\<sigma\>(t)> so that
<math|g<rsup|V<rprime|'>>(t)=Dg<rsup|V>(t)(1)=Df(\<sigma\>(t))\<circ\>D\<sigma\>(t)(1)=Df(\<sigma\>(t))(y-x)>.
Define now <math|h:[0,1]\<rightarrow\>X> by
<math|h=g<rsup|V><rsub|\|<rsup|>[0,1]>> then using
<reference|differentiability on [a,b]> we have <math|h<rprime|'>> is
defined on <math|[0,1]> and <math|h<rprime|'>=(g<rsup|V<rprime|'>>)<rsub|\|[0,1]>>,
now using the classical fundamental theorem of calculus (see
<reference|fundamental theorem of calculus>) we have\
<\eqnarray*>
<tformat|<table|<row|<cell|f(y)-f(x)>|<cell|=>|<cell|f(x+1(y-x))-f(x+0(y-x))>>|<row|<cell|>|<cell|=>|<cell|(f\<circ\>\<sigma\>)(1)-(f\<circ\>\<sigma\>)(0)>>|<row|<cell|>|<cell|\<equallim\><rsub|1,0\<in\>[0,1]>>|<cell|h(1)-h(0)>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|fundamental
theorem of calculus>>>|<cell|<big|int><rsub|0><rsup|1>h<rprime|'>>>>>
</eqnarray*>
Next using <reference|property of integral> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>f(y)-f(x)\<shortparallel\><rsub|Y>>|<cell|=>|<cell|\<shortparallel\><big|int><rsub|0><rsup|1>h<rprime|'>\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<big|int><rsub|0><rsup|1>\<shortparallel\>h<rprime|'>\<shortparallel\>>>>>
</eqnarray*>
Now if <math|t\<in\>[0,1]> we have <math|h<rprime|'>(t)=g<rsup|V<rprime|'>>(t)=Df(\<sigma\>(t))(y-x)\<Rightarrow\>\<shortparallel\>h<rprime|'>(t)\<shortparallel\>\<leqslant\>\<shortparallel\>Df(\<sigma\>(t))\<shortparallel\>\<shortparallel\>y-x\<shortparallel\><rsub|X>\<leqslant\>k\<shortparallel\>y-x\<shortparallel\><rsub|X>>
and thus
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>f(y)-f(x)\<shortparallel\><rsub|Y>>|<cell|\<leqslant\>>|<cell|<big|int><rsub|0><rsup|1>k\<shortparallel\>y-x\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(1-0)k\<shortparallel\>y-x\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|=>|<cell|k\<shortparallel\>y-x\<shortparallel\><rsub|X>>>>>
</eqnarray*>
proving our little theorem
\;
</proof>
<\definition>
<index|toplinear isomorphism>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>
and <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be a metric space
then a linear continuous isomorphism <math|L:X\<rightarrow\>Y> between
<math|X> and <math|Y> is a toplinear isomorphism if also
<math|L<rsup|-1>> is continuous (see definition <reference|toplinear
isomorphism> for a general definition)
</definition>
<\remark>
<label|in Banach spaces linear continuous mappings are toplinear>If
<math|X,Y> are Banach spaces then if <math|L:X\<rightarrow\>Y> is a
linear continuous isomorphism we have by <reference|inverse of continuous
linear mappings between Banach spaces is continuous> that <math|L> is a
toplinear isomorphism
</remark>
<\definition>
<index|<math|Gl(X)>>If <math|X,\<shortparallel\>\<shortparallel\>> is a
Banach space then we know by <reference|space of linear continuous maps
to a Banach space is Banach> that <math|L(X,X)> is a Banach space then
<math|Gl(X)\<subseteq\>L(X,X)> is defined by
<math|Gl(X)={L\<in\>L(X,X)\|L is toplinear}>
</definition>
<\theorem>
If <math|X,\<shortparallel\>\<shortparallel\>> is a Banach space then
<math|Gl(X)> is a group with the composition operator as the group
operator
</theorem>
<\proof>
\;
First consider <math|I:X\<rightarrow\>X> defined by <math|I(x)=x> (the
identity linear operator or <math|i<rsub|X>>) which is obviously linear,
continuous and open and if <math|L\<in\>Gl(X)> we have
<math|L\<circ\>I=I\<circ\>L=L> so <math|I> is the identity element.
Second if <math|L<rsub|1>,L<rsub|2>\<in\>Gl(X)> then
<math|L<rsub|1>\<circ\>L<rsub|2>> is also a isomorphism, open and
continuous and by <reference|inverse of continuous linear mappings
between Banach spaces is continuous> we have also that
<math|(L<rsub|1>\<circ\>L<rsub|2>)<rsup|-1>> is continuous, proving that
<math|L<rsub|1>\<circ\>L<rsub|2>\<in\>Gl(x)>
Third the composition is associative <math|[(L<rsub|1>\<circ\>(L<rsub|2>\<circ\>L<rsub|3>))(x)=L<rsub|1>((L<rsub|2>\<circ\>L<rsub|3>)(x))=L<rsub|1>(L<rsub|2>(L<rsub|3>(x)))=(L<rsub|1>\<circ\>L<rsub|2>)(L<rsub|3>(x))=((L<rsub|1>\<circ\>L<rsub|2>)\<circ\>L<rsub|3>)(x)]>\
Finally if <math|L\<in\>Gl(X)> then by definition
<math|L<rsup|-1>\<in\>Gl(X)> and <math|L\<circ\>L<rsup|-1>=L<rsup|-1>\<circ\>L=I>
</proof>
<\definition>
<index|<math|H<rsup|n>>>Let <math|X,\<shortparallel\>\<shortparallel\>>
be a normed vector space, <math|H:X\<rightarrow\>X> is a continuous
linear function and <math|n\<in\>\<bbb-N\><rsub|0>> then we define
<math|H<rsup|n>:X\<rightarrow\>X> recursively as follows\
<\eqnarray*>
<tformat|<table|<row|<cell|H<rsup|1>>|<cell|=>|<cell|H>>|<row|<cell|H<rsup|n+1>>|<cell|=>|<cell|H<rsup|>\<circ\>H<rsup|n><rsup|>>>>>
</eqnarray*>
</definition>
<\note>
If <math|H:X\<rightarrow\>X> is a continuous linear function, then
<math|-H:X\<rightarrow\>X> is obviously a continuous linear function
(<math|L(X,X)> is a vector space) and <math|(-H)<rsup|n>=(-1)<rsup|n>H>
</note>
<\proof>
The proof by induction on n is trivial as for every <math|x\<in\>X> we
have
<\eqnarray*>
<tformat|<table|<row|<cell|(-H)<rsup|1>(x)>|<cell|=>|<cell|(-H)(x)=(-1).H(x)>>|<row|<cell|(-H)<rsup|n+1>(x)>|<cell|=>|<cell|(-H)((-H)<rsup|n>(x))>>|<row|<cell|>|<cell|=>|<cell|(-1)(-1)<rsup|n>H(H<rsup|n>(x))>>|<row|<cell|>|<cell|=>|<cell|(-1)<rsup|n+1>H<rsup|n>(x)>>>>
</eqnarray*>
</proof>
<\lemma>
<label|power of a continuous linear map is linear and continuous>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a normed vector space and
<math|H:X\<rightarrow\>X> a continuous linear map then
<math|H<rsup|n>:X\<rightarrow\>X> is also linear and continuous and
<math|\<shortparallel\>H<rsup|n>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>>
so <math|\<shortparallel\>H<rsup|n>\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\><rsup|n>>.
In addition we have that <math|<big|sum><rsub|i=1><rsup|n>H<rsup|i>> is
continuous.
</lemma>
<\proof>
\;
<\enumerate>
<item>We prove linearity by induction
<\eqnarray*>
<tformat|<table|<row|<cell|H<rsup|1>(\<alpha\>x+\<beta\>y)>|<cell|=>|<cell|H(\<alpha\>x+\<beta\>y)=\<alpha\>H(x)+\<beta\>H(y)>>|<row|<cell|H<rsup|n+1>(\<alpha\>x+\<beta\>y)>|<cell|=>|<cell|H(H<rsup|n>(\<alpha\>x+\<beta\>y))>>|<row|<cell|>|<cell|\<equallim\><rsub|induction
hypothese>>|<cell|H(\<alpha\>H<rsup|n>(x)+\<beta\>H<rsup|n>(y))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>H(H<rsup|n>(x))+\<beta\>H(H<rsup|n>(y))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>H<rsup|n>(x)+\<beta\>H<rsup|n>(y)>>>>
</eqnarray*>
<item>We prove by induction that <math|\<shortparallel\>H<rsup|n>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>>
form which follows continuity\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>H<rsup|1>(x)\<shortparallel\>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H(x)\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\>\<shortparallel\>x\<shortparallel\>>>|<row|<cell|\<shortparallel\>H<rsup|n+1>(x)\<shortparallel\>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H(H<rsup|n>(x))\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H\<shortparallel\>\<shortparallel\>H<rsup|n>(x)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H\<shortparallel\>\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H\<shortparallel\><rsup|n+1>\<shortparallel\>x\<shortparallel\>>>>>
</eqnarray*>
<item>To prove continuity of <math|<big|sum><rsub|i=1><rsup|n>H<rsup|i>>
we prove by induction that <math|\<shortparallel\><big|sum><rsub|i=1><rsup|n>H<rsup|i>(x)\<shortparallel\>\<leqslant\>(<big|sum><rsub|i=1><rsup|n>\<shortparallel\>H\<shortparallel\>)\<shortparallel\>x\<shortparallel\>>
from which continuity follows
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\><big|sum><rsub|i=1><rsup|1>H<rsup|i>(x)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\>H(x)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H\<shortparallel\>\<shortparallel\>x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(<big|sum><rsub|i=1><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\>)\<shortparallel\>x\<shortparallel\>>>|<row|<cell|\<shortparallel\><big|sum><rsub|i=1><rsup|n+1>H<rsup|i>(x)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\><big|sum><rsub|i=1><rsup|n>H<rsup|i>(x)+H<rsup|n+1>(x)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\><big|sum><rsub|i=1><rsup|n>H<rsup|i>(x)\<shortparallel\>+\<shortparallel\>H\<shortparallel\><rsup|n+1>\<shortparallel\>x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\><rsub|induction
nypothese>>|<cell|(<big|sum><rsub|i=1><rsup|n>\<shortparallel\>H\<shortparallel\><rsup|i>)\<\|\|\>x\<shortparallel\>+\<shortparallel\>H\<shortparallel\><rsup|n+1>\<shortparallel\>x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(<big|sum><rsub|i=1><rsup|n+1>\<shortparallel\>H\<shortparallel\><rsup|i>)\<shortparallel\>x\<shortparallel\>>>>>
</eqnarray*>
</enumerate>
\;
</proof>
<\definition>
If <math|H:X\<rightarrow\>X> is a continuous linear function and
<math|\<forall\>x\<in\>X> we have <math|<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)>
converges then we say that <math|<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>>
converges and is defined by <math|(<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(x)=<big|sum><rsub|n=1><rsup|\<infty\>>(H<rsup|n>)(x)>
</definition>
<\lemma>
<label|I-H has inverse if norm of H is lower then 1>Let
<math|X,\<shortparallel\>\<shortparallel\>>be a Banach space and
<math|H:X\<rightarrow\>X> a continuous linear function with
<math|\<shortparallel\>H\<shortparallel\>\<less\>1> then
<math|I-H\<in\>Gl(E)>. Further <math|<big|sum><rsub|i=1><rsup|\<infty\>>H>
converges and <math|(I-H)<rsup|-1>=I+<big|sum><rsub|i=1><rsup|\<infty\>>H<rsup|n>>
</lemma>
<\proof>
Let <math|x\<in\>E> then we have by on the previous lemmas
(<reference|power of a continuous linear map is linear and continuous>)
that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>H<rsup|n>(x)\<shortparallel\>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>,
n\<in\>\<bbb-N\><rsub|0>>>>>
</eqnarray*>
now using <reference|example of a convergent serie> we have that
<math|<big|sum><rsub|i=1><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\><rsup|i>>
is convergent and equal to <math|<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>>
and using <reference|linear combination of convergent series> we have
also that <math|<big|sum><rsub|i=1><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\><rsup|i>\<shortparallel\>x\<shortparallel\>>
is convergent and equal to <math|\<shortparallel\>x\<shortparallel\><big|sum><rsub|i=1><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\><rsup|i>=<frac|\<shortparallel\>x\<shortparallel\>\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>>.
Using then <reference|convergent criteria of a serie in a Banach space>
we have that <math|<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)> is
convergent and\
<\eqnarray*>
<tformat|<table|<row|<cell|(1) \ \<shortparallel\><big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)\<shortparallel\>>|<cell|\<leqslant\>>|<cell|<big|sum><rsub|i=1><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>=<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>\<shortparallel\>x\<shortparallel\>>>>>
</eqnarray*>
Also by <reference|terms of a convergent serie goes to 0> we have that
<math|lim<rsub|n\<rightarrow\>\<infty\>>H<rsup|n>(x)=0> and thus also
<\eqnarray*>
<tformat|<table|<row|<cell|lim<rsub|n\<rightarrow\>\<infty\>>H<rsup|n+1>(x)>|<cell|=>|<cell|0>>>>
</eqnarray*>
So we can define the function <math|(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>):X\<rightarrow\>X>
by setting <math|\<forall\>x\<in\>X>
<\eqnarray*>
<tformat|<table|<row|<cell|(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(x)>|<cell|=>|<cell|I(x)+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)=x+<big|sum><rsub|n=1><rsup|\<infty\>>(H<rsup|n>)(x)>>>>
</eqnarray*>
We show now that <math|(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)>
is linear
<\eqnarray*>
<tformat|<table|<row|<cell|(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(\<alpha\>x+\<beta\>y)>|<cell|=>|<cell|\<alpha\>x+\<beta\>y+<big|sum><rsub|n=1><rsup|\<infty\>>(H<rsup|n>)(\<alpha\>x+\<beta\>y)>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|power
of a continuous linear map is linear and
continuous>>>|<cell|\<alpha\>x+\<beta\>y+<big|sum><rsub|n=1><rsup|\<infty\>>(\<alpha\>H<rsup|n>(x)+\<beta\>H<rsup|n>(y))>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|linear
combination of convergent series>>>|<cell|\<alpha\>x+\<beta\>y+\<alpha\><big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)+\<beta\><big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(y)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(x)+\<beta\>(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(y)>>>>
</eqnarray*>
Now we have using (1) that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(x)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\>I(x)+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>x\<shortparallel\>+\<shortparallel\><big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>(x)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|(1+<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>)\<shortparallel\>x\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|1|1-\<shortparallel\>H\<shortparallel\>>\<shortparallel\>x\<shortparallel\>>>>>
</eqnarray*>
proving continuity of <math|I+<big|sum><rsub|i=1><rsup|\<infty\>>H<rsup|n>>
and that <math|\<shortparallel\>I+<big|sum><rsub|i=1><rsup|\<infty\>>H\<shortparallel\>\<leqslant\><frac|1|1-\<shortparallel\>H\<shortparallel\>>>.\
We prove now by induction on <math|n> that\
<\eqnarray*>
<tformat|<table|<row|<cell|(I-H)(x+<big|sum><rsub|i=1><rsup|n>H<rsup|i>(x))>|<cell|=>|<cell|x-H<rsup|n+1>>>>>
</eqnarray*>
<\eqnarray*>
<tformat|<table|<row|<cell|(I-H)(x+<big|sum><rsub|i=1><rsup|1>H<rsup|i>(x))>|<cell|=>|<cell|(I-H)(x+H(x))>>|<row|<cell|>|<cell|=>|<cell|I(x)+H(x)-H(x)-H(H(x))>>|<row|<cell|>|<cell|=>|<cell|x-H<rsup|2>(x)>>|<row|<cell|(I-H)(x+<big|sum><rsub|i=1><rsup|n+1>H<rsup|i>(x))>|<cell|\<equallim\>>|<cell|(I-H)(x+<big|sum><rsub|i=1><rsup|n>H<rsup|i>(x)+H<rsup|n+1>(x))>>|<row|<cell|>|<cell|=>|<cell|(I-H)(x+<big|sum><rsub|i=1><rsup|n>H<rsup|i>(x))+(I-H)(H<rsup|n+1>(x))>>|<row|<cell|>|<cell|\<equallim\><rsub|induction
hypothese>>|<cell|x-H<rsup|n+1>(x)+H<rsup|n+1>(x)-H(H<rsup|n+1>(x))>>|<row|<cell|>|<cell|=>|<cell|x-H<rsup|n+2>(x)>>>>
</eqnarray*>
Again by induction we prove that\
<\eqnarray*>
<tformat|<table|<row|<cell|(I+<big|sum><rsub|i=1><rsup|n>H<rsup|i>)(I-H)(x)>|<cell|=>|<cell|x-H<rsup|n+1>(x)>>>>
</eqnarray*>
<\eqnarray*>
<tformat|<table|<row|<cell|(I+<big|sum><rsub|i=1><rsup|1>H<rsup|i>)(I-H)(x)>|<cell|=>|<cell|(I+H)(x-H(x))>>|<row|<cell|>|<cell|=>|<cell|x-H(x)+H(x)-H(H(x))>>|<row|<cell|>|<cell|=>|<cell|x-H<rsup|2>(x)>>|<row|<cell|(I+<big|sum><rsub|i=1><rsup|n+1>H<rsup|i>)(I-H)(x)>|<cell|=>|<cell|I((I-H)(x))+<big|sum><rsub|i=1><rsup|n+1>H<rsup|i>((I-H)(x))>>|<row|<cell|>|<cell|=>|<cell|I((I-H)(x))+<big|sum><rsub|i=1><rsup|n>H<rsup|i>((I-H)(x))+H<rsup|n+1>((I-H)(x))>>|<row|<cell|>|<cell|=>|<cell|(I+<big|sum><rsub|i=1><rsup|n>H<rsup|i>)(I-H)(x)+H<rsup|n+1>(x)-H<rsup|n+2>(x)>>|<row|<cell|>|<cell|\<equallim\><rsub|induction>>|<cell|x-H<rsup|n+1>(x)+H<rsup|n+1>(x)-H<rsup|n+2>(x)>>|<row|<cell|>|<cell|=>|<cell|x-H<rsup|n+2>(x)>>>>
</eqnarray*>
From the above it follows that\
<\eqnarray*>
<tformat|<table|<row|<cell|((I-H)\<circ\>(I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)(x)<rsup|>>|<cell|\<equallim\><rsub|<reference|limit
of a sum>>>|<cell|(I-H)(lim<rsub|n\<rightarrow\>\<infty\>>(x+<big|sum><rsub|i=1><rsup|n>H<rsup|i>(x))>>|<row|<cell|>|<cell|\<equallim\><rsub|continuity
of I-H and <reference|limit and continuity>>>|<cell|lim<rsub|n\<rightarrow\>\<infty\>>((I-H)(x+<big|sum><rsub|i=1><rsup|n>H<rsup|i>(x))>>|<row|<cell|>|<cell|=>|<cell|lim<rsub|n\<rightarrow\>\<infty\>>(x-H<rsup|n+1>(x))>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|limit
of a sum>>>|<cell|x-lim<rsub|n\<rightarrow\>\<infty\>>H<rsup|n+1>(x)=x-0=x>>|<row|<cell|>|<cell|=>|<cell|I(x)>>|<row|<cell|((I+<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>)\<circ\>(I-H))(x)>|<cell|=>|<cell|I((I-H)(x))-<big|sum><rsub|n=1><rsup|\<infty\>>H<rsup|n>((I-H)(x))>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|limit
of a sum>>>|<cell|lim<rsub|n\<rightarrow\>\<infty\>>(I((I-H)(x)-<big|sum><rsub|i=1><rsup|n>H<rsup|i>((I-H)(x)))>>|<row|<cell|>|<cell|=>|<cell|lim<rsub|n\<rightarrow\>\<infty\>>((I+<big|sum><rsub|i=1><rsup|n>H<rsup|i>)(I-H)(x))>>|<row|<cell|>|<cell|=>|<cell|lim<rsub|n<rsub|\<rightarrow\>\<infty\>>>(x-H<rsup|n+1>(x))>>|<row|<cell|>|<cell|=>|<cell|x-lim<rsub|n\<rightarrow\>\<infty\>>H<rsup|n+1>(x)=x-0=x>>|<row|<cell|>|<cell|=>|<cell|I(x)>>>>
</eqnarray*>
So we have proved that for the continuous linear function
<math|I+<big|sum><rsub|i=1><rsup|\<infty\>>H<rsup|i>> we have
<math|(I-H)\<circ\>(I+<big|sum><rsub|i=1><rsup|\<infty\>>H<rsup|i>)=I=(I+<big|sum><rsub|i=1><rsup|\<infty\>>H<rsup|i>)\<circ\>(I-H)>
\;
</proof>
<\corollary>
<label|I+H has inverse if norm of H is less then 1>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a Banach space and
<math|H:X\<rightarrow\>X> a continuous linear function with
<math|\<shortparallel\>\<shortparallel\>\<less\>1> then
<math|I+H\<in\>Gl(E)>. Further <math|<big|sum><rsub|i=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>>
converges and <math|(I+H)<rsup|-1>=I+<big|sum><rsub|i=1><rsup|\<infty\>>(-1)H<rsup|n>>
</corollary>
<\proof>
If <math|\<shortparallel\>H\<shortparallel\>\<less\>1\<Rightarrow\>\<shortparallel\>-H\<shortparallel\>\<less\>1\<Rightarrowlim\><rsub|previous
theorem><big|sum><rsub|i=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>=<big|sum><rsub|i=1><rsup|\<infty\>>(-H)<rsup|n>>
is convergent and <math|(I+H)=(I-(-H))\<in\>Gl(E)> and
<math|I+H=I-(-H))=I+<big|sum><rsub|i=1><rsup|\<infty\>>(-H)<rsup|n>=I+<big|sum><rsub|i=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>>
</proof>
<\theorem>
<label|inverse mapping is c-inifinite>Let
<math|X,\<shortparallel\>\<shortparallel\>> be a Banach space then we
have\
<\enumerate>
<item><math|Gl(X)> is a open set in <math|X>
<item>The function <math|\<tau\>:Gl(x)\<rightarrow\>Gl(X)> defined by
<math|\<tau\>(L)=L<rsup|-1>> is differentiable of class
<math|C<rsup|\<infty\>>> and <math|D\<tau\>(L)> is defined by
<math|D\<tau\>(L)(H)=-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>> where
<math|L\<in\>Gl(X)> and <math|H\<in\>L(X,X)>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|L\<in\>Gl(X)> then for any <math|H\<in\>L(X,X)> we can
write
<\eqnarray*>
<tformat|<table|<row|<cell|H>|<cell|=>|<cell|L+(H-L)>>|<row|<cell|>|<cell|=>|<cell|L\<circ\>I+L\<circ\>L<rsup|-1>\<circ\>(H-L)>>|<row|<cell|>|<cell|=>|<cell|L\<circ\>(I+L<rsup|-1>\<circ\>(H-L))>>>>
</eqnarray*>
so if <math|\<shortparallel\>L<rsup|-1>\<circ\>(H-L)\<shortparallel\>\<less\>1>
then <reference|I+H has inverse if norm of H is less then 1> proves
that <math|I+L<rsup|-1>\<circ\>(H-L)\<in\>Gl(X)> and thus
<math|H=L\<circ\>(I+L<rsup|-1>\<circ\>(H-L))\<in\>Gl(x)>. Now according
to <reference|composition of continuous linear map is continuous> we
have <math|\<shortparallel\>L<rsup|-1>\<circ\>(H-L)\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsup|-1>\<shortparallel\>\<shortparallel\>L-H\<shortparallel\>>.
Take now <math|\<delta\>=max(1,\<shortparallel\>L<rsup|-1>\<shortparallel\>)>
then if <math|\<shortparallel\>H-L\<shortparallel\>\<less\><frac|1|\<delta\>>>
we have <math|\<delta\>\<shortparallel\>H-L\<shortparallel\>\<less\>1>
and thus <math|\<shortparallel\>L<rsup|-1>\<circ\>(H-L)\<shortparallel\>\<leqslant\>\<shortparallel\>L<rsup|-1>\<shortparallel\>\<shortparallel\>H-L\<shortparallel\>\<less\>\<delta\>\<shortparallel\>H-L\<shortparallel\>\<less\>1>
and thus <math|H\<in\>Gl(X)>. We have thus proved that
<math|B<rsub|\<shortparallel\>\<shortparallel\>>(L,<frac|\<delta\>|2>)={H\<in\>L(X,X)\|\<shortparallel\>H-L\<shortparallel\>\<less\><frac|1|\<delta\>>}\<subseteq\>Gl(X)>
proving that <math|Gl(X)> is open
<item>Next we prove that <math|\<tau\>> is differentiable at every
<math|L\<in\>Gl(X)>
<\enumerate>
<item>First we prove that <math|\<tau\>> is differentiable at
<math|I\<in\>Gl(X)>, to do this note that if
<math|\<shortparallel\>H\<shortparallel\>\<less\>1> then using
<reference|I+H has inverse if norm of H is less then 1>
<math|I+H\<in\>Gl(X)> and <math|(I+H)<rsup|-1>=I+<big|sum><rsub|n=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>>.
So we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<tau\>(I+H)-\<tau\>(I)>|<cell|=>|<cell|(I+H)<rsup|-1>-I>>|<row|<cell|>|<cell|=>|<cell|I+<big|sum><rsub|n=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>-I>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|n=1><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|subserie
property>>>|<cell|-H+<big|sum><rsub|n=2><rsup|\<infty\>>(-1)<rsup|n>H<rsup|n>>>>>
</eqnarray*>
Define now <math|\<varepsilon\>(H)\<in\>L(E,E)> by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(0)>|<cell|=>|<cell|0,H=0>>|<row|<cell|\<varepsilon\>(H)>|<cell|=>|<cell|<big|sum><rsub|n=2><rsup|\<infty\>>(-1)<rsup|n><frac|H<rsup|n>|\<shortparallel\>H\<shortparallel\>>,H\<neq\>0>>>>
</eqnarray*>
So we have defined <math|\<varepsilon\>:B<rsub|\<shortparallel\>\<shortparallel\>>(I,1)<rsub|I>=B<rsub|\<shortparallel\>\<shortparallel\>>(0,1)\<rightarrow\>L(X,X)>\
Now if <math|0\<less\>\<shortparallel\>H\<shortparallel\>\<less\>1>
then we have as <math|n\<in\>{1,\<ldots\>,\<infty\>}>
<math|\<shortparallel\>(-1)<rsup|n>H<rsup|n>(x)\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>>
we can use <reference|example of a convergent serie> and
<reference|convergent criteria of a serie in a Banach space> and
<reference|subserie property> to prove that
<math|\<shortparallel\><big|sum><rsub|i=2><rsup|\<infty\>>(-1)<rsup|i>H<rsup|i>(x)\<shortparallel\>\<leqslant\><big|sum><rsub|i=2><rsup|\<infty\>>(\<shortparallel\>H\<shortparallel\><rsup|n>\<shortparallel\>x\<shortparallel\>)\<Rightarrowlim\><rsub|<reference|linear
combination of convergent series>>\<shortparallel\><big|sum><rsub|i=2><rsup|\<infty\>>(-1)<rsup|i><frac|H<rsup|i>(x)|\<shortparallel\>H\<shortparallel\>>\<leqslant\><frac|\<shortparallel\>x\<shortparallel\>|\<shortparallel\>H\<shortparallel\>><big|sum><rsub|i=2><rsup|\<infty\>>\<shortparallel\>H\<shortparallel\><rsup|n>=<frac|1|\<shortparallel\>H\<shortparallel\>>(<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>-\<shortparallel\>H\<shortparallel\>)\<shortparallel\>x\<shortparallel\>=<frac|\<shortparallel\>H\<shortparallel\>-\<shortparallel\>H\<shortparallel\>+\<shortparallel\>H\<shortparallel\><rsup|2>|\<shortparallel\>H\<shortparallel\>(1-\<shortparallel\>H\<shortparallel\>)>\<shortparallel\>x\<shortparallel\>=<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>\<shortparallel\>x\<shortparallel\>\<Rightarrow\>\<shortparallel\><big|sum><rsub|i=1><rsup|\<infty\>>(-1)<rsup|i><frac|H<rsup|i>|\<shortparallel\>H\<shortparallel\>>\<shortparallel\>\<leqslant\><frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>>
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varepsilon\>(H)\<shortparallel\>>|<cell|\<leqslant\>>|<cell|<frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>>>>>
</eqnarray*>
and given <math|\<varepsilon\>\<gtr\>0> choose
<math|0\<less\>\<shortparallel\>H\<shortparallel\>\<less\>min(<frac|\<varepsilon\>|2>,<frac|1|2>)>
then <math|\<shortparallel\>\<varepsilon\>(H)-0\<shortparallel\>=\<shortparallel\>\<varepsilon\>(H)\<shortparallel\>\<leqslant\><frac|\<shortparallel\>H\<shortparallel\>|1-\<shortparallel\>H\<shortparallel\>>\<leqslant\><frac|<frac|<frac|\<varepsilon\>|2>|>|1-<frac|1|2>>=\<varepsilon\>>
proving that <math|\<varepsilon\>(H)> is continuous
<math|0\<in\>L(E,E)> and we have also\
<\eqnarray*>
<tformat|<table|<row|<cell|\<tau\>(I+H)-\<tau\>(I)-(-H)>|<cell|=>|<cell|\<varepsilon\>(H)\<shortparallel\>H\<shortparallel\>>>>>
</eqnarray*>
proving that <math|D\<tau\>(I)(H)=-H=-(I<rsup|-1>\<circ\>H\<circ\>I<rsup|-1>)>
(this is trivial a linear continuous function)
<item>Next we prove that <math|\<tau\>> is differentiable at a
arbitrary <math|L\<in\>Gl(X)> so take <math|H\<in\>L(X,X)> such that
<math|><math|\<shortparallel\>H\<shortparallel\>\<less\><frac|1|max(1,\<shortparallel\>L<rsup|-1>\<\|\|\>)>=\<delta\>>
then as
<\eqnarray*>
<tformat|<table|<row|<cell|(1) L+H>|<cell|=>|<cell|(I+H\<circ\>L<rsup|-1>)\<circ\>L>>>>
</eqnarray*>
we have then <math|\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\>\<shortparallel\>L<rsup|-1>\<shortparallel\>\<less\>1>
and thus <math|I+H\<circ\>L<rsup|-1>\<in\>Gl(X)> and as
<math|L\<in\>Gl(X)> we have also <math|L+H\<in\>Gl(X)> and
<math|(L+H)<rsup|-1>> exists. So
<\eqnarray*>
<tformat|<table|<row|<cell|\<tau\>(L+H)-\<tau\>(H)>|<cell|=>|<cell|(L+H)<rsup|-1>-L<rsup|-1>>>|<row|<cell|>|<cell|\<equallim\><rsub|(1)>>|<cell|L<rsup|-1>\<circ\>[(I+H\<circ\>L<rsup|-1>)<rsup|-1>-I]>>|<row|<cell|>|<cell|=>|<cell|L<rsup|-1>\<circ\>[\<tau\>(I+H\<circ\>L<rsup|-1>)-\<tau\>(I)]>>|<row|<cell|>|<cell|\<equallim\><rsub|(b)>>|<cell|L<rsup|-1>\<circ\>[D\<tau\>(I)(H\<circ\>L<rsup|-1>)+\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>]>>|<row|<cell|>|<cell|\<equallim\><rsub|(b)>>|<cell|L<rsup|-1>\<circ\>[-(H\<circ\>L<rsup|-1>)+\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>]>>|<row|<cell|>|<cell|=>|<cell|-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>+L<rsup|-1>\<circ\>\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>>>>>
</eqnarray*>
This gives then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<tau\>(L+H)-\<tau\>(H)-(-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>)>|<cell|=>|<cell|L<rsup|-1>\<circ\>\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>>>>>
</eqnarray*>
Define now the function <math|\<varepsilon\><rprime|'>:B<rsub|\<shortparallel\>\<shortparallel\>>(L,\<delta\>)<rsub|L>=B<rsub|\<shortparallel\>\<shortparallel\>>(0,\<delta\>)\<rightarrow\>L(X,X)>
by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\><rprime|'>(H)>|<cell|=>|<cell|L<rsup|-1>\<circ\>\<varepsilon\>(H\<circ\>L<rsup|-1>)<frac|\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>|\<shortparallel\>H\<shortparallel\>>,
H\<neq\>0>>|<row|<cell|>|<cell|=>|<cell|0, H=0>>>>
</eqnarray*>
Note that <math|\<varepsilon\>\<in\>L(X,X)> and
<math|L<rsup|-1>\<in\>Gl(X)\<in\>L(X,X)> and also
<math|0\<in\>L(X,X)> we have <math|\<varepsilon\>(H)\<in\>L(X,x)> as
we should have
Then if <math|\<varepsilon\>\<gtr\>0> we can find a
<math|\<delta\><rsub|1>\<gtr\>0> such that if
<math|\<shortparallel\>H<rprime|'>\<shortparallel\>\<less\>\<delta\><rsub|1>>
then <math|\<shortparallel\>\<varepsilon\>(H<rprime|'>)\<shortparallel\>\<less\><frac|\<varepsilon\>|max(\<shortparallel\>L<rsup|-1>\<shortparallel\><rsup|2>,1)>>
then if <math|\<shortparallel\>H\<shortparallel\>\<less\><frac|\<delta\><rsub|1>|max(1,\<shortparallel\>L<rsup|-1>\<shortparallel\>>>
we have <math|\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>\<leqslant\>\<shortparallel\>H\<shortparallel\>\<shortparallel\>L<rsup|-1>\<shortparallel\>\<less\>\<delta\><rsub|1>>
and thus <math|\<shortparallel\>\<varepsilon\><rprime|'>(H)\<shortparallel\>=\<shortparallel\>L<rsup|-1>\<circ\>\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\><frac|\<shortparallel\>H\<circ\>L<rsup|-1>\<shortparallel\>|\<shortparallel\>H\<shortparallel\>>\<leqslant\>\<shortparallel\>L<rsup|-1>\<shortparallel\>\<shortparallel\>\<varepsilon\>(H\<circ\>L<rsup|-1>)\<shortparallel\><frac|\<shortparallel\>H\<shortparallel\>\<shortparallel\>L<rsup|-1>\<shortparallel\>|\<shortparallel\>H\<shortparallel\>>\<less\>\<shortparallel\>L<rsup|-1>\<shortparallel\><rsup|2><frac|\<varepsilon\>|max(\<shortparallel\>L<rsup|-1>\<shortparallel\><rsup|2>,1)>>
proving that <math|><math|\<varepsilon\><rprime|'>> is continue at
<math|0>. If we prove now that <math|-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>:L(X,X)\<rightarrow\>L(X,X)>
is linear and continue we have proved (b)
<\eqnarray*>
<tformat|<table|<row|<cell|(-L<rsup|-1>\<circ\>(\<alpha\>H+\<beta\>H<rprime|'>)\<circ\>L<rsup|-1>)(x)>|<cell|=>|<cell|(-L<rsup|-1>((\<alpha\>H+\<beta\>H<rprime|'>)(L<rsup|-1>(x)))>>|<row|<cell|>|<cell|=>|<cell|(-L<rsup|-1>(\<alpha\>H(L<rsup|-1>(x))+\<beta\>H<rprime|'>(L<rsup|-1>(x)))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(-L<rsup|-1>(H(L<rsup|-1>(x))))+\<beta\>(-L<rsup|-1>(H(L<rsup|-1>(x))))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>)(x)+\<beta\>(-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>)(x)>>>>
</eqnarray*>
So <math|-L<rsup|-1>\<circ\>(\<alpha\>H+\<beta\>H<rprime|'>)\<circ\>L<rsup|-1>=\<alpha\>(-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>)+\<beta\>(-L\<circ\>H<rprime|'>\<circ\>L<rsup|-1>)>
and thus we have proved linearity. Also
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>\<shortparallel\>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>L<rsup|-1>\<shortparallel\><rsup|2>\<shortparallel\>H\<shortparallel\>>>>>
</eqnarray*>
proving continuity. So we have <math|D\<tau\>(L)(H)=-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>>
</enumerate>
<item>Now to prove that <math|\<tau\>> is <math|C<rsup|\<infty\>>>
define the following functions (and prove that they are
<math|C<rsup|\<infty\>>>)
<\enumerate>
<item><math|><math|\<chi\>:Gl(X)\<rightarrow\>L(E,E)\<times\>L(E,E)>
by <math|\<chi\>(L)=(L,L)>. Then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<chi\>(\<alpha\>L+\<beta\>L<rprime|'>)>|<cell|=>|<cell|(\<alpha\>L+\<beta\>L<rprime|'>,\<alpha\>L+\<beta\>L<rprime|'>)>>|<row|<cell|>|<cell|=>|<cell|(\<alpha\>L,\<alpha\>L)+(\<beta\>L<rprime|'>,\<beta\>L<rprime|'>)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(L,L)+\<beta\>(L<rprime|'>,L<rprime|'>)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(L)+\<beta\>(L<rprime|'>)>>>>
</eqnarray*>
proving that <math|\<chi\>> is linear. Also we have if we use the
maximum norm <math|\<shortparallel\>(x,y)\<shortparallel\><rsub|m>=max(\<shortparallel\>x\<shortparallel\>,\<shortparallel\>y\<shortparallel\>)>
on <math|L(E,E)\<times\>L(E,E)> where
<math|\<shortparallel\>\<shortparallel\>> is the operator norm
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<chi\>(L)\<shortparallel\><rsub|m>>|<cell|=>|<cell|max(\<shortparallel\>L\<shortparallel\>,\<shortparallel\>L\<shortparallel\>)>>|<row|<cell|>|<cell|=>|<cell|\<shortparallel\>L\<shortparallel\>>>>>
</eqnarray*>
proving that <math|\<chi\>> is continuous and (with
<math|\<shortparallel\>\<chi\>\<shortparallel\>\<leqslant\>1>). Using
<reference|linear continuous mappings are C infinity> we have then
that <math|\<chi\>> is <math|C<rsup|\<infty\>>>
<item><math|\<rho\>:L(E,E)\<times\>L(E,E)\<rightarrow\>L(L(E,E),L(E,E))>
defined by <math|\<rho\>(L,L<rprime|'>):L(E,E)\<rightarrow\>L(E,E)>
which is defined by <math|\<rho\>(L,L<rprime|'>)(H)=-L\<circ\>H\<circ\>L<rprime|'>>.
Then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<rho\>(\<alpha\>L+\<beta\>L<rprime|'>,L<rprime|''>)(H)>|<cell|=>|<cell|-(\<alpha\>L+\<beta\>L<rprime|'>)\<circ\>H\<circ\>L<rprime|''>>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(-L\<circ\>H\<circ\>L<rprime|''>)+\<beta\>(-L<rprime|'>\<circ\>H\<circ\>L<rprime|''>)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>\<rho\>(L,L<rprime|''>)+\<beta\>\<rho\>(L<rprime|'>,L<rprime|''>)>>>>
</eqnarray*>
proving that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<rho\>(\<alpha\>L+\<beta\>L<rprime|'>,L<rprime|'>)>|<cell|=>|<cell|\<alpha\>\<rho\>(L)+\<beta\>\<rho\>(L<rprime|'>,L)>>>>
</eqnarray*>
and similar we prove that
<\eqnarray*>
<tformat|<table|<row|<cell|\<rho\>(L,\<alpha\>L<rprime|'>+\<beta\>L<rprime|''>)>|<cell|=>|<cell|\<alpha\>\<rho\>(L,L<rprime|'>)+\<beta\>\<rho\>(L,L<rprime|''>)>>>>
</eqnarray*>
proving that <math|\<rho\>> is multilinear. Also we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<rho\>(L,L<rprime|'>)(H)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\>-L\<circ\>H\<circ\>L<rprime|'>\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>L\<shortparallel\>\<shortparallel\>L<rprime|'>\<shortparallel\>\<shortparallel\>H\<shortparallel\>>>>>
</eqnarray*>
proving that <math|\<rho\>(L,L<rprime|'>)> is a continuous linear
function and that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<rho\>(L,L<rprime|'>)>|<cell|\<leqslant\>>|<cell|\<shortparallel\>L\<shortparallel\>\<shortparallel\>L<rprime|'>\<shortparallel\>>>>>
</eqnarray*>
proving that <math|\<rho\>> is continuous (and
<math|\<shortparallel\>\<rho\>\<shortparallel\>\<leqslant\>1>) and
thus by <reference|billinear continuous mappings are C infinity> we
have that <math|\<rho\>> is <math|C<rsup|\<infty\>>>
<item>Now given a <math|L\<in\>Gl(E)> take
<math|(\<rho\>\<circ\>\<chi\>\<circ\>\<tau\>)(L)> and apply it on
<math|H\<in\>Gl(X)> then\
<\eqnarray*>
<tformat|<table|<row|<cell|[(\<rho\>\<circ\>\<chi\>\<circ\>\<tau\>)(L)](H)>|<cell|=>|<cell|[\<rho\>(\<chi\>(\<tau\>(L)))](H)>>|<row|<cell|>|<cell|=>|<cell|[\<rho\>(\<chi\>(L<rsup|-1>))](H)>>|<row|<cell|>|<cell|=>|<cell|\<rho\>(L<rsup|-1>,L<rsup|-1>)(H)>>|<row|<cell|>|<cell|=>|<cell|-L<rsup|-1>\<circ\>H\<circ\>L<rsup|-1>>>|<row|<cell|>|<cell|=>|<cell|D\<tau\>(L)(H)>>>>
</eqnarray*>
proving that <math|D\<tau\>(L)=(\<rho\>\<circ\>\<chi\>\<circ\>\<tau\>)(L)>
or that <math|D\<tau\>=\<rho\>\<circ\>\<chi\>\<circ\>\<tau\>> and
must be continuous because <math|\<rho\>,\<chi\>,\<tau\>> are
continuous
<item>Finally to prove that <math|\<tau\>> is
<math|C<rsup|\<infty\>>> we use induction to prove that for every
<math|n\<in\>\<bbb-N\><rsub|0>> we have that <math|\<tau\>> is
<math|C<rsup|n>>
<\enumerate>
<item>In <math|2, 3 (a-c)> we have actually proved that
<math|\<tau\>=C<rsup|1>> so the theorem is true for <math|n=1>
<item>Assume that <math|\<tau\>> is <math|C<rsup|n>> then as
<math|D\<tau\>=\<rho\>\<circ\>\<chi\>\<circ\>\<tau\>> where
<math|\<rho\>,\<chi\>> are <math|C<rsup|\<infty\>>> thus by
<reference|generalized chain rule> we have that <math|D\<tau\>> is
<math|C<rsup|n>> and then by <reference|proving f is Cn based on
Cn-1 of Df> we have that <math|\<tau\>> is <math|C<rsup|n+1>>\
</enumerate>
</enumerate>
</enumerate>
\;
</proof>
<\definition>
<index|diffeomorphism>Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>>
and <math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over <math|\<bbb-K\>>, <math|U\<subseteq\>X,V\<subseteq\>Y> open
sets then a bijective function <math|f:U\<rightarrow\>V> is a
diffeomorphism (of class <math|C<rsup|r>>) if and only if
<math|f,f<rsup|-1>> are differentiable (of class <math|C<rsup|r>>). If
such a function exists we say that <math|U> and <math|V> are diffeomorph
(of class <math|C<rsup|r>>)
</definition>
<\note>
Using <reference|characterization of a bijective mapping> we see that
<math|f> is bijective if and only if there exists a
<math|g:V\<rightarrow\>U> and <math|f\<circ\>g=i<rsub|V>,g\<circ\>f=i<rsub|U>>
and <math|g=f<rsup|-1>> so we can reformulate our definition and say that
<math|f> is diffeomorph (of class <math|C<rsup|r>>) if and only if
<math|f> is differentiable (of class <math|C<rsup|r>>) and there exists a
function <math|g:V\<rightarrow\>U> that is differentiable (of class
<math|C<rsup|r>>) such that <math|g\<circ\>f=i<rsub|U>> and
<math|f\<circ\>g=i<rsub|V>>
</note>
<\note>
<label|inverse of diffeomorphisme>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over <math|\<bbb-K\>>, <math|U\<subseteq\>X,V\<subseteq\>Y> open
sets and \ <math|f:U\<rightarrow\>V> a diffeomorphism and
<math|g=f<rsup|-1>:V\<rightarrow\>U> then if we take <math|x\<in\>U> with
<math|y=f(x)> then we have <math|Dg(y)\<circ\>Df(x)=1<rsub|X>> and
<math|Df(x)\<circ\>Dg(y)=1<rsub|Y>> proving that <math|Df(x)> and
<math|Df(y)> are toplinear isomorphisms and that
<math|Df(x)<rsup|-1>=Dg(y)>
</note>
<\proof>
\;
<\enumerate>
<item>From <math|f\<circ\>g=1<rsub|V>> and the fact that
<math|D1<rsub|V>=1<rsub|Y>> we have using the chain rule
<math|1<rsub|Y>=D1<rsub|V>(y)=D(f\<circ\>g)(y)=Df(g(y))\<circ\>Dg(y)=Df(x)\<circ\>Dg(y)>
<item>From <math|g\<circ\>f=1<rsub|U>> and the fact that
<math|1<rsub|X>=D1<rsub|U>(x)=D(g\<circ\>f)(x)=Dg(f(x))\<circ\>Df(x)=Dg(y)\<circ\>Df(x)><math|>
</enumerate>
</proof>
<\lemma>
Let <math|X,\<shortparallel\>\<shortparallel\>>,
<math|Y,\<shortparallel\>\<shortparallel\>,
Z,\<shortparallel\>\<shortparallel\>> be normed spaces over
<math|\<bbb-K\>,U\<subseteq\>X,V\<subseteq\>Y,W\<subseteq\>Z> and
<math|f:U\<rightarrow\>V,g:V\<rightarrow\>W> diffeomorphismes (of class
<math|C<rsup|r>>) then <math|f\<circ\>g:U\<rightarrow\>W> is a
diffeomorphism (of class <math|C<rsup|r>>
</lemma>
<\proof>
This follows trivial from <reference|composition of bijective mappings>
and the chain rule <reference|generalized chain rule>
</proof>
<\definition>
Let <math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be normed vector
spaces over <math|\<bbb-K\>>, <math|U\<subseteq\>X,V\<subseteq\>Y> open
then a differentiable function (of class <math|C<rsup|r>>) is said to be
a local diffeomorphism (of class <math|C<rsup|r>>) at <math|x\<in\>U> if
there exists a <math|U(x)> open with <math|x\<in\>U(x)\<subseteq\>U> and
<math|V(f(x))> open with <math|f(x)\<in\>V(f(x))\<subseteq\>V> such that
the restriction <math|f<rsub|\|U(x)>:U(x)\<rightarrow\>V(f(x))> is a
diffeomorphism (of class <math|C<rsup|r>>)
</definition>
<section|The inverse function theorem>
<\theorem>
<label|The inversion theorem><dueto|The Inverse Function
Theorem><index|Inverse function theorem>Let
<math|X,\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be Banach spaces over
<math|\<bbb-R\>>, <math|U\<subseteq\>X> a open set and
<math|f:U\<rightarrow\>Y> a differentiable function of class
<math|C<rsup|r>,r\<in\>\<bbb-N\><rsub|0>>. Assume that
<math|x<rsub|0>\<in\>U> is that <math|Df(x<rsub|0>):X\<rightarrow\>Y> is
a isomorphism. Then <math|f> is a local diffeomorphism of class
<math|C<rsup|r>> at <math|x<rsub|0>\<in\>U>.
In other words there exists open sets <math|U(x<rsub|0>),V(f(x<rsub|0>))>
such that <math|x<rsub|0>\<in\>U(x<rsub|0>)\<subseteq\>U,f(x<rsub|0>)\<in\>V(f(x<rsub|0>))\<subseteq\>Y>
such that <math|f<rsub|\|U(x<rsub|0>)>:U(x<rsub|0>)\<rightarrow\>V(f(x<rsub|0>))>
is a diffeomorphism of class <math|C<rsup|r>>
</theorem>
<\proof>
The proof is quite long and is done in different stages
First note that <math|Df(x<rsub|0>)> is linear and continuous because
<math|f> is of class <math|C<rsup|r>> so that by <reference|in Banach
spaces linear continuous mappings are toplinear> we have that
<math|Df(x<rsub|0>)> is a toplinear isomorphism
<\proposition>
First it is proved that if the theorem is true for <math|x<rsub|0>=0>
then it is true for general <math|x<rsub|0>>.
</proposition>
<\proof>
First given <math|x\<in\>X> consider the translation
<math|\<tau\><rsub|x>:X\<rightarrow\>X> defined by
<math|\<tau\><rsub|x>(t)=x+t> then by <reference|translation is
C-inifinity> we have that <math|\<tau\><rsub|x>> is differentiable on
<math|X> of class on <math|C<rsup|\<infty\>>> and as trivially
<math|\<tau\><rsub|x><rsup|-1>=\<tau\><rsub|\<um\>x>> we have that a
translation is actually a diffeomorphism. Note also that if
<math|W\<subseteq\>X>, <math|W> is open then the restriction (which is
bijective) <math|\<tau\><rsub|x<rsub|\|W>><rsub|>:W\<rightarrow\>x+W=\<tau\><rsub|x<rsub|\|W>>(W)>
is differentiable on <math|W> of class <math|C<rsup|\<infty\>>> (see
<reference|restriction of C-r mapping>) and thus a diffeomorphism, note
also that <math|x+W> is open (see <reference|normed space properties>).
Now <math|\<tau\><rsub|x<rsub|0>>(\<um\>x<rsub|0>+U)=U> so
<math|\<tau\><rsub|x<rsub|0\|>\<um\>x<rsub|0>+U>:\<um\>x<rsub|0>+U\<rightarrow\>U>
is a diffeomorphism of class <math|C<rsup|\<infty\>>>. Consider now
<math|h=f\<circ\>\<tau\><rsub|x<rsub|0<rsub|\|\<um\>x<rsub|0>+U>>>:\<um\>x<rsub|0>+U\<rightarrow\>Y>
then <math|h(0)=f(\<tau\><rsub|x<rsub|0\|-x<rsub|0>+U>>(0))=f(\<tau\><rsub|x<rsub|0>>(0))=f(x<rsub|0>)>
and <math|h> is is differentiable of class <math|C<rsup|r>> and by our
hypothesis there exists open sets <math|U<rprime|'>(0),V(f(x<rsub|0>))>
such that <math|h<rsub|\|U<rprime|'>(0)>:U<rprime|'>(0)\<rightarrow\>V(f(x<rsub|0>))>
is a diffeomorphism of class <math|C<rsup|r>>. As
<math|\<tau\><rsub|\<um\>x<rsub|0>>(x<rsub|0>+U<rprime|'>(0))=U<rprime|'>(0)>
we have that <math|><math|\<tau\><rsub|\<um\>x<rsub|0<rsub|\|x<rsub|0>+U<rprime|'>(0)>>>:x<rsub|0>+U<rprime|'>(0)\<rightarrow\>U<rprime|'>(0)>
is a diffeomorphism of class <math|C<rsup|r>>and thus
<math|g=h<rsub|\|U<rprime|'>(0)>\<circ\>\<tau\><rsub|\<um\>x<rsub|0\|x<rsub|0>+U<rprime|'>(0)>>:x<rsub|0>+U<rprime|'>(0)\<rightarrow\>V(f(x<rsub|0>))>
is a diffeomorphism of class <math|C<rsup|r>>. Now if
<math|x\<in\>x<rsub|0>+U<rprime|'>(0)> then
<math|g(x)=h<rsub|\|U<rprime|'>(0)>(\<tau\><rsub|\<um\>x<rsub|0\|x<rsub|0>+U<rprime|'>(0)>>(x))=h<rsub|\|U<rprime|'>(0)>(\<tau\><rsub|x<rsub|0>>(x))=h<rsub|\|U<rprime|'>(0)>(\<um\>x<rsub|0>+x)=h(\<um\>x<rsub|0>+x)=f(\<tau\><rsub|x<rsub|0\|<rsub|\<um\>x<rsub|0>+U>>>(\<um\>x<rsub|0>+x)=f(\<tau\><rsub|x<rsub|0>>(x<rsub|0>+x))=f(x)>
or <math|g=f<rsub|\|x<rsub|0>+U<rprime|'>(0)>> is a diffeomorphism of
class <math|C<rsup|r>> from the open set
<math|x<rsub|0>+U<rprime|'>(0)\<rightarrow\>V(f(x<rsub|0>))> proving
our proposition
</proof>
So we only have to consider in our prove the case of a
<math|f:U\<rightarrow\>Y> with <math|0\<in\>U> so that
<math|Df(0):X\<rightarrow\>Y> is a toplinear isomorphism
<\proposition>
Next if the theorem is proved for a <math|f:U\<rightarrow\>Y> with
<math|f(0)> then it is also proved for the case <math|f(0)\<neq\>0>
</proposition>
<\proof>
Let <math|f(0)\<neq\>0> consider then the diffeomorphism of class
<math|C<rsup|\<infty\>>> <math|\<tau\><rsub|\<um\>f(0)>:Y\<rightarrow\>Y>
then <math|h=\<tau\><rsub|\<um\>f(0)>\<circ\>f:U\<rightarrow\>Y> is a
diffeomorphism of class <math|C<rsup|r>> with
<math|h(0)=\<tau\><rsub|\<um\>f(0)>(f(0))=\<um\>f(0)+f(0)=0> so by our
hypothesis there exists open sets <math|U(0),V<rprime|'>(0)> such that
<math|h<rsub|\|U(0)>:U(0)\<rightarrow\>V<rprime|'>(0)> is a
diffeomorphism of class <math|C<rsup|r>>. Then as
<math|\<tau\><rsub|f(0)<rsub|\|V<rprime|'>(0)>>:V<rprime|'>(0)\<rightarrow\>f(0)+V<rprime|'>(0)>
is a diffeomorphism of class <math|C<rsup|\<infty\>>> we have that
<math|g=\<tau\><rsub|f(0)<rsub|\|V<rprime|'>(0)>>\<circ\>h<rsub|\|U(0)>:U(0)\<rightarrow\>f(0)+V<rprime|'>(0)>
is a diffeomorphism of class <math|C<rsup|r>>. Now if
<math|x\<in\>U(0)> then <math|g(x)=\<tau\><rsub|f(0)<rsub|\|V<rprime|'>(0)>>(h<rsub|\|U(0)>(x)=\<tau\><rsub|f(0)>(h(x))=\<tau\><rsub|f(0)>(\<tau\><rsub|\<um\>f(0)>(f(x))=f(x)>
or <math|g=f<rsub|\|U(0)>> proving that
<math|f<rsub|\|U(0)>:U(0)\<rightarrow\>f(0)+V<rprime|'>(0)> is a
<math|C<rsup|r>> diffeomorphism
</proof>
Now we only have to prove the the inverse function theorem for the case
of a <math|f:U\<rightarrow\>Y> with <math|f(0)=0>\
<\proposition>
If the theorem is proved for a <math|f:U\<rightarrow\>X> with
<math|Df(0)=1<rsub|X>> then it also true for the more general case
<math|f:U\<rightarrow\>Y> with general <math|Df(0)>
</proposition>
<\proof>
Let <math|f:U\<rightarrow\>Y> be differentiable on <math|U> of class
<math|C<rsup|r>> with differential <math|Df(0):X\<rightarrow\>Y> now
<math|Df(0)> is toplinear and so <math|Df(0)<rsup|-1>:Y\<rightarrow\>X>
is defined, linear and continuous and thus by <reference|linear
continuous mappings are C infinity> is <math|C<rsup|\<infty\>>>. Define
then <math|h=Df(0)<rsup|-1>\<circ\>f:U\<rightarrow\>X> which is
differentiable on <math|U> of class <math|C<rsup|r>>,
<math|h(0)=Df(0)<rsup|-1>(f(0)=Df(0)<rsup|-1>\<equallim\><rsub|linearity>0>
and <math|Dh(0)=D(Df(0)<rsup|-1>)(f(0))\<circ\>Df(0)=Df(0)<rsup|-1>\<circ\>Df(0)=i<rsub|X>>
so our hypothesis applies and we find open sets
<math|U(0),V<rprime|'>(0)> such that
<math|h<rsub|\|U(0)>:U(0)\<rightarrow\>V<rprime|'>(0)> is a
diffeomorphism of class <math|C<rsup|r>>. Then as <math|Df(0)> is a
toplinear isomorphism it is by <reference|linear continuous mappings
are C infinity> differentiable on <math|X> of class <math|C<rsup|r>>
and thus also <math|Df(0)<rsub|\|V<rprime|'>(0)>:V<rprime|'>(0)\<rightarrow\>Df(0)(V<rprime|'>(0))=V(0)>
is a diffeomorphism of class <math|C<rsup|\<infty\>>>, also
<math|0\<in\>V<rprime|'>(0)\<Rightarrow\>Df(0)(0)=0\<Rightarrow\>0\<in\>V(0)>
and because <math|Df(0)> is toplinear we have also that <math|V(0)> is
open. Consider now <math|g=Df(0)<rsub|\|V<rprime|'>(0)>\<circ\>h<rsub|\|U(0)>:U(0)\<rightarrow\>Df(0)(V<rprime|'>(0))>
which is a diffeomorphism of class <math|C<rsup|r>>. Then if
<math|x\<in\>U(0)> then we have <math|g(x)=Df(0)<rsub|\|V<rprime|'>(0)>(h<rsub|\|U(0)>(x))=Df(0)(h(x)=Df(0)(Df(0)<rsup|-1>(f(x)))=f(x)>
so <math|f<rsub|\|U(0)>=h> and thus
<math|f<rsub|\|U(0)>:U(0)\<rightarrow\>V(0)> is a diffeomorphism of
class <math|C<rsup|r>>
</proof>
So to prove our theorem we only have to prove that if
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> is a Banach space and
<math|U\<subseteq\>X> a open set then if <math|f:U\<rightarrow\>X> is a
function that is differentiable on <math|U> of class
<math|C<rsup|r>,r\<geqslant\>1> with <math|f(0)=0> and
<math|Df(0)=i<rsub|X>> then there exists open sets <math|U(0),V(0)> such
that <math|f<rsub|\|U(0)>:U(0)\<rightarrow\>V(0)> is a diffeomorphism of
class <math|C<rsup|r>>\
\;
Define now <math|T:U\<rightarrow\>X> by <math|x\<rightarrow\>T(x)=x-f(x)>
then <math|T(0)=0> and <math|T> is of class <math|C<rsup|r>> by
<reference|sum of C-r functions is C-r>, so <math|DT> is continue and
<math|DT(0)=Dx(0)-Df(0)=i<rsub|X>-1<rsub|X>=0>. Hence we can choose a
<math|\<rho\><rprime|'>\<gtr\>0> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\><rprime|'>)\<subseteq\>U>
and if <math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\><rprime|'>)\<Rightarrow\>\<shortparallel\>DT(x)\<shortparallel\>\<less\><frac|1|2>>
(and thus also <math|\<leqslant\>>), take then
<math|\<delta\>=<frac|\<delta\><rprime|'>|2>> then
<math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)\<subseteq\>B<rsub|\<shortparallel\>>(0,\<rho\><rprime|'>)\<subseteq\>U
>
<\eqnarray*>
<tformat|<table|<row|<cell|(1) \<shortparallel\>DT(x)\<shortparallel\>\<leqslant\><frac|1|2>\<less\>1>|<cell|for>|<cell|\<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>\<rho\>>>>>
</eqnarray*>
Then as <math|DT(x)=i<rsub|X>-Df(x)\<Rightarrow\>Df(x)=i<rsub|X>-DT(x)>
we apply <reference|I-H has inverse if norm of H is lower then 1> to find
that\
<\eqnarray*>
<tformat|<table|<row|<cell|(2) Df(x) is a toplinear isomorphism if
x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)
or \<shortparallel\>x\<shortparallel\><rsub|X>\<leqslant\>\<rho\>>|<cell|>|<cell|>>>>
</eqnarray*>
<math|Df(x)> is a toplinear isomorphism if
<math|\<shortparallel\>x\<shortparallel\>\<leqslant\>\<rho\>>. Now using
the fact that <math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
is convex (<reference|balls are convex>), (1) and the fundamental theorem
of calculus (<reference|the fundamental theorem of calculus (3)>) we find
that if <math|x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
<\eqnarray*>
<tformat|<table|<row|<cell|(3) \<shortparallel\>T(x)\<shortparallel\>>|<cell|=>|<cell|\<shortparallel\>T(x)-T(0)\<shortparallel\>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|1|2>\<shortparallel\>x-0\<shortparallel\>>>|<row|<cell|>|<cell|=>|<cell|<frac|1|2>\<shortparallel\>x\<shortparallel\>>>>>
</eqnarray*>
we prove now the following proposition
<\proposition>
Let <math|0\<less\>\<sigma\>\<leqslant\>\<rho\>> then for every
<math|y\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<sigma\>|2>)
> there exists a uniquely determined
<math|x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
such that <math|f(x)=y>\
</proposition>
<\proof>
First note that <math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\>>(0,\<sigma\>)>
is closed and thus by (<reference|closed balls are closed>) is
complete, also <math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
\ Let <math|y\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<sigma\>|2>)\<Rightarrow\>\<shortparallel\>y\<shortparallel\>\<leqslant\><frac|\<sigma\>|2>>
consider now the function <math|T<rsub|y>:<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\>>(0,\<sigma\>)\<rightarrow\>X>
defined by <math|\<forall\>x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
we have
<\eqnarray*>
<tformat|<table|<row|<cell|(4) T<rsub|y>(x)<rsub|>>|<cell|=>|<cell|y+T(x)>>|<row|<cell|>|<cell|=>|<cell|y+x-f(x)>>>>
</eqnarray*>
Note then that <math|y=f(x)> if and only if <math|T<rsub|y>(x)=x>, so
if there exists a (uniquely determined) fixed point of <math|T<rsub|y>>
then <with|mode|math|we have >proved our proposition. Now for
<math|x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
we have by (4) and (5) that\
<\eqnarray*>
<tformat|<table|<row|<cell|(5) \<shortparallel\>T<rsub|y>(x)\<shortparallel\><rsub|X>>|<cell|=>|<cell|\<shortparallel\>y+T(x)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>y\<shortparallel\>+\<shortparallel\>T(x)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|\<sigma\>|2>+<frac|\<shortparallel\>x\<shortparallel\><rsub|X>|2>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|\<sigma\>|2>+<frac|\<sigma\>|2>>>|<row|<cell|>|<cell|=>|<cell|\<sigma\>>>>>
</eqnarray*>
from this it follows that T is a continuous function (because T and
constant functions are continuous) from the complete space
<math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
to the complete space <math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>.
Again using the fact that <math|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
is convex and the inverse function theorem and (1) we get for
<math|x<rsub|1>,x<rsub|2>\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
that
<\eqnarray*>
<tformat|<table|<row|<cell|(6) \<shortparallel\>T<rsub|y>(x<rsub|1>)-T<rsub|y>(x<rsub|2>)\<shortparallel\><rsub|X>>|<cell|=>|<cell|\<shortparallel\>T(x<rsub|1>)-T(x<rsub|2>)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|<frac|1|2>\<shortparallel\>x<rsub|1>-x<rsub|2>\<shortparallel\><rsub|X>>>>>
</eqnarray*>
proving that <math|T<rsub|y>> is a contraction and using the
contraction function theorem (<reference|the Banach Fixed Point
Theorem>) we have the existence of a
<math|x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<sigma\>)>
such that <math|T<rsub|y>(x)=x> proving our proposition.
</proof>
Let now given a <math|\<delta\>\<gtr\>0> define the open set
<math|U<rsub|\<delta\>>> with <math|0\<in\>U<rsub|\<delta\>>> be defined
by\
<\eqnarray*>
<tformat|<table|<row|<cell|(7) U<rsub|\<delta\>>>|<cell|=>|<cell|B<rsub|\<shortparallel\>\<shortparallel\>>(0,\<delta\>)<big|cap>f<rsup|-1>(B<rsub|\<shortparallel\>\<shortparallel\>>(0,<frac|\<delta\>|2>)>>>>
</eqnarray*>
\ We can then prove the following:
<\proposition>
There exists a <math|\<chi\>> with <math|0\<less\>\<chi\>\<less\>\<rho\>>
such that <math|f<rsub|\|U<rsub|\<chi\>>>:U<rsub|\<chi\>>\<rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
is a bijective function
</proposition>
<\proof>
T<abbr|ake <math|\<chi\>=<frac|\<rho\>|2>> then
<math|0\<less\>\<chi\>\<less\>\<rho\>>> then we must prove surjectivity
and injectivity
<\enumerate>
<item>Surjectivity.By the previous proposition we have if
<math|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
then there exists a unique <math|x\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)>
(the closed ball) such that <math|f(x)=y>. We prove now by
contradiction that <math|\<shortparallel\>x\<shortparallel\><rsub|X>\<neq\>\<chi\>>
[So <math|>assume \ that <math|\<shortparallel\>x\<shortparallel\><rsub|X>=\<chi\>>
take then <math|\<varepsilon\>=<frac|\<chi\>|2>-\<shortparallel\>y\<shortparallel\>\<gtr\>0>
then by continuity of <math|f> at <math|x> there exists a
<math|\<delta\><rprime|'>> such that if
<math|\<shortparallel\>x-x<rprime|'>\<shortparallel\><rsub|X>\<less\>\<delta\><rprime|'>>
then <math|\<shortparallel\>f(x)-f(x<rprime|'>)\<shortparallel\>\<less\>\<varepsilon\>>
and thus <math|\<shortparallel\>f(x<rprime|'>)\<shortparallel\><rsub|X>\<leqslant\>\<shortparallel\>f(x<rprime|'>)-f(x)\<shortparallel\>+\<shortparallel\>f(x)\<shortparallel\>\<less\>\<varepsilon\>+\<shortparallel\>y\<shortparallel\>\<less\><frac|\<chi\>|2>-\<shortparallel\>y\<shortparallel\>+\<shortparallel\>y\<shortparallel\>=<frac|\<chi\>|2>>.
Now if we take <math|\<delta\>=min(\<delta\><rprime|'>,<frac|\<rho\>|4>)>
and <math|x<rsub|1>=x(1+<frac|\<delta\>|\<shortparallel\>x\<shortparallel\>>)>
then <math|\<shortparallel\>x-x<rsub|1>\<shortparallel\><rsub|X>=\<shortparallel\><frac|\<delta\>|\<shortparallel\>x\<shortparallel\><rsub|X>>\<shortparallel\>x\<shortparallel\><rsub|X>=\<delta\>\<less\>\<delta\><rprime|'>>
and thus <math|\<shortparallel\>f(x<rsub|1>)\<shortparallel\><rsub|X>\<less\><frac|\<chi\>|2>\<Rightarrow\>f(x<rsub|1>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
and also <math|\<chi\>=\<shortparallel\>x\<shortparallel\><rsub|X>\<less\>(1+<frac|\<delta\>|\<shortparallel\>x\<shortparallel\><rsub|X>>)\<shortparallel\>x\<shortparallel\><rsub|X>=\<shortparallel\>x<rsub|1>\<shortparallel\>=(1+<frac|\<delta\>|\<shortparallel\>x\<shortparallel\><rsub|X>>)\<shortparallel\>x\<shortparallel\><rsub|X>=\<shortparallel\>x\<shortparallel\>+\<delta\>=\<chi\>+\<delta\>\<leqslant\>\<chi\>+<frac|\<rho\>|4>=<frac|\<rho\>|2>+<frac|\<rho\>|4>=<frac|3|4>\<rho\>\<less\>\<rho\>\<Rightarrow\>\<chi\>\<less\>\<shortparallel\>x<rsub|1>\<shortparallel\>\<less\>\<rho\>>.
Now as <math|f(x<rsub|1>)\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
we can again use the previous proposition to find a
<math|x<rsub|2>\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)>
such that <math|f(x<rsub|2>)=f(x<rsub|1>)>. Then as
<math|x<rsub|2>\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
and <math|x<rsub|1>\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
we have using the fact that <math|f(x<rsub|1>)=f(x<rsub|2>)=y\<in\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)>
and the previous proposition that <math|x<rsub|1>=x<rsub|2>> but then
we have <math|\<shortparallel\>x<rsub|2>\<shortparallel\><rsub|X>\<leqslant\>\<chi\>\<less\>\<shortparallel\>x<rsub|1>\<shortparallel\><rsub|X>=\<shortparallel\>x<rsub|2>\<shortparallel\><rsub|X>>
a contradiction.] As we now have proved that
<math|\<shortparallel\>x\<shortparallel\><rsub|X>\<neq\>\<chi\>> we
must have that <math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)>
(the open ball) and as also <math|f(x)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
we have <math|x\<in\>U<rsub|\<chi\>>> and thus surjectivity.
<item>Injectivity assume that there exists a
<math|x<rsub|1>,x<rsub|2>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)>
with <math|f(x<rsub|1>)=f(x<rsub|2>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<subseteq\><wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
then by the previous proposition we have <math|x<rsub|1>=x<rsub|2>>
proving injectivity
</enumerate>
</proof>
Now assume that <math|0\<less\>\<chi\>\<less\>\<rho\>> is chosen such
that <math|f<rsub|\|U<rsub|\<chi\>><rsub|>>:U<rsub|\<chi\>>\<rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
is a bijective function. Then by bijectivity there exists a inverse
function <math|g=(f<rsub|\|U<rsub|\<chi\>>>)<rsup|-1>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<rightarrow\>U<rsub|\<chi\>>>
then we have\
<\proposition>
<math|g> is continuous on <math|B<rsub|<next-line>\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
</proposition>
<\proof>
If <math|x\<in\>U<rsub|\<chi\>>\<subseteq\><below|<wide|B|\<bar\>><rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)|>>
a convex set then <math|x=x-f(x)+f(x)=T(x)+f(x)> so if
<math|x<rsub|1>,x<rsub|2>\<in\>U<rsub|\<chi\>>> then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>x<rsub|1>-x<rsub|2>\<shortparallel\><rsub|X>>|<cell|=>|<cell|\<shortparallel\>T(x<rsub|1>)+f(x<rsub|1>)-T(x<rsub|2>)-f(x<rsub|2>)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\>>|<cell|\<shortparallel\>T(x<rsub|1>)-T(x<rsub|2>)\<shortparallel\><rsub|X>+\<shortparallel\>f(x<rsub|1>)-f(x<rsub|2>)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<leqslant\><rsub|(1)
and <reference|the fundamental theorem of calculus
(3)>>>|<cell|<frac|1|2>\<shortparallel\>x<rsub|1>-x<rsub|2>\<shortparallel\><rsub|X>+\<shortparallel\>f(x<rsub|1>)-f(x<rsub|2>)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|>>|<row|<cell|<frac|1|2>\<shortparallel\>x<rsub|1>-x<rsub|2>\<shortparallel\>-x>|<cell|\<leqslant\>>|<cell|\<shortparallel\>f(x<rsub|1>)-f(x<rsub|2>)\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|>>|<row|<cell|(8)
\ \<shortparallel\>x<rsub|1>-x<rsub|2>\<shortparallel\><rsub|X>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>f(x<rsub|1>)-f(x<rsub|2>)\<shortparallel\><rsub|X>>>>>
</eqnarray*>
So if <math|y<rsub|1>,y<rsub|2>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
then by bijectivity of <math|g> we have that
<math|y<rsub|1>=f(g(y<rsub|1>)),y<rsub|2>=f(g(y<rsub|2>))> and thus by
(8) we have\
<\eqnarray*>
<tformat|<table|<row|<cell|(9) \<shortparallel\>g(y<rsub|1>)-g*y<rsub|2>)\<shortparallel\><rsub|X>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>y<rsub|1>-y<rsub|2>\<shortparallel\><rsub|X>>>>>
</eqnarray*>
now if <math|\<varepsilon\>\<gtr\>0> and
<math|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
then for all <math|y<rprime|'>\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
with <math|\<shortparallel\>y-y<rprime|'>\<shortparallel\><rsub|X>\<less\>\<varepsilon\>>
we have <math|\<shortparallel\>g(y)-g(y<rprime|'>)\<shortparallel\><rsub|X>\<leqslant\>\<shortparallel\>y-y<rprime|'>\<shortparallel\><rsub|X>\<less\>\<varepsilon\>>
proving continuity of <math|g> on <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
</proof>
Next we prove that <math|g> is differentiable
<\proposition>
<math|g> is differentiable on <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
and <math|Dg(y)=(Df(g(y))<rsup|-1>>
</proposition>
<\proof>
Now if <math|y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
we have that <math|g(y)\<in\>U<rsub|\<chi\>>\<Rightarrowlim\><rsub|(7)>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)\<subseteq\>U>
we have by (2) that <math|Df(g(y))> is a toplinear isomorphism and thus
<math|Df(g(y))<rsup|-1>> is continuous and linear. Then we need to show
that if <math|h\<in\>(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>))<rsub|y>={h\<in\>E\|y+h\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)}>
then there exists a <math|\<varepsilon\>-mapping>
<math|\<varepsilon\>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<rightarrow\>X>
such that\
<\eqnarray*>
<tformat|<table|<row|<cell|g(y+h)-g(y)-Df(g(y))<rsup|-1>(h)>|<cell|=>|<cell|\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\><rsub|X>>>>>
</eqnarray*>
So set <math|k=g(y+h)-g(y)> then <math|k+g(y)=g(y+h)\<in\>g(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>))=U<rsub|\<chi\>>\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<chi\>)\<subseteq\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>)\<Rightarrow\>k\<in\>(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,\<rho\>))<rsub|g(y)>>
then as <math|f> is differentiable at <math|g(y)> there exists a
<math|\<varepsilon\>-mapping> <math|\<varepsilon\><rprime|'>> such that
<math|f(g(y)+k)-f(g(y))-Df(g(y))(k)=\<varepsilon\><rprime|'>(k)\<shortparallel\>k\<shortparallel\><rsub|X>>
and thus
<\eqnarray*>
<tformat|<table|<row|<cell|f(g(y)+k)-f(g(y))>|<cell|>|<cell|Df(g(y))(k)+\<varepsilon\><rprime|'>(k)\<shortparallel\>k\<shortparallel\><rsub|X>>>>>
</eqnarray*>
then using <math|f(g(y)+k)=f(g(y+h))=y+h> and <math|f(g(y))=y> and thus
<math|f(g(y)+k)-f(g(y))=h> we find that\
<\eqnarray*>
<tformat|<table|<row|<cell|h>|<cell|=>|<cell|Df(g(y))(k)+\<varepsilon\><rprime|'>(k)\<shortparallel\>k\<shortparallel\>>>>>
</eqnarray*>
Now we apply <math|Df(g(y))<rsup|-1>> to both sides of the equation to
find (using linearity)
<\eqnarray*>
<tformat|<table|<row|<cell|Df(g(y))<rsup|-1>(h)>|<cell|=>|<cell|k+Df(g(y))<rsup|-1>(\<varepsilon\><rprime|'>(k))\<shortparallel\>k\<shortparallel\>>>|<row|<cell|>|<cell|=>|<cell|g(y+h)-g(y)+Df(g(y))<rsup|-1>(\<varepsilon\><rprime|'>(k))\<shortparallel\>k\<shortparallel\>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|>>|<row|<cell|g(y+h)-g(y)-Df(g(y))<rsup|-1>(h)>|<cell|=>|<cell|-Df(g(y))<rsup|-1>(\<varepsilon\><rprime|'>(k))\<shortparallel\>k\<shortparallel\><rsub|X>>>|<row|<cell|>|<cell|=>|<cell|-Df(g(y))<rsup|-1>(\<varepsilon\><rprime|'>(g(y+h)-g(y)))\<shortparallel\>g(y+h)-g(y)\<shortparallel\><rsub|X>>>>>
</eqnarray*>
Define now <math|\<varepsilon\>:(B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>))<rsub|y>\<rightarrow\>X>
defined by
<\eqnarray*>
<tformat|<table|<row|<cell|\<varepsilon\>(h)>|<cell|=>|<cell|-Df(g(y))<rsup|-1>(\<varepsilon\><rprime|'>(g(y+h)-g(y)))<frac|\<shortparallel\>g(y+h)-g(y)\<shortparallel\><rsub|X>|\<shortparallel\>h\<shortparallel\><rsub|X>>,
if h\<neq\>0>>|<row|<cell|>|<cell|=>|<cell|0 \ , if h=0>>>>
</eqnarray*>
Then we have that\
<\eqnarray*>
<tformat|<table|<row|<cell|g(y+h)-g(y)-Df(g(y))<rsup|-1>(h)>|<cell|=>|<cell|\<varepsilon\>(h)\<shortparallel\>h\<shortparallel\>>>>>
</eqnarray*>
So if we prove that <math|\<varepsilon\>> is continue at <math|0> then
we are done. Now to prove continuity of <math|h> at <math|0> note that
in the proof of the previous proposition we have by (9) that
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>g(y+h)-g(y)\<shortparallel\><rsub|X>>|<cell|\<leqslant\>>|<cell|2\<shortparallel\>y+h-y\<shortparallel\><rsub|X>=2\<shortparallel\>h\<shortparallel\>>>>>
</eqnarray*>
Hence if <math|h\<neq\>0> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|<frac|\<shortparallel\>g(y+h)-g(y)\<shortparallel\><rsub|X>|\<shortparallel\>h\<shortparallel\><rsub|X>>>|<cell|\<leqslant\>>|<cell|2>>>>
</eqnarray*>
So let <math|\<varepsilon\>\<less\>0> (mind you do not confuse
<math|\<varepsilon\>> with the <math|\<varepsilon\>-mapping>
<math|\<varepsilon\>>)
Now by continuity of <math|\<varepsilon\><rprime|'>> at <math|0> there
exists a <math|\<delta\><rprime|'>\<gtr\>0> such that if
<math|\<shortparallel\>z\<shortparallel\><rsub|X>\<less\>\<delta\><rprime|'>>
then <math|\<shortparallel\>\<varepsilon\><rprime|'>(z)\<shortparallel\><rsub|X>\<less\><frac|\<varepsilon\>|2>>.
By continuity of <math|g> at <math|y> we can find a <math|\<delta\>>
such that if <math|\<shortparallel\>h\<shortparallel\><rsub|X>\<less\>\<delta\>>
\ then<math|\<shortparallel\>g(y+h)-g(y)\<shortparallel\>\<less\>\<delta\><rprime|'>>
and thus if <math|\<shortparallel\>h\<shortparallel\>\<less\>\<delta\>>
we have <math|\<shortparallel\>\<varepsilon\><rprime|'>(g(y+h)-g(y))\<shortparallel\><rsub|X>\<less\><frac|\<varepsilon\>|2>>.
from this if follows that if <math|\<shortparallel\>h\<shortparallel\>\<less\>\<delta\>>
we have\
<\eqnarray*>
<tformat|<table|<row|<cell|\<shortparallel\>\<varepsilon\>(h)\<shortparallel\><rsub|X>>|<cell|\<less\>>|<cell|\<varepsilon\>>>>>
</eqnarray*>
proving differentiability of <math|g> at <math|y> and that the
differential is <math|Df(g(y))<rsup|-1>>
</proof>
To summarize we have proved now that <math|f<rsub|\|U<rsub|\<chi\>>>:U<rsub|\<chi\>>\<rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
has a differentiable inverse <math|g=(f<rsub|\|U<rsub|\<chi\>>>)<rsup|-1>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<rightarrow\>U<rsub|\<chi\>>>
and that <math|\<forall\>y\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
we have <math|Dg(y)=Df(g(y))<rsup|-1>> thus the first derivative
<math|Dg:B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)\<rightarrow\>L(E,E)>
is defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|(10) \ Dg>|<cell|=>|<cell|\<tau\>\<circ\>Df\<circ\>g>>>>
</eqnarray*>
Where <math|\<tau\>> is the inverse function which is by
<reference|inverse mapping is c-inifinite> of class
<math|C<rsup|\<infty\>>> and thus continue, <math|Df> is continue
(because it is of class <math|C<rsup|r>>) and <math|g> has been proved to
be continue so <math|Dg> is continue proving that <math|g> is
differentiable on <math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
of class <math|C<rsup|1>> and <math|f<rsub|\|U<rsub|\<chi\>>>> is a
diffeomorphism on <math|U<rsub|\<chi\>>> of class <math|C<rsup|1>>.\
Now to prove the general case with <math|C<rsup|r>> we proceed by
induction on r to prove that <math|g> is differentiable of class
<math|C<rsup|r>> if <math|f> is differentiable on <math|U> of class
<math|C<rsup|r>>\
<\enumerate>
<item><math|r=1> this we have just proved
<item>Assume it is true for <math|r> and prove it for <math|r+1> if
<math|f> is differentiable on <math|U> of class <math|C<rsup|r+1>> we
have that <math|Df> is differentiable on <math|U> of class
<math|C<rsup|r>> then we have by the induction hypothesis that <math|g>
is differentiable of class <math|C<rsup|r>> and because
<math|Dg=\<tau\>\<circ\>Df\<circ\>g> we have that <math|Dg> is
differentiable of class <math|C<rsup|r>> proving by <reference|proving
f is Cn based on Cn-1 of Df> that <math|g> is differentiable on
<math|B<rsub|\<shortparallel\>\<shortparallel\><rsub|X>>(0,<frac|\<chi\>|2>)>
of class <math|C<rsup|r+1>>
</enumerate>
This finally proves the inverse function theorem
</proof>
<\theorem>
<label|implicit function theorem><dueto|Implicit function
theorem><index|Implicit function theorem>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>,Y,\<shortparallel\>\<shortparallel\><rsub|Y>>
and <math|Z,\<shortparallel\>\<shortparallel\><rsub|Z>> be Banach spaces
over <math|\<bbb-R\>> and <math|U\<subseteq\>X,V\<subseteq\>Y> open sets
and <math|f:U\<times\>V\<rightarrow\>Z> be a function differentiable on
<math|U\<times\>V> of class <math|C<rsup|r>> . Now for some
<math|(x<rsub|0>,y<rsub|0>)\<in\>U\<times\>V> assume that
<math|D<rsub|2>f(x<rsub|0>,y<rsub|0>):Y\<rightarrow\>Z> is a isomorphism.
Then there exists open sets <math|U<rsub|0>,W<rsub|0>> with
<math|x<rsub|0>\<subseteq\>U<rsub|0>\<subseteq\>U> and
<math|f(x<rsub|0>,y<rsub|0>)\<in\>W<rsub|0>> and a function
<math|g:U<rsub|0>\<times\>W<rsub|0>\<rightarrow\>Y> that is
differentiable on <math|U<rsub|0>\<times\>W<rsub|0>> of class
<math|C<rsup|r>> such that <math|f(x,g(x,w))=w><math|>\
</theorem>
<\proof>
Define <math|\<Phi\>:U\<times\>V\<rightarrow\>X\<times\>Z> by
<math|(x,y)\<rightarrow\>(x,f(x,y))> then <math|\<Phi\>> is
<math|C<rsup|r>> because of <reference|c-r of mapping to product >, the
fact that <math|\<pi\><rsub|1>> is <math|C<rsup|\<infty\>>> and <math|f>
is <math|C<rsup|r>>. Also by \ <math|D\<Phi\>(x<rsub|0>,y)(h,k)=(D\<pi\><rsub|1>(x<rsub|0>,y<rsub|0>)(h,k),Df(x<rsub|0>,y<rsub|0>)(h,k))\<equallim\><rsub|\<pi\><rsub|1>
is linear>(\<pi\><rsub|1>(h,k),Df(x<rsub|0>,y<rsub|0>)(h,k))=(h,Df(x<rsub|0>,y<rsub|0>)(h,k))>
which is a isomorphism (because <math|Df(x<rsub|0>,y<rsub|0>)> is a
isomorphism so using the inverse function theorem <reference|The
inversion theorem> there exists a open
<math|U<rsub|0><rprime|'>\<subseteq\>U\<times\>V> and
<math|W<rprime|'><rsub|0>> open in <math|X\<times\>Z> with
<math|(x<rsub|0>,y<rsub|0>)\<in\>U<rprime|'><rsub|0>> and
<math|(x<rsub|0>,f(x<rsub|0>,y<rsub|0>))\<in\>W<rprime|'><rsub|0>> such
that <math|\<Phi\><rsub|\|U<rprime|'><rsub|0>>:U<rprime|'><rsub|0>\<rightarrow\>W<rprime|'><rsub|0>>
is a diffeomorphism so there exists a
<math|\<Phi\><rsup|-1>:W<rprime|'><rsub|0>\<rightarrow\>U<rprime|'><rsub|0>\<subseteq\>U\<times\>V>
now <math|\<Phi\><rsup|-1>(x,z)=((\<Phi\><rsup|-1>)<rsub|1>(x,z),(\<Phi\><rsup|-1>)<rsub|2>(x,z))>
and we have now by <reference|c-r of mapping to product > that
<math|(\<Phi\><rsup|-1>)<rsub|2>:W<rprime|'><rsub|0>\<rightarrow\>V> is
<math|C<rsup|r>>. Finally by definition of the product topology there
exists a <math|U<rsub|0>,W<rsub|0>> with
<math|x<rsub|0><rsub|>\<in\>U<rsub|0>,f(x<rsub|0>,y<rsub|0>)\<in\>W<rsub|0>>
and <math|U<rsub|0>\<times\>W<rsub|0>\<subseteq\>W<rprime|'><rsub|0>>
then <math|g=((\<Phi\><rsup|-1>)<rsub|2>)<rsub|\|U<rsub|0>\<times\>V<rsub|0>>:U<rsub|0>\<times\>V<rsub|0>\<rightarrow\>V>
\ is <math|C<rsup|r>> (see <reference|restriction of C-r mapping>). Also
<math|if (x,w)\<in\>U<rsub|0>\<times\>W<rsub|0>> then
<math|(x,w)=\<Phi\>((\<Phi\><rsup|-1>(x,w))<rsub|1>,g(x,w))=((\<Phi\><rsup|-1>(x,w))<rsub|1>,f((\<Phi\><rsup|-1>(x,w),g(x,w)))\<Rightarrow\>x=(\<Phi\><rsup|-1>(x,w))<rsub|1>\<Rightarrow\>(x,w)=(x,f(x,g(x,w))\<Rightarrow\>f(x,g(x,w))=w>
and we have found the desired <math|g>
</proof>
\;
\;
<chapter|Differentiable manifolds>
<\convention>
If we say that something is smooth if it is of class
<math|C<rsup|\<infty\>>> (meaning that it is <math|C<rsup|r>> for all
<math|r\<in\>\<bbb-N\>>). Also if we say that something is of class
<math|C<rsup|r>,r\<geqslant\>1(0)> then we mean a specific number
<math|r> that can include the smooth case <math|r=\<infty\>>.\
</convention>
<section|Manifolds>
<\definition>
<index|topological manifold><math|M> is a topological manifold modelled
on <math|X> if and only if <math|M> is a Hausdorff topological space,
<math|X> is a normed vector space and <math|\<forall\>m\<in\>M> there
exists a open set <math|U> with <math|x\<in\>U> which is homeomorphic
with a open set in <math|X> (see <reference|definition of a
homeomorphism>)
</definition>
<\remark>
If \ the <math|X> has two equivalent norms
<math|\<shortparallel\>\<shortparallel\><rsub|1>,\<shortparallel\>\<shortparallel\><rsub|2>>
(so the generated topological space is identical) we have obviously that
<math|M> is a topological manifold modelled on
<math|X,\<shortparallel\>\<shortparallel\><rsub|1>> if and only if
<math|M> is a topological manifold modelled on
<math|X,\<shortparallel\>\<shortparallel\><rsub|2>>
</remark>
<\remark>
If in the previous definition <math|M> is a topological space modelled on
<math|\<bbb-R\><rsup|n>> (with any of its equivalent norms) then we say
that <math|M> is a <math|n\<um\>>dimensional manifold and is often noted
as <math|M<rsup|n>>
</remark>
<\definition>
<index|local chart><index|local coordinate system><index|atlas>If
<math|M> is a topological space, <math|X> a normed space then a pair
<math|(U,\<varphi\>)> is called a local chart (or local coordinate
system) of <math|M> modelled on X if <math|U> is a open set in <math|M>
and <math|\<varphi\>> a homeomorphism
<math|\<varphi\>:U\<rightarrow\>\<varphi\>(U),><math|\<varphi\>(U)> a
open set in <math|X>. A atlas of <math|M> modelled on X is a set
<math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I}> of local
charts (or local coordinate systems) such that
<math|><math|<big|cup><rsub|i><rsub|>U<rsub|i>=M>. Here we have indexed
the set by itself (see <reference|set indexed by itself>)
</definition>
Clearly from the definition of a atlas and topological manifold we have the
following trivial theorem
<\theorem>
Let <math|M> be a topological Hausdorff space, <math|X> a normed vector
space. Then <math|M> is a topological manifold modelled on <math|X> if
and only there exists at least one atlas on <math|M> modelled on
<math|X>.\
</theorem>
<\proof>
\;
<math|\<Rightarrow\>> If <math|m\<in\>M> then by definition of a
topological space there exists a open sets <math|U<rsub|m>,V<rsub|m>> in
<math|M,X> such that <math|m\<in\>U<rsub|m>> and there exists a
homeomorphism <math|\<varphi\><rsub|m>:U<rsub|m>\<rightarrow\>V<rsub|m>\<equallim\><rsub|\<varphi\><rsub|m>
is surjective>\<varphi\><rsub|m>(U<rsub|m>)> , then
<math|<big|cup><rsub|m\<in\>M>U<rsub|m>=X> and
<math|\<cal-A\>={(U<rsub|m>,\<varphi\><rsub|m>)\|m\<in\>M}> is the atlas
we are searching for.
<math|\<Leftarrow\>> If <math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I}>
is a atlas then as <math|M=<big|cup><rsub|i\<in\>I>U<rsub|i>> we have
that <math|\<forall\>m\<in\>M> there exists a <math|i\<in\>I> such that
<math|m\<in\>U<rsub|i>> and <math|\<varphi\><rsub|i>:U\<rightarrow\>\<varphi\><rsub|i>(U)
(open)> is a homeomorphism, proving that <math|M> is a topological
manifold modelled on <math|X>\
</proof>
<\notation>
From now on if we have a function partial function <math|f> from
<math|X\<rightarrow\>Y> with <math|dom(f)=A> then we note
<math|f:A\<rightarrow\>B> as <math|f> if it is clear what its domain is.
\ For example if we have the homeomorphisms
<math|\<varphi\><rsub|1>:U<rsub|1>\<rightarrow\>\<varphi\><rsub|1>(U),\<varphi\><rsub|2>:U<rsub|2>\<rightarrow\>\<varphi\><rsub|2>(U<rsub|2>)>
with <math|U<rsub|1><big|cap>U<rsub|2>\<neq\>\<emptyset\>> then as
<math|\<varphi\><rsub|2><rsup|-1>:\<varphi\><rsub|2>(U<rsub|2>)\<rightarrow\>U<rsub|2>>
is also a homeomorphism we have that the domain of the partial function
<math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>> is
<math|U<rsub|1><big|cap>U<rsub|2>> \ and we can thus note
<math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>:\<varphi\><rsub|2>(U<rsub|1><big|cap>U<rsub|2>)\<rightarrow\>\<varphi\><rsub|1>(U<rsub|1><big|cap>U<rsub|2>)>
\ by <math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>>.\
As another example if <math|f:A\<rightarrow\>B> ,
<math|g:C\<rightarrow\>D> with <math|B<big|cap>C\<neq\>\<emptyset\>> then
the domain of the partial function <math|g\<circ\>f> is
<math|f<rsup|-1>(C)> and we name the function
<math|g\<circ\>f:f<rsup|-1>(C)\<rightarrow\>D> by <math|g\<circ\>f>.\
This is a convention that we assume from now.
</notation>
<\theorem>
<label|coordinate transform is a homeormorphism>Let <math|M> be a
topological space, <math|X> a normed vector space and
<math|(U,\<varphi\>),(V,\<psi\>)> be two local charts modelled on
<math|X> with <math|U<big|cap>V\<neq\>\<emptyset\>> then
<math|\<varphi\>\<circ\>\<psi\><rsup|-1><rsub|>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
is a homeomorphism with inverse <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
</theorem>
<\proof>
This is trivial using <reference|continuity of restricted
maps>,<reference|restriction of a bijective mapping> and
<reference|composition of bijective function and inverse bijective
function>.
</proof>
Note that if <math|(U,\<varphi\>),(V,\<psi\>)> be two local charts then
<math|U<big|cap>V> is open and thus open in the subspace topology of
<math|U> and <math|V>. Then from the fact that <math|\<varphi\>,\<psi\>>
are homeomorphism we have that <math|\<varphi\>(U<big|cap>V),\<psi\>(U<big|cap>V)>
\ are open in the subspace topology of <math|\<varphi\>(U),\<psi\>(V)>.
Finally from the fact that <math|\<varphi\>(U),\<psi\>(V)> are open we have
that <math|\<varphi\>(U<big|cap>V),\<psi\>(U<big|cap>V)> are open sets. So
we can consider differentiability of <math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
and <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>.
<\definition>
<index|<math|C<rsup|r>> compatible charts>Let
<math|r\<in\>{1,\<ldots\>,\<infty\>}> (in future we note this as
<math|r\<geqslant\>1>), <math|M> be a topological space, X a normed
vector space and <math|(U,\<varphi\>),(V,\<psi\>)> be two local charts
modelled on <math|X>. Then <math|(U,\<varphi\>),(V,\<psi\>)> are
<math|C<rsup|r>>compatible if and only if <math|U<big|cap>V=\<emptyset\>>
or <math|U<big|cap>V\<neq\>\<emptyset\>> and
<math|(\<psi\>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is a diffeomorphism of class <math|C<rsup|r>>.
</definition>
<\definition>
<index|atlas of class <math|C<rsup|r>>>Let <math|><math|M> be a
topological space, <math|X> a normed space, then a atlas
<math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I}> on <math|M>
modelled on X is of class <math|C<rsup|r>>(<math|r\<geqslant\>1)> if
<math|\<forall\>(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>> with
<math|U<big|cap>V\<neq\>\<emptyset\>> we have that
\ <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\><rsub|>(U<big|cap>V)\<rightarrow\>\<psi\>(U<rsup|><big|cap>V)>
is a diffeomorphism of class <math|C<rsup|r>>. In other words if every
two local charts in <math|\<cal-A\>> are <math|C<rsup|r>> compatible.
</definition>
<\theorem>
<label|characterization of a atlas of class C-r>Let <math|M> be a
topological space, <math|X> normed space, then a atlas
<math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I}> is of class
<math|C<rsup|r>> if and only if <math|\<forall\>(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>>
with <math|U<big|cap>V\<neq\>\<emptyset\>> we have that
<math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\><rsub|>(U<big|cap>V)\<rightarrow\>\<psi\>(U<rsup|><big|cap>V)>
is differentiable of class <math|C<rsup|r>>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>> is trivially as if follows from the definition of a
diffeomorphism of class <math|C<rsup|r>>
<math|\<Leftarrow\>> If <math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>>
and <math|U<big|cap>V\<neq\>\<emptyset\>> then
<with|mode|math|\<psi\><rsub|\<varphi\>>=\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>> by the hypothesis, then using
<reference|coordinate transform is a homeomorphism> we have that
<math|(\<psi\><rsub|\<varphi\>>)<rsup|-1>=\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
which is by hypothesis also differentiable of class <math|C<rsup|r>>
proving that <math|\<psi\><rsub|\<varphi\>>> is a diffeomorphism of class
<math|C<rsup|r>>.
</proof>
<\note>
<label|image of open set by chart is open>If
<math|(U,\<varphi\>)\<in\>\<cal-A\>> then because <math|\<varphi\>> is a
homeomorphism and <math|\<varphi\>(U)> is open we have that if <math|W>
is a open set in <math|M> then <math|\<varphi\>(W)> is open in <math|X>
</note>
<\example>
<label|atlas in a normed space>Let <math|X> be a normed space and
<math|U\<subseteq\>X> a open set then <math|{(U,i<rsub|U>)}> is a atlas
of class <math|C<rsup|\<infty\>>> on <math|U> modelled on <math|X>
</example>
<\proof>
First we prove that <math|{(U,i<rsub|U>)}> is a atlas on <math|U>. This
is trivial as <math|i<rsub|U>:U\<rightarrow\>i<rsub|U>(U)=U> is a
homeomorphism (as it is trivially bijective and continuous) and
<math|U=<big|cup><rsub|(U,i<rsub|U>)\<in\>{(U,i<rsub|U>)}>U>. Finally
<math|i<rsub|U>\<circ\>i<rsub|U><rsup|-1>:i<rsub|U>(U)\<rightarrow\>i<rsub|U>(U)=i<rsub|U>>
which is differentiable of class <math|C<rsup|\<infty\>>> because it is
linear.
</proof>
<\definition>
<index|<math|C<rsup|r>> compatable atlasses>Let <math|M> be a topological
space, <math|X> a normed space and <math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>):i\<in\>I}<rsub|>,\<cal-A\><rprime|'>={(V<rsub|j>,\<psi\><rsub|j>)\|j\<in\>J}<rsub|>>
be two atlases of class <math|C<rsup|r>> on <math|X> modelled on <math|X>
then <math|\<cal-A\>,\<cal-A\><rprime|'>> are called <math|C<rsup|r>>
compatible if <math|\<cal-A\><big|cup>\<cal-A\><rprime|'>> is also a
atlas of class <math|C<rsup|r>>
</definition>
<\lemma>
<label|characterization of compatibility>Let <math|M> be a topological
space, <math|X> a normed space and <math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>):i\<in\>I},\<cal-A\><rprime|'>={(V<rsub|j>,\<psi\><rsub|j>):j\<in\>J}>
be two atlases of class <math|C<rsup|r>> modelled on X then
<math|\<cal-A\>,\<cal-A\><rprime|'>> are <math|C<rsup|r>> compatible
<math|\<Leftrightarrow\>> if <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>V\<neq\>\<emptyset\>> we have that
<math|\<psi\>\<circ\>\<varphi\><rsub|><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
and <math|\<varphi\>\<circ\>\<psi\><rsub|><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>are
differentiable of class <math|C<rsup|r>>.
</lemma>
<\proof>
\;
<math|\<Rightarrow\>> Is trivial by the definition of a atlas of class
<math|C<rsup|r>>
<math|\<Leftarrow\>> Let <math|\<cal-A\><big|cup>\<cal-A\><rprime|'>={(W<rsub|i>,\<tau\><rsub|i>):i\<in\>K}>
then :
First if <math|m\<in\>M> then there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>\<subseteq\>\<cal-A\><big|cup>\<cal-A\><rprime|'>>
with <math|m\<in\>U\<Rightarrow\>M=<big|cup><rsub|i\<in\>K>W<rsub|i>><math|>
Second we have the following cases for
<math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\><big|cup>\<cal-A\><rprime|'>>
with <math|U<big|cap>V\<neq\>\<emptyset\>>
<\enumerate>
<item><math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>> then
<math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is of class <math|C<rsup|r>> because <math|\<cal-A\>> is a atlas of
class <math|C<rsup|r>>
<item><math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\><rprime|'>> then
<math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is of class <math|C<rsup|r>> because <math|\<cal-A\><rprime|'>> is a
atlas of class <math|C<rsup|r>>
<item><math|(U,\<varphi\>)\<in\>\<cal-A\>,(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
then <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is of class <math|C<rsup|r>> by hypothesis
<item><math|(U,\<varphi\>)\<in\>\<cal-A\><rprime|'>,(V,\<psi\>)\<in\>\<cal-A\>>
then <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is of class <math|C<rsup|r>> by hypothesis
</enumerate>
So in all cases we have <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is of class <math|C<rsup|r>> proving that
<math|\<cal-A\><big|cup>\<cal-A\><rprime|'>> is a atlas of class
<math|C<rsup|r>>
</proof>
<\lemma>
Let <math|M> be a topological space, <math|X> a normed space and
<math|\<Delta\><rsup|r>={\<cal-A\>:\<cal-A\>> is a atlas of class
<math|C<rsup|r> on M modelled on X}> then <math|<rprime|'>C<rsup|r>>
compatibility' is a equivalence relation. We note this relation as
<math|\<cal-C\><rsup|(r)>> see <reference|equivalence relation>.
</lemma>
<\proof>
\;
<\enumerate>
<item>(reflectivity) Let <math|\<cal-A\>> be a atlas then
<math|\<cal-A\>=\<cal-A\><big|cup>\<cal-A\>> is trivially a atlas
<item>(symmetricity) This is trivial as
<math|\<cal-A\><big|cup>\<cal-A\><rprime|'>=\<cal-A\><rprime|'><big|cup>\<cal-A\>>
<item>(transitivity) Let <math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I},\<cal-A\><rprime|'>={(V<rsub|i>,\<psi\><rsub|i>)\|i\<in\>J},\<cal-A\><rprime|''>={(W<rsub|i>,\<tau\><rsub|i>)\|i\<in\>K}>
be atlases of class <math|C<rsup|r>> such that
<math|\<cal-A\>,\<cal-A\><rprime|'>> are compatible and
<math|\<cal-A\><rprime|'>,\<cal-A\><rprime|''>> are compatible. Now if
<math|(U,\<varphi\>)\<in\>\<cal-A\>> and
<math|(W,\<tau\>)\<in\>\<cal-A\><rprime|''>> is such that
<math|U<big|cap>W\<neq\>\<emptyset\>> then we have to prove the
following two assertions :
<\enumerate>
<item><math|\<varphi\><rsub|\<tau\>>=\<varphi\>\<circ\>\<tau\><rsup|-1>:\<tau\>(U<big|cap>W)\<rightarrow\>\<varphi\>(U<big|cap>W)>
is <math|C<rsup|r>>
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>\<tau\>(U<big|cap>W)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>U<big|cap>W\<vdash\>\<tau\>(x)=y>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\><rprime|'>
is a atlas>>|<cell|\<exists\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>\<vdash\>x\<in\>V>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>U<big|cap>V,x\<in\>V<big|cap>W,x\<in\>U<big|cap>V<big|cap>W>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\>,\<cal-A\><rprime|'>
are C<rsup|r> compatable>>|<cell|\<varphi\><rsub|\<psi\>>=\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\><rprime|'>,\<cal-A\><rprime|''>
are C<rsup|r> compatable>>|<cell|\<psi\><rsub|\<tau\>>=\<psi\>\<circ\>\<tau\><rsup|-1>:\<tau\>(V<big|cap>W)\<rightarrow\>\<psi\>(V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|restriction
of C-r mapping> and \<psi\>(<big|cap>V<big|cap>W) is
open>>|<cell|\<varphi\><rsub|\<psi\><rsub|\|\<psi\>(U<big|cap>V<big|cap>W)>>=\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<varphi\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|restriction
of C-r mapping> and \<tau\>(U<big|cap>V<big|cap>W) is
open>>|<cell|\<psi\><rsub|\<tau\><rsub|\|\<tau\>(U<big|cap>V<big|cap>W)>>=\<psi\>\<circ\>\<tau\><rsup|-1>:\<tau\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<psi\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|generalized
chain rule>>>|<cell|\<varphi\><rsub|\<psi\><rsub|\|\<psi\>(U<big|cap>V<big|cap>W)>>\<circ\>\<psi\><rsub|\<tau\><rsub|\|\<tau\>(U<big|cap>V<big|cap>W)>>=\<varphi\>\<circ\>\<psi\><rsup|-1>\<circ\>\<psi\>\<circ\>\<tau\><rsup|-1>\|\<tau\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<varphi\>(U<big|cap>V<big|cap>W)\<equallim\><rsub|<reference|intermediate
bijective function>>\<varphi\>\<circ\>\<tau\><rsup|-1>:\<tau\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<varphi\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>>>
</eqnarray*>
now we can use <reference|C-r is a local property> and
<math|y=\<tau\>(x)\<in\>t(U<big|cap>V<big|cap>W) (open) >to prove
that <math|\<varphi\><rsub|t>> is <math|C<rsup|r>>\
<item><math|\<tau\><rsub|\<varphi\>>=\<tau\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>W)\<rightarrow\>\<tau\>(U<big|cap>W)>
is <math|C<rsup|r>>
<\eqnarray*>
<tformat|<table|<row|<cell|y\<in\>\<varphi\>(U<big|cap>W)>|<cell|\<Rightarrow\>>|<cell|\<exists\>x\<in\>U<big|cap>W\<vdash\>\<varphi\>(x)=y>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\><rprime|'>
is a atlas>>|<cell|\<exists\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>\<vdash\>x\<in\>V>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|x\<in\>U<big|cap>V,x\<in\>V<big|cap>W,x\<in\>U<big|cap>V<big|cap>W>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\>,\<cal-A\><rprime|'>
are C<rsup|r> compatable>>|<cell|\<psi\><rsub|\<varphi\>>=\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<cal-A\><rprime|'>,\<cal-A\><rprime|''>
are C<rsup|r> compatable>>|<cell|\<tau\><rsub|\<psi\>>=\<tau\>\<circ\>\<psi\><rsup|-1>:\<psi\>(V<big|cap>W)\<rightarrow\>\<tau\>(V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|restriction
of C-r mapping> and \<varphi\>(<big|cap>V<big|cap>W) is
open>>|<cell|\<psi\><rsub|\<varphi\><rsub|\|\<varphi\>(U<big|cap>V<big|cap>W)>>=\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<psi\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|restriction
of C-r mapping> and \<psi\>(U<big|cap>V<big|cap>W) is
open>>|<cell|\<tau\><rsub|\<psi\><rsub|\|\<psi\>(U<big|cap>V<big|cap>W)>>=\<tau\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<tau\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|<reference|generalized
chain rule>>>|<cell|\<tau\><rsub|\<psi\><rsub|\|\<psi\>(U<big|cap>V<big|cap>W)>>\<circ\>\<psi\><rsub|\<varphi\><rsub|\|\<varphi\>(U<big|cap>V<big|cap>W)>>=\<tau\>\<circ\>\<psi\><rsup|-1>\<circ\>\<psi\>\<circ\>\<varphi\><rsup|-1>\|\<varphi\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<tau\>(U<big|cap>V<big|cap>W)\<equallim\><rsub|<reference|intermediate
bijective function>>\<tau\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V<big|cap>W)\<rightarrow\>\<tau\>(U<big|cap>V<big|cap>W)
is C<rsup|r>>>>>
</eqnarray*>
now we can use <reference|C-r is a local property> and
<math|y=\<varphi\>(x)\<in\>\<varphi\>(U<big|cap>V<big|cap>W) (open)>
to prove that <math|\<tau\><rsub|\<varphi\>>> is <math|C<rsup|r>>
</enumerate>
From (a) and (b) it follows then that
<math|\<cal-A\>,\<cal-A\><rprime|''>> are compatible and hence
transitivity
</enumerate>
</proof>
<\definition>
<index|differentiable structure><index|<math|\<cal-C\><rsup|(r)>[\<cal-A\>]>>Let
<math|M> be a topological space, <math|X> a normed space and
<math|\<cal-A\>> a atlas of class <math|C<rsup|r>> on <math|M> modelled
on <math|X> then the equivalence class
<math|\<cal-C\><rsup|(r)><rsup|>[\<cal-A\>]={\<cal-A\><rprime|'>\|\<cal-A\>,\<cal-A\><rprime|'>
are compatible}\<in\>\<Delta\><rsup|r>> is called a differentiable
structure of class <math|C<rsup|r>> on <math|M> modelled on <math|X>
</definition>
<\lemma>
<label|maximal atlas of a differential structure
(1)><index|<math|\<cal-M\>(\<cal-A\>)>>Let <math|M> be a topological
space, <math|X> a vector space and <math|\<cal-A\>> a atlas of class
<math|C<rsup|r>> and <math|\<cal-C\><rsup|(r)>[\<cal-A\>]> the
differential structure on <math|M> defined by <math|\<cal-A\>>. Then
there exist a unique maximal (defined by the inclusion relation) atlas
<math|\<cal-M\>(\<cal-A\>)<rsup|>> on <math|M> modelled on <math|X> in
<math|\<cal-C\><rsup|(r)>[\<cal-A\>]>. Further we have that
<math|\<cal-M\>(\<cal-A\>)=<big|cup><rsub|\<cal-B\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]>\<cal-B\>>.
Note that <math|\<cal-C\><rsup|(r)>[\<cal-A\>]=\<cal-C\><rsup|(r)>[\<cal-M\>(\<cal-A\>)]>
as by definition <math|\<cal-M\>(\<cal-A\>)\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]>
</lemma>
<\proof>
Define <math|\<cal-M\>(\<cal-A\>)=<big|cup><rsub|\<cal-B\>\<in\>\<cal-C\>[\<cal-A\>]>\<cal-B\>>
then we prove the following
<\enumerate>
<item><math|\<cal-M\>(\<cal-A\>)> is a atlas of class <math|C<rsup|r>>.
First <math|if ><math|m\<in\>M> then as <math|\<cal-A\>> is a atlas
there exists a <math|(U,\<varphi\>)\<in\>\<cal-A\>\<subseteq\>\<cal-M\>(\<cal-A\>)>
such that <math|m\<in\>U> proving that <math|\<cal-M\>(\<cal-A\>)> is a
atlas. Further if <math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-M\>(\<cal-A\>)>
such that <math|U<big|cap>V\<neq\>\<emptyset\>> then there exists
<math|\<cal-A\><rprime|'>,\<cal-A\><rprime|''>> which are compatible
with <math|\<cal-A\>> such that <math|(U,\<varphi\>)\<in\>\<cal-A\><rprime|'>,(V,\<psi\>)\<in\>\<cal-A\><rprime|''>>
now by transitivity we have that <math|\<cal-A\><rprime|'>,\<cal-A\><rprime|''>>
are compatible and thus <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is <math|C<rsup|r>> proving that <math|\<cal-M\>(\<cal-A\>)> is a atlas
of class <math|C<rsup|r>> (see <reference|characterization of a atlas
of class C-r>).
<item><math|\<cal-M\>(\<cal-A\>)> is compatible with <math|\<cal-A\>>
thus a element of <math|\<cal-C\><rsup|(r)>[\<cal-A\>]>. If
<math|(U,\<varphi\>)\<in\>\<cal-A\>> and
<math|(V,\<psi\>)\<in\>\<cal-M\>(\<cal-A\>)> then there exists a
<math|\<cal-A\><rprime|'>> compatible with <math|\<cal-A\>> such that
<math|(V,\<psi\>)\<in\>\<cal-A\><rprime|'>> and thus
<math|><math|\<psi\>\<circ\>\<varphi\><rsub|><rsup|-1>\|\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
and <math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
are of class <math|C<rsup|r>> and by <reference|characterization of
compatibility> we have that <math|\<cal-A\>,\<cal-M\>(\<cal-A\>)> are
compatible.
<item><math|\<cal-M\>(\<cal-A\>)> is the unique maximal element in
<math|\<cal-C\><rsup|(r)>[\<cal-A\>]>. First maximality is trivially
for if <math|\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]\<Rightarrow\>\<cal-A\><rprime|'>\<subseteq\><big|cup><rsub|\<cal-B\>\<in\>\<cal-C\><rsup|r>[\<cal-A\>]>\<cal-B\>=\<cal-M\>(\<cal-A\>)>.
Second if there is another maximal atlas <math|\<cal-A\><rprime|''>> of
class <math|C<rsup|r>> in <math|\<cal-C\><rsup|(r)>[\<cal-A\>]> then we
have <math|\<cal-A\><rprime|''>\<subseteq\>\<cal-M\>(\<cal-A\>)\<subseteq\>\<cal-A\><rprime|''>>
because of maximality of <math|\<cal-A\><rprime|''>,\<cal-M\>(\<cal-A\>)>
proving that <math|\<cal-A\><rprime|''>=\<cal-M\>(\<cal-A\>)> and thus
uniqueness
</enumerate>
</proof>
<\definition>
<index|differentiable manifold>A differentiable manifold
<math|(M,\<cal-A\>)> of class <math|C<rsup|r>,r\<geqslant\>1> modelled on
a normed vector space <math|X>, is a pair of a topological Hausdorff
space together with the maximal atlas <math|\<cal-A\>> for a
differentiable structure of class <math|C<rsup|r>> on <math|M> modelled
on <math|X> (see <reference|maximal atlas of a differential structure
(1)>). Note that the differentiable structure is given by
<math|\<cal-C\><rsup|[r]>[\<cal-A\>]> (so we don't need the original
atlas once we know the maximal atlas). A differentiable manifold of class
<math|C<rsup|\<infty\>>> is called a smooth manifold.
</definition>
<\remark>
If <math|\<forall\>r\<in\>{1,\<ldots\>,\<infty\>}> we have that
<math|(M,\<cal-A\>)> is differentiable manifold of class <math|C<rsup|r>>
modelled on <math|X> then we say <math|(M,\<cal-A\>>) is a differentiable
manifold of class <math|C<rsup|\<infty\>>> (or a smooth manifold)
modelled on <math|X>. If <math|(M,\<cal-A\>)> is a topological manifold
mapped on <math|X> then we say that <math|(M,\<cal-A\>)> of class
<math|C<rsup|0>> modelled on <math|X>. Note that in the future if we want
to avoid speaking about <math|C<rsup|0>> manifolds we mention
<math|r\<geqslant\>1> which includes also the case <math|r=\<infty\>>
</remark>
<\theorem>
<label|characterization of differentiable manifolds>Let <math|M> be a
Hausdorff space, <math|X> a normed vector space then <math|(M,\<cal-A\>)>
is a differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on <math|X> if and only if <math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)\|i\<in\>I}>
is a set of pairs of open sets of <math|M> and homeomorphism from
<math|U<rsub|i>\<rightarrow\>\<varphi\>(U<rsub|i>)\<subseteq\>X,\<varphi\>(U<rsub|i>)>
open such that\
<\enumerate>
<item><math|<big|cup><rsub|i\<in\>I>U<rsub|i>=M>
<item><math|\<forall\>(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>> with
<math|U<big|cap>V\<neq\>\<emptyset\>> we have that
<math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>> on
<math|\<varphi\>(U<big|cap>V)>
<item>If <math|(V,\<psi\>)> is a pair of a open set in <math|M> and a
homeomorphism <math|\<psi\>:V\<rightarrow\>\<psi\>(V)> such that
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> we have that if
<math|U<big|cap>V\<neq\>\<emptyset\>>
<\enumerate>
<item><math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>>on
<math|\<psi\>(U<big|cap>V)>
<item><math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>> on
<math|\<varphi\>(U<big|cap>V)>
</enumerate>
then <math|(V,\<psi\>)\<in\>\<cal-A\>>
</enumerate>
</theorem>
<\proof>
\;
<math|\<Rightarrow\>>If <math|(M,\<cal-A\>)> is a differentiable manifold
of class <math|C<rsup|r>> then there exists a differentiable structure
<math|\<cal-C\><rsup|(r)>[\<cal-A\>]> such that
<math|\<cal-A\>=\<cal-M\>(\<cal-A\>)> is the maximal atlas of class
<math|C<rsup|r>> so (1) and (2) follows by the definition of a atlas of
class <math|C<rsup|r>>. For <math|(3)> if <math|(V,\<psi\>)> fulfills
(a),(b) then <math|\<cal-A\><big|cup>{(V,\<psi\>)}> is compatible with
<math|\<cal-A\>> and thus by maximality we have
<math|\<cal-A\><big|cup>{(V,\<psi\>)}\<subseteq\>\<cal-A\>\<Rightarrow\>(V,\<psi\>)\<in\>\<cal-A\>>.
<math|\<Leftarrow\>> If <math|(M,\<cal-A\>)> is such that
<math|\<cal-A\>> fulfils (1) and (2) then by <reference|characterization
of a atlas of class C-r> we have that <math|\<cal-A\>> is a atlas of
class <math|C<rsup|r>> so <math|M> is a topological manifold. Now if
<math|(V,\<psi\>)\<in\>\<cal-M\>(\<cal-A\>)\<Rightarrow\>\<exists\>\<cal-A\><rprime|'>\<in\>\<cal-C\>[\<cal-A\>]>
such that <math|(V,\<psi\>)\<in\>\<cal-A\><rprime|'>> which is compatible
with <math|\<cal-A\>> and thus <math|(a),(b)> of <math|(3)> applies (see
<reference|characterization of compatibility>) and thus
<math|(V,\<psi\>)\<in\>\<cal-A\>> so <math|\<cal-M\>(\<cal-A\>)\<subseteq\>\<cal-A\>\<Rightarrow\>\<cal-A\>=\<cal-M\>(\<cal-A\>)>
and thus <math|(M,\<cal-A\>)> is a differentiable manifold of class
<math|C<rsup|r>>
</proof>
<\note>
<label|restriction of local chart in differentiable manifold>If
<math|(M,\<cal-A\>)> is a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on <math|X> then if
<math|(U,\<varphi\>)\<in\>\<cal-A\>> and <math|W> open in <math|M> such
that <math|\<emptyset\>\<neq\>W\<subseteq\>U> then
<math|(W,\<varphi\><rsub|\|W>)\<in\>\<cal-A\>>
</note>
<\proof>
Take any <math|(V,\<psi\>)\<in\>\<cal-A\>> with
<math|V<big|cap>W\<neq\>\<emptyset\>\<Rightarrowlim\><rsub|W\<subseteq\>U>V<big|cap>U\<neq\>\<emptyset\>>
so we have that <math|\<psi\><rsub|\<varphi\>>=\<psi\>\<circ\>\<varphi\><rsup|-1>\|\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V),\<varphi\><rsub|\<psi\>>=\<varphi\>\<circ\>\<psi\><rsup|-1>\|\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
are differentiable of class <math|C<rsup|r>>. From this and the fact that
the open set <math|\<varphi\>(V<big|cap>W)\<subseteq\>\<varphi\>(V<big|cap>U)
and the open set \<psi\>(V<big|cap>W)\<subseteq\>\<psi\>(V<big|cap>U)> we
find by <reference|restriction of C-r mapping> that
<math|\<psi\><rsub|\<varphi\><rsub|\|\<varphi\>(V<big|cap>W)>>=\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(V<big|cap>W)\<rightarrow\>\<psi\>(V<big|cap>W)\<equallim\><rsub|<reference|restriction
of a bijective mapping>,<reference|restriction and composition of
functions>,<reference|image of a restricted
function>>\<psi\>\<circ\>(\<varphi\><rsub|\|W>)<rsup|-1>:\<varphi\><rsub|\|W>(V<big|cap>W)\<rightarrow\>\<psi\>(V<big|cap>W)>
is differentiable of class <math|C<rsup|r>.> Again using
<reference|restriction of C-r mapping> we have that
<math|\<varphi\><rsub|\<psi\><rsub|\|\<psi\>(V<big|cap>W)>>=\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(V<big|cap>W)\<rightarrow\>\<varphi\>(V<big|cap>W)\<equallim\><rsub|<reference|restriction
and composition of functions>,<reference|image of a restricted
function>>\<varphi\><rsub|\|W>\<circ\>\<psi\><rsup|-1>:\<psi\>(V<big|cap>W)\<rightarrow\>\<varphi\><rsub|\|W>(V<big|cap>W)>.
Then by (3) in <reference|characterization of differentiable manifolds>
we have that <math|(W,\<varphi\><rsub|\|W>)\<in\>\<cal-A\>>
</proof>
<\theorem>
<label|manifold defined on a open subset of a manifold>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed set <math|X> then
if <math|W\<subseteq\>M> is a non empty open subset we can define the
differential manifold of class <math|C<rsup|r>,r\<geqslant\>1>
<math|(W,\<cal-A\><rsub|\|W>)> where <math|\<cal-A\><rsub|\|W>={(W<big|cap>V,\<varphi\><rsub|\|V>):(V,\<varphi\>)\<in\>\<cal-A\>}>,
furthermore using <reference|restriction of local chart in differentiable
manifold> we have that <math|\<cal-A\><rsub|\|W>\<subseteq\>\<cal-A\>>
</theorem>
<\proof>
We use <reference|characterization of differentiable manifolds> to proof
this.
<\enumerate>
<item><math|<big|cup><rsub|(V,\<varphi\>)\<in\>\<cal-A\>>(W<big|cap>V)=(<big|cup><rsub|(V,\<varphi\>)\<in\>\<cal-A\>>V)<big|cap>W=M<big|cap>W=W>
<item>Let <math|(U<rprime|'>,\<varphi\><rprime|'>),(V<rprime|'>,\<psi\><rprime|'>)\<in\>\<cal-A\><rsub|\|W>>
with <math|U<rprime|'><big|cap>V<rprime|'>\<neq\>\<emptyset\>> then
there exists <math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-A\>> such that
<math|U<rprime|'>=W<big|cap>U\<subseteq\>U,V<rprime|'>=W<big|cap>V\<subseteq\>V>
and <math|\<varphi\><rprime|'>=\<varphi\><rsub|\|W>,\<psi\><rprime|'>=\<psi\><rsub|\|W>>
then from <reference|restriction of local chart in differentiable
manifold> we have that <math|(\<varphi\><rprime|'>,U<rprime|'>),(\<psi\><rprime|'>,V<rprime|'>)\<in\>\<cal-A\>\<Rightarrow\>\<psi\><rprime|'>\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(U<rprime|'><big|cap>V<rprime|'>)\<rightarrow\>\<psi\><rprime|'>(U<rprime|'><big|cap>V<rprime|'>)>
is differentiable of class <math|C<rsup|r>>.
<item>If <math|(V,\<psi\>)> is a pair of a open set in <math|W> and a
homeomorphism <math|\<psi\>:V\<rightarrow\>\<psi\>(V)> such that
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|\|W>> we have that
if <math|U<big|cap>V\<neq\>\<emptyset\>>
<\enumerate>
<item><math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>>on
<math|\<psi\>(U<big|cap>V)>
<item><math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>> on
<math|\<varphi\>(U<big|cap>V)>
</enumerate>
If we proof that <math|(V,\<psi\>)\<in\>\<cal-A\>> then as
<math|V\<subseteq\>W> we have that <math|(V,\<psi\>)=(V<big|cap>W,\<psi\><rsub|\|W>)\<in\>\<cal-A\><rsub|\|W>>.
So to proceed let <math|(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|\<emptyset\>\<neq\>U<big|cap>V\<equallim\><rsub|V\<subseteq\>W>(U<big|cap>W)<big|cap>V>
and thus if <math|\<varphi\><rprime|'>=\<varphi\><rsub|\|W>> and
<math|U<rprime|'>=U<big|cap>W> then we have
<math|(U<rprime|'>,\<varphi\><rprime|'>)\<in\>\<cal-A\><rsub|\|W>> and
as <math|U<rprime|'><big|cap>V\<neq\>\<emptyset\>> we have by the
hypothesis that <math|\<varphi\><rprime|'>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<rprime|'><big|cap>V)\<rightarrow\>\<varphi\><rprime|'>(U<rprime|'><big|cap>V)\<equallim\><rsub|U<rprime|'><big|cap>V=U<big|cap>V,<reference|restriction
and composition of functions>,<reference|image of a restricted
function>>\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)=\<varphi\>(U<big|cap>V)>
and <math|\<psi\>\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(U<rprime|'><big|cap>V)\<rightarrow\>\<psi\>(U<rprime|'><big|cap>V)\<equallim\><rsub|U<rprime|'><big|cap>V=U<big|cap>V,<reference|restriction
and composition of functions>,<reference|restriction of a bijective
mapping>,<reference|image of a restricted
function>>\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
are differentiable of class <math|C<rsup|r>>. And then by
<reference|characterization of differentiable manifolds> we have
<math|(V,\<psi\>)\<in\>\<cal-A\>> as must be proved.
</enumerate>
</proof>
<\note>
<label|atlas on open subset of manifold>If \ <math|(M,\<cal-A\><rsub|M>)>
is a differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on <math|X,W\<subseteq\>M> a open set then if
<math|\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>]\<Rightarrow\>\<cal-A\><rsub|\|W>\<in\>\<cal-C\><rsup|(r)>[(\<cal-A\><rsub|M>)<rsub|\|W>]>
</note>
<\proof>
Now as <math|\<cal-A\><rsub|\|W>\<subseteq\>\<cal-A\>,(\<cal-A\><rsub|M>)<rsub|\|W>\<subseteq\>\<cal-A\><rsub|M>>
we have <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|\|W>,\<forall\>(V,\<psi\>)\<in\>(\<cal-A\><rsub|M>)<rsub|\|W>>
(as <math|\<cal-A\>> is <math|C<rsup|r>> compatible with
<math|\<cal-A\><rsub|M>>) that <math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
and <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
are of class <math|C<rsup|r>> so that <math|\<cal-A\><rsub|\|W>> and
<math|(\<cal-A\><rsub|M>)<rsub|\|W>> are compatible so
<math|\<cal-A\><rsub|\|W>\<in\>\<cal-C\><rsup|(r)>[(\<cal-A\><rsub|M>)<rsub|\|W>]>
</proof>
<\example>
<label|normed space is a manifold>Let <math|X> be a normed space then let
<math|\<cal-M\><rsup|r><rsub|X>={(U,\<psi\>)\|U is open in X,
\<psi\>:U\<rightarrow\>\<psi\>(U) is a diffeomorphisme of class
C<rsup|r>}> then <math|(X,\<cal-M\><rsup|r><rsub|X>)> is a differentiable
manifold of class <math|C<rsup|r>>. Note that
<math|(X,i<rsub|X>)\<in\>\<cal-M\><rsup|r><rsub|X>> and
<math|{(X,i<rsub|X>)}> is compatible with
<math|\<cal-M\><rsub|X><rsup|r>> so <math|\<cal-M\><rsub|X><rsup|r>=\<cal-M\>({(X,i<rsub|X>)})>
</example>
<\proof>
We use <reference|characterization of differentiable manifolds> to proof
the assertion.
<\enumerate>
<item>Clearly <math|<big|cup><rsub|(U,\<psi\><rsub|U>)\<in\>\<cal-M\><rsup|r><rsub|X>>U=X
> as <math|X> is open and <math|i<rsub|X>> is a diffeomorphism of class
<math|C<rsup|r>> (<math|C<rsup|\<infty\>> actally)>
<item>If <math|(U,\<varphi\>),(V,\<psi\>)\<in\>\<cal-M\><rsup|r><rsub|X>>
with <math|U<big|cap>V\<neq\>\<emptyset\>> then as
<math|\<varphi\>,\<psi\>> are diffeomorphism's of class
<math|C<rsup|r>> we have that <math|\<psi\>:U<big|cap>V\<rightarrow\>\<psi\>(U<big|cap>V)>
and <math|\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>U<big|cap>V>
are diffeomorphism's of class <math|C<rsup|r>> and thus using the chain
rule \ <math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is a diffeomorphism of class <math|C<rsup|r>>
<item><math|>If <math|(V,\<psi\>)> is a pair of a open set in <math|M>
and a homeomorphism <math|\<psi\>:V\<rightarrow\>\<psi\>(V)> such that
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-M\><rsup|r><rsub|X>> we have
that if <math|U<big|cap>V\<neq\>\<emptyset\>>
<\enumerate>
<item><math|\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>\<varphi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>>on
<math|\<psi\>(U<big|cap>V)>
<item><math|\<psi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
is differentiable of class <math|C<rsup|r>> on
<math|\<varphi\>(U<big|cap>V)>
</enumerate>
Now take <math|x\<in\>V> then there exists a
<math|(U,\<varphi\>)\<in\>\<cal-M\><rsup|r><rsub|X>> such that
<math|x\<in\>U\<Rightarrow\>x\<in\>U<big|cap>V>(a open set) and
<math|\<psi\>(x)\<in\>\<psi\>(U<big|cap>V)> a open set then
<math|\<varphi\>:U<big|cap>V\<rightarrow\>\<varphi\>(U<big|cap>V)> and
<math|\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>V)\<rightarrow\>U<big|cap>V>
are of class <math|C<rsup|r>> (even diffeomorphism's) and thus
<math|\<varphi\><rsup|-1>\<circ\>\<varphi\>\<circ\>\<psi\><rsup|-1>:\<psi\>(U<big|cap>V)\<rightarrow\>U<big|cap>V>
and <math|\<psi\>\<circ\>\<varphi\><rsup|-1>\<circ\>\<varphi\>:U<big|cap>V\<rightarrow\>\<psi\>(U<big|cap>V)>
are of class <math|C<rsup|r>>. Then from <reference|C-r is a local
property> we have that <math|\<psi\>,\<psi\><rsup|-1>> are of class
<math|C<rsup|r>> and thus <math|\<psi\>:U\<rightarrow\>\<psi\>(U)> is a
diffeomorphism or <math|(V,\<psi\>)\<in\>\<cal-M\><rsup|r><rsub|X>>
</enumerate>
</proof>
<\example>
<label|open subset of metric space is a differentiable manifold>Let
<math|X> be a normed space then if <math|U> is a open set we have using
<reference|atlas in a normed space> that <math|{(U,i<rsub|U>)}> is a
atlas of class <math|C<rsup|\<infty\>>> on <math|U> modelled on <math|X>
\ (meaning that is a atlas of class <math|C<rsup|r>> for every
<math|r\<geqslant\>1>). So for every <math|r\<geqslant\>1> if
<math|\<cal-M\><rsup|>({(U,i<rsub|U>)})> is the maximal atlas on the
differential structure <math|\<cal-C\><rsup|(r)>[{(U,i<rsub|U>)}]> then
<math|(U,\<cal-M\><rsup|>({(U,i<rsub|U>)}))> is a differentiable manifold
of class <math|C<rsup|r>>.\
</example>
<section|Differentiable mappings between manifolds>
<\definition>
<index|differentiable mapping between
manifolds><index|<math|C<rsup|r>>>Let <math|(M,\<cal-A\><rsub|M>)> and
<math|(N,\<cal-A\><rsub|N>)> be differentiable manifolds of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on normed spaces <math|X,Y>
respectively then a continuous function <math|f:M\<rightarrow\>N> is
differentiable (of class <math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r)> if
for any <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>> we have that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (differentiable of class <math|C<rsup|k>>).\
</definition>
<\note>
From this definition it automatically follows that if <math|f> is
differentiable of class <math|C<rsup|r>>, that <math|f> is differentiable
of class <math|C<rsup|l>,1\<leqslant\>l\<leqslant\>r>
</note>
<\note>
The function <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is noted by <math|f<rsub|[\<varphi\>,\<psi\>]>>
</note>
<\theorem>
<label|definition of C-r between manifolds is independent of atlas>Let
<math|(M,\<cal-A\><rsub|M>)> and <math|(N,\<cal-A\><rsub|N>)> be
differentiable manifolds of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on normed spaces <math|X,Y> respectively then for a continuous
function <math|f:M\<rightarrow\>N> the following are equivalent
<\enumerate>
<item><math|f> is differentiable (of class
<math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r>)
<item><math|\<forall\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<forall\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
we have that <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>> we have
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<exists\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<exists\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
such that <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>> we have
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k><rsup|>>)
</enumerate>
</theorem>
<\proof>
\;
<math|(1)\<Rightarrow\>(2)>\
This follows from the fact that <math|\<cal-A\>\<subseteq\>\<cal-A\><rsub|M>,\<cal-A\><rprime|'>\<subseteq\>\<cal-A\><rsub|N>>
<math|(2)\<Rightarrow\>(3)>\
This follows by taking <math|\<cal-A\>=\<cal-A\><rsub|M>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<cal-A\><rprime|'>=\<cal-A\><rsub|N>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
<math|(3)\<Rightarrow\>(1)>
Assume <math|\<cal-A\>,\<cal-A\><rprime|'>> as in <math|(3)> and let
<math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
such that <math|f<rsup|-1>(V)<big|cap>U\<neq\>\<emptyset\>> and consider
<math|f<rsub|[\<varphi\>,\<psi\>]>=\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>.
To proof that <math|f<rsub|[\<varphi\>,\<psi\>]>> is differentiable (of
class <math|C<rsup|k>>) it suffices by (<reference|differentiability and
restricted mappings>, <reference|C-r is a local property>) to prove that
<math|\<forall\>x\<in\>\<varphi\>(f<rsup|-1>(V))> there exists a open
<math|W> with <math|x\<in\>W\<subseteq\>\<varphi\>(f<rsup|-1>(V))> such
that <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:W\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>).
So let <math|x\<in\>\<varphi\>(f<rsup|-1>(V))> then
<math|\<exists\>y\<in\>f<rsup|-1>(V)<big|cap>U> such that
<math|x=\<varphi\>(y)> and <math|f(y)\<in\>V>. By the definition of a
atlas there exists a <math|(U<rprime|'>,\<varphi\><rprime|'>)\<in\>\<cal-A\>\<subseteq\>\<cal-A\><rsub|M>,(V<rprime|'>,\<psi\><rprime|'>)\<in\>\<cal-A\><rprime|'>\<subseteq\>\<cal-A\><rsub|N>>
such that <math|y\<in\>U<rprime|'>> and <math|f(y)\<in\>V<rprime|'>>.\
As <math|y\<in\>U<big|cap>U<rprime|'>\<Rightarrow\>U<big|cap>U<rprime|'>\<neq\>\<emptyset\>>
and <math|f(y)\<in\>V<big|cap>V<rprime|'>\<Rightarrow\>V<big|cap>V<rprime|'>\<neq\>\<emptyset\>>
we have again by the definition of a atlas that
<math|\<varphi\><rprime|'>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>U<rprime|'>)\<rightarrow\>\<varphi\><rprime|'>(U<big|cap>U<rprime|'>)>
and <math|\<psi\>\<circ\>\<psi\><rprime|'><rsup|-1>:\<psi\><rprime|'>(U<big|cap>V)\<rightarrow\>\<psi\>(U<big|cap>V)>
are differentiable of class <math|C<rsup|r>>.\
By (3) and <math|y\<in\>f<rsup|-1>(V<rprime|'>)<big|cap>U<rprime|'>> we
have that <math|f<rsub|[\<varphi\><rprime|'>,\<psi\><rprime|'>]>=\<psi\><rprime|'>\<circ\>f\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(f<rsup|-1>(V<rprime|'>))\<rightarrow\>\<psi\><rprime|'>(V<rprime|'>)>
is differentiable of class <math|C<rsup|k>>.
From the two above points and the (generalized) chainrule
\ <math|><reference|differentiability of composition of differentiable
mappings>(<reference|generalized chain rule>) we have then that
<math|(\<psi\>\<circ\>\<psi\><rprime|'><rsup|-1>)\<circ\>f<rsub|[\<varphi\><rprime|'>,\<psi\><rprime|'>]>\<circ\>(\<varphi\><rprime|'>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(f<rsup|-1>(V<rprime|'><big|cap>V)<big|cap>U<big|cap>U<rprime|'>)\<rightarrow\>\<psi\>(V<rprime|'><big|cap>V)>
is differentiable (of class <math|C<rsup|k>>).\
Also <with|mode|math|(\<psi\>\<circ\>\<psi\><rprime|'><rsup|-1>)\<circ\>f<rsub|[\<varphi\><rprime|'>,\<psi\><rprime|'>]>\<circ\>(\<varphi\><rprime|'>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(f<rsup|-1>(V<rprime|'><big|cap>V)<big|cap>U<big|cap>U<rprime|'>)\<rightarrow\>\<psi\>(V<rprime|'><big|cap>V)=\<psi\>\<circ\>\<psi\><rprime|'><rsup|-1>\<circ\>\<psi\><rprime|'>\<circ\>f\<circ\>\<varphi\><rprime|'><rsup|-1>\<circ\>\<varphi\><rprime|'>\<circ\>\<psi\>:\<varphi\>(f<rsup|-1>(V<big|cap>V<rprime|'>)<big|cap>U<big|cap>U<rprime|'>)\<rightarrow\>\<psi\><rprime|'>(V<big|cap>V<rprime|'>)=\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V<big|cap>V<rprime|'>)<big|cap>U<big|cap>U<rprime|'>)\<rightarrow\>\<psi\>(V<big|cap>V<rprime|'>)>
is differentiable (of class <math|C<rsup|k>>)
Finally note that <math|x\<in\>\<varphi\>(y)\<in\>\<varphi\>(f<rsup|-1>(V<big|cap>V<rprime|'>)<big|cap>U<big|cap>U<rprime|'>)=W>
and that <math|W> is open by remark <reference|image of open set by chart
is open> and the fact that <math|f> is continuous.
</proof>
<\theorem>
<label|characterization of differential mapping between manifolds>Let
<math|(M,\<cal-A\><rsub|M>)> and <math|(N,\<cal-A\><rsub|N>)> be
differentiable manifolds of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on normed spaces <math|X,Y> respectively and
<math|f:M\<rightarrow\>N> a continuous function. Then the following
statements are equivalent
<\enumerate>
<item><math|f> is differentiable (of class
<math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r)>
<item><math|\<forall\>p\<in\>M,\<forall\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>>,<math|\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
with <math|p\<in\>U,f(p)\<in\>V> we have
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<forall\>p\<in\>M> <math|\<exists\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>>,
<math|\<exists\>(V,\<psi\>)\<in\>\<cal-A\><rsub|N>> such that
<math|p\<in\>U> and <math|f(p)\<in\>V> and such that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (differentiable of class <math|C<rsup|k>>)
(Essentially the definition of differentiable mappings is independent
of the choosen chart)
<item><math|\<exists\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<exists\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
such that <math|\<forall\>p\<in\>M>
<math|\<exists\>(U,\<varphi\>)\<in\>\<cal-A\>,\<exists\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|p\<in\>U,f(p)\<in\>V> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>)>
</enumerate>
</theorem>
<\proof>
\;
<\math>
(1)\<Rightarrow\>(2)
</math>
If <math|p\<in\>U,f(p)\<in\>V\<Rightarrow\>p\<in\>f<rsup|-1>(V)<big|cap>U>
and from the differentiability (of class <math|C<rsup|k>>) we have then
the desired result.
<math|(2)\<Rightarrow\>(1)>\
Let <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> and
<math|(V,\<psi\>)\<in\>\<cal-A\><rsub|N>> with
<math|f<rsup|-1>(V)<big|cap>U\<neq\>\<emptyset\>> then
<math|\<exists\>p\<in\>U> with <math|f(p)\<in\>V> and from (2) we have
then that <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
proving differentiability (of class <math|C<rsup|k>>) as
<math|(U,\<varphi\>),(V,\<psi\>)> was choosen arbitrarely.
<math|(1)\<Rightarrow\>(3)>
If <math|p\<in\>M> then by the definition of a atlas there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> with <math|p\<in\>U> and a
<math|(V,\<psi\>)\<in\>\<cal-A\><rsub|N>> with <math|f(p)\<in\>V>. From
this it follows that <math|p\<in\>f<rsup|-1>(V)\<Rightarrow\>U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>>
and from the hypothese we have then that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<math|(3)\<Rightarrow\>(1)>\
Let <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>>. Let
<math|y\<in\>\<varphi\>(f<rsup|-1>(V))> and let
<math|p=\<varphi\><rsup|-1>*(y)\<in\>U<big|cap>f<rsup|-1>(V)\<Rightarrow\>p\<in\>U>
and <math|f(p)\<in\>V>.\
From the hypothese (3) there exists a
<math|(U<rsub|p>,\<varphi\><rsub|p>)\<in\>\<cal-A\><rsub|M>,(V<rsub|f(p)>,\<psi\><rsub|f(p)>)\<in\>\<cal-A\><rsub|N>>
such that <math|p\<in\>U<rsub|p>,f(p)\<in\>V<rsub|f(p)>> and
<math|f<rsub|[\<varphi\><rsub|p>,\<psi\><rsub|f(p)>]>=\<psi\><rsub|f(p)>\<circ\>f\<circ\>\<varphi\><rsub|p><rsup|-1>:\<varphi\><rsub|p>(f<rsup|-1>(V))\<rightarrow\>\<psi\><rsub|f(p)>(V)>
is differentiable (of class <math|C<rsup|k>)>, note that by
<reference|differentiability and restricted mappings>
(<reference|restriction of C-r mapping>) we also have that
<math|\<psi\><rsub|f(p)>\<circ\>f\<circ\>\<varphi\><rsub|p><rsup|-1>:\<varphi\><rsub|p>(f<rsup|-1>(V<rsub|f(p)><big|cap>V))\<rightarrow\>\<psi\><rsub|f(p)>(V<rsub|f(p)><big|cap>V)>
is differentiable (of class <math|C<rsup|k>>).\
By the defnition of a atlas and the fact that
<math|p\<in\>U<big|cap>U<rsub|p>\<neq\>\<emptyset\>,f(p)\<in\>V,V<rsub|f(p)>>
we have that <math|\<varphi\><rsub|p>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>U<rsub|p>)\<rightarrow\>\<varphi\><rsub|p>(U<big|cap>U<rsub|p>)>
and <math|\<psi\>\<circ\>\<psi\><rsub|f(p)><rsup|-1>:\<psi\><rsub|f(p)>(V<big|cap>V<rsub|f(p)>)\<rightarrow\>\<psi\>(V<big|cap>V<rsub|f(p)>)>
are differentiable of class <math|C<rsup|r>>.
Using the (generalized) chain rule (<reference|generalized chain rule>)
<reference|differentiability of composition of differentiable mappings>
and differentiability of restricted mappings <reference|differentiability
and restricted mappings> (<reference|restriction of C-r mapping>) we have
that <math|(\<psi\>\<circ\>\<psi\><rsub|f(p)><rsup|-1>)\<circ\>f<rsub|[\<varphi\><rsub|p>,\<psi\><rsub|f(p)>]>\<circ\>(\<varphi\><rsub|p>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)>)<big|cap>U<big|cap>U<rsub|p>)\<rightarrow\>\<psi\>(V<big|cap>V<rsub|p>)>
is differentiable (of class <math|C<rsup|k>>).\
Note also that <math|><math|(\<psi\>\<circ\>\<psi\><rsub|f(p)><rsup|-1>)\<circ\>f<rsub|[\<varphi\><rsub|p>,\<psi\><rsub|f(p)>]>\<circ\>(\<varphi\><rsub|p>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)>)<big|cap>U<big|cap>U<rsub|p>)\<rightarrow\>\<psi\>(V<big|cap>V<rsub|p>)=<with|mode|text|<math|(\<psi\>\<circ\>\<psi\><rsub|f(p)><rsup|-1>)\<circ\>\<psi\><rsub|f(p)>\<circ\>f\<circ\>\<varphi\><rsub|p><rsup|-1>\<circ\>(\<varphi\><rsub|p>\<circ\>\<varphi\><rsup|-1>):\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)>)<big|cap>U<big|cap>U<rsub|p>)\<rightarrow\>\<psi\>(V<big|cap>V<rsub|p>)>>=<with|mode|text|<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)>)<big|cap>U<big|cap>U<rsub|p>)\<rightarrow\>\<psi\>(V<big|cap>V<rsub|p>)>>>,
<math|W=\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)>)<big|cap>U<big|cap>U<rsub|p>)>
is open by continuity of <math|f> and remark <reference|image of open set
by chart is open> and that <math|y=\<varphi\>(p)\<in\>\<varphi\>(f<rsup|-1>(V<big|cap>V<rsub|f(p)><big|cap>U<big|cap>U<rsub|p>)>.
So we have found a open <math|W> with <math|p\<in\>W> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:W\<rightarrow\>\<psi\>(V<big|cap>V<rsub|p>)>
is differentiable (of class <math|C<rsup|k>)> and thus using
<reference|differentiability and restricted mappings> (<reference|C-r is
a local property>) we have that <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>) proving that <math|f> is
differentiable (of class <math|C<rsup|k>>).
<\math>
(3)\<Rightarrow\>(4)
</math>
This is trivial by taking <math|\<cal-A\>=\<cal-A\><rsub|M>,\<cal-A\><rprime|'>=\<cal-A\><rsub|N>>
<math|(4)\<Rightarrow\>(3)>
This follows from the fact that <math|\<forall\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<forall\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
we have <math|\<cal-A\>\<subseteq\>\<cal-A\><rsub|M>,\<cal-A\><rprime|'>\<subseteq\>\<cal-A\><rsub|N>>
</proof>
<\corollary>
<label|differentiability of real functions and manifolds>Let
<math|(M,\<cal-A\><rsub|M>)> be a differentiable manifold of class
<math|C<rsup|r>> and <math|f:M\<rightarrow\>\<bbb-R\>> a continuous
function (of class <math|C<rsup|r>) > (where we consider <math|\<bbb-R\>>
to be the manifold <math|(\<bbb-R\>,\<cal-M\>({(\<bbb-R\>,i<rsub|\<bbb-R\>>)}))>then
the following are equivalent
<\enumerate>
<item><math|f> is differentiable of (class <math|C<rsup|k>>),
<math|1\<leqslant\>k\<leqslant\>r>
<item><math|><math|\<forall\>p\<in\>U there
\<exists\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> with <math|p\<in\>U>
for which we have <math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<bbb-R\>>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<exists\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>]>
such that <math|\<forall\>p\<in\>M> there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>> for which
<math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<bbb-R\>>
is differentiable (of class <math|C<rsup|k>>)
</enumerate>
</corollary>
<\proof>
\;
<math|(1)\<Rightarrow\>(2)>
Let <math|p\<in\>M> then by definition of a atlas there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> with <math|p\<in\>U> and of
course for <math|(i<rsub|\<bbb-R\>>,\<bbb-R\>)> we have
<math|f(p)\<in\>\<bbb-R\>>. Using <reference|definition of C-r between
manifolds is independent of atlas> (2) we have then that
<math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<bbb-R\>=i<rsub|\<bbb-R\>>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(\<bbb-R\>)\<rightarrow\>\<bbb-R\>>
is differentiable (of class <math|C<rsup|k>>)
<math|(2)\<Rightarrow\>(1)>
This follows from <reference|characterization of differentiability of a
mapping on a open subset of a manifold> (4) and the fact that
<math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<bbb-R\>=i<rsub|\<bbb-R\>>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(\<bbb-R\>)\<rightarrow\>\<bbb-R\>>
<math|(2)\<Rightarrow\>(3)> This follows by taking
<math|\<cal-A\>=\<cal-A\><rsub|M>>
<math|(3)\<Rightarrow\>(2)> This follows from
<math|\<cal-A\>\<subseteq\>\<cal-A\><rsub|M>>
</proof>
<\definition>
<label|differentiable mappings on a open subset of a maifold>Let
<math|(M,\<cal-A\><rsub|M>)> and <math|(N,\<cal-A\><rsub|N>)> be
differentiable manifolds of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on normed spaces <math|X,Y>. <math|U\<subseteq\>M> a open set
then the function <math|f:U\<rightarrow\>N> is differentiable (of class
<math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r>). If it is continuous and
differentiable of class <math|C<rsup|k>> considered of a function from
the manifold <math|(U,(\<cal-A\><rsub|M>)<rsub|\|U>)> to the manifold
<math|(N,\<cal-A\><rsub|N>)> (see <reference|manifold defined on a open
subset of a manifold>).
</definition>
<\theorem>
<label|characterization of differentiability of a mapping on a open
subset of a manifold>Let <math|(M,\<cal-A\><rsub|M>)> and
<math|(N,\<cal-A\><rsub|N>)> be differentiable manifolds of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on normed spaces <math|X,Y>.
<math|W\<subseteq\>M> a open set and a continuous function
<math|f:W\<rightarrow\>N> then the following is equivalent,
<math|1\<leqslant\>k\<leqslant\>r>
<\enumerate>
<item><math|f> is differentiable (of class <math|C<rsup|k>>)
<item><math|\<forall\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<forall\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
we have <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>f<rsup|-1>(V)=\<emptyset\>> we have that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<exists\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<exists\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
we have <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,\<forall\>(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>f<rsup|-1>(V)=\<emptyset\>> we have that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<forall\>p\<in\>W> there
<math|\<exists\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,\<exists\>(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
such that <math|p\<in\>U,f(p)\<in\>V> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<item><math|\<exists\>\<cal-A\>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|M>],\<exists\>\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\><rsub|N>]>
such that <math|\<forall\>p\<in\>W> there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>,(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
such that \ <math|p\<in\>U,f(p)\<in\>V> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
<item><math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,\<forall\>(W,\<psi\>)\<in\>\<cal-A\><rsub|N>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>> we have that
<math|>
</enumerate>
</theorem>
<\proof>
First note <math|f<rsup|-1>(V)\<subseteq\>W>. Now start the proof
<math|(1)\<Rightarrow\>(2)> Let <math|\<cal-A\>,\<cal-A\><rprime|'>> as
mentioned in (2) and let <math|(U,\<varphi\>)\<in\>\<cal-A\>,(V,\<psi\>)\<in\>\<cal-A\><rprime|'>>
with <math|U<big|cap>f<rsup|-1>(V)\<neq\>\<emptyset\>> then
<math|(U<rprime|'>,\<varphi\><rprime|'>)=(U<big|cap>W,\<varphi\><rsub|\|W>)\<in\>\<cal-A\><rsub|\|W>\<subseteq\>\<cal-A\><rsub|M>>
then from <reference|definition of C-r between manifolds is independent
of atlas> we have <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsub|\|W><rsup|-1>:\<varphi\><rsub|\|W>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>. As
<math|\<varphi\><rsub|\|W>(f<rsup|-1>(V))=\<varphi\>(f<rsup|-1>(V)<big|cap>W)=\<varphi\>(f<rsup|-1>(V))>
so <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)=\<psi\>\<circ\>f\<circ\>\<varphi\><rsub|\|W><rsup|-1>:\<varphi\><rsub|\|W>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable of class <math|C<rsup|k>>
<math|(2)\<Rightarrow\>(3)>
This follows by taking <math|\<cal-A\>=\<cal-A\><rsub|M>,\<cal-A\><rprime|'>=\<cal-A\><rsub|N>>
<math|(3)\<Rightarrow\>(1)>
Let <math|\<cal-A\>,\<cal-A\><rprime|'>,><math|(U,\<varphi\>),(V,\<psi\>)>
defined as in <math|(3)> then by <reference|atlas on open subset of
manifold> <math|\<cal-A\><rsub|\|W>\<in\>\<cal-C\><rsup|(3)>[\<cal-A\><rsub|M\|W>]>
and using <reference|definition of C-r between manifolds is independent
of atlas> and <math|><math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsub|\|W><rsup|-1>:\<varphi\><rsub|\|W>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)\<equallim\><rsub|See
proof (1)\<Rightarrow\>(2)>\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
we have that <math|f> is differentiable (of class <math|C<rsup|k>>)
<math|(1)\<Rightarrow\>(4)>\
Let <math|p\<in\>W> then by differentiability of <math|f> there exists a
<math|(U<rprime|'>,\<varphi\><rprime|'>)\<in\>(\<cal-A\><rsub|M>)<rsub|\|U>>
and <math|(V,\<psi\>)\<in\>\<cal-A\><rsub|N>> such that
<math|><math|p\<in\>U<rprime|'>,f(p)\<in\>V> such that
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>). Now as
<math|(U<rprime|'>,\<varphi\><rprime|'>)\<in\>(\<cal-A\><rsub|M>)<rsub|\|U>>
there exists a <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> such that
<math|U<rprime|'>=W<big|cap>U,\<varphi\><rprime|'>=\<varphi\><rsub|\|U>>.
Also <math|U<rprime|'><big|cap>f<rsup|-1>(V)=W<big|cap>U<big|cap>f<rsup|-1>(V)=U<big|cap>f<rsup|-1>(V)\<Rightarrow\>\<varphi\><rprime|'>(f<rsup|-1>(V))=\<varphi\><rprime|'>(U<rprime|'><big|cap>f<rsup|-1>(V))=\<varphi\><rprime|'>(U<big|cap>f<rsup|-1>(V))=\<varphi\>(f<rsup|-1>(V))>
so <math|\<varphi\><rprime|'>:f<rsup|-1>(V)<big|cap>U<rprime|'>\<rightarrow\>\<varphi\><rprime|'>(f<rsup|-1>(V)<big|cap>U<rprime|'>)\<equallim\><rsub|<reference|image
of a restricted function>>\<varphi\>:f<rsup|-1>(V)<big|cap>U<rprime|'>\<rightarrow\>\<varphi\>(f<rsup|-1>(V)<big|cap>U<rprime|'>)=\<varphi\>:f<rsup|-1>(V)<big|cap>U\<rightarrow\>\<varphi\>(f<rsup|-1>(V)<big|cap>U)=\<varphi\>(f<rsup|-1>(V))>
which has inverse <math|\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>f<rsup|-1>(V)<big|cap>U>
and thus <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)=\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
proving that <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
<math|(4)\<Rightarrow\>(1)> Let <math|p\<in\>W> then by the hypothese
there exists <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
such that <math|p\<in\>U,f(p)\<in\>V> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>) then
<math|(U<big|cap>W,\<varphi\><rsub|\|U>)\<in\>(\<cal-A\><rsub|M>)<rsub|\|U>>
and <math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V)<big|cap>U)\<rightarrow\>\<psi\>(V)=\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\><rsub|\|U>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)=\<psi\>\<circ\>f\<circ\>(\<varphi\><rsub|\|U>)<rsup|-1>:\<varphi\><rsub|\|U>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>) proving by
<reference|characterization of differential mapping between manifolds>
that is differentiable on <math|W>.
<math|(2)\<Rightarrow\>(3)> This is trivial by using
<math|\<cal-A\>=\<cal-A\><rsub|M>,\<cal-A\><rprime|'>=\<cal-A\><rsub|N>>
<math|(3)\<Rightarrow\>(2)> Follows from
<math|\<cal-A\>\<subseteq\>\<cal-A\><rsub|M>,\<cal-A\>\<subseteq\>\<cal-A\><rsub|N>>
</proof>
<\theorem>
<label|a chart is differentiable>Let <math|(M,\<cal-A\><rsub|M>)> be a
differentiable manifold of class <math|C<rsup|r>> modelled on <math|X>
then <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> we have that
<math|\<varphi\>:U\<rightarrow\>X> is a differentiable mapping of class
<math|C<rsup|r>> (where we consider <math|X> a differentiable manifold of
class <math|C<rsup|r>> defined as in <reference|normed space is a
manifold>).
</theorem>
<\proof>
\;
Let <math|p\<in\>U> then <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>> and
<math|(X,i<rsub|X>)\<in\>\<cal-M\>({X,i<rsub|X>})> are such that
<math|p\<in\>U>,<math|\<varphi\>(p)\<in\>X> and
<math|i<rsub|X>\<circ\>\<varphi\>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(\<varphi\><rsup|-1>(X))\<rightarrow\>X=i<rsub|X>:\<varphi\>(U)\<rightarrow\>X>
is differentiable of class <math|C<rsup|r>> proving by the previous
theorem that <math|\<varphi\>> is differentiable of class
<math|C<rsup|r>>.
</proof>
<\corollary>
<label|restriction of a differential mapping between manifolds>Let
<math|(M,\<cal-A\><rsub|M>)> and <math|(N,\<cal-A\><rsub|N>)> be
differentiable manifolds of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on normed spaces <math|X,Y>. <math|W\<subseteq\>M> a open set
and a differential continuous function (of class
<math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r)> <math|f:M\<rightarrow\>N>
then <math|f<rsub|\|W>:W\<rightarrow\>N> is a differentiable function (of
class <math|C<rsup|k>)>
</corollary>
<\proof>
This follows from (4) in (<reference|characterization of
differentiability of a mapping on a open subset of a manifold>). For if
<math|p\<in\>W> then there exists by differentiability of <math|f> a
<math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M>,(V,\<psi\>)\<in\>\<cal-A\><rsub|N>>
such that <math|p\<in\>U,f(p)\<in\>V> and
<math|\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>). Now
<math|\<psi\>\<circ\>f<rsub|\|W>\<circ\>\<varphi\><rsup|-1>:\<varphi\>((f<rsub|\|W>)<rsup|-1>(V))\<rightarrow\>\<psi\>(V)\<equallim\><rsub|<reference|restriction
and preimage>>\<psi\>\<circ\>f<rsub|\|W>\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V)<big|cap>W)\<rightarrow\>\<psi\>(V)\<equallim\><rsub|<reference|restriction
and composition of functions>>\<psi\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(V)<big|cap>W)\<rightarrow\>\<psi\>(V)>
which is differentiable (of class <math|C<rsup|k>>) because of
<reference|restriction of C-r mapping> and the fact that
<math|\<varphi\>(f<rsup|-1>(V)<big|cap>W)> is open because
<math|\<varphi\>> is a homeomorphism and <math|f> is continuous.
</proof>
<\theorem>
<label|differentiable function to a normed space>Let <math|(M,\<cal-A\>)>
be a differentiable manifold of class <math|C<rsup|r>> modelled on
<math|X>, <math|Y> a normed space then if
<math|1\<leqslant\>k\<leqslant\>r, W\<subseteq\>M> is a open set then for
a continuous function <math|f:W\<rightarrow\>Y> we have the following
equivalences
<\enumerate>
<item><math|f> is differentiable (of class <math|C<rsup|k>>)
<item><math|\<forall\>p\<in\>W> there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>> such that <math|p\<in\>\<varphi\>>
and <math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(W)\<rightarrow\>Y> is
differentiable (of class <math|C<rsup|k>>)
</enumerate>
</theorem>
<\proof>
\;
<math|(1)\<Rightarrow\>(2)>\
If <math|f> is differentiable (of class <math|C<rsup|k>)> and
<math|p\<in\>W> then using the definition of a atlas there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>> such that <math|><math|p\<in\>U> and
of course for <math|(i<rsub|Y>,Y)\<in\>\<cal-M\>({(i<rsub|Y>,Y)})> we
have that <math|f(p)\<in\>Y>. Then using <reference|characterization of
differentiability of a mapping on a open subset of a manifold> (2) we
have the differentiability (of class <math|C<rsup|k>>) of
<math|i<rsub|Y>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(Y))-Y=f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(W)\<rightarrow\>Y>
<math|(2)\<Rightarrow\>(1)>
This follows again from \ <math|i<rsub|Y>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(Y))-Y=f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(W)\<rightarrow\>Y>
and <reference|characterization of differentiability of a mapping on a
open subset of a manifold> (5)
</proof>
<\theorem>
<label|sum(product) of differentiable functions between manifolds>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>> modelled on <math|X,Y> a normed space,
<math|W\<subseteq\>X> \ and <math|f,g:W\<rightarrow\>Y> is differentiable
(of class <math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r)> then <math|f+g>
is differentiable of class <math|C<rsup|k>>, also if
<math|\<alpha\>\<in\>\<bbb-K\>>(where <math|\<bbb-K\>> is the field of
<math|Y>) then <math|\<alpha\>.f> furthermore if <math|Y=\<bbb-K\>> then
also <math|f.g> is differentiable of class <math|C<rsup|k>> is
differentiable (of class <math|C<rsup|r>>)
</theorem>
<\proof>
This follows trivially as if <math|(U,\<varphi\>)\<in\>\<cal-A\>> then
<math|(f+g)\<circ\>\<varphi\><rsup|-1>=f\<circ\>\<varphi\><rsup|-1>+g\<circ\>\<varphi\><rsup|-1>>,
<math|(\<alpha\>.f)\<circ\>\<varphi\><rsup|-1>=\<alpha\>.(f\<circ\>\<varphi\><rsup|-1>)>
and <math|(f.g)\<circ\>\<varphi\><rsup|-1>=(f\<circ\>\<varphi\><rsup|-1>).(g\<circ\>\<varphi\><rsup|-1>),>
and the properties of differentiability (of class <math|C<rsup|k>>)
between normed spaces.
</proof>
<\theorem>
<label|differentiable mappings between manifolds and composition>Let
<math|(M<rsub|1>,\<cal-A\><rsub|1>),(M<rsub|2>,\<cal-A\><rsub|2>)> and
<math|(M<rsub|3>,\<cal-A\><rsub|3>)> be differentiable manifolds of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on normed spaces <math|X,Y,Z>.
\ And <math|f:M<rsub|1>\<rightarrow\>M<rsub|2>>,
<math|g:M<rsub|2>\<rightarrow\>M<rsub|3>> be continuous functions which
are differentiable (of class <math|C<rsup|k>,1\<leqslant\>k\<leqslant\>r>)
then <math|g\<circ\>f:M<rsub|1>\<rightarrow\>M<rsub|3>> is differentiable
(of class <math|C<rsup|k>>)
</theorem>
<\proof>
\;
Of course <math|g\<circ\>f> is continuous.
Using <reference|characterization of differential mapping between
manifolds> (3) it is enough to prove that <math|\<forall\>p\<in\>M> there
exists a <math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M<rsub|1>>,(V,\<psi\>)\<in\>\<cal-A\><rsub|M<rsub|2>>>
with <math|p\<in\>U,g\<circ\>f(p)\<in\>W> such that
<math|\<psi\>\<circ\>(g\<circ\>f)\<circ\>\<varphi\><rsup|-1>:\<varphi\>((g\<circ\>f)<rsup|-1>(V))\<rightarrow\>\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>).\
First note that by differentiability of <math|f,g> there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\><rsub|M<rsub|1>>,(W,\<tau\>)\<in\>\<cal-A\><rsub|M<rsub|2>>,(V,\<psi\>)\<in\>\<cal-A\><rsub|M<rsub|3>>>
such that <math|x\<in\>U<rprime|'>,f(x)\<in\>>
<\eqnarray*>
<tformat|<table|<row|<cell|\<tau\>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(W))\<rightarrow\>\<tau\>(W)>|<cell|is
differentiable (of class C<rsup|k>)>|<cell|>>|<row|<cell|\<psi\>\<circ\>g\<circ\>\<tau\><rsup|-1>:\<tau\>(g<rsup|-1>(V))\<rightarrow\>\<psi\>(V)
>|<cell|is differentiable (of class C<rsup|k>)>|<cell|>>>>
</eqnarray*>
As <math|(g\<circ\>f)<rsup|-1>(V)=f<rsup|-1>(g<rsup|-1>(V))> is open, we
have that by remark <reference|image of open set by chart is open>
<math|\<varphi\>((g\<circ\>f)<rsup|-1>(V))> is open. Also
<math|\<tau\>\<circ\>f\<circ\>\<varphi\><rsup|-1>(\<varphi\>((g\<circ\>f)<rsup|-1>(V)))=\<tau\>(f(\<varphi\><rsup|-1>(\<varphi\>(f<rsup|-1>(g<rsup|-1>(V))))))=\<tau\>(f(f<rsup|-1>(g<rsup|-1>(V)))))\<subseteq\>\<tau\>(g<rsup|-1>(V))>
so thiat we can apply the (generalized) chain rule
(<reference|generalized chain rule>) <reference|differentiability of
composition of differentiable mappings> to proof that
<math|(\<psi\>\<circ\>g\<circ\>\<tau\><rsup|-1>)\<circ\>(\<tau\>\<circ\>f\<circ\>\<varphi\><rsup|-1>):\<varphi\>((g\<circ\>f)<rsup|-1>(V))\<rightarrow\>\<psi\>(V)=\<psi\>\<circ\>g\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>((g\<circ\>f)<rsup|-1>(V))-\<psi\>(V)>
is differentiable (of class <math|C<rsup|k>>)
</proof>
<\theorem>
<label|Cr between normed spaces os Cr between manifolds>Let <math|X,Y> be
normed vector spaces and <math|U\<subset\>X> a open set,
<math|f:U\<rightarrow\>Y> a function then <math|>f \ is differentiable of
class <math|C<rsup|r>> between the normed spaces if and only if <math|f>
is differentiable of class <math|C<rsup|r>> between the manifolds
<math|U,\<cal-M\>({(U,i<rsub|U>)})> and <math|Y,{(Y,i<rsub|Y>)}>
</theorem>
<\proof>
The proof is trivial as <math|i<rsub|Y>\<circ\>f\<circ\>(i<rsub|U>)<rsup|-1>=i<rsub|Y>\<circ\>f\<circ\>i<rsub|U>=f>
is\
</proof>
<section|Tangent spaces>
<\definition>
<label|differentiable curve><index|differential curve>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on a normed space <math|X>.
Given <math|\<varepsilon\>\<gtr\>0> consider then the open interval
<math|I<rsub|\<varepsilon\>>=]-\<varepsilon\>,\<varepsilon\>[\<subseteq\>\<bbb-R\>>
centered around <math|0>. A curve is a pair
<math|(I<rsub|\<varepsilon\>>,c)> where
<math|c:I<rsub|\<varepsilon\>>\<rightarrow\>M> is a continuous function.
A curve <math|(I<rsub|\<varepsilon\>>,c)> is called a differentiable
curve if <math|c> is differentiable (using
<math|\<cal-M\>({(I<rsub|\<varepsilon\>>,i<rsub|I<rsub|\<varepsilon\>>>)})>
for the differentiable structure (see <reference|open subset of metric
space is a differentiable manifold>)\
</definition>
<\note>
If <math|(I<rsub|\<varepsilon\>>,c)> is a differentiable curve then we
note this by <math|c> if <math|I<rsub|\<varepsilon\>>> is not important.
</note>
<\theorem>
<label|characterization of a differential curve>Let <math|(M,\<cal-A\>)>
be a differential manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on a normed space <math|X> and <math|\<varepsilon\>\<gtr\>0>
then if <math|c:I<rsub|\<varepsilon\>>\<rightarrow\>M> is a curve then we
have the following equivalence
<\enumerate>
<item><math|(I<rsub|\<varepsilon\>>,c)> is a differentiable curve
<item><math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|c<rsup|-1>(U)\<neq\>\<emptyset\>> we have that
<math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)> is
differentiable
<item><math|\<forall\>x\<in\>I<rsub|\<varepsilon\>>> we have
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with <math|c(x)\<in\>U>
that <math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable
<item><math|\<forall\>x\<in\>I<rsub|\<varepsilon\>>> we have
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with <math|c(x)\<in\>U>
that <math|<frac|d(\<varphi\>\<circ\>c)(t)|dt>(x)> exists
<item><math|\<forall\>x\<in\>I<rsub|\<varepsilon\>>> we have
<math|\<exists\>(U,\<varphi\>)\<in\>\<cal-A\>> with <math|c(x)\<in\>U>
so that <math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable.
<item><math|\<forall\>x\<in\>I<rsub|\<varepsilon\>>> we have that
<math|\<exists\>(U,\<varphi\>)\<in\>\<cal-A\>> such that
<math|c(x)\<in\>U> and that <math|<frac|d(\<varphi\>\<circ\>c)(t)|dt>(x)>
exists.
</enumerate>
Note that if <math|(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|c<rsup|-1>(U)> we have that <math|dom(\<varphi\>\<circ\>c)=c<rsup|-1>(U)>
so that <math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is indeed a function. from now on if we write <math|\<varphi\>\<circ\>c>
then we actually mean <math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
and not the partial function <math|\<varphi\>\<circ\>c:I<rsub|\<varepsilon\>>\<rightarrow\>\<varphi\>(U)>
<math|>
</theorem>
<\proof>
\;
<math|(1\<Rightarrow\>2)>
Using <reference|definition of C-r between manifolds is independent of
atlas> (2) we have that <math|\<forall\>(U,\<varphi\>)\<in\>U,(I<rsub|\<varepsilon\>>,i<rsub|I<rsub|\<varepsilon\>>>)\<in\>\<cal-M\>({I<rsub|\<varepsilon\>>,i<rsub|I<rsub|\<varepsilon\>>>)})>
that <math|\<varphi\>\<circ\>c\<circ\>i<rsub|I<rsub|\<varepsilon\>>>:i<rsub|I<rsub|\<varepsilon\>>><rsup|-1>(c<rsup|-1>(U))\<rightarrow\>\<varphi\>(U)=\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable.
<math|(2\<Rightarrow\>1)>
If <math|x\<in\>I<rsub|\<varepsilon\>>> then by the definition of a atlas
there exists a <math|(U,\<varphi\>)\<in\>\<cal-A\>> such that
<math|c(x)\<in\>U> we have by (2) that
<math|\<varphi\>\<circ\>c\<circ\>i<rsub|I\<varepsilon\>>:i<rsub|I<rsub|\<varepsilon\>>><rsup|-1>(c<rsup|-1>(U))\<rightarrow\>\<varphi\>(U)=\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable. Using <reference|definition of C-r between manifolds
is independent of atlas> (3) we have then that <math|c> is
differentiable.
<math|(1\<Rightarrow\>3)>
Using <reference|characterization of differential mapping between
manifolds> (2) we have that <math|(\<forall\>(U,\<varphi\>)\<in\>U,(I<rsub|\<varepsilon\>>,i<rsub|I<rsub|\<varepsilon\>>>)\<in\>\<cal-M\>>
such that <math|c(p)\<in\>U> we have that
<math|><math|\<varphi\>\<circ\>c\<circ\>i<rsub|I<rsub|\<varepsilon\>>>:i<rsub|I<rsub|\<varepsilon\>>><rsup|-1>(c<rsup|-1>(U))\<rightarrow\>\<varphi\>(U)=\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable.
<math|(3\<Rightarrow\>1)>\
If <math|x\<in\>I<rsub|\<varepsilon\>>> then by the definition of a atlas
we have a <math|(U,\<varphi\>)\<in\>\<cal-A\>> such that
<math|c(x)\<in\>U> and then by (3) we have
<math|\<varphi\>\<circ\>c\<circ\>i<rsub|I\<varepsilon\>>:i<rsub|I<rsub|\<varepsilon\>>><rsup|-1>(c<rsup|-1>(U))\<rightarrow\>\<varphi\>(U)=\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable. Using <reference|characterization of differential
mapping between manifolds> (3) we have that <math|c> is differentiable.
<math|(3\<Leftrightarrow\>4)>
This follow from <reference|differentiability of real or complex
functions>\
<math|(1\<Leftrightarrow\>5)>
Using <math|\<varphi\>\<circ\>c\<circ\>i<rsub|I\<varepsilon\>>:i<rsub|I<rsub|\<varepsilon\>>><rsup|-1>(c<rsup|-1>(U))\<rightarrow\>\<varphi\>(U)=\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
this follows from <reference|characterization of differential mapping
between manifolds> (3)
<math|(5\<Leftrightarrow\>6)>
This follow from <reference|differentiability of real or complex
functions>\
</proof>
<\definition>
<index|tangential curves>Let <math|(M,\<cal-A\>)> be a differentiable
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled on the normed
space <math|X>. Consider then a point <math|p\<in\>M> and let
<math|(I<rsub|\<varepsilon\><rsub|1>>,c<rsub|1>),(I<rsub|\<varepsilon\><rsub|2>>,c<rsub|2>)>
be two differentiable curves on <math|M> then these curves are called
tangential at p if\
<\enumerate>
<item><math|c<rsub|1>(0)=c<rsub|2>(0)=p>
<item><math|\<forall\>*(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|p\<in\>U> we have <math|<frac|d(\<varphi\>\<circ\>c<rsub|1>)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rsub|2>)(t)|dt>(0)>
</enumerate>
</definition>
<\theorem>
<label|criteria for tangentiality>Let <math|(M,\<cal-A\>)> be a
differential manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|X.> Consider then a point <math|p\<in\>M> and
let <math|(I<rsub|\<varepsilon\><rsub|>>,c)> be a differentiable curve on
<math|M> then if <math|(U<rsub|1>,\<varphi\><rsub|1>),(U<rsub|2>,\<varphi\><rsub|2>)\<in\>\<cal-A\>>
are two charts \ at <math|p> (<math|p\<in\>U<rsub|1><big|cap>U<rsub|2>)>
then we have <math|<frac|d(\<varphi\><rsub|1>\<circ\>c)(t)|dt>(0)=D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsup|-1><rsub|2>)(\<varphi\><rsub|2>(p))(<frac|d(\<varphi\><rsub|2>\<circ\>c)(t)|dt>(0))>.
From this it follows that to prove that two differential curves are
tangential at p, it is enough to prove that there exists a
<math|(\<varphi\>,U)\<in\>\<cal-A\>> with <math|p\<in\>U> and
<math|<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)>.
Note that we have noted here <math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>:\<varphi\><rsub|2>(U<rsub|1><big|cap>U<rsub|2>)\<rightarrow\>\<varphi\><rsub|1>(U<rsub|1><big|cap>U<rsub|2>)>
by <math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>>.
</theorem>
<\proof>
First note that by definition a manifold
<math|\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>:\<varphi\><rsub|2>(U<rsub|1><big|cap>U<rsub|2>)\<rightarrow\>\<varphi\><rsub|1>(U<rsub|1><big|cap>U<rsub|2>)>
is differentiable of class <math|C<rsup|r>>. Then as <math|c> is a
differentiable curve we have that <math|\<varphi\><rsub|1>\<circ\>c:c<rsup|-1>(U<rsub|1>)\<rightarrow\>\<varphi\><rsub|1>(U<rsub|1>)>
is differentiable and by <reference|restriction of C-r mapping>
<math|\<varphi\><rsub|1>\<circ\>c:c<rsup|-1>(U<rsub|1><big|cap>U<rsub|2>)\<rightarrow\>\<varphi\><rsub|1>(U<rsub|1><big|cap>U<rsub|2>)>
is differentiable and has the same differential on every
<math|x\<in\>c<rsup|-1>(U<rsub|1><big|cap>U<rsub|2>)> using the chain
rule we have then <math|<frac|d(\<varphi\><rsub|1>\<circ\>c)(t)|dt>(0)=D(\<varphi\><rsub|1>\<circ\>c)(0)(1)=D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>\<circ\>\<varphi\><rsub|2>\<circ\>c)(0)(1)=(D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(c(0)))\<circ\>D(\<varphi\><rsub|2>\<circ\>c)(0))(1)=D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))(<frac|d(\<varphi\><rsub|2>\<circ\>c)(t)|dt>(0))>
</proof>
<\lemma>
Let <math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X> and
let <math|p\<in\>X> then the relation 'tangential' between differential
curves at <math|p> is a equivalence relation. Lets call this relation
<math|\<bbb-T\>> and given <math|c> a differential curve in M at <math|p>
then we can form the equivalence class <math|\<bbb-T\>[c]>
</lemma>
<\proof>
\;
<\enumerate>
<item>Reflexivity, this is trivial as
<math|<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)>
for every <math|(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U>
<item>Symmetry if <math|c<rsub|1>> is tangential with <math|c<rsub|2>>
then for every <math|(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> then
<math|<frac|d(\<varphi\>\<circ\>c<rsub|1>)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rsub|2>)(t)|dt>(0)\<Rightarrowlim\><rsub|reverse
the equality>c<rsub|2>> is tangential with <math|c<rsub|1>>
<item>Transitivity, again this is trivial as if <math|c<rsub|1>> is
tangential to <math|c<rsub|2>> and <math|c<rsub|2>> is tangential to
<math|c<rsub|3>\<Rightarrow\><frac|d(\<varphi\>\<circ\>c<rsub|1>)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rsub|2>)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rsub|3>)(t)|dt>(0)\<Rightarrow\>c<rsub|1>>
is tangential to <math|c<rsub|3>>
</enumerate>
</proof>
<\definition>
<index|tangent vector><index|tangent space><index|<math|\<bbb-T\>[c]>><index|<math|T<rsub|p>M>>Let
<math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X> and
let <math|p\<in\>X>. A tangent vector to M at p is an equivalence class
of of tangential curves in M through p. So if <math|v<rsub|p>> is a
tangent vector to M at p then there exists a differential curve c in M at
p so that <math|v<rsub|p>=\<bbb-T\>[c]>. We call <math|c> a
representative for the tangent vector <math|v<rsub|p>>. The set of all
the tangent vectors to M at p is noted by <math|T<rsub|p>M> and is called
the tangent space on <math|M> at <math|x>.
</definition>
<\definition>
<index|<math|T(\<varphi\>,p)>>Let <math|(M,\<cal-A\>)> be a differential
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled on the normed
space <math|X> and let <math|p\<in\>X,(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U>
then we define the function <math|T(\<varphi\>,p):T<rsub|p>M\<rightarrow\>X>
by <math|T(\<varphi\>,p)(v<rsub|p>)=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0),
where c\<in\>v<rsub|2>> is a representative for the vector
<math|v<rsub|p>>
</definition>
<\proof>
Of course we must prove that this definition is independent of the choice
of <math|c>, which is trivial for if <math|c<rprime|'>\<in\>v<rsub|p>=\<bbb-T\>[c]\<Rightarrow\>T(\<varphi\>,p)(v<rsub|p>)=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rprime|'>)(t)|dt>(0)>
</proof>
<\theorem>
<label|T(f,p) is bijective>Let <math|(M,\<cal-A\>)> be a differential
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled on the normed
space <math|X> and let <math|p\<in\>X> then we have\
<\enumerate>
<item><math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> we have
that the function <math|T(\<varphi\>,p):T<rsub|p>M\<rightarrow\>X> is
bijective
<item><math|\<forall\>(U<rsub|1>,\<varphi\><rsub|1>),(U<rsub|2>,\<varphi\><rsub|2>)\<in\>\<cal-A\>,p\<in\>U<rsub|1><big|cap>U<rsub|2>>
we have <math|D(\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1><rsup|-1>)(\<varphi\><rsub|1>(p))\<circ\>T(\<varphi\><rsub|1>,p)=T(\<varphi\><rsub|2>,p)>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>
<\enumerate>
<item>Injectivity. If <math|T(\<varphi\>,p)(v<rsub|p>)=T(\<varphi\>,p)(v<rprime|'><rsub|p>)>
then for <math|c\<in\>v<rsub|p>,c<rprime|'>\<in\>v<rprime|'><rsub|p>>
we have <math|T(\<varphi\>,p)(v<rsub|p>)=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>c<rprime|'>)(t)|dt>(0)>
and then from <reference|criteria for tangentiality> we have that
<math|c,c<rprime|'>> are tangential as p and thus
<math|v<rsub|p>=\<bbb-T\>[c]=\<bbb-T\>[c<rprime|'>]=v<rprime|'><rsub|p>>
<item>Surjectivity. Let <math|h\<in\>X> and define and take
<math|\<varphi\>(p)\<in\>\<varphi\>(U)> which because
<math|\<varphi\>> is a homeomorphism is open in <math|X> so there
exists a <math|\<delta\>\<gtr\>0> such that
<math|B<rsub|\<shortparallel\>\<shortparallel\>>(\<varphi\>(p),\<delta\>)\<subseteq\>\<varphi\>(U)>,
now choose <math|\<varepsilon\>\<gtr\>0> such that
<math|\<varepsilon\>\<shortparallel\>h\<shortparallel\>\<leqslant\>\<delta\>>
(if <math|\<shortparallel\>h\<shortparallel\>=0> you can take
<math|\<varepsilon\>=1>, otherwise take
<math|\<varepsilon\>=<frac|\<delta\>|\<shortparallel\>h\<shortparallel\>>>)<math|>.
Then if <math|t\<in\>I<rsub|\<varepsilon\>>> we have that
<math|\<shortparallel\>\<varphi\>(p)-(\<varphi\>(p)+t.h)\<shortparallel\>=\|t\|\<shortparallel\>h\<shortparallel\>\<less\>\<varepsilon\>\<shortparallel\>h\<shortparallel\>\<leqslant\>\<delta\>\<Rightarrow\>\<varphi\>(p)+t.h\<in\>\<varphi\>(U)>.
So if we define <math|\<lambda\>:I<rsub|\<varepsilon\>>\<rightarrow\>\<varphi\>(U)>
by <math|\<lambda\>(t)=\<varphi\>(p)+t.h> then we have
<math|\<lambda\>(I<rsub|\<varepsilon\>>)\<subseteq\>\<varphi\>(U)>.
As <math|\<varphi\>:U\<rightarrow\>\<varphi\>(U)> is by definition a
bijection <math|\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>U> is
defined so we can define <math|c:I<rsub|\<varepsilon\>>\<rightarrow\>U>
by <math|c=\<varphi\><rsup|-1>\<circ\>\<lambda\>>. Note now
<\enumerate>
<item><math|c> is continuous beeing the composition of continuous
functions (<math|\<varphi\>> is a homeomorphism, and
<math|\<varphi\>> is continuous.
<item>If <math|x\<in\>I<rsub|\<varepsilon\>>> then
<math|c(x)\<in\>U> or <math|x\<in\>c<rsup|-1>(U)> and as
<math|\<varphi\>\<circ\>c=\<lambda\>> is differentiable (being the
sum of a constant function and a linear function) and then using
(1) and <reference|characterization of a differential curve> (5) we
have that <math|c> is a diffential curve.
<item><math|c(0)=\<varphi\><rsup|-1>(\<varphi\>(p)+0.h)=\<varphi\><rsup|-1>(\<varphi\>(p))>
, so we have finally proved that <math|c> is a differential curve
at <math|p>.
</enumerate>
\ \ Take now <math|v<rsub|p>=\<bbb-T\>[c]> then
<math|T(\<varphi\>,p)[v<rsub|p>]=<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0)=<frac|d(\<varphi\>\<circ\>(\<varphi\><rsup|-1>\<circ\>\<lambda\>))(t)|dt>(x)=<frac|d\<lambda\>(t)|dt>(0)=h>
proving surjectivity.
</enumerate>
<item>Let <math|v<rsub|p>=\<bbb-T\>[c]> then
<math|T(\<varphi\><rsub|2>,p)(v<rsub|p>)=<frac|d(\<varphi\><rsub|2>\<circ\>c)(t)|dt>(0)\<equallim\><rsub|<reference|criteria
for tangentiality>>D(\<varphi\><rsub|2>\<circ\>\<varphi\><rsub|1><rsup|-1>)(\<varphi\><rsub|1>(p))(<frac|d(\<varphi\><rsub|1>\<circ\>c)(t)<rsub|>|dt>(0))=D(\<varphi\><rsub|2>\<circ\>\<varphi\><rsup|-1><rsub|1>)(\<varphi\><rsub|1>(p))(T(\<varphi\><rsub|1>,p)(v<rsub|p>))\<Rightarrow\>T(\<varphi\><rsub|2>,p)=D(\<varphi\><rsub|2>\<circ\>\<varphi\><rsup|-1><rsub|1>)(\<varphi\><rsub|1>(p))\<circ\>T(\<varphi\><rsub|1>,p)>
</enumerate>
</proof>
<\theorem>
<label|TpM is a vector space>Let <math|(M,\<cal-A\>)> be a differentiable
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled on the normed
space <math|X, p\<in\>M>. Then we can equip the tangent space
<math|T<rsub|p>M> to <math|M> at p with the
<with|font-series|bold|unique> structure of a topological vector space
(meaning that the vector operations and topology is independent of the
choice of <math|\<varphi\>>) such that
<math|T(\<varphi\>,p):T<rsub|p>M\<rightarrow\>X> is a toplinear
isomorphism <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|p\<in\>U>.
</theorem>
<\proof>
\;
First we define the vector structure on <math|T<rsub|p>M>. As
<math|T(\<varphi\>,p)> is bijective by <reference|T(f,p) is bijective> we
have that <math|T(\<varphi\>,p)<rsup|-1>:X\<rightarrow\>T<rsub|p>M> is a
bijection and thus we can use <reference|induced vector space structure>
to define a vector space structure on <math|T<rsub|p>M> such that
<math|f=T(\<varphi\>,p)<rsup|-1>> is a isomorphism and from
<reference|inverse of isomorphisme> we have
<math|T(\<varphi\>,p)=f<rsup|-1>> is a isomorphism.
Second we prove that this vector structure is independent of the choice
of the <math|\<varphi\>>. So take <math|(U<rsub|1>,\<varphi\><rsub|1>),(U<rsub|2>,\<varphi\><rsub|2>)\<in\>\<cal-A\>>
and let <math|+<rsub|1>,.<rsub|1>> defined by
<math|f<rsub|1>=T(\<varphi\><rsub|1>,p)<rsup|-1>> and
<math|+<rsub|2>,.<rsub|2>> defined by
<math|f<rsub|2>=T(\<varphi\><rsub|2>,p)<rsup|-1>>. Then
\ <math|v<rsub|p>+<rsub|1>w<rsub|p>=f<rsub|1>(f<rsup|-1><rsub|1>(v<rsub|p>)+f<rsub|1><rsup|-1>(w<rsub|p>))=T(\<varphi\><rsub|1>,p)<rsup|-1>(T(\<varphi\><rsub|1>,p)(v<rsub|p>)+T(\<varphi\><rsub|1>,p)(w<rsub|p>))=T(\<varphi\><rsub|1>,p)<rsup|-1>{D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsup|-1><rsub|2>)(\<varphi\><rsub|2>(p))(T(\<varphi\><rsub|2>,p)(v<rsub|p>))+D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>)(T(\<varphi\><rsub|2>,p)(w<rsub|p>))}=(w<rsub|p>))\<equallim\><rsub|Linearity
of the differential>T(\<varphi\><rsub|1>,p)<rsup|-1>(D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p)){T(\<varphi\><rsub|2>,p)(v<rsub|p>)+T(\<varphi\><rsub|2>,p)(\<omega\><rsub|p>)}={[D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))\<circ\>T(\<varphi\><rsub|2>,p)]<rsup|-1>\<circ\>D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p)}{T(\<varphi\><rsub|2>,p)(v<rsub|p>)+T(\<varphi\><rsub|2>,p)}\<equallim\><rsub|<reference|inverse
of diffeomorphisme>>T(\<varphi\><rsub|2>,p)<rsup|-1>{T(\<varphi\><rsub|2>,p)(v<rsub|p>)+T(\<varphi\><rsub|2>,p)(w<rsub|p>)]=v<rsub|p>+<rsub|2>w<rsub|p>>.
In a similar fashion we have that <math|\<alpha\>.<rsub|1>v<rsub|p>=f<rsub|1>(\<alpha\>.f<rsub|1><rsup|-1>(v<rsub|p>))=T(\<varphi\><rsub|1>,p)<rsup|-1>(\<alpha\>.T(\<varphi\><rsub|1>,p)(v<rsub|p>))={T(\<varphi\><rsub|2>,p)<rsup|-1>\<circ\>[D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))]<rsup|-1>}(\<alpha\>.D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))(T(\<varphi\><rsub|2>,p)(v<rsub|p>))={T(\<varphi\><rsub|2>,p)<rsup|-1>\<circ\>D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))}(D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))(\<alpha\>.T(\<varphi\><rsub|2>,p)(v<rsub|p>))=T(\<varphi\><rsub|2>,p)<rsup|-1>(\<alpha\>.T(\<varphi\><rsub|2>,p)(v<rsub|p>))=\<alpha\>.<rsub|2>v<rsub|p>>.\
Finally, <math|f<rsub|1>=T(\<varphi\><rsub|1>,p)<rsup|-1>:X\<rightarrow\>T<rsub|p>M,f<rsub|2>=T(\<varphi\><rsub|2>,p)<rsup|-1>>
are bijection's and <math|f<rsub|1>=[D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>)\<circ\>T(\<varphi\><rsub|2>,p)]<rsup|-1>=T(\<varphi\><rsub|2>,p)<rsup|-1>\<circ\>[D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))]<rsup|-1>=f<rsub|2>\<circ\>h>
where <math|h=[D(\<varphi\><rsub|1>\<circ\>\<varphi\><rsub|2><rsup|-1>)(\<varphi\><rsub|2>(p))]<rsup|-1>>
is a homeomorphism (<reference|inverse of diffeomorphisme>). So we can
use <reference|induced topology> to prove the existence of a topology
independent of the choice of <math|(U,\<varphi\>)> so that
<math|f<rsub|1>> is a homeomorphism and thus
<math|f<rsub|1><rsup|-1>=T(\<varphi\>,p)> is a homeomorphism and is thus
a toplinear isomorphism.\
</proof>
For the rest of the text we always consider <math|T<rsub|p>M> endowed with
the topology and vector space operations as defined in the previous
theorem.
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>, r\<geqslant\>1> modelled on the normed space <math|X>,
let <math|p\<in\>M,> <math|W> a open set with <math|p\<in\>W> and
<math|f:W\<rightarrow\>\<bbb-R\>> a continuous function differentiable on
<math|W> (see <reference|differentiable mappings on a open subset of a
maifold>). If then <math|v<rsub|p>=\<bbb-T\>[c]\<in\>T<rsub|p>M> then
<math|<frac|d(f\<circ\>c)(t)|dt>(0)> is defined and if
<math|(U,\<varphi\>)\<in\>\<cal-A\>> is such that <math|p\<in\>U> then
<math|<frac|d(f\<circ\>c)(t)|dt>(0)=[D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))\<circ\>T(\<varphi\>,p)](v<rsub|p>)>
</theorem>
<\proof>
Using <reference|characterization of differentiability of a mapping on a
open subset of a manifold> (3) (2) we have that
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with
<math|\<emptyset\>\<neq\>f<rsup|-1>(\<bbb-R\>)<big|cap>U=W<big|cap>U>
that <math|i<rsub|\<bbb-R\>>\<circ\>f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(f<rsup|-1>(\<bbb-R\>))\<rightarrow\>\<bbb-R\>=f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>W)\<rightarrow\>\<bbb-R\>>
is differentiable. Now if <math|p\<in\>U\<Rightarrow\>p\<in\>W<big|cap>U\<neq\>\<emptyset\>>
so <math|f\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U<big|cap>W)\<rightarrow\>\<bbb-R\>>
is differentiable. Also from <math|c(0)=p\<in\>U\<Rightarrow\>c<rsup|-1>(U)\<neq\>\<emptyset\>>
so using <reference|characterization of a differential curve> (2) we have
that <math|\<varphi\>\<circ\>c:c<rsup|-1>(U)\<rightarrow\>\<varphi\>(U)>
is differentiable and thus also <math|f\<circ\>\<varphi\><rsup|-1>:c<rsup|-1>(U<big|cap>W)\<rightarrow\>\<varphi\>(U<big|cap>W)>
is differentiable. So we have by the chain rule the differentiability of
<math|(f\<circ\>\<varphi\><rsup|-1>)\<circ\>(\<varphi\>\<circ\>c):c<rsup|-1>(U<big|cap>W)\<rightarrow\>\<bbb-R\>=f\<circ\>c:c<rsup|-1>(U<big|cap>W)\<rightarrow\>\<bbb-R\>>
and thus as <math|0=c<rsup|-1>(p)\<in\>c<rsup|-1>(U<big|cap>W)> we have
the existance of <math|D(f\<circ\>c)(0)(1)=<frac|d(f\<circ\>c)(t)|dt>(0)>
(see <reference|differentiability of real or complex functions>).\
Finally we have by the above that <math|<frac|d(f\<circ\>c)(t)|dt>(0)=D((f\<circ\>\<varphi\><rsup|-1>)\<circ\>(\<varphi\>\<circ\>c))(0)(1)\<equallim\><rsub|<reference|differentiability
of composition of differentiable mappings> (Chain
rule)>[D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))\<circ\>D(\<varphi\>\<circ\>c)(0)](1)=D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))(D(\<varphi\>\<circ\>c)(0)(1))=D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))(<frac|d(\<varphi\>\<circ\>c)(t)|dt>(0))=D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))(T(\<varphi\>,p)[v<rsub|p>])=[D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))\<circ\>T(\<varphi\>,p)](v<rsub|p>)>
</proof>
<\definition>
<label|directional derivative><index|directional
derivative><index|<math|v<rsub|p>[f]>>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|X, p\<in\>M>,<math| p\<in\>W\<subseteq\>M> a
open set, <math|f:W\<rightarrow\>\<bbb-R\>> a differentiable function and
<math|v<rsub|p>\<in\>T<rsub|p>M,v=\<bbb-T\>[c]> then the directional
derivative <math|v<rsub|p>[f]> of <math|f> along <math|v<rsub|p>> at
<math|p\<in\>M> is defined by <math|v<rsub|p>[f]=<frac|d(f\<circ\>c)(t)|dt>(0)>.
Using the previous theorem and the fact that <math|T(\<varphi\>,p)> is
independent of the representation of <math|\<bbb-T\>[c]> of
<math|v<rsub|p>> we see that <math|v<rsub|p>[f]> is indeed well defined
</definition>
<\proof>
<math|<frac|d(f\<circ\>c)|dx>(0)> is independent of <math|(U,\<varphi\>)>
and <math|[D(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))\<circ\>T(\<varphi\>,p)](v<rsub|p>)>
is independent of <math|c> so we have that the definition of
<math|v<rsub|p>[f]> is independent of the choice of <math|(U,\<varphi\>)>
and <math|v<rsub|p>=\<bbb-T\>[c]>
</proof>
<\note>
<label|directional derivative is linear>Let <math|(M,\<cal-A\>)> be a
differential manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|X, p\<in\>M,f:M\<rightarrow\>\<bbb-R\>> a
differential function<math|, (U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> and
<math|v<rsub|p>,w<rsub|p>\<in\>T<rsub|p>M,\<lambda\>,\<mu\>\<in\>\<bbb-R\>>
then using the previous theorem and the linearity of
<math|T(\<varphi\>,p)> we have <strong|<math|(\<lambda\>v<rsub|p>+\<mu\>w<rsub|p>)[f]=\<lambda\>v<rsub|p>[f]+\<mu\>w<rsub|p>[f]>>.
Also if <math|f,g\<in\>:M\<rightarrow\>\<bbb-R\>> and
<math|v<rsub|p>=\<bbb-T\>[c]> then <math|v<rsub|p>[\<lambda\>f+\<mu\>g]=<frac|d((\<lambda\>f+\<mu\>g)\<circ\>c)(t)|dt>(0)=<frac|d(\<lambda\>(f\<circ\>c)+\<mu\>(g\<circ\>c))(t)|dt>(0)=\<lambda\><frac|d(f\<circ\>c)(t)|dt>(0)+\<mu\><frac|d(g\<circ\>c)(t)|dt>(0)=\<lambda\>v<rsub|p>[f]+\<mu\>v<rsub|p>[g]>
or <strong|<math|v<rsub|p>[\<lambda\>f+\<mu\>g]=\<lambda\>v<rsub|p>[f]+\<mu\>v<rsub|p>[g]>>
</note>
<\notation>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space
<math|\<bbb-R\><rsup|n>> then if <math|(U,\<varphi\>)\<in\>\<cal-A\>> we
have <math|\<varphi\>(p)=(\<varphi\><rsub|1>(p),\<ldots\>,\<varphi\><rsub|n>(p))>
where <math|\<varphi\><rsub|i>=\<pi\><rsub|i>\<circ\>\<varphi\>>. Instead
of using the symbols <math|\<varphi\>,\<psi\>,\<ldots\>> it is customary
to use instead <math|x,y> so we have then for example
<math|<math|>x(p)=(x<rsub|1>(p),\<ldots\>,x<rsub|n>(p))>.\
</notation>
<\example>
<label|basis of tangent space in finite dimensional
case><index|<math|\<partial\><rsub|i>(p)>>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|\<bbb-R\><rsup|n>> and <math|p\<in\>M>, if
<math|(U,x)\<in\>\<cal-A\>> with <math|p\<in\>U> then as <math|x(U)> is
open by the definition of a chart there exists a
<math|\<varepsilon\>\<gtr\>0> such that
<math|(x<rsub|1><rsup|0>,\<ldots\>,x<rsub|n><rsup|0>)=x(p)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(x(p),\<varepsilon\>)>.
We can then define <math|\<forall\>i\<in\>{1,\<ldots\>,n}> for
<math|\<tau\><rsub|i>:I<rsub|\<varepsilon\>>=]\<um\>\<varepsilon\>,\<varepsilon\>[\<rightarrow\>x(U):\<tau\><rsub|i>(t)\<rightarrow\>(x<rsub|1><rsup|0>,\<ldots\>,x<rsub|i-1><rsup|0>,x<rsub|1><rsup|0>+t,x<rsub|i+1><rsup|0>,\<ldots\>,x<rsub|n><rsup|0>)>
define now <math|\<forall\>i\<in\>{1,\<ldots\>,n}> the function
<math|><math|x<rsub|[i]>:I<rsub|\<varepsilon\>>\<rightarrow\>U\<subseteq\>M>
by <math|x<rsub|[i]>=x<rsup|-1>\<circ\>\<tau\><rsub|i>>.\
We proof now that <math|x<rsub|[i]>> is a differential curve, the vector
<math|\<bbb-T\>[x<rsub|[i]>]> is noted as <math|\<partial\><rsub|i>(x,p)>
<\proof>
\;
First as <math|\<tau\><rsub|i>=x(p)+(0,\<ldots\>,t,\<ldots\>0)> is the
sum of a constant function and a linear function it is differentiable
(and thus continuous), from the fact that <math|x> is a homeomorphism
we have that <math|x<rsub|[i]>> must be continuous.
Second <math|\<forall\>t\<in\>I<rsub|\<varepsilon\>>> we have
<math|x\<circ\>x<rsub|[i]>:I<rsub|\<varepsilon\>>\<rightarrow\>x(U)=x\<circ\>x<rsup|-1>\<circ\>\<tau\><rsub|i>:I<rsub|\<varepsilon\>>\<rightarrow\>x(U)=\<tau\><rsub|i>:I<rsub|\<varepsilon\>>\<rightarrow\>x(U)>
is differentiable, so using <reference|characterization of a
differential curve> (5) we have that <math|x<rsub|i]>> is a
differentiable curve.
</proof>
\;
<\theorem>
<label|apply basis vector on function>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on the normed space <math|\<bbb-R\><rsup|n>> and
<math|W\<subseteq\>M> a open set and <math|f:W\<rightarrow\>\<bbb-R\>>
a continuous differentiable function, <math|p\<in\>W> then
<math|\<forall\>(U,x)\<in\>\<cal-A\>> with <math|p\<in\>U> we have
<math|\<partial\><rsub|i>(x,p)[f]=<frac|d(f\<circ\>x<rsup|-1>)(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x(p))>
</theorem>
</example>
<\proof>
<math|\<partial\><rsub|i>(x,p)[f]=<frac|d(f\<circ\>x<rsub|[i]>)(t)|dt>(0)=<frac|d((f\<circ\>x<rsup|-1>)\<circ\>\<tau\><rsub|i>)(t)|dt>(0)=D((f\<circ\>x<rsup|-1>)\<circ\>\<tau\><rsub|i>)(0)(1)\<equallim\><rsub|chain
rule>D(f\<circ\>x<rsup|-1>)(\<tau\><rsub|i>(0))(D\<tau\><rsub|i>(0)(1))=D(f\<circ\>x<rsup|-1>)(x(p))(0,\<ldots\>,0<rsub|i-1>,1,0<rsub|i+1>,\<ldots\>,0)\<equallim\><rsub|<reference|differential
of multiparameter function>><frac|d(f\<circ\>x<rsup|-1>)(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x(p))<rsub|>>
</proof>
<\theorem>
<index|coordinate vector fields><label|coordinate vector fields>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1,p\<in\>M> modelled on the normed space
<math|\<bbb-R\><rsup|n>> and <math|(U,x)\<in\>\<cal-A\>,p\<in\>U> then
<math|T<rsub|p>M> <math|{\<partial\><rsub|1>(x,p),\<ldots\>,\<partial\><rsub|n>(x,p)}>
is a basis for <math|T<rsub|p>M>. The <math|\<partial\><rsub|i>(x,p)> are
called the coordinate vector fields. Further we have that
<math|\<forall\>i\<in\>{1,\<ldots\>,n}> and <math|><math|\<forall\>
(U,x)\<in\>\<cal-A\>> with <math|p\<in\>U> that
<math|T(x,p)(\<partial\><rsub|i>(x,p))=e<rsub|i>> (<math|e<rsub|i>> i-the
base vector in <math|R<rsup|n>>). Finally if <math|v\<in\>T<rsub|p>M>
then <math|v=<big|sum><rsub|i=1><rsup|n>h<rsub|i>\<partial\><rsub|i>(x,p)>
where <math|h<rsub|i>> are the coordinates of <math|v> in the basis
<math|\<partial\><rsub|i>(x,p)> then <math|T(x,p)(v)=(h<rsub|1>,\<ldots\>,h<rsub|n>)>
so that <math|\<pi\><rsub|i>\<circ\>T(x,p)(v)=h<rsub|i>>
</theorem>
<\proof>
Using <reference|induced basis on induced vector space> and
<reference|TpM is a vector space> we have proved our theorem if we can
prove that given a <math|(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> we have
that <math|\<partial\><rsub|i>(x,p)=T(x,p)<rsup|-1>(e<rsub|i>)> where
<math|e<rsub|i>=(\<delta\><rsub|1,i>,\<ldots\>,\<delta\><rsub|n,i>)\<in\>\<bbb-R\><rsup|n>>
(the basis for <math|\<bbb-R\><rsup|n>)>. To prove this we must prove
that <math|T(x,p)(\<partial\><rsub|i>(x,p))=e<rsub|i>>,
<math|T(x,p)(\<partial\><rsub|i>(x,p))\<equallim\><rsub|\<partial\><rsub|i>(x,p)=\<bbb-T\>[x<rsub|[i]>]><frac|d(x\<circ\>x<rsub|[i]>)(t)|dt>(0)=<frac|d\<tau\><rsub|i>(t)<rsub|>|dt>(0)=e<rsub|i>>.
The fact that if <math|v\<in\>T<rsub|p>M> that then
<math|v=<big|sum><rsub|i=1><rsup|n>h<rsub|i>\<partial\><rsub|i>(x,p)>
follows from the fact that <math|\<partial\><rsub|i>(x,p),i\<in\>{1,\<ldots\>,n}>
forms a basis of <math|T<rsub|p>M>. Form this we have that
<math|T(x,p)v=<big|sum><rsub|i=1><rsup|n>h<rsub|i>T(x,p)(\<partial\><rsub|i>(x,p))=<big|sum><rsub|i=1><rsup|n>h<rsub|i>e<rsub|i>=(h<rsub|1>,\<ldots\>,h<rsub|n>)>
</proof>
<\theorem>
<label|transformation rule of the components of a
vector><index|transformation of vector components>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1,p\<in\>M> modelled on the normed space
<math|\<bbb-R\><rsup|n>>, <math|p\<in\>M> and
<math|(U,x),(V,y)\<in\>\<cal-A\>> with <math|p\<in\>U<big|cap>V> then we
have by the previous theorem for any <math|v\<in\>T<rsub|p>M> that
<math|v=<big|sum><rsub|i=1><rsup|n>h<rsub|i>\<partial\><rsub|i>(x,p)=<big|sum><rsub|i=1><rsup|n>k<rsub|i>\<partial\><rsub|i>(y,p)>.
We have then that <math|k<rsub|i>=<big|sum><rsub|j=1><rsup|n><frac|d(y\<circ\>x<rsup|-1>(x(p))(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|j>>h<rsub|j>>
</theorem>
<\proof>
From the previous theorem we have that
<math|h<rsub|i>=\<pi\><rsub|i>\<circ\>T(x,p)(v),k<rsub|i>=\<pi\><rsub|i>\<circ\>T(y,p)(v)>.
From <reference|T(f,p) is bijective> it follows that
<math|T(y,p)=D(y\<circ\>x<rsup|-1>)(x(p))\<circ\>T(x,p)> and thus
<math|k<rsub|i>=\<pi\><rsub|i>\<circ\>D(y\<circ\>x<rsup|-1>)(x(p))\<circ\>T(x,p)(v)=\<pi\><rsub|i>\<circ\>D(y\<circ\>x<rsup|-1>)(x(p))(h<rsub|i>)>.
As <math|y\<circ\>x<rsup|-1>:\<bbb-R\><rsup|n>\<rightarrow\>\<bbb-R\><rsup|n>>
is differentiable of class <math|C<rsup|r>> we can use
<reference|Jacobian> to have that <math|k<rsub|i>=<big|sum><rsub|j=1><rsup|n><frac|d(y\<circ\>x<rsup|-1>)(x(p))(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|j>>h<rsub|j>>.
</proof>
<\definition>
<index|<math|\<equiv\><rsub|p>>><index|<math|\<cal-D\>(M,p)>>Let
(<math|M,\<cal-A\>)> be a differential manifold of class <math|C<rsup|r>,
r\<geqslant\>1> modelled on the normed space <math|X,p\<in\>M> consider
then <math|\<cal-D\>(M,p)={f\| \<exists\>U\<subseteq\>M,p\<in\>U,U open
and f:U\<rightarrow\>\<bbb-R\>> a differentiable function} then we define
the relation <math|\<equiv\><rsub|p>>on <math|\<cal-D\>(M,p)> as follows
<math|f<rsub|1>\<equiv\><rsub|p>f<rsub|2>\<Leftrightarrow\>\<exists\>open
U\<in\>M,p\<in\>U\<vdash\>\<forall\>x\<in\>U> we have
<math|f<rsub|1>(x)=f<rsub|2>(x)>. In other ways we say that
<math|f<rsub|1>\<equiv\><rsub|2>f<rsub|2>> if both functions agree on a
open set around <math|p>
</definition>
<\theorem>
Let (<math|M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>, r\<geqslant\>1> modelled on the normed space
<math|X,p\<in\>M> then <math|\<equiv\><rsub|p>> forms a equivalence
relation
</theorem>
<\proof>
\;
<\enumerate>
<item>Reflexivity, this is trivial as we have for every open set
<math|U\<ni\>p> that <math|\<forall\>x\<in\>U\<vdash\>f(x)=f(x)\<Rightarrow\>f\<equiv\><rsub|p>f>
<item>Symmetry, let <math|f<rsub|1>\<equiv\><rsub|p>f<rsub|2>\<Rightarrow\>\<exists\>open
set U\<ni\>p\<vdash\>\<forall\>x\<in\>U> we have
<math|f<rsub|1>(x)=f<rsub|2>(y)\<Rightarrow\>f<rsub|2>\<equiv\><rsub|p>f<rsub|1>>
<item>Transitivity, let <math|f<rsub|1>\<equiv\><rsub|p>f<rsub|2>> and
<math|f<rsub|2>\<equiv\><rsub|p>f<rsub|3>> then there exists open sets
<math|U<rsub|1>\<ni\>p,U<rsub|2>\<ni\>p> such that
<math|\<forall\>x\<in\>U<rsub|1>\<vdash\>f<rsub|1>(x)=f<rsub|2>(x),\<forall\>x\<in\>U<rsub|2>\<vdash\>f<rsub|2>(x)=f<rsub|3>(x)\<Rightarrow\>\<forall\>x\<in\>U=U<rsub|1><big|cap>U<rsub|2>\<vdash\>f<rsub|1>(x)=f<rsub|3>(x)>
and also <math|p\<in\>U<rsub|1><big|cap>U<rsub|2>=U> (which is open)
<math|\<Rightarrow\>f<rsub|1>\<equiv\><rsub|p>f<rsub|3>>
</enumerate>
</proof>
<\definition>
<label|definition of germs><index|germs><index|<math|\<varepsilon\>(M,p)>>Let
(<math|M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X,
p\<in\>M> then the space of germs <math|\<varepsilon\>(M,p)> is the set
of equivalence classes defined by <math|\<equiv\><rsub|p>>. In other
words <math|\<varepsilon\>(M,p)= {\<equiv\><rsub|p>[f]:
f\<in\>\<cal-D\>(M,p)><math|}>
</definition>
The idea of a germ is that for local properties at <math|p> (like
differentiability) two functions can be considered equal if they agree on a
chosen open set around <math|p>.
<\definition>
Let (<math|M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X,
p\<in\>M> and let <math|f\<in\>\<varepsilon\>(M,p)> where
<math|f=\<equiv\><rsub|p>[f<rprime|'>]> then we define
<math|f(p)=f<rprime|'>(p)> which is trivial by the definition from
<math|\<equiv\><rsub|p>>to be independent of the choice of the
representative <math|f<rprime|'>>
</definition>
<\theorem>
<index|algebra of germs><label|algebra of germs>Let <math|(M,\<cal-A\>)>
be a differential manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on the normed space <math|X,p\<in\>M> then we can define the
following operations on <math|\<varepsilon\>(M,p)>
<\enumerate>
<item><math|+:\<varepsilon\>(M,p)\<times\>\<varepsilon\>(M,p)\<rightarrow\>\<varepsilon\>(M,p)>
where if <math|f= \<equiv\><rsub|p>[f<rprime|'>]> and
<math|g=\<equiv\><rsub|p>[g<rprime|'>],f<rprime|'>:U\<rightarrow\>\<bbb-R\>,g<rprime|'>:V\<rightarrow\>\<bbb-R\>>
then <math|f+g=\<equiv\><rsub|p>[h]> where
<math|h:W=U<big|cap>V\<rightarrow\>\<bbb-R\>> is defined by
<math|h=(f<rprime|'>)<rsub|\|W>+(g<rprime|'>)<rsub|\|W>>
<item><math|,:\<bbb-R\>\<times\>\<varepsilon\>(M,p)\<rightarrow\>\<varepsilon\>(M,p)>,
\ where if <math|\<alpha\>\<in\>\<bbb-R\>> and
<math|f=\<equiv\><rsub|p>[f<rprime|'>],
f<rprime|'>:U\<rightarrow\>\<bbb-R\>> then
<math|\<alpha\>.f=\<equiv\><rsub|p>[h]> where
<math|h:U\<rightarrow\>\<bbb-R\> \ is defined by
h=\<alpha\>.f<rprime|'>>
<item><math|\<bullet\>:\<varepsilon\>(M,p)\<times\>\<varepsilon\>(M,p)\<rightarrow\>\<varepsilon\>(M,p)>
where where if <math|f= \<equiv\><rsub|p>[f<rprime|'>]> and
<math|g=\<equiv\><rsub|p>[g<rprime|'>],f<rprime|'>:U\<rightarrow\>\<bbb-R\>,g<rprime|'>:V\<rightarrow\>\<bbb-R\>>
then <math|f\<bullet\>g=\<equiv\><rsub|p>[h]> where
<math|h:W=U<big|cap>V\<rightarrow\>\<bbb-R\>> is defined by
<math|h=(f<rprime|'>)<rsub|\|W>.(g<rprime|'>)<rsub|\|W>>
</enumerate>
Then <math|\<varepsilon\>(M,p),+,.> forms a vector space over
<math|\<bbb-R\>> and <math|\<varepsilon\>(M,p),+,.,\<bullet\>> is a
commutative algebra
</theorem>
<\proof>
\;
First we must proof that <math|f+g,f\<bullet\>g > and <math|\<alpha\>.f>
are in <math|\<varepsilon\>(M,p)>. So if
<math|f,g\<in\>\<varepsilon\>(M,p)> then
<math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>]>
where <math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>,g<rprime|'>:V<rprime|'>\<rightarrow\>\<bbb-R\>>
are differentiable functions and thus
<math|(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>,(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>>
are differentiable functions on the open set
<math|U<rprime|'><big|cap>V<rprime|'>> containing <math|p> and thus
<math|(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>,(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>,\<alpha\>.(f<rprime|'>)<rsub|\|U<rprime|'>>>
are differentiable functions and thus
<math|f+g=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>],f.g=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]>
and <math|\<alpha\>.f=\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>]> are in
<math|\<varepsilon\>(M,p)>
Second we must proof that the definition of <math|+,.> and
<math|\<bullet\>> is independent of the chosen representation. So let
<math|f = \<equiv\><rsub|p>[f<rprime|'>]=
\<equiv\><rsub|p>[f<rprime|''>]> and <math|g=
\<equiv\><rsub|p>[g<rprime|'>]=\<equiv\><rsub|p>[g<rprime|''>]> where
<math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>,g<rprime|'>:V<rprime|'>\<rightarrow\>\<bbb-R\>,f<rprime|''>:U<rprime|''>\<rightarrow\>\<bbb-R\>,g<rprime|''>:V<rprime|''>\<rightarrow\>\<bbb-R\>>
where <math|p\<in\>U<rprime|'>,U<rprime|''>,V<rprime|'>,V<rprime|''>>
(open sets) then there exists a <math|U,V> open with <math|p\<in\>U,V>
such that <math|\<forall\>x\<in\>U\<vdash\>f<rprime|'>(x)=f<rprime|''>(x),\<forall\>x\<in\>V\<vdash\>g<rprime|'>(x)=g<rprime|''>(x)>
so we have <math|\<forall\>x\<in\>W=U<big|cap>U<rprime|'><big|cap>U<rprime|''><big|cap>V<big|cap>V<rprime|'><big|cap>V<rprime|''>\<ni\>p>
we have <math|f<rprime|'>(x)+g<rprime|'>(x)=f<rprime|''>(x)+g<rprime|''>(x),f<rprime|'>(x).g<rprime|'>(x)=f<rprime|''>(x).g<rprime|''>(x),\<alpha\>.f<rprime|'>(x)=\<alpha\>.f<rprime|''>(x)>
so <math|f<rprime|'>+g<rprime|'>\<equiv\><rsub|p>f<rprime|''>+g<rprime|''>,f<rprime|'>.g<rprime|'>\<equiv\>f<rprime|''>.g<rprime|''>,\<alpha\>.f<rprime|'>\<equiv\>a.f<rprime|''>>
and thus <math|\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|''>)<rsub|\|U<rprime|''><big|cap>V<rprime|''>>+(g<rprime|''>)<rsub|\|U<rprime|''><big|cap>V<rprime|''>>],\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|''>)<rsub|\|U<rprime|''><big|cap>V<rprime|''>>.(g<rprime|''>)<rsub|\|U<rprime|''><big|cap>V<rprime|''>>],\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>]=\<equiv\><rsub|p>[\<alpha\>.f<rprime|''>]>
proving the independence of the chosen representative of the equivalence
class.\
Third we must prove that we have the axioms of vectorspace and algebra
fulfilled for the defined operators, so let
<math|f,g,h\<in\>\<varepsilon\>(M,p),\<alpha\>,\<beta\>\<in\>\<bbb-R\>>
then <math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>],h=\<equiv\><rsub|p>[h<rprime|'>]>
where <math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>,g<rprime|'>:V<rprime|'>\<rightarrow\>\<bbb-R\>,h<rprime|'>:W<rprime|'>\<rightarrow\>\<bbb-R\>>
are differential functions
<\enumerate>
<item><math|\<varepsilon\>(M,p),+> is a abelian group
<\enumerate>
<item>(Associativity of <math|+>)
<math|><math|(f+g)+h=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]+h=\<equiv\><rsub|p>[((f<rprime|'>)<rsub|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>)<rsub|\|U<rprime|'><big|cap>V<big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>)+(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+((g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>)]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+((g<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=f+\<equiv\><rsub|p>[(g<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>]=f+(g+h)>
<item>(Neutral element) Consider <math|0:M\<rightarrow\>\<bbb-R\>>
with <math|0(m)=0> (which as a constant function is differentiable)
then <math|0+f=\<equiv\><rsub|M>[0<rsub|\|M<big|cap>U<rprime|'>>+(f<rprime|'>)<rsub|\|M<big|cap>U<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'>>]=\<equiv\><rsub|p>[f<rprime|'>]=f>
<item>(Inverse element) <math|>Take
<math|\<um\>f=\<equiv\><rsub|p>[\<um\>f<rprime|'>]> then
<math|f+(\<um\>f)= \<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>U<rprime|'>>+(\<um\>f<rprime|'>)<rsub|\|U<rprime|'><big|cap>U<rprime|'>>]=\<equiv\><rsub|p>[f<rprime|'>+(-f<rprime|'>)]=\<equiv\><rsub|p>[0<rsub|\|U<rprime|'>>]=\<equiv\><rsub|p>[0]>
<item>(Commutativity) <math|f+g=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=g+f>
</enumerate>
<item><math|\<varepsilon\>(M,p),+,.> is a real vector space
<\enumerate>
<item><math|(\<alpha\>+\<beta\>).f=\<equiv\><rsub|p>[(\<alpha\>+\<beta\>)f<rprime|'>]=\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>+\<beta\>.f<rprime|'>]=\<equiv\><rsub|p>[\<alpha\>.f]+\<equiv\><rsub|p>[\<beta\>.f<rprime|'>]=\<alpha\>.f+\<beta\>.g>
<item><math|\<alpha\>.(f+g)=\<alpha\>.[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[\<alpha\>.((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>)]=\<equiv\><rsub|p>[\<alpha\>.(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+\<alpha\>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>]+\<equiv\><rsub|p>[\<alpha\>.g<rprime|'>]=a.f+\<alpha\>.g<rsub|>>
<item><math|(\<alpha\>.\<beta\>).f=\<equiv\><rsub|p>[(\<alpha\>.\<beta\>).f<rprime|'>]=\<equiv\><rsub|p>[\<alpha\>.(\<beta\>.f<rprime|'>)]=\<alpha\>.\<equiv\><rsub|p>[\<beta\>.f<rprime|'>]=\<alpha\>.(\<beta\>.f)>
<item><math|1.f=\<equiv\><rsub|p>[1.f<rprime|'>]=\<equiv\><rsub|p>[f]=f>
</enumerate>
<item><math|\<varepsilon\>(m,p),+,.\<bullet\>> is a commutative algebra
<\enumerate>
<item><math|(f+g)\<bullet\>h=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]\<bullet\>h=\<equiv\><rsub|p>[((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>.(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>.(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>.(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>W<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>W<rprime|'>>]+\<equiv\><rsub|p>[(g<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>.(h<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>]=f\<bullet\>g+g\<bullet\>h>
<item><math|f\<bullet\>(g+h)=f\<bullet\>\<equiv\><rsub|p>[(g<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>.((g<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>+(h<rprime|'>)<rsub|\|V<rprime|'><big|cap>W<rprime|'>>)<rsub|\|U<rprime|'><big|cap>V<big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>+(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'><big|cap>W<rprime|'>>]=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]+\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>W<rprime|'>>.(h<rprime|'>)<rsub|\|U<rprime|'><big|cap>W<rprime|'>>]=f\<bullet\>g+f\<bullet\>h>
<item><math|(\<alpha\>.f)\<bullet\>(\<beta\>.g)=\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>]\<bullet\>\<equiv\><rsub|p>[\<beta\>.g<rprime|'>]=\<equiv\><rsub|p>[(\<alpha\>.f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(\<beta\>.g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[(\<alpha\>.\<beta\>).((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=(\<alpha\>.\<beta\>).\<equiv\><rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=(\<alpha\>.\<beta\>).(f\<bullet\>g)>
<item><math|f\<bullet\>g=\<equiv\><rsub|p>[f<rprime|'>]\<bullet\>\<equiv\><rsub|p>[g<rprime|'>]=\<equiv\><rsub|p>[f<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>.g<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=\<equiv\><rsub|p>[g<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>.f<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=g\<bullet\>f>
</enumerate>
</enumerate>
</proof>
<\definition>
<index|<math|\<Delta\><rsub|v<rsub|p>>>><label|local vector as a
function>Let <math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space
<math|X,p\<in\>M> then given a <math|v<rsub|p>\<in\>T<rsub|p>M> we define
a function <math|\<Delta\><rsub|v<rsub|p>>:\<varepsilon\>(M,p)\<rightarrow\>\<bbb-R\>>
by <math|\<Delta\><rsub|v<rsub|p>>(f)=\<Delta\><rsub|v<rsub|p>>(\<equiv\><rsub|p>[f<rprime|'>])=v<rsub|p>[f<rprime|'>]>
where <math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>> is a
representative of <math|f>.
</definition>
<\proof>
We must of course prove that this definition is independent of the chosen
representative <math|f<rprime|'>>. So if
<math|f=\<equiv\><rsub|p>[f<rprime|'>]=\<equiv\><rsub|p>[f<rprime|''>]>
then there exists \ open set <math|U\<ni\>p> such that
<math|\<forall\>x\<in\>U> we have <math|f<rprime|'>(x)=f<rprime|''>(x)>
then if <math|v<rsub|p>=\<bbb-T\>[c]> we have as
<math|c(0)=p\<Rightarrow\>0\<in\>c<rsup|-1>(U)> we have that
<math|v<rsub|p>[f<rprime|'>]=<frac|d(f<rprime|'>\<circ\>c)(t)|dt>(0)\<equallim\>D*(f\<circ\>c)(0)(1)\<equallim\><rsub|<reference|differentiability
and restricted mappings>>D(f<rprime|''>\<circ\>c)(0)(1)=
\ <frac|d(f<rprime|''>\<circ\>c)(t)|dt>(0)=v<rsub|p>[f<rprime|''>]>
</proof>
<\definition>
<index|derivation><label|derivation><index|<math|\<Delta\><rsub|v<rsub|p>>>>Let
<math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space
<math|X,p\<in\>M> then a function <math|\<Delta\>:\<varepsilon\>(M,p)\<rightarrow\>\<bbb-R\>>
which is such that
<\enumerate>
<item><math|\<Delta\>(\<alpha\>.f+\<beta\>.g)=\<alpha\>\<Delta\>(f)+\<beta\>\<Delta\>(g)>
<item><math|\<Delta\>(f\<bullet\>g)=\<Delta\>(f).g(p)+f(p).\<Delta\>(g)>
</enumerate>
is called a derivation
</definition>
<\theorem>
<index|<math|\<Delta\><rsub|v<rsub|p>>>><label|local vector as a function
is a derivation>Let <math|(M,\<cal-A\>)> be a differential manifold of
class <math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space
<math|X,p\<in\>M> then given a <math|v<rsub|p>\<in\>T<rsub|p>M> we have
that <math|\<Delta\><rsub|v<rsub|p>>> is a derivation
</theorem>
<\proof>
Let <math|f,g\<in\>\<varepsilon\>(M,p)> with
<math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>]>
where <math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>,g<rprime|'>:V<rprime|'>\<rightarrow\>\<bbb-R\>>
with <math|f<rprime|'>,g<rprime|'>> differentiable and let
<math|v<rsub|p>=\<bbb-T\>[c]> then\
<\enumerate>
<item><math|(\<alpha\>.f+\<beta\>.g)=
\<equiv\><rsub|p>[(\<alpha\>.f<rprime|'>+b.\<gamma\><rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]\<Rightarrow\>\<Delta\><rsub|v<rsub|p>>(\<alpha\>.f+\<beta\>.g)=v<rsub|p>[(\<alpha\>f<rprime|'>+\<beta\>g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=v<rsub|p>[(\<alpha\>.(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+\<beta\>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=<frac|d((\<alpha\>.(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>+\<beta\>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>)\<circ\>c)(t)|dt>(0)=<frac|d(\<alpha\>.(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c+\<beta\>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c(t)|dt>(0)=\<alpha\><frac|d*(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c(t)|dt>(0)+\<beta\><frac|d(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c(t)|dt>(0)=\<alpha\><frac|d(f<rprime|'>\<circ\>c)(t)|dt>(0)+\<beta\><frac|d(g<rprime|'>\<circ\>c)(t)|dt>(0)=\<alpha\>v<rsub|p>[f]+\<beta\>v<rsub|p>[g]=\<alpha\>\<Delta\><rsub|v<rsub|p>>(f)+\<beta\>\<Delta\><rsub|v<rsub|p>>(g)>
<item><math|(f\<bullet\>g)=\<equiv\><rsub|p>[(f<rprime|'>)<rsub|U<rprime|'><big|cap>V<rprime|'>>(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]\<Rightarrow\>\<Delta\><rsub|v<rsub|p>>(f\<bullet\>g)=v<rsub|p>[(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>]=<frac|d(((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>.(g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>))\<circ\>c(t)|dt>(0)=<frac|d((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c).(g<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c)(t)|dt>(0)=(f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>(c(0))<frac|d((g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c(t)|dt>(0)+<frac|d((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c)(t)|dt>(0)(g<rprime|'><rsub|\|U<rprime|'><big|cap>V<rprime|'>>(c(0))=f<rprime|'>(p)<frac|d((g<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c(t)|dt>(0)+<frac|d((f<rprime|'>)<rsub|\|U<rprime|'><big|cap>V<rprime|'>>\<circ\>c)(t)|dt>(0)g<rprime|'>(p)=f(p)v<rsub|p>[g]+v<rsub|p>[f]g(p)=\<Delta\><rsub|v<rsub|p>>(f).g(p)+f(p).\<Delta\><rsub|v<rsub|p>>(p)>
</enumerate>
</proof>
\;
<section|The tangent bundle>
<\definition>
<index|<math|T<rsub|M>>><label|TM definition>Let <math|(M,\<cal-A\>)> be
a differential manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|X> define then
<math|T<rsub|M>=<big|cup><rsub|p\<in\>M>{p}\<times\>T<rsub|p>M>. We
define the surjective mapping <math|\<pi\>:T<rsub|M>\<rightarrow\>M> by
<math|\<pi\>(p,v<rsub|p>)\<equiv\>\<pi\>((p,v<rsub|p>))=p> (or
<math|\<pi\>=\<pi\><rsub|1>> the projection operator on the first
coordinate) and note <math|\<pi\><rsup|-1>(A)=<wide|A|~>> for every
<math|A\<subseteq\>T<rsub|M>>
</definition>
<\note>
<label|topology and vectorspace on {p}XTpM>Let <math|(M,\<cal-A\>)> be a
differential manifold of class <math|C<rsup|r>,r\<geqslant\>1> modelled
on the normed space <math|X> then if <math|p\<in\>M> we can endown
<math|{p}\<times\>T<rsub|p>M> on a trivial way with a topology and vector
space structure using <reference|induced vector space
structure>,<reference|induced topology> and the bijection
<math|f<rsub|p>:T<rsub|p>M\<rightarrow\>{p}\<times\>T<rsub|p>M> where
<math|f<rsub|p>(v)=(p,v)>. Using the induced structure and topology we
have <math|\<alpha\>.(p,v<rsub|1>)+\<beta\>.(p,v<rsub|2>)=(p,\<alpha\>.v<rsub|1>+\<beta\>.v<rsub|2>)>
and the topology of <math|{p}\<times\>T<rsub|p>M> consist of the sets
\ <math|{p}\<times\>U>.\
</note>
<\definition>
Let <math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X>
then given a <math|\<cal-A\><rprime|'>={(U<rsub|i>,\<varphi\><rsub|i>)}<rsub|i\<in\>I>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]>
define then <math|<wide|\<cal-A\><rprime|'>|~>={(<wide|U<rsub|i>|~>,<wide|\<varphi\><rsub|i>|~>)}<rsub|i\<in\>I>>
where
<\itemize>
<item><math|<wide|U<rsub|i>|~>=\<pi\><rsup|-1>(U<rsub|i>)>
<item><math|<wide|\<varphi\><rsub|i>|~>:<wide|U<rsub|i>|~>\<rightarrow\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
by <math|<wide|\<varphi\><rsub|i>|~>((p,v<rsub|p>))\<equiv\><wide|\<varphi\><rsub|i>|~>(p,v<rsub|p>)=(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p)(v<rsub|p>))>
</itemize>
We prove now the following
</definition>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X>
then given a <math|\<cal-A\><rprime|'>={(U<rsub|i>,\<varphi\><rsub|i>)}<rsub|i\<in\>I>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]>
we have for <math|<wide|\<cal-A\><rprime|'>|~>={(<wide|U<rsub|i>|~>,<wide|\<varphi\><rsub|i>|~>)}<rsub|i\<in\>I>>
the following
<\enumerate>
<item><math|\<forall\>i\<in\>I,U\<subseteq\>U<rsub|i>> we have
<math|<wide|\<varphi\><rsub|i>|~>(<wide|U|~>)=\<varphi\><rsub|i>(U)\<times\>X>
<item><math|\<forall\>i\<in\>I> we have that
<math|<wide|\<varphi\><rsub|i>|~>:<wide|U<rsub|i>|~>\<rightarrow\><wide|\<varphi\><rsub|i>|~>(<wide|U<rsub|i>|~>)=\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
is a bijection with inverse <math|<wide|\<varphi\><rsub|i>|~><rsup|-1>:\<varphi\><rsub|i>(U<rsub|i>)\<times\>X\<rightarrow\><wide|\<varphi\><rsub|i>|~>(<wide|U<rsub|>|~>)>
defined by <math|<wide|\<varphi\><rsub|i>|~><rsup|-1>(x,y)=(\<varphi\><rsub|i><rsup|-1>(x),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))<rsup|-1>(y))>
<item><math|<big|cup><rsub|i\<in\>I><wide|U<rsub|i>|~>=T<rsub|M>>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>I\<vdash\><wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|b>|~>\<neq\>\<emptyset\>>
we have that <math|>
<\enumerate>
<item><math|<wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>:\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X\<rightarrow\>\<varphi\><rsub|\<beta\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
is a bijection.
<item><math|><math|\<forall\>(x,y)\<in\>\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X
\ we have <wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>(x,y)=(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsup|-1><rsub|\<alpha\>>(x),D(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>)(x)(y))>.
Note that because <math|\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<beta\>>>
are homeomorphisms we have that <math|\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
is open in the product topology of <math|X\<times\>X>
<item><math|<wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>>
is of class <math|C<rsup|r-1>> (from this and the fact that
<math|<wide|\<varphi\><rsub|\<alpha\>>|~>\<circ\><wide|\<varphi\><rsub|b>|~><rsup|-1>>
is the inverse of <math|<wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>>
we have then that <math|<wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>>
is a diffeomorphism of class <math|C<rsup|r-1>>
</enumerate>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>Let <math|U\<subseteq\>U<rsub|i>> then we prove that
<math|<wide|\<varphi\><rsub|i>|~>(<wide|U|~>)=\<varphi\><rsub|i>(U)\<times\>X>
<\enumerate>
<item>If <math|(x,h)\<in\><wide|\<varphi\>|~><rsub|i>(<wide|U|~>)>
then <math|\<exists\>(p,v)\<in\><wide|U|~> > such that
<math|(x,h)=<wide|\<varphi\><rsub|i>|~>(p,v)=(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p)(v))\<Rightarrow\>x=\<varphi\><rsub|i>(p),h=T(\<varphi\><rsub|i>,p)(v)\<in\>X>
now as <math|(p,v)\<in\><wide|U|~>=\<pi\><rsup|-1>(U)\<Rightarrow\>p=\<pi\>(p,v)\<in\>U\<Rightarrow\>(x,h)\<in\>\<varphi\><rsub|i>(U)\<times\>X>
<item>If <math|(x,h)\<in\>\<varphi\><rsub|i>(U)\<times\>X> then there
exists a <math|p\<in\>U> such that <math|x=\<varphi\><rsub|i>(p)> and
as <math|T(\<varphi\><rsub|i>,p)> is surjective a
<math|v\<in\>T<rsub|p>M> such that
<math|h=T(\<varphi\><rsub|i>,p)(v)\<Rightarrow\>(p,v)\<in\><wide|U|~>>
and <math|(x,h)=<wide|\<varphi\><rsub|i>|~>(p,v)\<in\><wide|\<varphi\><rsub|i>|~>(<wide|U|~>)>
</enumerate>
<item>We only have to prove injectivity so let
<math|<wide|\<varphi\><rsub|i>|~>(p,v)=<wide|\<varphi\><rsub|i>|~>(p<rprime|'>,v<rprime|'>)\<Rightarrow\>(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p)(v))=(\<varphi\><rsub|i>(p<rprime|'>),T(\<varphi\><rsub|i>,p<rprime|'>)(v<rprime|'>))\<Rightarrow\>\<varphi\><rsub|i>(p)=\<varphi\><rsub|i>(p<rprime|'>)\<wedge\>T(\<varphi\><rsub|i>,p)(v)=T(\<varphi\><rsub|i>,p<rprime|'>)(v<rprime|'>)\<Rightarrowlim\><rsub|\<varphi\><rsub|i>
is a homeomorphism>p=p<rprime|'>\<wedge\>T(\<varphi\><rsub|i>,p)(v)=T(\<varphi\><rsub|i>,p<rprime|'>)(v<rprime|'>)\<Rightarrow\>p=p<rprime|'>\<wedge\>T(\<varphi\><rsub|i>,p)(v)=T(\<varphi\><rsub|i>,p)(v<rprime|'>)\<Rightarrow\>p=p<rprime|'>\<wedge\>v=v<rprime|'>\<Rightarrow\>(p,v)=(p<rprime|'>,v<rprime|'>)>.
For the inverse consider <math|(x,y)\<in\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
and define <math|<wide|\<varphi\><rsub|i>|\<bar\>>> by
<math|<wide|\<varphi\><rsub|i>|\<bar\>>(x,y)=(\<varphi\><rsub|i><rsup|-1>(x),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))<rsup|-1>(y))>
then
<\enumerate>
<item><math|p=\<varphi\><rsub|i><rsup|-1>(x)\<in\>U<rsub|i>\<Rightarrow\>T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))<rsup|-1>(y)=T(\<varphi\><rsub|i>,p)<rsup|-1>(y)\<in\>T<rsub|p>M
> and thus <math|<wide|\<varphi\><rsub|i>|\<bar\>>(x,y)\<in\>{p}\<times\>T<rsub|p>M\<Rightarrow\><wide|\<varphi\><rsub|i>|\<bar\>>(x,y)\<in\>T<rsub|M>>.
Furthermore <math|\<pi\>(<wide|\<varphi\><rsub|i>|\<bar\>>(x,y))=p\<in\>U<rsub|i>>
proving that <math|<wide|\<varphi\><rsub|i>|\<bar\>>(x,y)\<in\>\<pi\><rsup|-1>(U<rsub|i>)=<wide|U<rsub|i>|~>>.
So we conclude that <math|<wide|\<varphi\><rsub|i>|\<bar\>>:\<varphi\><rsub|i>(U<rsub|i>)\<times\>X\<rightarrow\><wide|U<rsub|i>|~>>
<item>If <math|(p,v)\<in\><wide|U|~>> then
<math|<wide|\<varphi\><rsub|i>|\<bar\>>(<wide|\<varphi\><rsub|i>|~>(p,v))=<wide|\<varphi\><rsub|i>|\<bar\>>(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p)(v))=(\<varphi\><rsub|i><rsup|-1>(\<varphi\><rsub|i>(p)),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(\<varphi\><rsub|i>(p)))<rsup|-1>(T(\<varphi\><rsub|i>,p)(v)))=(p,T(\<varphi\><rsub|i>,p)<rsup|-1>(T(\<varphi\><rsub|i>,p)(v)))=(p,v)>
<item>If <math|(x,y)\<in\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
then <math|<wide|\<varphi\><rsub|i>|~>(<wide|\<varphi\><rsub|i>|\<bar\>>(x,y))=<wide|\<varphi\><rsub|i>|~>(\<varphi\><rsub|i><rsup|-1>(x),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))<rsup|-1>(y))=(\<varphi\><rsub|i>(\<varphi\><rsub|i><rsup|-1>(x)),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))(T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(x))(y)))=(x,y)>
</enumerate>
From <math|a,b,c> if follows that <math|<wide|\<varphi\><rsub|i>|\<bar\>>>
is the inverse of <math|<wide|\<varphi\><rsub|i>|~>>
<item>If <math|(p,v)\<in\>T<rsub|M>\<Rightarrow\>p\<in\>M,v\<in\>T<rsub|p>M\<Rightarrowlim\><rsub|\<cal-A\><rprime|'>
is a atlas>\<exists\>i\<in\>I\<succ\>p\<in\>U<rsub|i>\<Rightarrow\>\<pi\>(p,v)\<in\>U<rsub|i>\<Rightarrow\>(p,v)\<in\><wide|U<rsub|i>|~>\<Rightarrow\>T<rsub|M>\<subseteq\><big|cup><rsub|i\<in\>I><wide|U<rsub|i>|~>\<subseteq\>T<rsub|M>>
<item>Let <math|\<alpha\>,\<beta\>\<subset\>I> with
<math|<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>\<neq\>\<emptyset\>>
then
<\enumerate>
<item><math|<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>=\<pi\><rsup|-1>(U<rsub|\<alpha\>>)<big|cap>\<pi\><rsup|\<um\>1>(U<rsub|\<beta\>>)=\<pi\><rsup|-1>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)=<wide|U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>|~>>
and so <math|<wide|\<varphi\><rsub|\<alpha\>>|~>(<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>)=<wide|\<varphi\><rsub|\<alpha\>>|~>(<wide|U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>|~>)\<equallim\><rsub|(1)>\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>,
<math|<wide|\<varphi\><rsub|\<beta\>>|~>(<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>)=<wide|\<varphi\><rsub|\<beta\>>|~>(<wide|U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>|~>)\<equallim\><rsub|(1)>\<varphi\><rsub|\<beta\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>.
From (2) it follows then that <math|<wide|\<varphi\><rsub|\<beta\>>|~>:<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>\<rightarrow\>\<varphi\><rsub|\<beta\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
and <math|<wide|\<varphi\><rsub|\<alpha\>>|~>:<wide|U<rsub|\<alpha\>>|~><big|cap><wide|U<rsub|\<beta\>>|~>\<rightarrow\>\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
are bijective so that <math|<wide|\<varphi\><rsub|b>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>:\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X\<rightarrow\>\<varphi\><rsub|\<beta\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
is a bijection.
<item>If <math|(x,y)\<in\>\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>><big|cap>U<rsub|\<beta\>>)\<times\>X>
then <math|<wide|\<varphi\><rsub|\<beta\>>|~>\<circ\><wide|\<varphi\><rsub|\<alpha\>>|~><rsup|-1>(x,y)=<wide|\<varphi\><rsub|\<beta\>>|~>(\<varphi\><rsub|\<alpha\>><rsup|-1>(x),T(\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))<rsup|-1>(y))=(\<varphi\><rsub|\<beta\>>(\<varphi\><rsub|\<alpha\>><rsup|-1>(x)),T(\<varphi\><rsub|\<beta\>>,\<varphi\><rsup|-1><rsub|\<alpha\>>(x))(T(\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))(y))=(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>(x),T(\<varphi\><rsub|\<beta\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))\<circ\>T(\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))<rsup|-1>(y))\<equallim\><rsub|<reference|T(f,p)
is bijective>>(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>(x),D(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>)(\<varphi\><rsub|\<alpha\>>(\<varphi\><rsub|\<alpha\>><rsup|-1>(x))\<circ\>T(\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))\<circ\>T(\<varphi\><rsub|\<alpha\>>,\<varphi\><rsub|\<alpha\>><rsup|-1>(x))<rsup|-1>(y))=(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>(x),D(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>)(x)(y))>
<item>From (4 b) and <reference|c-r of mapping to product > we have
proved (c) if we can proof that <math|f<rsub|1>=(x,y)\<rightarrow\>\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>(x)>
and <math|f<rsub|2>=(x,y)\<rightarrow\>D(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>)(x)(y)>
are of class <math|C<rsup|r-1>>
<\enumerate>
<item><math|f<rsub|1>> is <math|C<rsup|r>> and thus
<math|C<rsup|r-1>>. For if we define
<math|h:\<varphi\><rsub|i>(U)\<times\>X:(x,v)\<rightarrow\>x>
(which is linear and thus <math|C<rsup|\<infty\>>>) then
<math|f<rsub|1>=(\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>)\<circ\>h>
which is of class <math|C<rsup|r>> because
<math|\<varphi\><rsub|b>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>>
is of class <math|C<rsup|r>> (definition of a manifold of class
<math|C<rsup|r>>) and the generalized chain rule
<reference|generalized chain rule>.
<item><math|f<rsub|2>> is of class <math|C<rsup|r-1>> because of
<reference|(x,h)-\<gtr\>Df(x)(h) differentiability> and the fact
that <math|\<varphi\><rsub|\<beta\>>\<circ\>\<varphi\><rsub|\<alpha\>><rsup|-1>>
is of class <math|C<rsup|r>>
</enumerate>
</enumerate>
\
</enumerate>
</proof>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differential manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on the normed space <math|X>
then there exists a topology on <math|T<rsub|M>> such that <math|\<pi\>>
is continuous and such that <math|T<rsub|M>> is a topological manifold
modelled on <math|X\<times\>X>. <math|T<rsub|M>> has a uniquely defined
differentiable structure <math|\<cal-C\><rsup|(r)>[<wide|\<cal-A\>|~>]>of
class <math|C<rsup|r-1>> such that if
<math|\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]> then
<math|<wide|\<cal-A\><rprime|'>|~>\<in\>\<cal-C\><rsup|(r-1)>[<wide|\<cal-A\>|~>]>
</theorem>
<\proof>
\;
<\enumerate>
<item><strong|Topology on <math|T<rsub|M>>> Let
<math|\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)}<rsub|i\<in\>I>> be the
maximal atlas of <math|M><math| then define
\ \<cal-S\>={(<wide|\<varphi\><rsub|i>|~>)<rsup|-1>(V)\|i\<in\>I,V>
open in <math|<wide|\<varphi\><rsub|i>|~>(<wide|U<rsub|i>|~>)=\<varphi\><rsub|i>(U<rsub|i>)\<times\>X}><math|>.
Using <reference|sub basis topology> this defines a topology on
<math|T<rsub|M>>
<item><strong|<math|\<pi\>> <strong|is continue using the above
topology and the topology on <math|M>>>. \ If <math|U> is open in
<math|M> then <math|\<forall\>x\<in\>U> there exists a <math|i<rsub|x>>
with <math|(U<rsub|i<rsub|x>>,\<varphi\><rsub|i<rsub|x>>)\<in\>\<cal-A\>>
and <math|x\<in\>U<rsub|i<rsub|x>>> if we note then the open set
<math|V<rsub|i<rsub|x>>=U<rsub|i<rsub|x>><big|cap>U> in <math|M> then
<math|U=<big|cup><rsub|x\<in\>U>V<rsub|i<rsub|x>>>. So
<math|\<pi\><rsup|-1>(U)=\<pi\><rsup|-1>(<big|cup><rsub|x\<in\>U>V<rsub|i<rsub|x>>)=<big|cup><rsub|x\<in\>U>\<pi\><rsup|-1>(V<rsub|i<rsub|x>>)=<big|cup><rsub|x\<in\>U><wide|V<rsub|i<rsub|x>>|~>>
. By the previous theorem we have <math|<wide|\<varphi\><rsub|i<rsub|x>>|~>(<wide|V<rsub|i<rsub|x>>|~>)=\<varphi\><rsub|i<rsub|x>>(V<rsub|i<rsub|x>>)\<times\>X>
which because <math|\<varphi\><rsub|i<rsub|x>>> is a homeomorphism (so
<math|\<varphi\><rsub|i<rsub|x>>(V<rsub|x>)> is open in
<math|\<varphi\><rsub|i<rsub|x>>(U<rsub|i<rsub|x>>)>) is open in
<math|\<varphi\><rsub|i<rsub|x>>(U<rsub|i<rsub|x>>)\<times\>X>. From
the fact that <math|<wide|\<varphi\><rsub|i<rsub|x>>|~>> is a bijective
function we have <math|<wide|V<rsub|i<rsub|x>>|~>=<wide|(\<varphi\><rsub|i<rsub|x>>|~><rsub|<rsup|>>)<rsup|-1>(<wide|\<varphi\><rsub|i<rsub|x>(<wide|V<rsub|i<rsub|x>>|~>)>|~>)=<wide|(\<varphi\><rsub|i<rsub|x>>|~>)<rsup|-1>(\<varphi\><rsub|i<rsub|x>>(V<rsub|i<rsub|x>>)\<times\>X)>
\ and thus by the definition of our topology
<math|<wide|V<rsub|i<rsub|x>>|~>\<in\>\<cal-S\>> is open, so we have
that <math|\<pi\><rsup|-1>(U)=<big|cup><rsub|x\<in\>U><wide|V<rsub|i<rsub|x>>|~>>
is open.
<item><strong|The topology is Hausdorf> Let
<math|(p,v),(p<rprime|'>,v<rprime|'>)\<in\>T<rsub|M>> with
<math|(p,v)\<neq\>(p.v<rprime|'>)>. Consider now the following cases
<\enumerate>
<item><math|p=p<rprime|'>>. In this case we must have
<math|v\<neq\>v<rprime|'>>. As <math|\<cal-A\>> is a atlas there
exists a <math|i\<in\>I> such that
<math|p\<in\>U<rsub|i>,(U<rsub|i>,\<varphi\><rsub|i>)\<in\>\<cal-A\>>
then as <math|v\<neq\>v<rprime|'>\<Rightarrowlim\><rsub|T(\<varphi\><rsub|i>,p)
is bijective>T(\<varphi\><rsub|i>,p)(v)\<neq\>T(\<varphi\><rsub|i>,p)(v<rprime|'>)\<Rightarrow\>\<exists\>V<rsub|1>,V<rsub|2>>
open in <math|X> with <math|T(\<varphi\><rsub|i>,p)(v)\<in\>V<rsub|1>,T(\<varphi\><rsub|i>,p)(v<rprime|'>)\<in\>V<rsub|2>>
and <math|V<rsub|1><big|cap>V<rsub|2>=\<emptyset\>>. We have then
that <math|\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|1>,\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|2>>
are open in <math|\<varphi\><rsub|i>(U<rsub|i>)\<times\>X> and thus
<math|W<rsub|1>=<wide|\<varphi\><rsub|i>|~><rsup|-1>(\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|1>)\<in\>\<cal-S\>,W<rsub|2>=<wide|\<varphi\><rsub|i>|~><rsup|-1>(\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|2>)\<in\>\<cal-S\>>.
Now <math|<wide|\<varphi\><rsub|i>|~>(p,v)=(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p)(v))\<in\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|1>\<Rightarrow\>(p,v)\<in\>W<rsub|1>>
and <math|<wide|\<varphi\><rsub|i>|~>(p,v<rprime|'>)=(\<varphi\><rsub|i>(p),T(\<varphi\><rsub|i>,p))\<in\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|2>\<Rightarrow\>(p,v)\<in\>W<rsub|2>>.
Finally if <math|(q,w)\<in\>W<rsub|1>*<big|cap>W<rsub|2>\<Rightarrow\>(\<varphi\><rsub|i>(q),T(\<varphi\><rsub|i>,p)(w))\<in\>(\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|1>)<big|cap>(\<varphi\><rsub|i>(U<rsub|i>)\<times\>V<rsub|2>)\<Rightarrow\>T(\<varphi\><rsub|i>,p)(w)\<in\>V<rsub|1><big|cap>V<rsub|2>=\<emptyset\>>
a contradiction so we must conclude that
<math|W<rsub|1><big|cap>W<rsub|2>=\<emptyset\>>
<item><math|p\<neq\>p<rprime|'>> We proceed by contradiction. So
assume that there exists <math|(p,v),(p<rprime|'>,v<rprime|'>)> with
<math|(p,v)\<neq\>(p<rprime|'>,v<rprime|'>)> such that
<math|\<forall\>U,V> open with <math|(p,v)\<in\>U,(p<rprime|'>,v<rprime|'>)\<in\>V>
we have <math|U<big|cap>V\<neq\>\<emptyset\>>. As <math|\<cal-A\>> is
a atlas there exists a <math|k,l\<in\>I> such that
<math|p\<in\>U<rsub|k>,p<rprime|'>\<in\>U<rsub|l>> and as <math|M> is
Hausdorff we find a <math|W,W<rprime|'>> open such that
<math|p\<in\>W,p<rprime|'>\<in\>W<rprime|'>> and
<math|W<big|cap>W<rprime|'>=\<emptyset\>> then using
<reference|restriction of local chart in differentiable manifold> we
find a <math|i,j\<in\>I> such that
<math|p\<in\>U<rsub|i>=W<big|cap>U<rsub|k>,p<rprime|'>\<in\>U<rsub|j>=W<rprime|'><big|cap>U<rsub|l>>
and thus <math|U<rsub|i><big|cap>U<rsub|j>=\<emptyset\>>. Then as
<math|p=\<pi\>(p,v),p<rprime|'>=\<pi\>(p<rprime|'>,v<rprime|'>)\<Rightarrow\>(p,v)\<in\>\<pi\><rsup|-1>(U<rsub|i>)=<wide|U<rsub|i>|~>>
and <math|(p<rprime|'>,v<rprime|'>)\<in\>\<pi\><rsup|-1>(U<rsub|j>)=<wide|U<rsub|j>|~>>.
As <math|\<pi\>> is continuous we have that
<math|<wide|U<rsub|i>|~>,<wide|U<rsub|j>|~>> is open and thus there
exists by hypothese a <math|(p<rprime|''>,v<rprime|''>)\<in\><wide|U<rsub|i>|~><big|cap><wide|U<rsub|j>|~>>
and thus <math|p<rprime|''>=\<pi\>(p<rprime|''>,v<rprime|''>)\<in\>\<pi\>(<wide|U<rsub|i>|~>)<big|cap>\<pi\>(<wide|U<rsub|j>|~>)=\<pi\>(\<pi\><rsup|-1>(U<rsub|i>))<big|cap>\<pi\>(\<pi\><rsup|-1>(U<rsub|j>))\<subseteq\>U<rsub|i><big|cap>U<rsub|j>\<neq\>\<emptyset\>>
which is a contradiction.
</enumerate>
<item><strong|<math|\<forall\>i\<in\>I> we have that
<math|<wide|\<varphi\><rsub|i>|~>:<wide|U<rsub|i>|~>\<rightarrow\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
is a homeomorphism to a open set in <math|X\<times\>X>.> From the
previous theorem we already know that
<math|<wide|\<varphi\><rsub|i>|~>> is a bijection, so we must just
proof that it is continuous and open.\
<\enumerate>
<item>Continuity is trivial because by our definition of a subbase we
have if <math|V> is open in <math|<wide|\<varphi\><rsub|i>|~>(<wide|U<rsub|i>|~>)>
then <math|<wide|\<varphi\><rsub|i>|~><rsup|-1>(V)> is a element of
the subbase and thus open.
<item>Openeness. Let <math|U\<subseteq\><wide|U<rsub|k>|~>> be open.
Then <math|U=<big|cup><rsub|a\<in\>A>(<big|cap><rsub|b\<in\>B<rsub|a>><wide|\<varphi\><rsub|b>|~><rsup|-1>(V<rsub|b>)>
where <math|B<rsub|\<alpha\>>> is finite and
<math|V<rsub|b>\<subseteq\><wide|\<varphi\><rsub|b>|~>(<wide|U<rsub|b>|~>)>
is open. Then noting <math|W<rsub|a>=<big|cap><rsub|b\<in\>B<rsub|\<alpha\>>><wide|\<varphi\><rsub|b>|~><rsup|-1>(V<rsub|b>)\<subseteq\>U>
we have <math|<wide|\<varphi\><rsub|k>|~>(U)=<big|cup><rsub|a\<in\>A><wide|\<varphi\><rsub|k>|~>(W<rsub|a>)>.
So if we proof that <math|\<forall\>a\<in\>A\<succ\><wide|
\<varphi\><rsub|k>|~>(W<rsub|\<alpha\>>)> is open we are done. Now if
<math|W<rsub|a>=\<emptyset\>> then
<math|<wide|\<varphi\><rsub|k>|~>(W<rsub|a>)=\<emptyset\>> which is
open. Lets consider the remaining case
<math|W<rsub|\<alpha\>>\<neq\>\<emptyset\>> then there exists a
<math|u\<in\>U\<subseteq\><wide|U<rsub|k>|~>> such that
<math|\<forall\>b\<in\>B<rsub|a>> we have
<math|u\<in\><wide|\<varphi\><rsub|b>|~><rsup|-1>(V<rsub|b>)\<subseteq\><wide|U<rsub|b>|~>>
so <math|\<emptyset\>\<neq\><wide|U<rsub|k>|~><big|cap><wide|U<rsub|b>|~>>
and from the fact that then by the previous theorem
<math|<wide|\<varphi\><rsub|k>|~>\<circ\><wide|\<varphi\><rsub|b>|~><rsup|-1>>
is of class <math|C<rsup|r-1>> we have that
<math|<wide|\<varphi\><rsub|k>|~>(<wide|\<varphi\><rsub|b>|~><rsup|-1>(V<rsub|b>))>
is open and then from the finiteness of <math|B<rsub|a>> we have that
<math|<wide|\<varphi\><rsub|k>|~>(W<rsub|a>)=<big|cap><rsub|b\<in\>B<rsub|a>><wide|\<varphi\><rsub|k>|~>(<wide|\<varphi\><rsub|b>|~><rsup|-1>(V<rsub|b>))>
is open. This completes our proof of openess. <math|>
</enumerate>
<item><math|<wide|\<cal-A\>|~>={(<wide|\<varphi\><rsub|i>|~>,<wide|U<rsub|i>|~>)}<rsub|i\<in\>I>>
is a atlas modelled on <math|X\<times\>X>. We have by <math|(b)>
already proved that <math|<wide|\<varphi\><rsub|\<alpha\>>|~>:<wide|U<rsub|\<alpha\>>|~>\<rightarrow\>\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>>)\<times\>X>
is a homeomorphism, by the definition of the topology we have
<math|<wide|U<rsub|\<alpha\>>|~>> is open and we have also
<math|\<varphi\><rsub|\<alpha\>>(U<rsub|\<alpha\>>)\<times\>X> is open
in <math|X\<times\>X> and finally we have by the previous theorem that
<math|<big|cup><rsub|i\<in\>I><wide|U<rsub|i>|~>=T<rsub|M>>
<item>Again by the previous theorem we have that
<math|<wide|\<cal-A\>|~>> is not just a atlas modelled on <math|X> but
also of class <math|C<rsup|r-1>>.
<item>Finally if <math|\<cal-A\><rprime|'>\<in\>\<cal-C\><rsup|(r)>[\<cal-A\>]>
then as <math|\<cal-A\>> is the maximal atlas we have
<math|\<cal-A\><rprime|'>\<subseteq\>\<cal-A\>> and thus
<math|<wide|\<cal-A\><rprime|'>|~>\<subseteq\><wide|\<cal-A\>|~>> from
which it follows easely that they are <math|C<rsup|r-1>> compatable and
thus that <math|<wide|\<cal-A\><rprime|'>|~>\<in\>\<cal-C\><rsup|(r-1)>[\<cal-A\>]>
</enumerate>
\
</proof>
\;
From now on we consider that <math|T<rsub|M>> as a differentiable manifold
of class <math|C<rsup|r-1>> modelled on <math|X\<times\>X> as defined in
the previous theorem.
Note also that if <math|(<wide|U|~>,<wide|\<varphi\>|~>)\<in\><wide|\<cal-A\>|~>>
then if <math|(p,v)\<in\><wide|U|~>> then we have
<math|\<pi\><rsub|1>(<wide|\<varphi\>|~>(p,v))=\<pi\><rsub|1>(\<varphi\>(p),T(\<varphi\>,p)(v))=\<varphi\>(p)=\<varphi\>(\<pi\>(p,v))>
or <math|\<pi\><rsub|1>\<circ\><wide|\<varphi\>|~>=\<varphi\>\<circ\>\<pi\>>.
we can summarize this in the following diagram which is valid for all
<math|(U,\<varphi\>)\<in\>\<cal-A\>\<Rightarrow\>(<wide|U|~>,<wide|\<varphi\>|~>)\<in\><wide|\<cal-A\>|~>>
<\equation*>
<tabular|<tformat|<table|<row|<cell|T<rsub|M>>|<cell|\<supseteq\>>|<cell|<wide|U|~>>|<cell|<huge|\<rightarrowlim\><rsup|<small|<wide|\<varphi\>|~>>>>>|<cell|<huge|>\<varphi\>(U)\<times\>X>|<cell|\<subseteq\>>|<cell|X\<times\>X>>|<row|<cell|<huge|\<downarrow\><small|\<pi\>>>>|<cell|>|<cell|<huge|\<downarrow\><small|\<pi\>>>>|<cell|>|<cell|<huge|\<downarrow\><small|\<pi\><rsub|1>>>>|<cell|>|<cell|<huge|\<downarrow\><small|\<pi\><rsub|1>>>>>|<row|<cell|M>|<cell|\<supseteq\>>|<cell|U>|<cell|<huge|\<rightarrowlim\><rsup|<small|\<varphi\>>>>>|<cell|\<varphi\>(U)>|<cell|\<subseteq\>>|<cell|X>>>>>
</equation*>
\;
<\theorem>
Let <math|(M,\<cal-A\>={(U<rsub|i>,\<varphi\><rsub|i>)}<rsub|i\<in\>I>)>
be a differentiable manifold of class <math|C<rsup|r>> modelled on
<math|X> then <math|><math|T<rsub|M>,M,\<pi\>,X,\<cal-T\>> where
<math|\<cal-T\>={(U<rsub|i>,<wide|\<varphi\><rsub|i>|^><rsup|-1>)}<rsub|i\<in\>I>>
where <math|<wide|\<varphi\><rsub|i>|^>:U<rsub|i>\<times\>X\<rightarrow\>\<pi\><rsup|-1>(U<rsub|i>)=<wide|U<rsub|i>|~>>
is defined by <math|(p,x)\<rightarrow\><wide|\<varphi\><rsub|i>|^>(p,x)=(p,T(\<varphi\><rsub|i>,p)<rsup|-1>(x))>
forms a vector bundle called the tangent bundle. So we have in this
vector bundle using the terminology for a vector bundle
(<reference|vector bundle>)
<\enumerate>
<item><math|T<rsub|M>> is the total space
<item>M is the base space
<item><math|\<pi\>> is the projection map of the bundle
<item><math|\<cal-T\>> is the trivialization and
<math|(<wide|U<rsub|i>|~>,<wide|\<varphi\><rsub|i>|^>)> a trivializing
neighborhood
<item><math|\<pi\><rsup|-1>({p})=T<rsub|p>M> is called a fiber
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|><math|\<pi\>:T<rsub|M>\<rightarrow\>M> is indeed a
continuous surjection (as proved before)
<item><math|\<pi\><rsup|-1>({p})={p}\<times\>T<rsub|p>M>
<\eqnarray*>
<tformat|<table|<row|<cell|(p<rprime|'>,v<rprime|'>)\<in\>\<pi\><rsup|-1>({p})>|<cell|\<Rightarrow\>>|<cell|p=\<pi\>(p<rprime|'>,v<rprime|'>)=p<rprime|'>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(p<rprime|'>,v<rprime|'>)\<in\>{p}\<times\>T<rsub|p>M>>|<row|<cell|(p<rprime|'>,v<rprime|'>)\<in\>{p}\<times\>T<rsub|p>M>|<cell|\<Rightarrow\>>|<cell|p=p<rprime|'>>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<pi\>(p<rprime|'>,v<rprime|'>)=p>>>>
</eqnarray*>
Using the note <reference|topology and vectorspace on {p}XTpM> we have
that <math|{p}\<times\>T<rsub|p>M=\<pi\><rsup|-1>({p})> has the
structure of a vector space.\
<item>Because <math|\<cal-A\>> is a atlas we have
<math|M=<big|cup><rsub|i\<in\>I>U<rsub|i>>
<item><math|\<forall\>i\<in\>I> we have
<math|\<forall\>p\<in\>U<rsub|i>> and <math|\<forall\>x\<in\>X> we have
<math|\<pi\>(<wide|\<varphi\><rsub|i>|^>(p,x))=\<pi\>(p,T(\<varphi\><rsub|i>,p)<rsup|-1>(x))=p>
<item>If we define <math|\<varphi\><rsub|i><rsup|\<circ\>>:U<rsub|i>\<times\>X\<rightarrow\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X
by (p,y)\<rightarrow\>(\<varphi\><rsub|i>(p),y)> then we can prove that
<math|\<varphi\><rsub|i><rsup|\<circ\>>> is a homeomorphism
<\enumerate>
<item><math|Injectivity.> If <math|\<varphi\><rsub|i><rsup|\<circ\>>(p,y)=\<varphi\><rsub|i><rsup|\<circ\>>(p<rprime|'>,y<rprime|'>)\<Rightarrow\>y=y<rprime|'>>
and <math|\<varphi\><rsub|i>(p)=\<varphi\><rsub|i>(p<rprime|'>)\<Rightarrowlim\><rsub|\<varphi\><rsub|i>
is a homeomorphism>p=p<rprime|'>\<Rightarrow\>(p,y)=(p<rprime|'>,y<rprime|'>)>
<item>Surjectivity. If <math|(x,y)\<in\>\<varphi\><rsub|i>(U<rsub|i>)\<times\>X>
then by surjectivity of <math|\<varphi\><rsub|i>> there exists a
<math|p\<in\>U<rsub|i>> such that <math|\<varphi\><rsub|i>(p)=x> and
thus <math|\<varphi\><rsub|i><rsup|\<circ\>>(p,y)=(\<varphi\><rsub|i>(p),y)=(x,y)>
<item>Continuity. As <math|\<pi\><rsub|1>\<circ\>\<varphi\><rsub|i><rsup|\<circ\>>>
is <math|(p,y)\<rightarrow\>\<varphi\><rsub|i>(p)> and is thus equal
to <math|\<varphi\><rsub|i>\<circ\>\<pi\><rsub|1>> the composition of
two continuous mappings we have that
<math|\<pi\><rsub|1>\<circ\>\<varphi\><rsub|i><rsup|\<circ\>>> is
continuous. As <math|\<pi\><rsub|2>\<circ\>\<varphi\><rsub|i><rsup|\<circ\>>=\<pi\><rsub|2>>
is also continuous we have that <math|\<varphi\><rsub|i><rsup|\<circ\>>>
is continuous (see <reference|continuous mappings to a product
space>).
<item>The inverse <math|(\<varphi\><rsub|i><rsup|\<circ\>>)<rsup|-1>>
of <math|\<varphi\><rsub|i><rsup|\<circ\>>> is defined by
<math|(\<varphi\><rsub|i><rsup|\<circ\>>)<rsup|-1>(x,y)=(\<varphi\><rsub|i><rsup|-1>(x),y)>
[<math|(\<varphi\><rsub|i><rsup|\<circ\>>)<rsup|-1>(\<varphi\><rsub|i><rsup|\<circ\>>(p,y))=(\<varphi\><rsub|i><rsup|\<circ\>>)<rsup|-1>(\<varphi\><rsub|i>(p),y)=(\<varphi\><rsub|i><rsup|-1>(\<varphi\><rsub|i>(p)),y)=(p,y)>]
and this is also continuous because (see <reference|continuous
mappings to a product space>)
<\enumerate>
<item><math|\<pi\><rsub|1>\<circ\>(\<varphi\><rsup|\<circ\>><rsub|i>)<rsup|-1>=\<varphi\><rsub|i><rsup|-1>\<circ\>\<pi\><rsub|1>>
is continuous beeing the composition of two continuos functions.
<item><math|\<pi\><rsub|2>\<circ\>(\<varphi\><rsub|i><rsup|\<circ\>>)<rsup|-1>=\<pi\><rsub|2>>
is also continuous.
</enumerate>
</enumerate>
<item>Now as <math|\<forall\>(p,x)\<in\>U<rsub|i>\<times\>X> we have
<math|(<wide|\<varphi\><rsub|i>|~>)<rsup|-1>\<circ\>\<varphi\><rsub|i><rsup|\<circ\>>(p,x)=(<wide|\<varphi\><rsub|i>|~>)<rsup|-1>(\<varphi\><rsub|i>(p),x)=(\<varphi\><rsub|i><rsup|-1>(\<varphi\><rsub|i>(p)),T(\<varphi\><rsub|i>,\<varphi\><rsub|i><rsup|-1>(\<varphi\><rsub|i>(p)))<rsup|-1>(x))=(p,T(\<varphi\><rsub|i>,p)<rsup|-1>(x)=<wide|\<varphi\><rsub|i>|^>(p,x)>
proving that <math|<wide|\<varphi\><rsub|i>|^>=(<wide|\<varphi\><rsub|i>|~>)<rsup|-1>\<circ\>\<varphi\><rsub|i><rsup|\<circ\>>>
which is the composition of two homeomorphism. From this it follows
that <math|\<forall\>i\<in\>I> we have that
<math|<wide|\<varphi\><rsub|i>|^>> is a homeomorphism as required by a
vector bundle.
<item>Finally if we define <math|<wide|\<varphi\><rsub|i>|^><rsup|(p)>:X\<rightarrow\>\<pi\><rsup|-1>({p})=T<rsub|p>M>
by <math|<wide|\<varphi\><rsub|i>|^><rsup|(p)>(x)=<wide|\<varphi\><rsub|i>|^>(p,x)=(p,T<rsub|>(\<varphi\><rsub|i>,p)(x))>
then <math|<wide|\<varphi\><rsub|i>|^><rsup|(p)>(\<alpha\>.x+\<beta\>.y)=(p,T(\<varphi\><rsub|i>,p)(\<alpha\>.x+\<beta\>.y))\<equallim\><rsub|linearity
of T(\<varphi\><rsub|i>,p)>(p,\<alpha\>.T(\<varphi\><rsub|i>,p)(x)+\<beta\>.T(\<varphi\><rsub|i>,p))\<equallim\><rsub|<reference|topology
and vectorspace on {p}XTpM>>\<alpha\>.(p,T(\<varphi\><rsub|i>,p)(x))+\<beta\>.(p,T(\<varphi\><rsub|i>,p)(y))>
proving linearity
</enumerate>
</proof>
\;
<\example>
Let <math|(M,\<cal-A\>)> be a manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modeled on <math|\<bbb-R\><rsup|n>> (a n
dimensional manifold) and let <math|(U,\<varphi\>)\<in\>\<cal-A\>> then
we have <math|\<varphi\>:M\<rightarrow\>\<bbb-R\><rsup|n>>. We note
<math|\<varphi\><rsub|i>=\<pi\><rsub|i>\<circ\>\<varphi\>\<equiv\>x<rsub|i>>.
We have then\
<\enumerate>
<item><math|T<rsub|M>> is then a manifold of class <math|C<rsup|r-1>>
modeled on <math|\<bbb-R\><rsup|n>\<times\>\<bbb-R\><rsup|n>=\<bbb-R\><rsup|2n>>
so <math|T<rsub|M>> is a <math|2n> dimensional manifold
<item><math|(<wide|U|~>,<wide|\<varphi\>|~>)> where
<math|<wide|\<varphi\>|~>:<wide|U|~>=\<pi\><rsup|-1>(U)\<rightarrow\>\<varphi\>(U)\<times\>\<bbb-R\><rsup|n>\<subseteq\>\<bbb-R\><rsup|n>\<times\>\<bbb-R\><rsup|n>=\<bbb-R\><rsup|2n>>
where
<\equation*>
<wide|\<varphi\>|~>(p,v)=(\<varphi\>(p),T(\<varphi\>,p)(v))
</equation*>
<item>Since <math|(\<partial\><rsub|1>(\<varphi\>,p),\<ldots\>,\<partial\><rsub|n>(\<varphi\>,p))>
is a basis for <math|T<rsub|p>M> we have
<math|\<forall\>v\<in\>T<rsub|p>M> that\
<\equation*>
v=<big|sum><rsub|i=1><rsup|n>h<rsub|i>\<partial\><rsub|i>(\<varphi\>,p)
</equation*>
and as we have proved in the previous example (<reference|coordinate
vector fields>) that
<\equation*>
T(\<varphi\>,p)(\<partial\><rsub|i>(\<varphi\>,p))=e<rsub|i>
</equation*>
we have that \ \ \
<\eqnarray*>
<tformat|<table|<row|<cell|<wide|\<varphi\>|~>(p,v)>|<cell|=>|<cell|(x<rsub|1>(p),\<ldots\>,x<rsub|n>(p),T(\<varphi\>,p)(<big|sum><rsub|i=1><rsup|n>h<rsub|i>\<partial\><rsub|i>(\<varphi\>,p))>>|<row|<cell|>|<cell|=>|<cell|(x<rsub|1>(p),\<ldots\>.,x<rsub|n>(p),<big|sum><rsub|i=1><rsup|n>h<rsub|i>e<rsub|i>)>>|<row|<cell|>|<cell|\<equallim\><rsub|e<rsub|i>=(0,\<ldots\>,1,\<ldots\>0)>>|<cell|(x<rsub|1>(p),\<ldots\>,x<rsub|n>(p),h<rsub|1>,\<ldots\>,h<rsub|n>)>>>>
</eqnarray*>
<item>Let <math|(V,\<psi\>)\<in\>\<cal-A\>> with
<math|V<big|cap>U\<neq\>\<emptyset\>> and then we have if
<math|x,h\<in\>\<varphi\>(U)\<times\>\<bbb-R\><rsup|n>\<subseteq\>\<bbb-R\><rsup|n>\<times\>\<bbb-R\><rsup|n>=\<bbb-R\><rsup|2n>>
<\eqnarray*>
<tformat|<table|<row|<cell|<wide|\<psi\>|~>\<circ\><wide|\<varphi\>|~><rsup|-1>(x,h)>|<cell|=>|<cell|(\<psi\>\<circ\>\<varphi\><rsup|-1>(x),D(\<psi\>\<circ\>\<varphi\><rsup|-1>)(x)(h))>>|<row|<cell|<wide|\<psi\>|~>\<circ\><wide|\<varphi\>|~><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>,h<rsub|1>,\<ldots\>,h<rsub|n>)>|<cell|\<equallim\><rsub|<reference|Jacobian>>>|<cell|(y<rsub|1>(x),\<ldots\>,y<rsub|n>(x),<big|sum><rsub|i=1><rsup|n><frac|dy<rsub|1>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x)h<rsub|i>,\<ldots\>,<big|sum><rsub|i=1><rsup|n><frac|dy<rsub|n>(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>h<rsub|i>)>>>>
</eqnarray*>
where we have <math|y<rsub|i>(x)=\<pi\><rsub|i>\<circ\>(\<psi\>\<circ\>\<varphi\><rsup|-1>)(x)>
</enumerate>
</example>
<section|Vector fields>
<\definition>
<label|vector field><index|vector field>Let <math|(M,\<cal-A\>)> be a
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modeled on the normed
space <math|X> then a vector field <math|V> is a function
<math|V:M\<rightarrow\>T<rsub|M>> such that
<math|\<pi\>\<circ\>V=i<rsub|M>>. If the function is differentiable of
class <math|C<rsup|r-1>> then <math|V> is said to be a differentiable
vector field of class <math|C<rsup|r-1>> on <math|M>
</definition>
The condition <math|\<pi\>\<circ\>V=i<rsub|M>> ensures that if
<math|p\<in\>M> that <math|\<pi\><rsub|2>\<circ\>V(p)\<in\>T<rsub|p>M> is a
vector tangent at <math|p>
<\lemma>
<label|characterization of differentiable vector
fields><index|<math|V<rsup|\<varphi\>>>>Let <math|(M,\<cal-A\>)> be a
manifold of class <math|C<rsup|r>,r\<geqslant\>1> modeled on the normed
space <math|X> and <math|(U,\<varphi\>)\<in\>\<cal-A\>> then
\ <math|V<rsup|\<varphi\>>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<varphi\>(U)*\<times\>X>
is defined and the following diagram \ commutes
<\equation*>
<tabular|<tformat|<table|<row|<cell|T<rsub|M>>|<cell|\<supseteq\>>|<cell|<wide|U|~>>|<cell|<huge|\<rightarrowlim\><rsup|<small|<wide|\<varphi\>|~>>>>>|<cell|\<varphi\>(U)\<times\>X>|<cell|\<subseteq\>>|<cell|X\<times\>X>>|<row|<cell|V<huge|\<uparrow\>\<downarrow\>>\<pi\>>|<cell|>|<cell|>|<cell|>|<cell|\<pi\><rsub|1><huge|\<downarrow\>\<uparrow\>>V<rsup|\<varphi\>>>|<cell|>|<cell|<huge|\<downarrow\>>\<pi\><rsub|1>>>|<row|<cell|M>|<cell|\<supseteq\>>|<cell|U>|<cell|<huge|\<rightarrowlim\><rsub|<small|\<varphi\>>>>>|<cell|\<varphi\>(U)>|<cell|\<subseteq\>>|<cell|X>>>>>
</equation*>
If <math|V<rsup|[\<varphi\>]>:U\<rightarrow\>X> is defined by
<math|p\<rightarrow\>V<rsup|[\<varphi\>]>(p)=T(\<varphi\>,p)(\<pi\><rsub|2>(V(p)))>
then we have that <math|V<rsup|\<varphi\>>(x)=(x,V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>(x))>.
<math|h<rsup|(V,\<varphi\>)>> is called the local representive of
<math|V>.
Finally <math|V> is differentiable of class <math|C<rsup|r-1>> if and
only if <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> we have that
<math|V<rsup|\<varphi\>>> (or <math|V<rsup|[\<varphi\>]>>) is
differentiable of class <math|C<rsup|r-1>>
</lemma>
<\proof>
We have for the homeomorphism <math|\<varphi\>> that
<math|\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>U> and thus
<math|\<forall\>y\<in\>\<varphi\>(U)> we have that
<math|\<pi\>(V(\<varphi\><rsup|-1>(y)))=\<varphi\><rsup|-1>(y)\<in\>U\<Rightarrow\>V(\<varphi\><rsup|-1>(y))\<in\>\<pi\><rsup|-1>(U)=<wide|U|~>>
and thus <math|\<forall\>y\<in\>\<varphi\>(U)> we have that
<math|<wide|\<varphi\>|~>(V(\<varphi\><rsup|-1>(y)))\<in\>\<varphi\>(U)\<times\>X>
is defined and thus <math|V<rsup|\<varphi\>>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>>
is a function <math|\<varphi\>(U)\<rightarrow\>\<varphi\>(U)\<times\>X>.
From the definition from <math|V<rsup|\<varphi\>>> it follows that
<math|V<rsup|\<varphi\>>\<circ\>\<varphi\>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>\<circ\>\<varphi\>=<wide|\<varphi\>|~>\<circ\>V>
so we have proved that the diagram commutes. Now if
<math|V(\<varphi\><rsup|-1>(x))=(\<pi\><rsub|1>(V(\<varphi\><rsup|-1>(x))),\<pi\><rsub|2>(V(\<varphi\><rsup|-1>(x))))=(p,v)\<in\>T<rsub|M>>
then <math|V<rsup|\<varphi\>>(x)=<wide|\<varphi\>|~>(V(\<varphi\><rsup|-1>(x)))=(\<varphi\>(\<pi\><rsub|1>(V(\<varphi\><rsup|-1>(x)),T(\<varphi\>,\<pi\><rsub|1>(V(\<varphi\><rsup|-1>(x))))(\<pi\><rsub|2>(V(\<varphi\><rsup|-1>(x))))\<equallim\><rsub|\<pi\><rsub|1>\<equiv\>\<pi\>,\<pi\><rsub|1>\<circ\>V=1<rsub|M>>(\<varphi\>(\<varphi\><rsup|-1>(x)),T(\<varphi\>,\<varphi\><rsup|-1>(x))(\<pi\><rsub|2>(V(\<varphi\><rsup|-1>(x)))))>.\
From this it follows that if we define
<math|V<rsup|[\<varphi\>]>:U\<rightarrow\>X> by
<math|y\<rightarrow\>T(\<varphi\>,y)(\<pi\><rsub|2>(V(y)))> then
<math|V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>(x)=T(\<varphi\>,\<varphi\><rsup|-1>(x))(\<pi\><rsub|2>(V(\<varphi\><rsup|-1>(x))))>
and thus <math|V<rsup|\<varphi\>>(x)=(x,V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>(x))>
If <math|V> is differentiable of class <math|C<rsup|r-1>> then if
<math|(U,\<varphi\>)\<in\>\<cal-A\>> and
<math|p\<in\>U\<Rightarrow\>p=\<pi\>(V(p))\<Rightarrow\>V(p)\<in\>\<pi\><rsup|-1>(U)=<wide|U|~>\<Rightarrow\>p\<in\>U<big|cap>V<rsup|-1>(<wide|U|~>)\<neq\>\<emptyset\>>
and thus we have that <math|V<rsup|\<varphi\>>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>>
is differentiable of class <math|C<rsup|r-1>> as
<math|(<wide|U|~>,<wide|\<varphi\>|~>)\<in\><wide|\<cal-A\>|~>> the
maximal atlas of <math|T<rsub|M>>
If <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> we have that
<math|V<rsup|\<varphi\>>> is differentiable of class <math|C<rsup|r-1>>
then if <math|p\<in\>M> and <math|(V,\<psi\>)\<in\><wide|\<cal-A\>|~>>
(the maximal atlas of <math|T<rsub|M>>) is such that <math|V(p)\<in\>V>
then there exists a <math|(U,\<varphi\>)\<in\>\<cal-A\>> such that
<math|V=<wide|U|~>>, <math|\<psi\>=<wide|\<varphi\>|~>> with
<math|\<psi\>\<circ\>V\<circ\>\<varphi\><rsup|-1>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>=V<rsup|\<varphi\>>>
is differentiable of class <math|C<rsup|r-1>>. From <reference|condition
for differentiability on manifolds> it then follows that <math|V> is
differentiable of class <math|C<rsup|r-1>>.\
Finally <math|V<rsup|\<varphi\>>> is differentiable of class
<math|C<rsup|r-1>> if and only if <math|\<pi\><rsub|1>\<circ\>V<rsup|\<varphi\>>=i<rsub|X>>
and <math|\<pi\><rsub|2>\<circ\>V<rsup|\<varphi\>>=V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>>
are differentiable of class <math|C<rsup|r-1>>. This is equivalent (as
<math|i<rsub|X>> is <math|C<rsup|\<infty\>>>) with
<math|V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>> is differentiable
of class <math|C<rsup|r-1>> and this by <reference|differentiable
function to a normed space> means that <math|h> is differentiable of
class <math|C<rsup|r-1>>.
</proof>
<\example>
<index|directional derivative><index|<math|V[f]>><label|vector field on a
finite differential manifold><dueto|Vector fields on finite dimensional
manifolds>Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|r>,r\<geqslant\>1> modelled on <math|\<bbb-R\><rsup|n>>. If
<math|(U,\<varphi\>)\<in\>\<cal-A\>, then \<forall\>p\<in\>U> \ as
<math|\<forall\>i\<in\>{1,\<ldots\>,n}>
<math|\<partial\><rsub|i>(\<varphi\>,p)> is a basis for <math|T<rsub|p>M>
(see example <reference|basis of tangent space in finite dimensional
case>,<reference|coordinate vector fields>) defined by
<math|\<partial\><rsub|i>(\<varphi\>,p)=\<bbb-T\>[c<rsub|i>]> where
<math|c<rsub|i>=\<varphi\><rsup|-1>\<circ\>\<tau\><rsub|i>,\<tau\><rsub|i>(t)=(x<rsup|0><rsub|1>,\<ldots\>,x<rsub|i-1><rsup|0>,x<rsub|i><rsup|0>+t,x<rsub|i+1><rsup|0>,\<ldots\>,x<rsub|n><rsup|0>)>
and <math|x<rsup|0><rsub|i>=\<pi\><rsub|i>(\<varphi\>(p))>. If now
<math|V> is a vector field on <math|M> then <math|\<forall\>p\<in\>U> we
have a unique decomposition
<\equation*>
\<pi\><rsub|2>\<circ\>V(p)=<big|sum><rsub|i=1><rsup|n>h<rsub|i>(p)\<partial\><rsub|i>(\<varphi\>,p)
</equation*>
where
<\equation*>
\<forall\>i\<in\>{1,\<ldots\>,n},h<rsub|i>(p):U\<rightarrow\>\<bbb-R\>
</equation*>
defines the local representative of <math|V> in the local coordinate
chart <math|(U,\<varphi\>)>. <math|h<rsub|i>> is called the <math|i>th
<em|coordinate function> of the vector field <math|V> in the chart
<math|(U,\<varphi\>)>.\
If now <math|\<Omega\>\<subseteq\>M> is a open set in <math|M> and
<math|f:\<Omega\>\<rightarrow\>\<bbb-R\>> a differentiable function, then
for every vector field <math|V> we define
<\equation*>
V[f]:\<Omega\>\<rightarrow\>\<bbb-R\> by
V[f](p)=(\<pi\><rsub|2>\<circ\>V(p))[f] (see <reference|directional
derivative>)
</equation*>
We call <math|V[f]> the directional derivate of <math|f> in the direction
<math|V> in <math|\<Omega\>>. We have then for
<math|(U,\<varphi\>)\<in\>\<cal-A\>> and
<math|p\<in\>U,\<varphi\><rsub|i>=\<pi\><rsub|i>\<circ\>\<varphi\>> that
<math|\<forall\>(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<varphi\>(U)\<subseteq\>\<bbb-R\><rsup|n>>
<\eqnarray*>
<tformat|<table|<row|<cell|\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>)>|<cell|=>|<cell|\<pi\><rsub|i>\<circ\>\<varphi\>\<circ\>\<varphi\><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>)>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|i>(x<rsub|1>,\<ldots\>,x<rsub|n>)>>|<row|<cell|>|<cell|=>|<cell|x<rsub|i>>>>>
</eqnarray*>
So we have that\
<\eqnarray*>
<tformat|<table|<row|<cell|<space|0.2spc><frac|d(\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>)(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|j>>(y)>|<cell|=>|<cell|\<delta\><rsup|i><rsub|j>>>>>
</eqnarray*>
from which it follows that
<\eqnarray*>
<tformat|<table|<row|<cell|V[\<varphi\><rsub|i>](p)>|<cell|=>|<cell|(\<pi\><rsub|2>\<circ\>V(p))[\<varphi\><rsub|i>]>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|j=1><rsup|n>h<rsub|j>(p)\<partial\><rsub|j>(\<varphi\>,p)[\<varphi\><rsub|i>]>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|apply
basis vector on function>>>|<cell|<big|sum><rsub|j=1><rsup|n>h<rsub|j>(p)<frac|d(\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>)(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|j>>(\<varphi\>(p))>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|j=1><rsup|n>h<rsub|j>(p)\<delta\><rsup|i><rsub|j>>>|<row|<cell|>|<cell|=>|<cell|h<rsub|i>(p)>>>>
</eqnarray*>
Or the <math|i>the coordinate of <math|V> is just
<math|V[\<varphi\><rsub|i>]>
</example>
<\theorem>
<label|differentiable vector fields property one>Let <math|(M,\<cal-A\>)>
be a differentiable manifold of class <math|C<rsup|r>,r\<geqslant\>1>
modelled on <math|\<bbb-R\><rsup|n>>. Then the following \ are equivalent
for a vector field <math|V:M\<rightarrow\>T<rsub|M>>
<\enumerate>
<item><math|V> is differentiable of class <math|C<rsup|r-1>>
<item><math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> the coordinate
functions <math|h<rsub|i>:U\<rightarrow\>\<bbb-R\>,i\<in\>{1,\<ldots\>,n}>
are differentiable of class <math|C<rsup|r-1>>
<item>If <math|\<Omega\>\<subseteq\>M> is a open set and
<math|f:\<Omega\>\<rightarrow\>\<bbb-R\>> is a differentiable function
of class <math|C<rsup|r>> then <math|V[f]:\<Omega\>\<rightarrow\>\<bbb-R\>>
is differentiable of class <math|C<rsup|r-1>>
</enumerate>
</theorem>
<\proof>
\;
<math|1\<Leftrightarrow\>2>
Not that <math|V> is differentiable of class <math|C<rsup|r-1>> if and
only if for <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>>
<math|V<rsup|\<varphi\>>=<wide|\<varphi\>|~>\<circ\>V\<circ\>\<varphi\><rsup|-1>:\<varphi\>(U)\<rightarrow\>\<varphi\>(U)\<times\>\<bbb-R\><rsup|n>:x\<rightarrow\>(x,V<rsup|[\<varphi\>]>\<circ\>\<varphi\><rsup|-1>(x))>
we have that <math|v<rsup|[\<varphi\>]>> is differentiable of class
<math|C<rsup|r-1>>. As we have for <math|p\<in\>U> that
<math|p=\<varphi\><rsup|-1>(x),><math|x=\<varphi\>(p)=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<bbb-R\><rsup|n>>
and as <math|V(p)=(p,v)\<in\>{p}\<times\>T<rsub|p>M>
<\eqnarray*>
<tformat|<table|<row|<cell|V(p)=V\<circ\>\<varphi\><rsup|-1>(x)>|<cell|=>|<cell|<wide|\<varphi\>|~><rsup|-1>\<circ\>V<rsup|\<varphi\>>(x)>>|<row|<cell|>|<cell|=>|<cell|<wide|\<varphi\>|~><rsup|-1>(V<rsup|\<varphi\>>(x))>>|<row|<cell|>|<cell|>|<cell|<wide|\<varphi\>|~><rsup|-1>(x,V<rsup|[\<varphi\>]>(\<varphi\><rsup|-1>(x))))>>|<row|<cell|>|<cell|=>|<cell|(\<varphi\><rsup|-1>(x),T(\<varphi\>,\<varphi\><rsup|-1>(x))<rsup|-1>(V<rsup|[\<varphi\>]>(\<varphi\><rsup|-1>(x))))>>|<row|<cell|>|<cell|\<equallim\>>|<cell|(p,T(\<varphi\>,p)<rsup|-1>(\<pi\><rsub|1>\<circ\>V<rsup|[\<varphi\>]<rsub|>>(p),\<ldots\>,\<pi\><rsub|n>\<circ\>V<rsup|[\<varphi\>]>(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,T(\<varphi\>,p)<rsup|-1>(<big|sum><rsub|i=1><rsup|n>(\<pi\><rsub|i>\<circ\>V<rsup|[\<varphi\>]>(p))e<rsub|i>))>>|<row|<cell|>|<cell|\<equallim\><rsub|T(\<varphi\>,p)<rsup|-1>
is linear>>|<cell|(p,<big|sum><rsub|i=1><rsup|n>(\<pi\><rsub|i>\<circ\>V<rsup|[\<varphi\>]>(p))T(\<varphi\>,p)<rsup|-1>(e<rsub|i>))>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|coordinate
vector fields>>>|<cell|(p,<big|sum><rsub|i=1><rsup|n>(\<pi\><rsub|i>\<circ\>V<rsup|[\<varphi\>]>(p))\<partial\><rsub|i>(\<varphi\>,p)>>>>
</eqnarray*>
so that\
<\eqnarray*>
<tformat|<table|<row|<cell|<big|sum><rsub|i=1><rsup|n>h<rsub|i>(p)\<partial\><rsub|i>(\<varphi\>,p)>|<cell|=>|<cell|\<pi\><rsub|2>\<circ\>V(p)>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|i=1><rsup|n>(\<pi\><rsub|i>\<circ\>V<rsup|[\<varphi\>]>(p))\<partial\><rsub|i>(\<varphi\>,p)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|\<partial\><rsub|i>(p)
forms a basis>\<forall\>i\<in\>{1\<ldots\>.n}>|<cell|h<rsub|i>=V<rsup|[\<varphi\>]>(p)>>>>
</eqnarray*>
so we have the equivalence
<\eqnarray*>
<tformat|<table|<row|<cell|V is differentiable of class
C<rsup|r-1>>|<cell|\<Leftrightarrow\>>|<cell|V<rsup|[\<varphi\>]> \ is
differentiable of class C<rsup|r-1>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|\<forall\>i\<in\>{1,\<ldots\>,n}
h<rsub|i>=\<pi\><rsub|i>\<circ\>V<rsup|[\<varphi\>]>(p) is
differentiable of class C<rsup|r-1>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|V<rsup|[\<varphi\>]>
is differentiable of class C<rsup|r-1>>>|<row|<cell|>|<cell|\<Leftrightarrow\>>|<cell|V
is differentiable of class C<rsup|r-1>>>>>
</eqnarray*>
\;
\ <math|3\<Rightarrow\>2>\
<math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> we have that
<math|\<varphi\><rsub|i>=\<pi\><rsub|i>\<circ\>\<varphi\>:U\<rightarrow\>\<bbb-R\>>
is differentiable of class <math|C<rsup|r>> as if
<math|(W,\<psi\>)\<in\>\<cal-A\>> with
<math|W<big|cap>U=W<big|cap>\<varphi\><rsub|i><rsup|-1>(U)\<neq\>\<emptyset\>
then we have that ><math|\<varphi\>\<circ\>\<psi\><rsup|-1>> is
differentiable of class <math|C<rsup|r>> and thus
<math|\<pi\><rsub|i>\<circ\>(\<varphi\>\<circ\>\<psi\><rsup|-1>)=(\<pi\><rsub|i>\<circ\>\<varphi\>)\<circ\>\<psi\><rsup|-1>>
is of class <math|C<rsup|r>>. From the previous example we have then that
<math|><math|V[\<varphi\><rsub|i>]=h<rsub|i>> and from the hypothese (3)
that <math|h<rsub|i>> is differentiable of class <math|C<rsup|r-1>>
<math|(2\<Rightarrow\>3)>
Given<math|f:\<Omega\>\<rightarrow\>\<bbb-R\>> a differentiable function
of class <math|C<rsup|r>> then <math|\<forall\>i\<in\>{1,\<ldots\>,n}>
and <math|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>> with <math|p\<in\>U>
we have that <math|V[f](p)=<big|sum><rsub|i=1><rsup|n>h<rsub|i>(p)\<partial\><rsub|i>(\<varphi\>,p)[f]>
where by (2) we have the hypothese that <math|h<rsub|i>(p)> is
differentiable of class <math|C<rsup|r-1>>, \ and
<math|\<partial\><rsub|i>(p)[f]=<frac|d(f\<circ\>\<varphi\><rsup|-1>)|dx<rsub|i>>(\<varphi\>(p))=D<rsub|i>(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))(1)>.
Now <math|(f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>:\<varphi\><rsub|i>(U)\<rightarrow\>\<bbb-R\>>
defined by <math|t\<rightarrow\>(f\<circ\>\<varphi\><rsup|-1>)(x<rsub|1>,\<ldots\>,x<rsub|i-1>,t,x<rsub|i+1>,\<ldots\>,x<rsub|n>),(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<varphi\>(U)>
is differentiable of class <math|C<rsup|r>> (because by hypothese
<math|f\<circ\>\<varphi\><rsup|-1>> is differentiable of class
<math|C<rsup|r>> and using <reference|Cr and partial derivates>). Also
using <reference|fv is differentiable> <math|\<forall\>v\<in\>\<bbb-R\>
((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)<rsub|v>:\<bbb-R\>\<rightarrow\>\<bbb-R\>>
defined by <math|x\<in\>\<varphi\><rsub|i>(U)\<rightarrow\>D((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)(x)(v)>
is differentiable of class <math|C<rsup|r-1>> (but also of class
<math|C<rsup|r-1>> considered as a mapping between manifolds (see
<reference|Cr between normed spaces os Cr between manifolds>). From
<math|[((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)<rsub|v>\<circ\>\<varphi\>](p)=((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)<rsub|v>(\<varphi\>(p))=D((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>(\<varphi\>(p))(v)=D<rsub|i>(f\<circ\>\<varphi\><rsup|-1>)(\<varphi\>(p))(v)>
if follows that the function <math|p\<rightarrow\>\<partial\><rsub|i>(\<varphi\>,p)[f]>
is equal to <math|((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)<rsub|1>\<circ\>\<varphi\>>
which is differentiable of class <math|C<rsup|r-1>> [because
<math|\<varphi\>> is differentiable of class <math|C<rsup|r-1>> (see
<reference|a chart is differentiable>), and we have just proved that
<math|((f\<circ\>\<varphi\><rsup|-1>)<rsup|(i)>)<rsub|1>> is
differentiable of class <math|C<rsup|r-1>> and thus the composition
should be also differentiable of class <math|C<rsup|r-1>>, see
<reference|differentiable mappings between manifolds and composition>)].
Finally <math|V[f]> is differentiable of class <math|C<rsup|r-1>> because
it is a sum of products of functions differentiable of class
<math|C<rsup|r-1>> (see <reference|sum(product) of differentiable
functions between manifolds>).
</proof>
<\definition>
<label|real smooth functions on smooth
manifolds><index|<math|C<rsup|\<infty\>>(M)>>Let <math|(M,\<cal-A\>)> be
a differentiable manifold modelled on <math|\<bbb-R\><rsup|n>> of class
<math|C<rsup|\<infty\>>> (as smooth manifold) then define
<math|C<rsup|\<infty\>>(M)={f:M\<rightarrow\>\<bbb-R\>,f> is of class
<math|C<rsup|\<infty\>>}> (the set of real smooth functions on <math|M>)
</definition>
<\theorem>
<math|C<rsup|\<infty\>>(M)> is a real commutative algebra defined \ by
the usual pointwise operations. So for
<math|f,g\<in\>C<rsup|\<infty\>>,\<alpha\>\<in\>\<bbb-R\>> define
<\enumerate>
<item><math|f+g:M\<rightarrow\>\<bbb-R\>> is defined by
<math|x\<rightarrow\>f(x)+g(x)>
<item><math|><math|\<alpha\>.f:M\<rightarrow\>\<bbb-R\>> is defined by
<math|x\<rightarrow\>\<alpha\>.f(x)>
<item><math|f\<bullet\>g:M\<rightarrow\>\<bbb-R\>> is defined by
<math|x\<rightarrow\>f(x).g(x)>
</enumerate>
</theorem>
<\proof>
First of all we must proof that <math|f+g,\<alpha\>.f> and
<math|f\<bullet\>g> are in <math|C<rsup|\<infty\>>(M)> but this is easy
because it follows from <reference|sum(product) of differentiable
functions between manifolds>. As <math|C<rsup|\<infty\>>(M)> is a
subspace of the vectorspace of mappings from
<math|M\<rightarrow\>\<bbb-R\>> (see <reference|subspace>) we use
<reference|vector space of mappings> to prove that
<math|C<rsup|\<infty\>>(M)> is a vectorspace.
To proof that it is a algebra (see <reference|algebra>) note that
<\enumerate>
<item><math|[(f+g)\<bullet\>h](x)=(f+g)(h(x))=f((h(x))+g(h(x))=(f\<bullet\>h)(x)+(g\<bullet\>h)(x)=(f\<bullet\>h+g\<bullet\>h)(x)\<Rightarrow\>(f+g)\<bullet\>h=f\<bullet\>h+g\<bullet\>h>
<item><math|[h\<bullet\>(f+g)](x)=h((f+g)(x))=h(f(x)+g(x))=h(f(x))+h(g(x))=(h\<bullet\>f)(x)+(h\<bullet\>g)(x)=(h\<bullet\>f+h\<bullet\>g)(x)\<Rightarrow\>h\<bullet\>(f+g)=h\<bullet\>f+h\<bullet\>g>
<item><math|[(\<alpha\>.f)\<bullet\>(\<beta\>.g)](x)=(\<alpha\>.f)(x)(\<beta\>.g)(x)=\<alpha\>.f(x).\<beta\>.g(x)=\<alpha\>.\<beta\>.f(x).g(x)=[\<alpha\>.\<beta\>(f\<bullet\>g)](x)\<Rightarrow\>(\<alpha\>.f)\<bullet\>(\<beta\>.g)>
<item><math|(f\<bullet\>g)(x)=f(x).g(x)=g(x).f(x)\<Rightarrow\>f\<bullet\>g=g\<bullet\>f>
</enumerate>
</proof>
<\definition>
<label|derivation in a algebra>Let <math|A> be a real algebra then a
function <math|\<Delta\>:A\<rightarrow\>A> is called a derivation on
<math|A> if it fullfils the following conditions
<\enumerate>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,x,y\<in\>A> we
have <math|\<Delta\>(\<alpha\>.x+\<beta\>.y)=\<alpha\>\<Delta\>(x)+\<beta\>\<Delta\>(y)>
(linearity)
<item><math|\<forall\>x,y\<in\>A> we have
<math|\<Delta\>(x.y)=x.\<Delta\>(y)+\<Delta\>(x).y> (Leibnitz)
</enumerate>
</definition>
<\theorem>
<label|Lie product of derivations is a derivation on a commutative
algebra>Let <math|A> be a real algebra and
<math|\<Delta\><rsub|1>,\<Delta\><rsub|2>> be two derivations on <math|A>
then if we define <math|[\<Delta\><rsub|1>,\<Delta\><rsub|2>]=\<Delta\><rsub|1>\<circ\>\<Delta\><rsub|2>-\<Delta\><rsub|2>\<circ\>\<Delta\><rsub|1>>.
<math|[\<Delta\><rsub|1>,\<Delta\><rsub|2>]> is then also a derivation if
<math|A> is commutative
</theorem>
<\proof>
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,x,y\<in\>A> we have
<\enumerate>
<item><math|[\<Delta\><rsub|1>,\<Delta\><rsub|2>](\<alpha\>.x+\<beta\>.y)=\<Delta\><rsub|1>(\<Delta\><rsub|2>(\<alpha\>.x+\<beta\>.y))-\<Delta\><rsub|2>(\<Delta\><rsub|1>(\<alpha\>.x+\<beta\>.y))=\<Delta\><rsub|1>(\<alpha\>.\<Delta\><rsub|2>(x)+\<beta\>.\<Delta\><rsub|2>(y))-\<Delta\><rsub|2>(\<alpha\>.\<Delta\><rsub|1>(x)+\<beta\>.\<Delta\><rsub|1>(y))=\<alpha\>.\<Delta\><rsub|1>(\<Delta\><rsub|2>(x))+\<beta\>.\<Delta\><rsub|1>(\<Delta\><rsub|2>(y))-\<alpha\>.\<Delta\><rsub|2>(\<Delta\><rsub|1>(x))-\<beta\>.\<Delta\><rsub|2>(\<Delta\><rsub|1>(y))=\<alpha\>(\<Delta\><rsub|1>(\<Delta\><rsub|2>(x))-\<Delta\><rsub|2>(\<Delta\><rsub|1>(x)))+\<beta\>(\<Delta\><rsub|1>(\<Delta\><rsub|2>(y))-\<Delta\><rsub|2>(\<Delta\><rsub|1>(y)))=\<alpha\>[\<Delta\><rsub|1>,\<Delta\><rsub|2>](x)+\<beta\>[\<Delta\><rsub|1>,\<Delta\><rsub|2>]>
<item><math|[\<Delta\><rsub|1>,\<Delta\><rsub|2>](x.y)=\<Delta\><rsub|1>(\<Delta\><rsub|2>(x.y))-\<Delta\><rsub|2>(\<Delta\><rsub|1>(x.y))=\<Delta\><rsub|1>(x.\<Delta\><rsub|2>(y)+\<Delta\><rsub|2>(x).y)-\<Delta\><rsub|2>(x.\<Delta\><rsub|1>(y)+\<Delta\><rsub|1>(x).y)=\<Delta\><rsub|1>(x).\<Delta\><rsub|2>(y)+x.\<Delta\><rsub|1>(\<Delta\><rsub|2>(y))+\<Delta\><rsub|2>(x).\<Delta\><rsub|1>(y)+\<Delta\><rsub|1>(\<Delta\><rsub|2>(x)).y-\<Delta\><rsub|2>(x).\<Delta\><rsub|1>(y)-x.\<Delta\><rsub|2>(\<Delta\><rsub|1>(y))-\<Delta\><rsub|1>(x).\<Delta\><rsub|2>(y)-\<Delta\><rsub|2>(\<Delta\><rsub|1>(x)).y=x.\<Delta\><rsub|1>(\<Delta\><rsub|2>(y))-x.\<Delta\><rsub|2>(\<Delta\><rsub|1>(y))+y.\<Delta\><rsub|1>(\<Delta\><rsub|2>(x))-y.\<Delta\><rsub|2>(\<Delta\><rsub|1>(x))=x.[\<Delta\><rsub|1>,\<Delta\><rsub|2>](y)+y.[\<Delta\><rsub|1>,\<Delta\><rsub|2>](x)>
(note that commutativity is essential here)
</enumerate>
</proof>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold modelled on
<math|\<bbb-R\><rsup|n>> of class <math|C<rsup|\<infty\>>> and let
<math|V:M\<rightarrow\>T<rsub|M>> be a vector field differentiable of
class <math|C<rsup|\<infty\>>> define then the function
<math|V:C<rsup|\<infty\>>(M)\<rightarrow\>C<rsup|\<infty\>>(M)> defined
by <math|f\<rightarrow\>V[f]> is well defined
(<math|V[f]\<in\>C<rsup|\<infty\>>>) and forms a derivation in
<math|C<rsup|\<infty\>>(M)>.<math|>\
</theorem>
<\proof>
Let <math|r\<in\>\<bbb-N\>> then <math|V> is differentiable of class
<math|C<rsup|r+1>> and from the previous theorem
<reference|differentiable vector fields property one> we have then that
<math|V[f]> is differentiable of class <math|C<rsup|r>> so <math|V[f]> is
differentiable of class <math|C<rsup|\<infty\>>>. To proof that it is a
derivation (see <reference|derivation>) note that
<\enumerate>
<item><math|\<forall\>p\<in\>M \ we have
V[\<alpha\>.f+\<beta\>.g](p)=\<pi\><rsub|2>\<circ\>V(p)[\<alpha\>.f+\<beta\>.g]\<equallim\><rsub|<reference|directional
derivative>,\<pi\><rsub|2>\<circ\>V(p)=\<bbb-T\>[c]><frac|d((\<alpha\>.f+\<beta\>.g)\<circ\>c)(t)|dt>(0)=\<alpha\><frac|d(f\<circ\>c)(t)|dt>+\<beta\><frac|d(g\<circ\>c)(t)|dt>=\<alpha\>.\<pi\><rsub|2>\<circ\><frac|d(f\<circ\>c)(t)|dt>+\<beta\>.\<pi\><rsub|2>\<circ\><frac|d(g\<circ\>c)(t)|dt>=\<alpha\>.V(p)[f]+\<beta\>.V(p)[g]=\<alpha\>.V[f](p)+\<beta\>.V[g](p)=(\<alpha\>.V[f]+\<beta\>.V[g])(p)\<Rightarrow\>V[\<alpha\>.f+\<beta\>.g]=\<alpha\>.V[f]+\<beta\>.V[g]>
<item><math|\<forall\>p\<in\>M> we have
<math|V[f\<bullet\>g](p)=\<pi\><rsub|2>\<circ\>V(p)[f\<bullet\>g]\<equallim\><rsub|\<pi\><rsub|2>\<circ\>V(p)=\<bbb-T\>[c]><frac|d((f.g)\<circ\>c)(t)|dt>(0)=<frac|d((f\<circ\>c).(g\<circ\>c)(t)|dt>(0)=(f\<circ\>c)(0)<frac|d(g\<circ\>c)(t)|dt>(0)+(g\<circ\>c)<frac|d(f\<circ\>c)(t)|dt>(0)=f(p)<frac|d(g\<circ\>c)(t)|dt>(0)+g(p)<frac|d(f\<circ\>c)(t)|dt>(0)=f(p)(\<pi\><rsub|2>\<circ\>V(p))[g]+g(p)(\<pi\><rsub|2>\<circ\>V(p))[f]=f(p)V[g](p)+g(p)V[f](p)=(f\<bullet\>V[g])(p)+(g\<bullet\>V[f])(p)=(f\<bullet\>V[g]+g\<bullet\>V[f])(p)\<Rightarrow\>V[f\<bullet\>g]=f\<bullet\>V[g]+g\<bullet\>V[f]>
</enumerate>
Concluding the proof of our theorem.<math|>
</proof>
<\note>
<label|vector field represents two functions>Given a vector field
<math|V:M\<rightarrow\>T<rsub|M>> then we can consider this as two
functions
<\enumerate>
<item><math|V:M\<rightarrow\>T<rsub|M>>
<item><math|V:C<rsup|\<infty\>>(M)\<rightarrow\>C<rsup|\<infty\>>(M)>\
</enumerate>
To avoid confusion we use <math|V[f]> in case (2) and <math|V(p)> in case
(1) but in both cases we can consider them as a function application.
</note>
<\definition>
<label|set of derivations on smooth functions><index|<math|\<cal-D\>(M)>>Let
\ <math|(M,\<cal-A\>)> be a differentiable manifold modelled on
<math|\<bbb-R\><rsup|n>> of class <math|C<rsup|\<infty\>>> then we define
<math|\<cal-D\><rsup|\<infty\>>(M)={\<Delta\>\|
\<Delta\>:C<rsup|\<infty\>>(M)\<rightarrow\>C<rsup|\<infty\>>(M) is a
derivation}\<subseteq\>Hom(C<rsup|\<infty\>>(M),C<rsup|\<infty\>>(M))>.
</definition>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold modelled on
<math|\<bbb-R\><rsup|n>> of class <math|C<rsup|\<infty\>>> then
<math|\<cal-D\><rsup|\<infty\>>(M),+,.> forms a real vector space. Here
<math|+,.> are defined on the usual way via pointwise operators (if
<math|\<Delta\><rsub|1>,\<Delta\><rsub|2>\<in\>\<cal-D\>(M)> then
<math|\<Delta\><rsub|1>+\<Delta\><rsub|2>:C<rsup|\<infty\>>(M)\<rightarrow\>C<rsup|\<infty\>>(M)>
is defined by <math|(\<Delta\><rsub|1>+\<Delta\><rsub|2>)(f)=\<Delta\><rsub|1>(f)+\<Delta\><rsub|1>(f)>
and if <math|\<alpha\>\<in\>\<bbb-R\>,\<Delta\>\<in\>D(M)> then
<math|\<alpha\>.\<Delta\>> is defined by
<math|*\<alpha\>.\<Delta\>)(f)=\<alpha\>.\<Delta\>(f)>)
</theorem>
<\proof>
\;
Second note that <math|\<cal-D\><rsup|\<infty\>>(M)\<subseteq\>Hom(C<rsup|\<infty\>>(M),C<rsup|\<infty\>>(M))>
and the operators defined on <math|\<cal-D\><rsup|\<infty\>>(M)> are the
restricted vector operators on <math|M(C<rsup|\<infty\>>(M),C<rsup|\<infty\>>(M))>
(see <reference|hom(X,Y)> and the fact that <math|C<rsup|\<infty\>>(M)>
beeing a algebra forms a vector space). Using <reference|subspace> we
have proved our theorem if we can prove that
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>> and
<math|\<Delta\><rsub|1>,\<Delta\><rsub|2>\<in\>\<cal-D\><rsup|\<infty\>>(M)>
that <math|\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>\<in\>\<cal-D\><rsup|\<infty\>>(M)>.
To do this note that linearity follows from the fact that
<math|\<cal-D\><rsup|\<infty\>>(M)\<subseteq\>Hom(C<rsup|\<infty\>>(M),C<rsup|\<infty\>>(M))>
(which is a vector space) and the Leibnitz condition from (given
<math|f,g\<in\>C<rsup|\<infty\>>(M))>
<\eqnarray*>
<tformat|<table|<row|<cell|(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(f\<bullet\>g)>|<cell|=>|<cell|\<alpha\>.\<Delta\><rsub|1>(f\<bullet\>g)+\<beta\>.\<Delta\><rsub|2>(f\<bullet\>g)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>.(f\<bullet\>\<Delta\><rsub|1>(g)+\<Delta\><rsub|1>(f)\<bullet\>g)+\<beta\>.(f\<bullet\>\<Delta\><rsub|2>(g)+\<Delta\><rsub|2>(f)\<bullet\>g)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>.(f\<bullet\>\<Delta\><rsub|1>(g))+\<beta\>.(f\<bullet\>\<Delta\><rsub|2>(g))+\<alpha\>.(\<Delta\><rsub|1>(f)\<bullet\>g)+\<beta\>.(\<Delta\><rsub|2>(f)\<bullet\>g)>>|<row|<cell|>|<cell|=>|<cell|f\<bullet\>(\<alpha\>.\<Delta\><rsub|1>(g))+f\<bullet\>(\<beta\>.\<Delta\><rsub|2>(g))+(\<alpha\>.\<Delta\><rsub|1>(f))\<bullet\>g+(\<beta\>.\<Delta\><rsub|2>(f))\<bullet\>g>>|<row|<cell|>|<cell|=>|<cell|f\<bullet\>(\<alpha\>\<Delta\><rsub|1>+\<beta\>\<Delta\><rsub|2>)(g)+((\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(f))\<bullet\>g>>>>
</eqnarray*>
</proof>
<\definition>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>.
<math|\<forall\>p\<in\>M> define <math|\<varepsilon\><rsup|\<infty\>>(M,p)={f\<in\>\<cal-D\>(M,p):f
is differentiable of class C<rsup|\<infty\>> on its
domain}\<subseteq\>\<varepsilon\>(M,p)> (see <reference|definition of
germs> )
</definition>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>. Then
<math|\<varepsilon\><rsup|\<infty\>>(M,p),+,\<times\>,\<bullet\>> (see
for the definition at <reference|algebra of germs>) forms a real
commutative algebra
</theorem>
<\proof>
As <math|\<varepsilon\><rsup|\<infty\>>(M,p)\<subseteq\>\<varepsilon\>(M,p)>
and uses the restricted operators we must only prove the following:
<\enumerate>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
we have <math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>]>
where <math|f<rprime|'>,g<rprime|'>\<in\>\<cal-D\>(M,p)> and are
differentiable of class <math|C<rsup|\<infty\>>> on their domain and so
<math|\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>> is differentiable of
class <math|C<rsup|\<infty\>>> on the intersection of their domains and
thus <math|\<alpha\>.f+\<beta\>.g=\<equiv\><rsub|p>[\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>]\<in\>\<varepsilon\><rsup|\<infty\>>(M)>.
<item>Also if <math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>]>
then on the intersection of their domains
<math|f<rprime|'>\<bullet\>g<rprime|'>> is differentiable of class
<math|C<rsup|\<infty\>>>(see <reference|sum(product) of differentiable
functions between manifolds>) so that
<math|f\<bullet\>g=\<equiv\><rsub|p>[f<rprime|'>\<bullet\>g<rprime|'>]\<in\>\<varepsilon\><rsup|\<infty\>>(M)>
</enumerate>
</proof>
<\definition>
<index|derivation (local)><index|local derivation><label|local
derivation>Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>,
<math|p\<in\>M>. Then a function <math|\<Delta\>:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<bbb-R\>>
that fullfills the following:
<\enumerate>
<item><math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
we have <math|\<Delta\>(\<alpha\>.f+\<beta\>.g)=\<alpha\>.\<Delta\>(f)+\<beta\>.\<Delta\>(g)>
<item><math|\<forall\>f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)> we
have <math|\<Delta\>(f\<bullet\>g)=\<Delta\>(f).g(p)+f(p).\<Delta\>(g)>
</enumerate>
is called a local derivation at p (this is similar with the definition in
<reference|derivation>). The set of all the local derivations at p is
called <math|\<cal-D\><rsup|\<infty\>>(M,p)>
</definition>
<\theorem>
<label|constructing local derivation from derivation>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>,
<math|p\<in\>M> and <math|\<Delta\>:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<varepsilon\><rsup|\<infty\>>(M,p)>
is a derivation on <math|\<varepsilon\><rsup|\<infty\>>(M,p)<rsup|>> then
<math|\<Delta\>{p}:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<bbb-R\>>
defined by <math|\<Delta\>{p}(f)=\<Delta\>(f)(p)> is a local derivation
at <math|p>
</theorem>
<\proof>
Let <math|\<alpha\>,\<beta\>\<in\>\<bbb-R\>,f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
then we have\
<\enumerate>
<item><math|\<Delta\>{p}(\<alpha\>.f+\<beta\>.g)=\<Delta\>(\<alpha\>.f+\<beta\>.g)(p)=(\<alpha\>.\<Delta\>(f)+\<beta\>.\<Delta\>(g))(p)=\<alpha\>.\<Delta\>(f)(p)+\<beta\>.\<Delta\>(g)(p)=\<alpha\>.\<Delta\>{p}(f)+\<beta\>.\<Delta\>{p}(g)>
<item><math|\<Delta\>{p}(f\<bullet\>g)=\<Delta\>(f\<bullet\>g)(p)=(f\<bullet\>\<Delta\>(g)+\<Delta\>(f)\<bullet\>g)(p)=f(p).\<Delta\>(g)(p)+\<Delta\>(f)(p).g(p)=f(p).\<Delta\>{p}(g)+\<Delta\>{p}(f).g(p)>
</enumerate>
</proof>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>,
<math|p\<in\>M>. Then as <math|\<cal-D\><rsup|\<infty\>>(M,p)\<subseteq\>M(\<varepsilon\><rsup|\<infty\>>(M,p),\<bbb-R\>)>
(see <reference|vector space of mappings>) then
<math|\<cal-D\><rsup|\<infty\>>(M,p),+,.> forms a vector space where we
define the <math|+,.> in the usal pointwise way (meaning that
<math|\<plusassign\>\<boxplus\><rsub|\|\<cal-D\><rsup|\<infty\>>(M,p)\<times\>\<cal-D\><rsup|\<infty\>>(M,p)>,.\<odot\><rsub|\|\<bbb-R\>\<times\>\<cal-D\><rsup|\<infty\>>(M,p)>>)
</theorem>
<\proof>
By <reference|subspace> we only have to prove that
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,\<forall\>\<Delta\><rsub|1>,\<Delta\><rsub|2>\<in\>\<cal-D\><rsup|\<infty\>>(M,p)>
<math|\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>\<in\>\<cal-D\><rsup|\<infty\>>(M,p)>.
Now if <math|f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p),\<lambda\>,\<delta\>\<in\>\<bbb-R\>>
then we have
<\eqnarray*>
<tformat|<table|<row|<cell|(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(\<lambda\>.f+\<delta\>.g)>|<cell|=>|<cell|\<alpha\>\<Delta\><rsub|1>(\<lambda\>.f+\<delta\>.g)+\<beta\>\<Delta\><rsub|2>(\<lambda\>.f+\<delta\>.g)>>|<row|<cell|>|<cell|=>|<cell|\<lambda\>.\<alpha\>.\<Delta\><rsub|1>(f)+\<delta\>.\<alpha\>.\<Delta\><rsub|1>(g)+\<lambda\>.\<beta\>.\<Delta\><rsub|2>(f)+\<delta\>.\<beta\>.\<Delta\><rsub|2>(g)>>|<row|<cell|>|<cell|=>|<cell|\<lambda\>.(\<alpha\>.\<Delta\><rsub|1>(f)+\<beta\>.\<Delta\><rsub|2>(f))+\<delta\>.(\<alpha\>.\<Delta\><rsub|1>(g)+\<beta\>.\<Delta\><rsub|2>(g))>>|<row|<cell|>|<cell|=>|<cell|\<lambda\>.(\<alpha\>.\<Delta\><rsub|1>+\<Beta\>.\<Delta\><rsub|2>)(f)+\<beta\>.\<Delta\>(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(g)>>|<row|<cell|(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(f\<bullet\>g)>|<cell|=>|<cell|\<alpha\>.\<Delta\><rsub|1>(f\<bullet\>g)+\<beta\>.\<Delta\><rsub|2>(f\<bullet\>g)>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>.[g(p).\<Delta\><rsub|1>(f)+f(p).\<Delta\><rsub|1>(g)]+\<beta\>.[g(p).\<Delta\><rsub|2>(f)+f(p).\<Delta\><rsub|2>(g)]>>|<row|<cell|>|<cell|=>|<cell|g(p).[\<alpha\>.\<Delta\><rsub|1>(f)+\<beta\>.\<Delta\><rsub|2>(f)]+f(p).[\<alpha\>.\<Delta\><rsub|1>(g)+\<beta\>.\<Delta\><rsub|2>(g)]>>|<row|<cell|>|<cell|=>|<cell|g(p).(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(f)+f(p).(\<alpha\>.\<Delta\><rsub|1>+\<beta\>.\<Delta\><rsub|2>)(g)>>>>
</eqnarray*>
proving linearity and the leibnitz rule.
\;
</proof>
<\lemma>
<label|Differential of composition of function and a+tx>Let
<math|X,\<shortparallel\>\<shortparallel\><rsub|X>> and
<math|Y,\<shortparallel\>\<shortparallel\><rsub|Y>> be real normed
spaces, <math|U\<subseteq\>X> open, <math|f:U\<rightarrow\>Y>
differentiable of class <math|C<rsup|\<infty\>>> then if
<math|\<forall\>t\<in\>\<bbb-R\>,a\<in\>X> we define <math|g(t)> by\
<\eqnarray*>
<tformat|<table|<row|<cell|g(t):U\<rightarrow\>Y>|<cell|where>|<cell|x\<rightarrow\>g(t)(x)=f(a+t.x)>>>>
</eqnarray*>
then <math|g(t)> is differentiable of class <math|C<rsup|\<infty\>>> and
<math|\<forall\>n\<geqslant\>1> we have
<math|D<rsup|n>g=t<rsup|n>.(Df\<circ\>\<lambda\>(t))> where
<math|\<lambda\>(t)> is defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<lambda\>(t):U\<rightarrow\>Y>|<cell|where>|<cell|x\<rightarrow\>\<lambda\>(t)(x)=a+t.x>>>>
</eqnarray*>
</lemma>
<\proof>
\;
Note first that <math|><math|\<lambda\>(t)> is differentiable of class
<math|C<rsup|\<infty\>>> with <math|D(\<lambda\>(t))(x)> is defined by
<math|h\<rightarrow\>t.h>. Then as <math|g(t)=f\<circ\>\<lambda\>(t)>
then we have <math|\<forall\>x\<in\>U,\<forall\>h\<in\>X>
<\eqnarray*>
<tformat|<table|<row|<cell|Dg(t)(x)(h)>|<cell|=>|<cell|D(f\<circ\>\<lambda\>(t))(x)(h)>>|<row|<cell|>|<cell|=>|<cell|Df(\<lambda\>(t)(x))\<circ\>D(\<lambda\>(t))(x)(h)>>|<row|<cell|>|<cell|=>|<cell|Df(\<lambda\>(t)(x))(D(\<lambda\>(t))(x)(h))>>|<row|<cell|>|<cell|=>|<cell|Df(\<lambda\>(t)(x))(t.h)>>|<row|<cell|>|<cell|\<equallim\><rsub|Differenetial
is a linear mapping>>|<cell|t.Df(\<lambda\>(t)(x))(h)>>>>
</eqnarray*>
and so\
<\eqnarray*>
<tformat|<table|<row|<cell|Dg(t)>|<cell|=>|<cell|t.Df\<circ\>\<lambda\>(t)>>>>
</eqnarray*>
We prove the rest of the theorem by induction
<\enumerate>
<item>Case <math|n=1> This is what we just have proved,
<item>Assume the theorem is true for <math|n> and prove it for
<math|n+1>. We have then\
<\eqnarray*>
<tformat|<table|<row|<cell|D<rsup|n+1>g(t)>|<cell|=>|<cell|D(D<rsup|n>g(t))>>|<row|<cell|>|<cell|>|<cell|D(t<rsup|n<rsub|>>.(D<rsup|n>g\<circ\>\<lambda\>(t)))>>>>
</eqnarray*>
</enumerate>
from which it follows that\
<\eqnarray*>
<tformat|<table|<row|<cell|D<rsup|n+1>g(t)(x)>|<cell|=>|<cell|D(t<rsup|n>.(D<rsup|n>(f\<circ\>\<lambda\>(t))(x))>>|<row|<cell|>|<cell|=>|<cell|t<rsup|n>.D<rsup|n+1>(f)(\<lambda\>(t)(x))\<circ\>D(\<lambda\>(t))(x)>>>>
</eqnarray*>
and thus <math|\<forall\>h\<in\>X> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|D<rsup|n+1>g(t)(x)(h)>|<cell|=>|<cell|t<rsup|n>.D<rsup|n+1>(f)(\<lambda\>(t)(x)(t.h)>>|<row|<cell|>|<cell|=>|<cell|t<rsup|n+1>.D<rsup|n+1>(f)(\<lambda\>(t)(x))(h)>>>>
</eqnarray*>
from which we have
<\eqnarray*>
<tformat|<table|<row|<cell|D<rsup|n+1>g(t)>|<cell|=>|<cell|t<rsup|n+1>.(D<rsup|n+1>f\<circ\>\<lambda\>(t))>>>>
</eqnarray*>
proving our lemma.
\;
</proof>
<\lemma>
<label|local derivation of constant is 0>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|\<infty\>>> modelled on
<math|\<bbb-R\><rsup|n>>,<math|p\<in\>M> and
<math|\<Delta\>:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<bbb-R\>>
a local derivation at p then if <math|f\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
is such that <math|\<exists\>a\<in\>\<bbb-R\>,\<exists\>U\<subseteq\>M,U>
open \ with <math|p\<in\>U> and <math|\<forall\>q\<in\>U> we have
<math|f(p)=a> then <math|\<Delta\>(f)=0>\
</lemma>
<\proof>
If <math|f\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)> fullfills the
condition of the lemma then <math|f=\<equiv\><rsub|p>[f<rprime|'>]> with
<math|\<forall\>q\<in\>U> we have that <math|f<rprime|'>(q)=f(q)=a>, then
if <math|1\<in\>C<rsup|\<infty\>>(M)> is defined by
<math|1<rsub|c>:M\<rightarrow\>\<bbb-R\>:x\<rightarrow\>1> we have
<math|\<forall\>q\<in\>U> that <math|f<rprime|'>(q)=a.1<rsub|c>(q)\<Rightarrow\>f<rprime|'>\<equiv\><rsub|p>a.1<rsub|c>\<Rightarrow\>f=\<equiv\><rsub|p>[f<rprime|'>]=\<equiv\><rsub|p>[a.1<rsub|c>]>
and thus <math|\<Delta\>(f)=\<Delta\>(a.1<rsub|c>)=a.\<Delta\>(1<rsub|c>)>.
Now <math|\<Delta\>(1<rsub|c>)=\<Delta\>(1<rsub|c>.1<rsub|c>)=1<rsub|c>(p)\<Delta\>(1<rsub|c>)+1<rsub|c>(p)\<Delta\>(1<rsub|c>)=\<Delta\><rsub|>(1<rsub|c>)+\<Delta\>(1<rsub|c>)\<Rightarrow\>\<Delta\>(1<rsub|c>)=0\<Rightarrow\>\<Delta\>(f)=0>
<math|>
</proof>
<\lemma>
<label|expansion of function that is infinite differentiable>Given
<math|\<bbb-R\><rsup|n>,a\<in\>\<bbb-R\><rsup|n>>, <math|U> open in
<math|\<bbb-R\><rsup|n>> such that <math|a=(a<rsub|1>,\<ldots\>,a<rsub|n>)\<in\>U>
and <math|f:U\<rightarrow\>\<bbb-R\>> a function that is differentiable
of class <math|C<rsup|\<infty\>>> then there exists a
<math|\<delta\>\<gtr\>0> such that <math|a\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<subseteq\>U>
and there exists a set of functions <math|g<rsub|i>:B<rsub|\<shortparallel\>\<shortparallel\>)max>(a,\<delta\>)\<rightarrow\>\<bbb-R\>>
differentiable of class <math|C<rsup|\<infty\>>> such that
<math|\<forall\>x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<subseteq\>U>
we have that
<\enumerate>
<item><math|g<rsub|i>(a)=<frac|df|dx<rsub|i>>(a)>
<item><math|f(x)=f(a)+<big|sum><rsub|i=1><rsup|n>(x<rsub|i>-a<rsub|i>)g<rsub|i>(x)>
</enumerate>
</lemma>
<\proof>
\;
First as <math|U> is open and <math|a\<in\>U> there exists a
<math|\<delta\>\<gtr\>0> such that <math|a\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<subseteq\>U>.
Given <math|x\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)>
define <math|\<lambda\><rsub|x>:[0,1]\<rightarrow\>\<bbb-R\><rsup|n>> by
<math|t\<rightarrow\>a+t(x-a)> then <math|\<lambda\><rsub|x>> is
differentiable of class <math|C<rsup|\<infty\>>> (sum of a constant and
linear function), <math|\<lambda\><rsub|x>(0)=a,\<lambda\><rsub|x>(1)=x>
and <math|><math|\<shortparallel\>\<lambda\><rsub|x>(t)-a)\<shortparallel\><rsub|max>=\|t\|\<shortparallel\>x-a\<shortparallel\><rsub|max>\<leqslant\>\<shortparallel\>x-a\<shortparallel\><rsub|max>\<Rightarrow\>\<forall\>t\<in\>[0,1]>
we have <math|\<lambda\><rsub|x>(t)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)>.
From this it follows that <math|h:[0,1]\<rightarrow\>\<bbb-R\>> defined
by <math|f\<circ\>\<lambda\><rsub|x>> is differentiable on <math|[0,1]>
of class <math|C<rsup|\<infty\>>> (see <reference|Cr on arbitrary sets>)
and thus using <reference|composition of derivates on arbitrary sets> and
<reference|fundamental theorem of calculus> we have
<math|h(1)-h(0)=<big|int><rsub|0><rsup|1>h<rprime|'>(t)dt\<Rightarrow\>f(\<lambda\><rsub|x>(1))-f(\<lambda\><rsub|x>(0))=<big|int><rsub|0><rsup|1>h<rprime|'>(t)\<Rightarrow\>f(x)-f(a)=<big|int><rsub|0><rsup|1>h<rprime|'>(t)dt>.
Now as <math|h=(f\<circ\>\<lambda\><rsub|x>)<rsub|\|[0,t]>> we have by
<reference|differentiability on [a,b]> (last remark) that
<math|h<rprime|'>=[(f\<circ\>\<lambda\><rsub|x>)<rprime|'><rsub|>]<rsub|\|[0,1]>>
and as <math|(f\<circ\>\<lambda\><rsub|x>)<rprime|'>(t)=<frac|d(f\<circ\>\<lambda\><rsub|x>)|dx>(t)=D(f\<circ\>\<lambda\><rsub|x>)(t)(1)\<equallim\><rsub|chain
rule>Df(\<lambda\><rsub|x>(t))(x-a)\<equallim\><rsub|<reference|differential
of multiparameter function>><big|sum><rsub|i=1><rsup|n>(x<rsub|i>-a<rsub|i>)<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(x-a))>.
So <math|f(x)-f(a)=<big|int><rsub|0><rsup|1>(<big|sum><rsub|i=1><rsup|n>(x<rsub|i>-a<rsub|i>)<frac|df(t<rsub|12>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(x-a))dt\<equallim\><rsub|<reference|linearity
of integral>><big|sum><rsub|i=1><rsup|n>(x<rsub|i>-a<rsub|i>)<big|int><rsub|0><rsup|1><frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(x-a))dt>.
If we define then
<\eqnarray*>
<tformat|<table|<row|<cell|g<rsub|i>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<rightarrow\>\<bbb-R\>>|<cell|as>|<cell|x\<rightarrow\>g<rsub|i>(x)=<big|int><rsub|0><rsup|1><frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(x-a))dt>>|<row|<cell|>|<cell|we
have then>|<cell|>>|<row|<cell|f(x)>|<cell|=>|<cell|f(a)+<big|sum><rsub|i=1><rsup|n>(x<rsub|i>-a<rsub|i>)g<rsub|i>(x)>>|<row|<cell|>|<cell|note
that>|<cell|>>|<row|<cell|g<rsub|i>(a)>|<cell|=>|<cell|<big|int><rsub|0><rsup|1><frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(a-a))dt>>|<row|<cell|>|<cell|=>|<cell|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(0)<big|int><rsub|0><rsup|1>dt>>|<row|<cell|>|<cell|=>|<cell|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(0)>>>>
</eqnarray*>
So the only thing left to prove is that <math|g<rsub|i>> is
<math|C<rsup|\<infty\>>> to do this consider the function
<math|h<rsub|i>> defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|h<rsub|i>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<times\>[0,1]\<rightarrow\>\<bbb-R\>>|<cell|where
>|<cell|(x,t)\<rightarrow\><frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(a+t(x-a))>>>>
</eqnarray*>
then we have\
<\eqnarray*>
<tformat|<table|<row|<cell|g<rsub|i>(a)>|<cell|=>|<cell|<big|int><rsub|0><rsup|1>h<rsub|i>(x,t)dt>>>>
</eqnarray*>
If we consider the function <math|l> defined by\
<\eqnarray*>
<tformat|<table|<row|<cell|l:B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<times\>[0,1]\<rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)>|<cell|where
>|<cell|(x,t)\<rightarrow\>a+t(x-a)>>|<row|<cell|so
that>|<cell|>|<cell|>>|<row|<cell|h<rsub|i>>|<cell|=>|<cell|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>\<circ\>l>>>>
</eqnarray*>
then <math|l> is continuous for if <math|(x,t),(x<rprime|'>,t<rprime|'>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<times\>[0,1]>
then
<\eqnarray*>
<tformat|<table|<row|<cell|l(h<rprime|'>,t<rprime|'>)-l(h,t)>|<cell|=>|<cell|a+t<rprime|'>(x<rprime|'>-a)-a-t(x-a)>>|<row|<cell|>|<cell|=>|<cell|t<rprime|'>x<rprime|'>-tx-a(t<rprime|'>-t)>>|<row|<cell|>|<cell|=>|<cell|t<rprime|'>x<rprime|'>-tx<rprime|'>+tx<rprime|'>-tx-a(t<rprime|'>-t)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|>>|<row|<cell|\<shortparallel\>l(h<rprime|'>,t<rprime|'>)-l(h,t)\<shortparallel\><rsub|max>>|<cell|\<leqslant\>>|<cell|\<shortparallel\>x<rprime|'><rsub|>\<shortparallel\><rsub|max>\|t<rprime|'>-t\|+\|t\|\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|max>+\<shortparallel\>a\<shortparallel\><rsub|max>\|t<rprime|'>-t\|>>|<row|<cell|>|<cell|as
\|t\|\<less\>1 and>|<cell|\<shortparallel\>x<rprime|'>\<shortparallel\><rsub|max>\<leqslant\>\<shortparallel\>x<rprime|'>-a\<shortparallel\><rsub|max>+\<shortparallel\>a\<shortparallel\>\<less\>\<delta\>+\<shortparallel\>a\<shortparallel\>
we have>>|<row|<cell|>|<cell|\<less\>>|<cell|(\<delta\>+2\<shortparallel\>a\<shortparallel\><rsub|max>)\|t<rprime|'>-t\|+\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|max>>>>>
</eqnarray*>
Given <math|\<varepsilon\>\<gtr\>0> <math|(x,t)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<times\>[0,1]>
then if we take <math|\<lambda\>=min(\<delta\>,<frac|\<varepsilon\>|2(\<shortparallel\>a\<shortparallel\><rsub|max>+1)>,<frac|\<varepsilon\>|2>)>
then we have if <math|max{\<shortparallel\>x<rprime|'>-x\<shortparallel\><rsub|max>\<rfloor\>t<rprime|'>-t\|}=\<shortparallel\>(x<rprime|'>,t<rprime|'>)-(x,t)\<shortparallel\><rsub|max>\<less\>\<lambda\>>
that <math|\<shortparallel\>l(h<rprime|'>,t<rprime|'>)-l(h,t)\<shortparallel\><rsub|max>\<less\>\<varepsilon\>>
proving continuity.
From <reference|Cr and partial derivates> and <math|C<rsup|\<infty\>>> of
<math|f> it follows that <math|D<rsub|i>f> is differentiable of class
<math|C<rsup|\<infty\>>> and as using <reference|fv is differentiable>
<math|f<rsub|1>> is differentiable of class <math|C<rsup|\<infty\>>>
where <math|f<rsub|1>(x)=D<rsub|i>f(x)(1)=<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>(x)>
we have that <math|f<rsub|1>=<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>>
proving that <math|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>>
is <math|C<rsup|\<infty\>>>. From this it follows that
<math|<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>> is continuous
and thus also that <strong|<math|h<rsub|i>> is continuous> (as it is the
composition of two continuous functions.
Furthermore if we define <math|\<forall\>t\<in\>[0,1]
\ \ \ \ (h<rsub|i>)<rsub|2>(t)> as follows\
<\eqnarray*>
<tformat|<table|<row|<cell|(h<rsub|i>)<rsub|2>(t):B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<rightarrow\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)>|<cell|where>|<cell|x\<rightarrow\>(h<rsub|i>)<rsub|2>(t)(x)=h<rsub|i>(x,t)=<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n)>|d<rsub|t<rsub|i>>>(a+t(x-a))>>>>
</eqnarray*>
Now if we define <math|\<forall\>t\<in\>[0,1]> the mapping
<math|\<lambda\>(t):B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>):x\<rightarrow\>a+t(x-a)=(a-ta)+tx>
we have <math|(h<rsub|i>)<rsub|2>(t)=<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>\<circ\>\<lambda\>(t)>
and by the lemma <reference|Differential of composition of function and
a+tx> we have then that <math|(h<rsub|i>)<rsub|2>(t)> is differentiable
of class <math|C<rsup|\<infty\>>>and if <math|j\<geqslant\>1> that
<\eqnarray*>
<tformat|<table|<row|<cell|D<rsup|j>((h<rsub|i>)<rsub|2>(t))>|<cell|=>|<cell|t<rsup|j>.(D<rsup|j>(<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>)\<circ\>\<lambda\>(t))>>>>
</eqnarray*>
So if we define <math|(h<rsub|i>)<rsup|(j)>> as\
<\eqnarray*>
<tformat|<table|<row|<cell|(h<rsub|i>)<rsup|j>:B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)\<times\>[0,1]\<rightarrow\>\<bbb-R\>>|<cell|where
>|<cell|(x,t)\<rightarrow\>(h<rsub|i>)<rsup|j>=D<rsup|j>((h<rsub|i>)<rsub|2>(t))(x)=t<rsup|j>.(D<rsup|j>(<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>)(\<lambda\>(t)(x))>>>>
</eqnarray*>
By <reference|power function is continuous> then function
<math|.<rsup|j>:t\<rightarrow\>t<rsup|j>> is continuous and by the
continuity of <math|\<pi\><rsub|2>> we have that the function
<math|k=.<rsup|j>\<circ\>\<pi\><rsub|2>> is continuous and
<math|k(x,t)=t<rsup|j>>. Also as <math|\<lambda\>(t)(x)=l(x,t)> then we
have that
<\eqnarray*>
<tformat|<table|<row|<cell|(h<rsub|i>)<rsup|j>=>|<cell|k.(D<rsup|j>(<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>)\<circ\>l)>|<cell|>>>>
</eqnarray*>
which as <math|l> and <math|D<rsup|j>(<frac|df(t<rsub|1>,\<ldots\>,t<rsub|n>)|dt<rsub|i>>)>
are continuous is a product of continuous functions and thus
<math|(h<rsub|i>)<rsup|j>> is continuous. So we can apply the theorem
about differentiation under the integral sign (see
<reference|differentiation under the integral sign>) to prove that
<math|g<rsub|i>> is differentiable of class <math|C<rsup|\<infty\>>>
proving our theorem.
\
</proof>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>,
<math|p\<in\>M> and <math|\<Delta\>:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<bbb-R\>>
a local derivation at p then there exists a
<math|(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> and a
<math|(\<lambda\><rsub|1>,\<ldots\>,\<lambda\><rsub|n>)\<in\>\<bbb-R\><rsup|n>>
such that <math|\<Delta\>=<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>>
(see <reference|local vector as a function>).
</theorem>
<\proof>
As <math|p\<in\>M> there exists <math|(U<rprime|'>,\<varphi\><rprime|'>)\<in\>\<cal-A\>>
with <math|p\<in\>U>. Given <math|f\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
there exists a <math|f<rprime|'>\<in\>C<rsup|\<infty\>>(M)> such that
<math|f=\<equiv\><rsub|p>[f<rprime|'>]>. Then
<math|f<rsup|\<circ\>>=f<rprime|'>\<circ\>\<varphi\><rprime|'><rsup|-1>:\<varphi\><rprime|'>(U)\<rightarrow\>\<bbb-R\>>
is a function differentiable of class <math|C<rsup|\<infty\>>>. By the
previous lemma <reference|expansion of function that is infinite
differentiable> there exists a <math|\<delta\>\<gtr\>0> such that
<math|\<varphi\>(p)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(\<varphi\>(p),\<delta\>)\<subseteq\>U>
and there exists a set of functions <math|g<rsub|i>:B<rsub|\<shortparallel\>\<shortparallel\>)max>(\<varphi\>(p),\<delta\>)\<rightarrow\>\<bbb-R\>>
differentiable of class <math|C<rsup|\<infty\>>> such that
<math|\<forall\>x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(\<varphi\>(p),\<delta\>)\<subseteq\>U>
we have that
<\enumerate>
<item><math|g<rsub|i>(\<varphi\>(p))=<frac|df<rsup|\<circ\>>(x<rsub|1>,\<ldots\>,x<rsub|n>)|dx<rsub|i>>(\<varphi\>(p))>
<item><math|f<rsup|\<circ\>>(x)=f<rsup|\<circ\>>(\<varphi\>(p))+<big|sum><rsub|i=1><rsup|n>(x<rsub|i>-\<varphi\><rsub|i>(p))g<rsub|i>(x)>
</enumerate>
\ because <math|\<varphi\>> is a homeomorphism we have that
<math|W=\<varphi\><rsup|-1>(B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>))>
is open in <math|U> and thus open in <math|M>.\
We take then <math|(U,\<varphi\>)=(W<big|cap>U<rprime|'>,\<varphi\><rprime|'><rsub|\|W<big|cap>U<rprime|'>>)>.
If we define then <math|g<rprime|'><rsub|i>=g<rsub|i>\<circ\>\<varphi\>>
then we can translate (1) and (2) as if <math|q\<in\>U> so that
<math|x=\<varphi\>(q)\<in\>B<rsub|\<shortparallel\>\<shortparallel\><rsub|max>>(a,\<delta\>)>
<\enumerate>
<item><math|g<rprime|'><rsub|i>(p)=<frac|d(f\<circ\>\<varphi\><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>)|dx<rsub|i>>(\<varphi\>(p))=\<partial\><rsub|i>(\<varphi\>,p)[f<rprime|'>]=\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>(f)>
<item><math|f<rprime|'>(q)=f<rprime|'>\<circ\>\<varphi\><rsup|-1>(\<varphi\>(p))=f<rsup|\<circ\>>(x)=f<rprime|'>\<circ\>\<varphi\><rsup|-1>(\<varphi\>(p))+<big|sum><rsub|i=1><rsup|n>(\<varphi\><rsub|i>(q)-\<varphi\><rsub|i>(p))g<rsub|i>(\<varphi\>(q))=f<rprime|'>(p)+<big|sum><rsub|i=1><rsup|n>(\<varphi\><rsub|i>(q)-\<varphi\><rsub|i>(p))g<rprime|'><rsub|i>(q)=f<rprime|'>(p)+<big|sum><rsub|i=1><rsup|n>[\<varphi\><rsub|i>(q)g<rprime|'><rsub|i>(q)-\<varphi\><rsub|i>(p)g<rprime|'><rsub|i>(q)]>
</enumerate>
\ then as we take <math|\<equiv\><rsub|p>[g<rprime|'><rsub|i>]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p),\<equiv\><rsub|p>[\<varphi\><rsub|i>]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
and given a <math|z\<in\>U> <math|[z]=\<equiv\><rsub|p>[y\<rightarrow\>z]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
we have\
<\eqnarray*>
<tformat|<table|<row|<cell|f>|<cell|=>|<cell|\<equiv\><rsub|p>[f<rprime|'>]>>|<row|<cell|>|<cell|=>|<cell|[f<rprime|'>(p)]+<big|sum><rsub|i=1><rsup|n>[(\<equiv\><rsub|p>[\<varphi\><rsub|i>])\<bullet\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])-\<varphi\><rsub|i>(p)(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])]>>>>
</eqnarray*>
and thus\
<\eqnarray*>
<tformat|<table|<row|<cell|\<Delta\>(f)>|<cell|=>|<cell|\<Delta\>[f<rprime|'>(p)]+<big|sum><rsub|i=1><rsup|n>{\<Delta\>((\<equiv\><rsub|p>[\<varphi\><rsub|i>])\<bullet\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>]))-\<varphi\><rsub|i>(p)\<Delta\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])}>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|local
derivation of constant is 0>>>|<cell|0+<big|sum><rsub|i=1><rsup|n>{\<Delta\>(\<equiv\><rsub|p>[\<varphi\><rsub|i>])(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])(p)+(\<equiv\><rsub|p>[\<varphi\><rsub|i>](p))\<Delta\>(\<equiv\><rsub|p>[g<rprime|'>])-\<varphi\><rsub|i>(p)\<Delta\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])}>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|i=1><rsup|n>{\<Delta\>(\<equiv\><rsub|p>[\<varphi\><rsub|i>]).g<rprime|'><rsub|i>(p)+\<varphi\><rsub|i>(p)\<Delta\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])-\<varphi\><rsub|i>(p)\<Delta\>(\<equiv\><rsub|p>[g<rprime|'><rsub|i>])}>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|i=1><rsup|n>\<Delta\>(\<equiv\><rsub|p>[\<varphi\><rsub|i>])\<Delta\><rsub|\<partial\><rsub|i>[\<varphi\>,p]>(f)>>|<row|<cell|>|<cell|=>|<cell|<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>(f)>>>>
</eqnarray*>
\ if we define <math|\<lambda\><rsub|i>=\<Delta\>(\<equiv\><rsub|p>[\<varphi\><rsub|i>])>
so that
<\eqnarray*>
<tformat|<table|<row|<cell|\<Delta\>>|<cell|=>|<cell|<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>>>>>
</eqnarray*>
as we must prove.
</proof>
<\definition>
<index|<math|\<Omega\><rsub|p>>>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|\<infty\>>> modelled on
<math|\<bbb-R\><rsup|n>>, <math|p\<in\>M> then we define the mapping
<math|\<Omega\><rsub|p>> by\
<\eqnarray*>
<tformat|<table|<row|<cell|\<Omega\><rsub|p>:T<rsub|p>M\<rightarrow\>\<cal-D\><rsup|\<infty\>>(M,p)>|<cell|where>|<cell|v<rsub|p>\<rightarrow\>\<Delta\><rsub|v<rsub|p>>
(see <reference|local vector as a function>,<reference|local vector as
a function is a derivation>)>>>>
</eqnarray*>
</definition>
<\theorem>
<index|<math|\<Omega\><rsub|p>>><label|set of local vectors is isomorph
with set of local derivations>Let <math|(M,\<cal-A\>)> be a
differentiable manifold of class <math|C<rsup|\<infty\>>> modelled on
<math|\<bbb-R\><rsup|n>>, <math|p\<in\>M> then
<math|\<Omega\><rsub|p>:T<rsub|p>M\<rightarrow\>\<cal-D\><rsup|\<infty\>>(M,p)>
is a isomorphism.
</theorem>
<\proof>
\;
First we prove that <math|\<Omega\><rsub|p>> is linear, as
<math|\<Omega\><rsub|p>(\<alpha\>.v<rsub|p>+\<beta\>.w<rsub|p>)=\<Delta\><rsub|\<alpha\>.v<rsub|p>+\<beta\>.w<rsub|p>>>
and thus if <math|f\<in\>\<cal-D\><rsup|\<infty\>>(M,p)> we have
<math|(\<alpha\>.v<rsub|p>+\<beta\>.\<omega\><rsub|p>)[f]\<equallim\><rsub|<reference|directional
derivative is linear>>\<alpha\>v<rsub|p>[f]+\<beta\>w<rsub|p>[f]=\<alpha\>\<Omega\><rsub|p>(v<rsub|p>)(f)+\<beta\>.\<Omega\><rsub|p>(w<rsub|p>)(f)>
from which it follows that <math|\<Omega\>(\<alpha\>,v<rsub|p>+\<beta\>.w<rsub|p>)=\<alpha\>.\<Omega\><rsub|p>(v<rsub|p>)+\<beta\>.\<Omega\><rsub|p>(w<rsub|p>)>.
Second we prove that <math|\<Omega\><rsub|p>> is a bijection
<\enumerate>
<item>(Injectivity) If <math|\<Omega\><rsub|p>(v<rsub|p>)=\<Omega\><rsub|p>(w<rsub|p>)\<Rightarrow\>\<Delta\><rsub|v<rsub|p>>=\<Delta\><rsub|w<rsub|p>>>
let <math|(U,\<varphi\>)\<in\>\<cal-A\>> with <math|p\<in\>U> then for
<math|\<varphi\><rsub|i>=\<pi\><rsub|i>\<circ\>\<varphi\>> we have if
<math|x=(x<rsub|1>,\<ldots\>,x<rsub|n>)\<in\>\<varphi\>(U)> we have
<math|\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>)=\<pi\><rsub|i>(\<varphi\>(\<varphi\><rsup|-1>(x<rsub|1>,\<ldots\>,x<rsub|n>)=x<rsub|i>\<Rightarrow\>\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>=\<pi\><rsub|i>>
which is linear and continuous so that it is differential of class
<math|C<rsup|\<infty\>> (and thus also and ><math|\<varphi\><rsub|i>>
is differential of class <math|C<rsup|\<infty\>>>,
<math|D(\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>)=D(\<pi\><rsub|i>)=\<pi\><rsub|i>>
take now <math|\<equiv\><rsub|p>[\<varphi\><rsub|i>]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
then <math|>\
<\eqnarray*>
<tformat|<table|<row|<cell|\<Delta\><rsub|v<rsub|p>>(\<equiv\><rsub|p>[\<varphi\><rsub|i>])=\<Delta\><rsub|w<rsub|p>>(\<equiv\><rsub|p>[\<varphi\><rsub|i>])>|<cell|\<Rightarrow\>>|<cell|v<rsub|p>[\<varphi\><rsub|i>]=w<rsub|p>[\<varphi\><rsub|i>]>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|[D(\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>)(\<varphi\><rsub|>(p))\<circ\>T(\<varphi\>,p)](v<rsub|p>)=[D(\<varphi\><rsub|i>\<circ\>\<varphi\><rsup|-1>)(\<varphi\><rsub|>(p))\<circ\>T(\<varphi\>,p)](w<rsub|p>)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<pi\><rsub|i>\<circ\>T(\<varphi\><rsub|>,p)(v<rsub|p>)=\<pi\><rsub|i>\<circ\>T(\<varphi\>,p)(w<rsub|p>)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|i\<in\>{1,\<ldots\>,n}>>|<cell|T(\<varphi\>,p)(v<rsub|p>)=T(\<varphi\>,p)(w<rsub|p>)>>|<row|<cell|>|<cell|\<Rightarrowlim\><rsub|T(\<varphi\>,p)
is isomorphism>>|<cell|v<rsub|p>=w<rsub|p>>>>>
</eqnarray*>
<item>(Surjectivity) Let <math|\<Delta\>\<in\>\<cal-D\><rsup|\<infty\>>(M,p)>
then by the previous theorem we have the existance of a
<math|(U,\<varphi\>)\<in\>\<cal-A\>,p\<in\>U> so that
<math|\<Delta\>=<big|sum><rsub|i=1><rsup|\<infty\>>\<lambda\><rsub|i>\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>>.
Take now <math|v<rsub|p>=<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<partial\><rsub|i>(\<varphi\>,p)>
then we have <math|\<Omega\><rsub|p>(v<rsub|p>)=<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<Omega\><rsub|p>(\<partial\><rsub|i>(\<varphi\>,p))=<big|sum><rsub|i=1><rsup|n>\<lambda\><rsub|i>\<Delta\><rsub|\<partial\><rsub|i>(\<varphi\>,p)>=\<Delta\>>
</enumerate>
\;
</proof>
<\definition>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>, and
<math|V:M\<rightarrow\>T<rsub|M>> be a vector field differentiable of
class <math|C<rsup|\<infty\>>>. Then if
<math|f\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)> where
<math|f=\<equiv\><rsub|p>[f<rprime|'>]> then <math|V[f]<rsub|p>> is
defined by <math|V[f]<rsub|p>=\<equiv\><rsub|p>[V[f<rprime|'>]]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>.
We define then <math|V[ ]<rsub|p>:\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<varepsilon\><rsup|\<infty\>>(M,p)>
by <math|f\<rightarrow\>V[ ]<rsub|p>(f)=V[f]<rsub|p>>
</definition>
<\proof>
We must prove that this definition is well defined and that is
independent of the choice of <math|f<rprime|'>>. First as
<math|f=\<equiv\><rsub|p>[f<rprime|'>]> we have that
<math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>> is differentiable
of class <math|C<rsup|\<infty\>>> and thus by <reference|differentiable
vector fields property one> <math|V[f<rprime|'>]:\<Omega\>\<rightarrow\>\<bbb-R\>>
is differentiable of class <math|C<rsup|\<infty\>>> from which it follows
that <math|\<equiv\><rsub|p>[V[f<rprime|'>]]> is defined and in
<math|\<varepsilon\><rsup|\<infty\>>(M,p)>. Now if
<math|f=\<equiv\><rsub|p>[f<rprime|'>]=\<equiv\><rsub|p>[f<rprime|''>]>
then <math|f<rprime|'>\<equiv\><rsub|p>f<rprime|''>> so there exists a
<math|W> open with <math|p\<in\>W> and <math|f<rprime|'>,f<rprime|''>>
have the same value in <math|W>. Then if <math|q\<in\>W> we have that
<math|V[f<rprime|'>](q)=(\<pi\><rsub|2>\<circ\>V(q))[f<rprime|'>]=\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(f)=(\<pi\><rsub|2>\<circ\>V(q))[f<rprime|'>]=V[f<rprime|''>](q)>
proving that <math|V[f<rprime|'>],V[f<rprime|''>]> has the same value in
<math|W> or that <math|V[f<rprime|'>]\<equiv\><rsub|p>V[f<rprime|''>]>
and thus <math|\<equiv\><rsub|p>[V[f<rprime|'>]]=\<equiv\><rsub|p>[V[f<rprime|''>]]>
</proof>
<\theorem>
<label|property of V1[V2]p]p>Let <math|(M,\<cal-A\>)> be a differentiable
manifold of class <math|C<rsup|\<infty\>>> modelled on
<math|\<bbb-R\><rsup|n>>, and <math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>>
be a vector field differentiable of class <math|C<rsup|\<infty\>>> then
we have <math|\<forall\>f=\<equiv\><rsub|p>[f<rprime|'>]\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
\ <math|V<rsub|1>[V<rsub|2>[f]<rsub|p>]<rsub|p>=V<rsub|1>[V<rsub|2>[f<rprime|'>]](p)>
</theorem>
<\proof>
If <math|f=\<equiv\><rsub|p>[f<rprime|'>]> then\
<\eqnarray*>
<tformat|<table|<row|<cell|V<rsub|1>[V<rsub|2>[f]<rsub|p>]<rsub|p>(p)>|<cell|=>|<cell|V<rsub|1>[\<equiv\><rsub|p>[V<rsub|2>[f<rprime|'>]]]<rsub|p>(p)>>|<row|<cell|>|<cell|=>|<cell|\<equiv\><rsub|p>[V<rsub|1>[V<rsub|2>[f<rprime|'>]]](p)>>|<row|<cell|>|<cell|=>|<cell|V<rsub|1>[V<rsub|2>[f<rprime|'>]](p)>>>>
</eqnarray*>
\;
</proof>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>, and
<math|V:M\<rightarrow\>T<rsub|M>> be a vector field differentiable of
class <math|C<rsup|\<infty\>>>. Then <math|V[ ]<rsub|p>> is a derivation
of <math|\<varepsilon\><rsup|\<infty\>>(M,p)>
</theorem>
<\proof>
<math|\<forall\>\<varepsilon\>,\<beta\>\<in\>\<bbb-R\>,f,g\<in\>\<varepsilon\><rsup|\<infty\>>(M,p)>
with <math|f=\<equiv\><rsub|p>[f<rprime|'>],g=\<equiv\><rsub|p>[g<rprime|'>]>
where <math|f<rprime|'>:U<rprime|'>\<rightarrow\>\<bbb-R\>,g<rprime|'>:V<rprime|'>\<rightarrow\>\<bbb-R\>>
with <math|p\<in\>U<rprime|'><big|cap>V<rprime|'>>
<\enumerate>
<item><math|\<forall\>q\<in\>U<rprime|'><big|cap>V<rprime|'>> we have
<math|V[\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>](q)=\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>)\<equallim\><rsub|<reference|local
vector as a function is a derivation>>\<alpha\>.\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(f<rprime|'>)+\<beta\>.\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(g<rprime|'>)=\<alpha\>.V[f<rprime|'>](q)+\<beta\>.V[g<rprime|'>](q)\<Rightarrow\>\<equiv\><rsub|p>[V[\<alpha\>.f<rprime|'>+\<beta\>.g<rprime|'>]]=\<equiv\><rsub|p>\<alpha\>.[V[f<rprime|'>]]+\<beta\>.[V[g<rprime|'>]]>
<item><math|\<forall\>q\<in\>U<rprime|'><big|cap>V<rprime|'>> we have
<math|V[f<rprime|'>\<bullet\>g<rprime|'>](q)=\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(f<rprime|'>\<bullet\>g<rprime|'>)\<equallim\><rsub|<reference|local
vector as a function is a derivation>>f<rprime|'>(q).\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(g<rprime|'>)+g<rprime|'>(q).\<Delta\><rsub|\<pi\><rsub|2>\<circ\>V(q)>(f<rprime|'>)=f<rprime|'>(q).V[g<rprime|'>](q)+g<rprime|'>(q).V[f<rprime|'>]\<Rightarrow\>\<equiv\><rsub|p>[V(f<rprime|'>\<bullet\>g<rprime|'>)]=f<rprime|'>\<bullet\>V[g<rprime|'>]+g<rprime|'>\<bullet\>V[f<rprime|'>]>
</enumerate>
</proof>
<\definition>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>, and
<math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>> be vector fields
differentiable of class <math|C<rsup|\<infty\>>> then we define
<math|[V<rsub|1>,V<rsub|2>]<rsub|p><rsub|>=V<rsub|1>[
]<rsub|p>\<circ\>V<rsub|2>[ ]<rsub|p>-V<rsub|2>[
]<rsub|p>\<circ\>V<rsub|1> []<rsub|p>> which is a function from
<math|\<varepsilon\><rsup|\<infty\>>(M,p)\<rightarrow\>\<varepsilon\><rsup|\<infty\>>(M,p)>
which by <reference|Lie product of derivations is a derivation on a
commutative algebra> is also a derivation.
</definition>
<\definition>
<index|Lie Product><label|Lie Product><dueto|Lie Product>Let
<math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>, and
<math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>> be vector fields
differentiable of class <math|C<rsup|\<infty\>>>. Then by
<reference|constructing local derivation from derivation> we have
<math|\<forall\>p\<in\>M> that <math|[V<rsub|1>,V<rsub|2>]<rsub|p>{p}> is
a local derivation and by <math|<reference|set of local vectors is
isomorph with set of local derivations>>
<math|\<Omega\><rsub|p><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|p>{p})> is a
vector element of <math|T<rsub|p>M>. This defines a new vector field
<math|[V<rsub|1>,V<rsub|2>]> as follows
<math|[V<rsub|1>,V<rsub|2>]:M\<rightarrow\>T<rsub|M>:p\<rightarrow\>(p,\<Omega\><rsub|p><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|p>{p})>,
called the Lie product of <math|V<rsub|1>,V<rsub|2>>
</definition>
<\theorem>
Let <math|(M,\<cal-A\>)> be a differentiable manifold of class
<math|C<rsup|\<infty\>>> modelled on <math|\<bbb-R\><rsup|n>>, and
<math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>> be vector fields
differentiable of class <math|C<rsup|\<infty\>>>, then
<\enumerate>
<item><math|\<forall\>f:U\<rightarrow\>\<bbb-R\>,U\<subseteq\>M> open
and <math|f> differentiable of class <math|C<rsup|\<infty\>>> we have
<math|[V<rsub|1>,V<rsub|2>][f]=V<rsub|1>[V<rsub|2>[f]]-V<rsub|2>[V<rsub|1>[f]]>
<item><math|[V<rsub|1>,V<rsub|2>]> is differentiable of class
<math|C<rsup|\<infty\>>>
</enumerate>
</theorem>
<\proof>
\;
<\enumerate>
<item>If <math|f:U\<rightarrow\>\<bbb-R\>, \ \ U> open in <math|M> a
differentiable function of class <math|C<rsup|\<infty\>>> then if
<math|q\<in\>U> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|[V<rsub|1>,V<rsub|2>][f](q)>|<cell|=>|<cell|(\<pi\><rsub|2>\<circ\>[V<rsub|1>,V<rsub|2>](q))[f]>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<Omega\><rsub|q><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|q>{q}))[f]>>|<row|<cell|>|<cell|=>|<cell|\<Omega\><rsub|q><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|q>{q}))[f]>>|<row|<cell|>|<cell|=>|<cell|v<rsub|q>[f]>>>>
</eqnarray*>
if we define <math|v<rsub|q>=\<Omega\><rsub|q><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|q>{q}\<in\>T<rsub|q>M>,
\ but then\
<\eqnarray*>
<tformat|<table|<row|<cell|v<rsub|q>[f]>|<cell|=>|<cell|\<Delta\><rsub|v<rsub|q>>(\<equiv\><rsub|q>[f])>>|<row|<cell|>|<cell|=>|<cell|\<Omega\><rsub|q>(v<rsub|q>)(\<equiv\><rsub|q>[f])>>|<row|<cell|>|<cell|=>|<cell|\<Omega\><rsub|q>(\<Omega\><rsub|q><rsup|-1>([V<rsub|1>,V<rsub|2>]<rsub|q>{q}))(\<equiv\><rsub|q>[f])>>|<row|<cell|>|<cell|=>|<cell|[V<rsub|1>,V<rsub|2>]<rsub|q>{q}(\<equiv\><rsub|q>[f])>>|<row|<cell|>|<cell|=>|<cell|((V<rsub|1>[
]<rsub|q> \<circ\>V<rsub|2>[ ]<rsub|q>-V<rsub|2>[
]<rsub|q>\<circ\>V<rsub|1>[ ]<rsub|q>)(\<equiv\><rsub|q>[f]))(q)>>|<row|<cell|>|<cell|=>|<cell|V<rsub|1>[V<rsub|2>[\<equiv\><rsub|q>[f]]<rsub|q>]<rsub|q>(q)-V<rsub|2>[V<rsub|1>[\<equiv\><rsub|q>[f]]<rsub|q>]<rsub|q>(q)>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|property
of V1[V2]p]p>>>|<cell|V<rsub|1>[V<rsub|2>[f]](q)-V<rsub|2>[V<rsub|1>[f]](q)<rsub|>>>|<row|<cell|>|<cell|=>|<cell|(V<rsub|1>[V<rsub|2>[f]]-V<rsub|2>[V<rsub|1>[f]])(q)>>>>
</eqnarray*>
proving as we have choosen <math|q> arbitrary in <math|> that
<math|[V<rsub|1>,V<rsub|2>][f]=V<rsub|1>[V<rsub|2>[f]]-V<rsub|2>[V<rsub|1>[f]]>
<item>If <math|f:U\<rightarrow\>\<bbb-R\>, \ \ U> open in <math|M> a
differentiable function of class <math|C<rsup|\<infty\>>> then we have
by <reference|differentiable vector fields property one> that
<math|V<rsub|1>[f],V<rsub|2>[f]> are differentiable of class
<math|C<rsup|\<infty\>>>, applying <reference|differentiable vector
fields property one> again proves that
<math|V<rsub|2>[V<rsub|1>[f]],V<rsub|1>[V<rsub|2>[f]]> are
differentiable of class <math|C<rsup|\<infty\>>> from which it follows
that <math|V<rsub|1>[V<rsub|2>[f]]-V<rsub|2>[V<rsub|1>[f]]> is
differentiable of class <math|C<rsup|\<infty\>>> so that using (1) we
have that <math|[V<rsub|1>,V<rsub|2>][f]> is differentiable of class
<math|C<rsup|\<infty\>>>. Then as we have choosen <math|f> arbitrarely
we can use <reference|differentiable vector fields property one> to
prove that <math|[V<rsub|1>,V<rsub|2>]> is differentiable of class
<math|C<rsup|\<infty\>>>.
</enumerate>
</proof>
<\theorem>
<label|Lie Product is a derivation>Let <math|(M,\<cal-A\>)> be a
differentiable manifold modelled on <math|\<bbb-R\><rsup|n>> of class
<math|C<rsup|\<infty\>>> and let <math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>>
be vector fields differentiable of class <math|C<rsup|\<infty\>>> then
<math|[V<rsub|1>,V<rsub|2>]> is a derivation in
<math|C<rsup|\<infty\>>(M)>
</theorem>
<\proof>
\;
<\enumerate>
<item><math|[V<rsub|1>,V<rsub|2>](\<alpha\>.f+\<beta\>.g)=V<rsub|1>[V<rsub|2>[\<alpha\>.f+\<beta\>.g]]-V<rsub|2>[V<rsub|1>[\<alpha\>.f+\<beta\>.g]]=V<rsub|1>[\<alpha\>.V<rsub|2>[f]+\<beta\>.V<rsub|2>[g]]-V<rsub|2>[\<alpha\>.V<rsub|1>[f]-\<beta\>.V<rsub|1>[g]]=\<alpha\>.V<rsub|1>[V<rsub|2>[f]]+\<beta\>.V<rsub|1>[V<rsub|2>[g]]-\<alpha\>.V<rsub|2>[V<rsub|1>[f]]-\<beta\>.V<rsub|2>[V<rsub|1>[g]]=\<alpha\>[V<rsub|1>,V<rsub|2>][f]+\<beta\>.[V<rsub|1>,V<rsub|2>][g]>
<item><math|[V<rsub|1>,V<rsub|2>][f\<bullet\>g]=V<rsub|1>[V<rsub|2>[f\<bullet\>g]]-V<rsub|2>[V<rsub|1>[f\<bullet\>g]]=V<rsub|1>[f\<bullet\>V<rsub|2>[g]+g\<bullet\>V<rsub|2>[f]]-V<rsub|2>[f\<bullet\>V<rsub|1>[g]+g\<bullet\>V<rsub|1>[f]]=f\<bullet\>V<rsub|1>[V<rsub|2>[g]]+V<rsub|2>[g]\<bullet\>V<rsub|1>[f]+g\<bullet\>V<rsub|1>[V<rsub|2>[f]]+V<rsub|2>[f]\<bullet\>V<rsub|1>[g]-f\<bullet\>V<rsub|2>[V<rsub|1>[g]]-V<rsub|1>[g]\<bullet\>V<rsub|2>[f]-g\<bullet\>V<rsub|2>[V<rsub|1>[f]]-V<rsub|1>[f]V<rsub|2>[g]=f\<bullet\>V<rsub|1>[V<rsub|2>[g]]+g\<bullet\>V<rsub|1>[V<rsub|2>[f]]-f\<bullet\>V<rsub|2>[V<rsub|1>[g]]-g\<bullet\>V<rsub|2>[V<rsub|1>[f]]=f\<bullet\>(V<rsub|1>[V<rsub|<rsub|2>>[g]]-V<rsub|2>[V<rsub|1>[g]])+g\<bullet\>(V<rsub|1>[V<rsub|2>[f]]-V<rsub|2>[V<rsub|1>[f]])=f\<bullet\>[V<rsub|1>,V<rsub|2>][g]+g\<bullet\>[V<rsub|1>,V<rsub|2>][f]>
</enumerate>
</proof>
<\definition>
<index|Lie algebra><label|Lie algebra><index|Jacobi's identity>Let
<math|A> be a real algebra together with a product
<math|[,]:A\<times\>A\<rightarrow\>A> such that <math|\<forall\>X,Y,Z>
and <math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>> we have
Bilinearity
<\enumerate>
<item><math|[\<alpha\>X+\<beta\>Y,Z]=\<alpha\>[X,Z]+\<beta\>[Y,Z]>
<item><math|[X,\<alpha\>Y+\<beta\>Z]=\<alpha\>[X,Y]+\<beta\>[X,Z]>
<item>Anti-commutativity: <math|[X,Y]=\<um\>[Y,X]>
<item><math|Jacobi<rprime|'>s identity:
[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0>
</enumerate>
</definition>
then <math|A> is called a Lie algebra
<\theorem>
<index|Jacobi's identity><label|Jacobi's identity>Let
<math|(M,\<cal-A\>)> be a differentiable manifold modelled on
<math|\<bbb-R\><rsup|n>> of class <math|C<rsup|\<infty\>>> and let
<math|V<rsub|1>,V<rsub|2>:M\<rightarrow\>T<rsub|M>> be vector fields
differentiable of class <math|C<rsup|\<infty\>>> then
<math|\<cal-X\>[M]={V:V \ is a smooth vector field><math|}> is a real
vector space using <math|V<rsub|1>+V<rsub|2>:M\<rightarrow\>T<rsub|M>>
defined by <math|(V<rsub|1>+V<rsub|2>)(p)=V<rsub|1>(p)+V<rsub|2>(p)> and
<math|\<alpha\>.V<rsub|1>:M\<rightarrow\>T<rsub|M>> defined by
<math|(\<alpha\>V<rsub|1>)(p)=\<alpha\>.V<rsub|1>(p)> [using the
sum,product defined in <reference|topology and vectorspace on {p}XTpM>,
so if <math|V<rsub|1>(p)=(p,v<rsub|1>),V<rsub|2>(p)=(p,v<rsub|2>)\<Rightarrow\>V<rsub|1>(p)+V<rsub|2>(p)=(p,v<rsub|1>+v<rsub|2>),\<alpha\>.V<rsub|1>(p)=(p,\<alpha\>.v<rsub|1>)>].\
Using the vector space operators we have
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,\<forall\>V<rsub|1>,V<rsub|2>,V\<in\>\<cal-X\>[M],\<forall\>f,g\<in\>C<rsup|\<infty\>>(M)>
we have\
<\enumerate>
<item><math|(V<rsub|1>+V<rsub|2>)[f]=V<rsub|1>[f]+V<rsub|2>[f]>
<item><math|(\<alpha\>.V)[f]=\<alpha\>.(V[f])>
</enumerate>
Furthermore the Lie product satisies the following and forms thus a Lie
Algebra:
<\enumerate>
<item>Bilinearity
<\enumerate>
<item><math|[\<alpha\>X+\<beta\>Y,Z]=\<alpha\>[X,Z]+\<beta\>[Y,Z]>
<item><math|[X,\<alpha\>Y+\<beta\>Z]=\<alpha\>[X,Y]+\<beta\>[X,Z]>
</enumerate>
<item>Anti-commutativity: <math|[X,Y]=\<um\>[Y,X]>
<item><math|Jacobi<rprime|'>s identity:
[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0>
</enumerate>
</theorem>
<\proof>
\;
Note that <math|\<forall\>p\<in\>M,V<rsub|1>,V<rsub|2>,V\<in\>\<cal-X\>[M]>
we have that\
<\eqnarray*>
<tformat|<table|<row|<cell|\<pi\><rsub|2>\<circ\>(V<rsub|1>+V<rsub|2>)(p)>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<pi\><rsub|2>\<circ\>(V<rsub|1>(p)+V<rsub|2>(p)))>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<pi\><rsub|2>(p,\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p)))>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p))>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p)>>|<row|<cell|\<pi\><rsub|2>\<circ\>(\<alpha\>.V)(p)>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<pi\><rsub|2>(\<alpha\>.V(p)))>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<pi\><rsub|2>(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>V(p))))>>|<row|<cell|>|<cell|=>|<cell|\<pi\><rsub|2>(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>V(p)))>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>.(\<pi\><rsub|2>\<circ\>V(p))>>>>
</eqnarray*>
First we proof that <math|\<chi\>[M]> is a real vector space. So let
<math|\<alpha\>\<in\>\<bbb-R\>,V,V<rsub|1>,V<rsub|2>\<in\>\<cal-X\>[M]>
then we have
<\enumerate>
<item><math|V<rsub|1>+V<rsub|2>\<in\>\<cal-X\>[M]>. To proof this note
first that if <math|V<rsub|1>(p)=(p,v<rsub|1>),v<rsub|1>=\<pi\><rsub|2>\<circ\>V<rsub|1>(p)\<in\>T<rsub|p>M,V<rsub|2>(p)=(p,v<rsub|2>),v<rsub|2>=\<pi\><rsub|2>\<circ\>V<rsub|2>(p)\<in\>T<rsub|p>M>
then\
<\eqnarray*>
<tformat|<table|<row|<cell|(V<rsub|1>+V<rsub|2>)(p)>|<cell|=>|<cell|V<rsub|1>(p)+V<rsub|2>(p)>>|<row|<cell|>|<cell|=>|<cell|(p,v<rsub|1>+v<rsub|2>),v<rsub|1>+v<rsub|2>\<in\>T<rsub|p>M>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(V<rsub|1>+V<rsub|2>)(p)\<in\>T<rsub|M>
and \<pi\><rsub|1>\<circ\>(V<rsub|1>+V<rsub|2>)=i<rsub|M>>>>>
</eqnarray*>
Proving that <math|V<rsub|1>+V<rsub|2>> is a vector field. Next we
prove property (1) for the <math|+> operator. Given <math|p\<in\>M> we
have then by the definition of <math|[]> (see <reference|vector field
on a finite differential manifold>)
<\eqnarray*>
<tformat|<table|<row|<cell|(V<rsub|1>+V<rsub|2>)[f](p)>|<cell|=>|<cell|(\<pi\><rsub|2>\<circ\>(V<rsub|1>+V<rsub|2>)(p))[f]>>|<row|<cell|>|<cell|=>|<cell|(\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p))[f]>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|directional
derivative is linear>>>|<cell|(\<pi\><rsub|2>\<circ\>V<rsub|1>(p))[f]+(\<pi\><rsub|2>\<circ\>V<rsub|2>(p))[f]>>|<row|<cell|>|<cell|=>|<cell|V<rsub|1>[f](p)+V<rsub|2>[f](p)>>>>
</eqnarray*>
proving that <math|(V<rsub|1>+V<rsub|2>)[f]=V<rsub|1>[f]+V<rsub|2>[f]>
Finally we have to prove that <math|V<rsub|1>+V<rsub|2>> is
differentiable of class <math|C<rsup|\<infty\>>>. So take
<math|f:U\<subseteq\>M\<rightarrow\>\<bbb-R\>> a differentiable
function of class <math|C<rsup|\<infty\>>> then by
<reference|differentiable vector fields property one> we have that
<math|V<rsub|1>[f],V<rsub|2>[f]> are differentiable of class
<math|C<rsup|\<infty\>>\<Rightarrow\>(V<rsub|1>+V<rsub|2>)f=V<rsub|1>[f]+V<rsub|2>[f]>
is differentiable of class <math|C<rsup|\<infty\>>> and thus by
<reference|differentiable vector fields property one> again we have
that <math|V<rsub|1>+V<rsub|2>> is a differentiable vector field of
class <math|C<rsup|\<infty\>>.>\
<item><math|\<alpha\>.V\<in\>\<cal-X\>[M]>. Note that if
<math|V(p)=(p,v),v=\<pi\><rsub|2>\<circ\>V(p)\<in\>T<rsub|p>M>
<\eqnarray*>
<tformat|<table|<row|<cell|(\<alpha\>.V(p))>|<cell|=>|<cell|(p,\<alpha\>.v),\<alpha\>.v\<in\>T<rsub|p>M>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|\<alpha\>.V(p)\<in\>T<rsub|M>
and \<pi\><rsub|1>\<circ\>(\<alpha\>.V)=i<rsub|M>>>>>
</eqnarray*>
Proving that <math|\<alpha\>.V> is a vector field. Next we prove
property (2) for the . operator. Given <math|p\<in\>M> we have then\
<\eqnarray*>
<tformat|<table|<row|<cell|(\<alpha\>.V)[f](p)>|<cell|=>|<cell|(\<pi\><rsub|2>\<circ\>(\<alpha\>.V)(p))[f]>>|<row|<cell|>|<cell|=>|<cell|(\<alpha\>.(\<pi\><rsub|2>\<circ\>V(p)))[f]>>|<row|<cell|>|<cell|\<equallim\><rsub|<reference|directional
derivative is linear>>>|<cell|\<alpha\>.((\<pi\><rsub|2>\<circ\>V(p))[f])>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>.(V[f](p))>>>>
</eqnarray*>
Proving that <math|(\<alpha\>.V)[f]=\<alpha\>.V[p]>. Finally we have to
prove that <math|\<alpha\>.V> is differentiable of class
<math|C<rsup|\<infty\>>>. So take <math|f:U\<subseteq\>M\<rightarrow\>\<bbb-R\>>
a differentiable function of class <math|C<rsup|\<infty\>>> then by
<reference|differentiable vector fields property one> we have that
<math|V<rsub|>[f]> is differentiable of class
<math|C<rsup|\<infty\>>\<Rightarrow\>(\<alpha\>V)f=\<alpha\>V[f]> is
differentiable of class <math|C<rsup|\<infty\>>> and thus by
<reference|differentiable vector fields property one> again we have
that <math|\<alpha\>.V> is a differentiable vector field of class
<math|C<rsup|\<infty\>>.>
</enumerate>
(1) and (2) proves that <math|+,.> is well defined on
<math|\<cal-X\>[M]>.\
Now to proof the rest of the vector space axioms.\
<\enumerate>
<item>Abelian Group axions of <math|+>
<\enumerate>
<item><math|\<forall\>V<rsub|1>,V<rsub|2>,V<rsub|3>\<in\>\<cal-X\>[M],
\<forall\>p\<in\>M> we have
<\eqnarray*>
<tformat|<table|<row|<cell|((V<rsub|1>+V<rsub|2>)+V<rsub|3>)(p)>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>((V<rsub|1>+V<rsub|2>)+V<rsub|3>)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(V<rsub|1>+V<rsub|2>)(p)+\<pi\><rsub|2>\<circ\>V<rsub|3>(p))>>|<row|<cell|>|<cell|=>|<cell|(p,(\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p))+\<pi\><rsub|2>\<circ\>V<rsub|3>(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+(\<pi\><rsub|2>\<circ\>V<rsub|2>(p)+\<pi\><rsub|2>\<circ\>V<rsub|3>(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>(V<rsub|2>+V<rsub|3>)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(V<rsub|1>+(V<rsub|2>+V<rsub|3>))(p))>>|<row|<cell|>|<cell|=>|<cell|((V<rsub|1>+V<rsub|2>)+V<rsub|3>)(p)>>|<row|<cell|>|<cell|\<Rightarrow\>>|<cell|(V<rsub|1>+V<rsub|2>)+V<rsub|3>=V<rsub|1>+(V<rsub|2>+V<rsub|3>)>>>>
</eqnarray*>
<item>Define the vector field <math|0> by
<math|p\<rightarrow\>0(p)=(p,0)> (last <math|0> is zero of
<math|T<rsub|p>M>) then\
<\eqnarray*>
<tformat|<table|<row|<cell|\<forall\>(U,\<varphi\>)\<in\>\<cal-A\>
we have>|<cell|0<rsup|[\<varphi\>]>(p)>|<cell|=T(\<varphi\>,p)0(p)>>|<row|<cell|>|<cell|>|<cell|=
T(\<varphi\>,p)(0)>>|<row|<cell|>|<cell|>|<cell|=0>>>>
</eqnarray*>
Proving that <math|0<rsup|[\<varphi\>]> is the constant function >0
which is <math|C<rsup|\<infty\>>> so <math|0\<in\>\<cal-X\>[M]> (see
<reference|characterization of differentiable vector fields>).
Finally <math|\<forall\>V\<in\>\<cal-X\>[M],(U,\<varphi\>)\<in\>\<cal-A\>>
<\eqnarray*>
<tformat|<table|<row|<cell|(V+0)(p)>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(V+0)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V(p)+\<pi\><rsub|2>\<circ\>0(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V(p))>>|<row|<cell|>|<cell|=>|<cell|V(p)>>>>
</eqnarray*>
Proving the <math|V+0> (<math|0+V> follows then from commutativity
that is proved later)
<item>Let <math|V\<in\>\<cal-X\>[M]> take then
<math|\<um\>V=(\<um\>1).V\<in\>\<cal-X\>[M]> then
<math|\<forall\>p\<in\>M>
<\eqnarray*>
<tformat|<table|<row|<cell|(V+(\<um\>V))(p)>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(V+(\<um\>1).V)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V(p)+\<pi\><rsub|2>\<circ\>((\<um\>1).V)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,1.(\<pi\><rsub|2>\<circ\>V(p))+(\<um\>1).(\<pi\><rsub|2>\<circ\>V(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,(1+(\<um\>1)).(\<pi\><rsub|2>\<circ\>V(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,0.(\<pi\><rsub|2>\<circ\>V(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,0)>>|<row|<cell|>|<cell|=>|<cell|0(p)>>>>
</eqnarray*>
So <math|V+(-V)=0> (<math|(-V)+V)> follows then from commutativity
that is proved later).
<item>Let <math|V<rsub|1>,V<rsub|2>> then <math|\<forall\>p\<in\>M>\
<\eqnarray*>
<tformat|<table|<row|<cell|(V<rsub|1>+V<rsub|2>)(p)>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>V<rsub|2>(p)+\<pi\><rsub|2>\<circ\>V<rsub|1>(p))>>|<row|<cell|>|<cell|=>|<cell|(V<rsub|2>+V<rsub|1>)(p)>>>>
</eqnarray*>
So it follows that <math|V<rsub|1>+V<rsub|2>=V<rsub|2>+V<rsub|1>>
</enumerate>
<item>Remaining vector space axioms. We have
<math|\<forall\>\<alpha\>,\<beta\>\<in\>\<bbb-R\>,\<forall\>V<rsub|1>,V<rsub|2>,V\<in\>\<cal-X\>[M]>\
<\enumerate>
<item><math|\<alpha\>.(V<rsub|1>+V<rsub|2>)=\<alpha\>.V<rsub|1>+\<alpha\>.V<rsub|2>>.
This follows from, given <math|p\<in\>M>,
<\eqnarray*>
<tformat|<table|<row|<cell|(\<alpha\>.(V<rsub|1>+V<rsub|2>))(p)>|<cell|=>|<cell|(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>(V<rsub|1>+V<rsub|2>))(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>V<rsub|1>(p)+\<pi\><rsub|2>\<circ\>V<rsub|2>(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>V<rsub|1>(p))+\<alpha\>.(\<pi\><rsub|2>\<circ\>V<rsub|2>(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(\<alpha\>.V<rsub|1>)(p)+\<pi\><rsub|2>\<circ\>(\<alpha\>.V<rsub|2>)(p))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(\<alpha\>.V<rsub|1>+\<alpha\>.V<rsub|2>)(p))>>|<row|<cell|>|<cell|=>|<cell|(\<alpha\>.V<rsub|1>+\<alpha\>.V<rsub|2>)(p)>>>>
</eqnarray*>
<item><math|(\<alpha\>+\<beta\>).V<rsub|>=\<alpha\>.V<rsub|1>+\<beta\>.V<rsub|2>>.
This follows from, given <math|p\<in\>M,>
<\eqnarray*>
<tformat|<table|<row|<cell|((\<alpha\>+\<beta\>).V)(p)>|<cell|=>|<cell|(p,(\<alpha\>+\<beta\>).((\<pi\><rsub|2>\<circ\>V)(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<alpha\>.((\<pi\><rsub|2>\<circ\>V)(p))+\<beta\>.((\<pi\><rsub|2>\<circ\>V)(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(\<alpha\>.V)(p)+\<pi\><rsub|2>\<circ\>(\<beta\>.V)(p))>>|<row|<cell|>|<cell|=>|<cell|(\<alpha\>.V+\<beta\>.V)(p)>>>>
</eqnarray*>
<item><math|(\<alpha\>.\<beta\>).V=\<alpha\>.(\<beta\>.V)>. This
given <math|p\<in\>M> follows from
<\eqnarray*>
<tformat|<table|<row|<cell|((\<alpha\>.\<beta\>).V)(p)>|<cell|=>|<cell|(p,(\<alpha\>.\<beta\>).(\<pi\><rsub|2>\<circ\>V(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<alpha\>.(\<beta\>.(\<pi\><rsub|2>\<circ\>V(p))))>>|<row|<cell|>|<cell|=>|<cell|(p,\<alpha\>.(\<pi\><rsub|2>\<circ\>(\<beta\>.V)(p)))>>|<row|<cell|>|<cell|=>|<cell|(p,\<pi\><rsub|2>\<circ\>(\<alpha\>.(\<beta\>.V))(p))>>|<row|<cell|>|<cell|=>|<cell|(\<alpha\>.(\<beta\>.V))(p)>>>>
</eqnarray*>
</enumerate>
</enumerate>
\;
This concludes our proof that <math|\<cal-X\>[M]> is a vector space.
\;
Now to proof Lie Algebra axions, take <math|X,Y,Z\<in\>\<cal-X\>[M]>\
<\enumerate>
<item>Bilinearity
<\enumerate>
<item><math|\<forall\>f\<in\>C<rsup|\<infty\>>(M)> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|[\<alpha\>X+\<beta\>Y,Z](f)>|<cell|=>|<cell|(\<alpha\>X+\<beta\>Y)[Z[f]]-Z[(\<alpha\>X+\<beta\>Y)[f])>>|<row|<cell|>|<cell|\<equallim\>>|<cell|\<alpha\>X[Z[f]]+\<beta\>Y[Z[f]]-Z[\<alpha\>X[f]+\<beta\>Y[f]]>>|<row|<cell|>|<cell|\<equallim\><rsub|Z[.]
is a derivation>>|<cell|\<alpha\>X[Z[f]]+\<beta\>Y[Z[f]]-\<alpha\>Z[X[f]]-\<beta\>Z[Y[f]]>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(X[Z[f]]-Z[X[f]])+\<beta\>(Y[Z[f]]=Z[Y[f]])>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>[X,Z](f)+\<beta\>[Y,Z](f)>>>>
</eqnarray*>
giving that <math|[\<alpha\>X+\<beta\>Y,Z]=\<alpha\>[X,Z]+\<beta\>[Y,Z]>
<item><math|\<forall\>f\<in\>C<rsup|\<infty\>>(M)> we have\
<\eqnarray*>
<tformat|<table|<row|<cell|[X,\<alpha\>Y+\<beta\>Z](f)>|<cell|=>|<cell|X[(\<alpha\>Y+\<beta\>Z)[f]]-(\<alpha\>Y+\<beta\>Z)[X[f]]>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>X[Y[f]]+\<beta\>X[Z[f]]-\<alpha\>Y[X[f]]-\<beta\>Z[X[f]]>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>(X[Y[f]]-Y[X[f]])+\<beta\>(X[Z[f]]-Z[X[f]])>>|<row|<cell|>|<cell|=>|<cell|\<alpha\>[X,Y](f)+\<beta\>[X,Z](f)>>>>
</eqnarray*>
giving <math|[X,\<alpha\>Y+\<beta\>Z]=\<alpha\>[X,Y]+\<beta\>[X,Z]>
</enumerate>
<item>Anti-commutativity. <math|\<forall\>f\<in\>C<rsup|\<infty\>>(M)>
we have\
<\eqnarray*>
<tformat|<table|<row|<cell|[X,Y](f)>|<cell|=>|<cell|X[Y[f]]-Y[X[f]]>>|<row|<cell|>|<cell|=>|<cell|\<um\>(Y[X[f]]-X[Y[f]])>>|<row|<cell|>|<cell|=>|<cell|-[Y,X](f)>>>>
</eqnarray*>
proving the <math|[X,Y]=\<um\>[Y,X]>
<item>Jacobi's identity. <math|\<forall\>f\<in\>C<rsup|\<infty\>(M)>>
we have\
<\eqnarray*>
<tformat|<table|<row|<cell|([X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]])(f)>|<cell|=>|<cell|[X,[Y,Z]](f)+[Z,[X,Y]](f)+[Y,[Z,X]](f)>>|<row|<cell|>|<cell|=>|<cell|X[[Y,Z][f]]-[Y,Z][X[f]]+Z[[X,Y][f]]-[X,Y][Z[f]]+Y[[Z,X][f]]-[Z,X][Y[f]]>>|<row|<cell|>|<cell|=>|<cell|X[Y[Z[f]]-Z[Y[f]]]-Y[Z[X[f]]]+Z[Y[X[f]]]+Z[X[Y[f]]-Y[X[f]]]-X[Y[Z[f]]]+Y[X[Z[f]]]+Y[Z[X[f]]-X[Z[f]]]-Z[X[Y[f]]]+X[Z[Y[f]]]>>|<row|<cell|>|<cell|=>|<cell|X[Y[Z[f]]]<rsub|(1)>-X[Z[Y[f]]]<rsub|(2)>-Y[Z[X[f]]]<rsub|(3)>+Z[Y[X[f]]]<rsub|(4)>+Z[X[Y[f]]<rsub|(5)>-Z[Y[X[f]]]<rsub|(4)>-X[Y[Z[f]]]<rsub|(1)>+Y[X[Z[f]]]<rsub|(6)><rsub|>+Y[Z[X[f]]]<rsub|(3)>-Y[X[Z[f]]]<rsub|(6)>-Z[X[Y[f]]]<rsub|(5)>+X[Z[Y[f]]]<rsub|(2)>>>|<row|<cell|>|<cell|=>|<cell|0
>>>>
</eqnarray*>
</enumerate>
\;
</proof>
\;
\;
\;
<chapter|Notations>
\;
<tabular|<tformat|<table|<row|<cell|<math|\<vdash\>>>|<cell|with,
where>>|<row|<cell|<math|\<succ\>>>|<cell|fulfilling
>>|<row|<cell|<math|\<in\>>>|<cell|element
of>>|<row|<cell|<math|<big|cup>>>|<cell|union>>|<row|<cell|<math|<big|cap>>>|<cell|intersection>>|<row|<cell|<math|<big|prod>>>|<cell|product>>|<row|<cell|<math|*\<times\>>>|<cell|set
product or product of two sets (depending on
context)>>|<row|<cell|<math|\<bbb-R\>>>|<cell|set of real
numbers>>|<row|<cell|<math|\<bbb-R\><rsub|+>>>|<cell|set of positive real
numbers>>|<row|<cell|<math|\<bbb-R\><rsub|+,0>>>|<cell|set of positive real
numbers including <math|0>>>|<row|<cell|<math|\<bbb-C\>>>|<cell|set of
complex numbers>>|<row|<cell|<math|\<bbb-K\>>>|<cell|either
<math|\<bbb-R\>> or <math|\<bbb-C\>>>>|<row|<cell|<math|\<bbb-N\><rsub|0>>>|<cell|set
of positive integer's>>|<row|<cell|<math|\<bbb-N\>>>|<cell|set of integers
including 0>>|<row|<cell|<math|\<bbb-Z\>>>|<cell|set of
integers>>|<row|<cell|<math|\<bbb-Q\>>>|<cell|set of rational
numbers>>|<row|<cell|<math|\<wedge\>>>|<cell|and>>|<row|<cell|<math|\<vee\>>>|<cell|or>>|<row|<cell|<math|\<sim\>>>|<cell|not>>>>>
\;
\;
<\the-index|idx>
<index-1|<with|mode|math|C<rsup|r>> compatable
atlasses|<pageref|auto-295>>
<index-1|<with|mode|math|C<rsup|r>> compatible charts|<pageref|auto-293>>
<index-1|<with|mode|math|C<rsup|r>> on set|<pageref|auto-269>>
<index-1|<with|mode|math|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)>|<pageref|auto-152>>
<index-1|<with|mode|math|<frac|df(x)|dx>(y)>|<pageref|auto-258>>
<index-1|<with|mode|math|\<bbb-C\>>|<pageref|auto-107>>
<index-1|<with|mode|math|\<bbb-N\><rsub|0>>|<pageref|auto-63>>
<index-1|<with|mode|math|\<bbb-R\>>|<pageref|auto-58>>
<index-1|<with|mode|math|\<bbb-T\>[c]>|<pageref|auto-308>>
<index-1|<with|mode|math|\<bbb-Z\>>|<pageref|auto-65>>
<index-1|<with|mode|math|\<cal-C\><rsup|(r)>[\<cal-A\>]>|<pageref|auto-297>>
<index-1|<with|mode|math|\<cal-D\>(M)>|<pageref|auto-333>>
<index-1|<with|mode|math|\<cal-D\>(M,p)>|<pageref|auto-317>>
<index-1|<with|mode|math|\<cal-M\>(\<cal-A\>)>|<pageref|auto-298>>
<index-1|<with|mode|math|\<cal-P\>>|<pageref|auto-192>>
<index-1|<with|mode|math|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>>|<pageref|auto-239>>
<index-1|<with|mode|math|\<cal-T\><rsup|-1><rsub|f,Y>>|<pageref|auto-177>>
<index-1|<with|mode|math|\<Delta\><rsub|v<rsub|p>>>|<pageref|auto-321>,
<pageref|auto-323>, <pageref|auto-324>>
<index-1|<with|mode|math|\<delta\>[\<Delta\>,p]>|<pageref|auto-334>>
<index-1|<with|mode|math|\<equiv\><rsub|p>>|<pageref|auto-316>>
<index-1|<with|mode|math|\<mu\>(\<cal-P\>)>|<pageref|auto-241>>
<index-1|<with|mode|math|\<partial\><rsub|i>(p)>|<pageref|auto-313>>
<index-1|<with|mode|math|\<varepsilon\>(M,p)>|<pageref|auto-319>>
<index-1|<with|mode|math|\<varepsilon\>-mapping>|<pageref|auto-252>>
<index-1|<with|mode|math|<big|sum><rsub|a\<in\>A>a>|<pageref|auto-89>,
<pageref|auto-91>>
<index-1|abelian group|<pageref|auto-86>>
<index-1|absolute value|<pageref|auto-60>>
<index-1|absolute value (complex)|<pageref|auto-110>>
<index-1|accumulate (filterbase)|<pageref|auto-219>>
<index-1|accumulation point|<pageref|auto-134>>
<index-1|algebra|<pageref|auto-122>>
<index-1|algebra of germs|<pageref|auto-320>>
<index-1|atlas|<pageref|auto-292>>
<index-1|atlas of class <with|mode|math|C<rsup|r>>|<pageref|auto-294>>
<index-1|axiom of choice|<pageref|auto-44>>
<index-1|<with|mode|math|B<rsub|d>(x,\<varepsilon\>)>|<pageref|auto-150>>
<index-1|Baire Category Theorem|<pageref|auto-235>>
<index-1|Baire set|<pageref|auto-143>>
<index-1|Banach fixed point theorem|<pageref|auto-234>>
<index-1|Banach space|<pageref|auto-230>, <pageref|auto-232>>
<index-1|basis|<pageref|auto-117>>
<index-1|basis of <with|mode|math|\<bbb-R\><rsup|n>>|<pageref|auto-118>>
<index-1|basis (topology)|<pageref|auto-137>>
<index-1*|bijection>
<index-2|function|<pageref|auto-24>>
<index-1|bijective sets|<pageref|auto-29>>
<index-1|bounded set|<pageref|auto-155>>
<index-1|box topology|<pageref|auto-140>>
<index-1|<with|mode|math|C<rsup|1>>|<pageref|auto-264>>
<index-1|<with|mode|math|C<rsup|n>>|<pageref|auto-266>>
<index-1|<with|mode|math|C<rsup|\<infty\>>>|<pageref|auto-267>>
<index-1|<with|mode|math|C<rsup|r>>|<pageref|auto-302>>
<index-1|<with|mode|math|C<rsup|\<infty\>>(M)>|<pageref|auto-332>>
<index-1|Cauchy property|<pageref|auto-227>>
<index-1|Cauchy sequence|<pageref|auto-163>>
<index-1|chain|<pageref|auto-40>>
<index-1|chain rule|<pageref|auto-259>>
<index-1|closed ball|<pageref|auto-151>>
<index-1|closed set|<pageref|auto-131>>
<index-1|closure|<pageref|auto-132>>
<index-1|compact set|<pageref|auto-213>>
<index-1|compact space|<pageref|auto-212>>
<index-1|complete metric space|<pageref|auto-229>>
<index-1|complex conjugate|<pageref|auto-109>>
<index-1|complex inner product space|<pageref|auto-167>>
<index-1|complex numbers|<pageref|auto-108>>
<index-1|composition of relations|<pageref|auto-13>>
<index-1|connected set|<pageref|auto-248>>
<index-1|connected topological space|<pageref|auto-247>>
<index-1|continuous mapping|<pageref|auto-171>>
<index-1|contraction|<pageref|auto-233>>
<index-1|Convergence property|<pageref|auto-228>>
<index-1|convergent filterbase|<pageref|auto-218>>
<index-1|convergent sequence|<pageref|auto-162>>
<index-1|convex set|<pageref|auto-278>>
<index-1|coordinate vector fields|<pageref|auto-314>>
<index-1|countable|<pageref|auto-73>>
<index-1|countable infinite set|<pageref|auto-78>>
<index-1|countable set|<pageref|auto-79>>
<index-1|dense subset|<pageref|auto-142>>
<index-1|derivate|<pageref|auto-257>>
<index-1|derivation|<pageref|auto-322>>
<index-1|derivation (local)|<pageref|auto-336>>
<index-1|derived set|<pageref|auto-135>>
<index-1|diameter|<pageref|auto-156>>
<index-1|diffeomorphism|<pageref|auto-283>>
<index-1|differentiable manifold|<pageref|auto-299>>
<index-1|differentiable mapping at x|<pageref|auto-253>>
<index-1|differentiable mapping between manifolds|<pageref|auto-301>>
<index-1|differentiable structure|<pageref|auto-296>>
<index-1|differential|<pageref|auto-254>>
<index-1|differential curve|<pageref|auto-304>>
<index-1|differential of constant|<pageref|auto-255>>
<index-1|differential of linear function|<pageref|auto-256>>
<index-1|directional derivative|<pageref|auto-311>, <pageref|auto-330>>
<index-1|domain|<pageref|auto-17>>
<index-1|equivalence class|<pageref|auto-15>>
<index-1|equivalence relation|<pageref|auto-14>>
<index-1|equivalent metrics|<pageref|auto-153>>
<index-1|equivalent norms|<pageref|auto-160>>
<index-1|even number|<pageref|auto-69>>
<index-1|<with|mode|math|f<rsup|\|C>>|<pageref|auto-25>>
<index-1|<with|mode|math|f<rsup|n>>|<pageref|auto-102>>
<index-1|<with|mode|math|f(x<rsub|m>:\<ldots\>:x<rsub|n>)>|<pageref|auto-191>>
<index-1|faithful left action|<pageref|auto-94>>
<index-1|faithful right action|<pageref|auto-97>>
<index-1|family|<pageref|auto-32>>
<index-1|family of sets|<pageref|auto-33>>
<index-1|fiber bundle|<pageref|auto-208>>
<index-1|field|<pageref|auto-101>>
<index-1|filterbase|<pageref|auto-216>>
<index-1|filterbase on a set|<pageref|auto-221>>
<index-1|finer topology|<pageref|auto-136>>
<index-1|finite set|<pageref|auto-71>>
<index-1|first countable set|<pageref|auto-200>>
<index-1|first derivative|<pageref|auto-263>>
<index-1|free left action|<pageref|auto-96>>
<index-1|free right action|<pageref|auto-99>>
<index-1|function|<pageref|auto-28>>
<index-1|Fundamental theorem of calculus|<pageref|auto-277>>
<index-1|Fundamental theorem of Calculus (classical)|<pageref|auto-276>>
<index-1|Fundamental theorem of claculus
(generalized)|<pageref|auto-279>>
<index-1|fundamentally system of neighborhoods|<pageref|auto-199>>
<index-1|general linear group|<pageref|auto-182>>
<index-1|generalized interval|<pageref|auto-56>>
<index-1|generation basis|<pageref|auto-138>>
<index-1|germs|<pageref|auto-318>>
<index-1|<with|mode|math|Gl(X)>|<pageref|auto-281>>
<index-1|group|<pageref|auto-85>>
<index-1|<with|mode|math|H<rsup|n>>|<pageref|auto-282>>
<index-1|Hausdorff space|<pageref|auto-194>>
<index-1|Heine-Borel theorem|<pageref|auto-225>>
<index-1|<with|mode|math|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>|<pageref|auto-187>>
<index-1|<with|mode|math|Hom(X,Y)>|<pageref|auto-180>>
<index-1|homeomorphism|<pageref|auto-174>>
<index-1|image|<pageref|auto-19>>
<index-1|imaginary part|<pageref|auto-112>>
<index-1|Implicit function theorem|<pageref|auto-286>>
<index-1|induced topology|<pageref|auto-175>>
<index-1|induced vector space structure|<pageref|auto-120>>
<index-1|inductive set|<pageref|auto-62>>
<index-1|infimum|<pageref|auto-53>>
<index-1|infinite set|<pageref|auto-72>>
<index-1*|injection>
<index-2|function|<pageref|auto-22>>
<index-1|inner product norm|<pageref|auto-169>>
<index-1|inner product space|<pageref|auto-166>, <pageref|auto-168>>
<index-1|inner set|<pageref|auto-130>>
<index-1|integer numbers|<pageref|auto-66>>
<index-1|Inverse function theorem|<pageref|auto-285>>
<index-1|inverse induced topology|<pageref|auto-176>>
<index-1|invertible linear transformation|<pageref|auto-181>>
<index-1|isometry|<pageref|auto-154>, <pageref|auto-161>>
<index-1|isomorphism|<pageref|auto-119>>
<index-1|Jacobian matrix|<pageref|auto-261>>
<index-1|Jacobi's identity|<pageref|auto-340>, <pageref|auto-341>>
<index-1|<with|mode|math|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y))>|<pageref|auto-190>>
<index-1|<with|mode|math|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>|<pageref|auto-188>>
<index-1|Lagrange's theorem|<pageref|auto-275>>
<index-1|left group action|<pageref|auto-92>>
<index-1|Lie algebra|<pageref|auto-339>>
<index-1|Lie product|<pageref|auto-338>>
<index-1|<with|mode|math|lim<rsub|h\<rightarrow\>x>f(x)>|<pageref|auto-148>>
<index-1|limit of a function|<pageref|auto-147>>
<index-1|limit of sequence|<pageref|auto-202>>
<index-1|limit point|<pageref|auto-133>>
<index-1|linear combination|<pageref|auto-113>>
<index-1|linear dependent set|<pageref|auto-115>>
<index-1|linear independent set|<pageref|auto-114>>
<index-1|linear mapping|<pageref|auto-179>>
<index-1|local chart|<pageref|auto-290>>
<index-1|local coordinate system|<pageref|auto-291>>
<index-1|local derivation|<pageref|auto-337>>
<index-1|local weak maximum|<pageref|auto-273>>
<index-1|local weak minimum|<pageref|auto-272>>
<index-1|localized derivation|<pageref|auto-335>>
<index-1|lower bound|<pageref|auto-51>>
<index-1|lower bound property|<pageref|auto-55>>
<index-1|L(X,Y)|<pageref|auto-183>>
<index-1|<with|mode|math|M(X,V)>|<pageref|auto-106>>
<index-1|mapping|<pageref|auto-27>>
<index-1|maximal element|<pageref|auto-38>>
<index-1|maximal filterbase|<pageref|auto-222>>
<index-1|maximum|<pageref|auto-48>>
<index-1|metric space|<pageref|auto-146>>
<index-1|minimum|<pageref|auto-49>>
<index-1|multilinear continuous mapping|<pageref|auto-189>>
<index-1|multilinear mapping|<pageref|auto-186>>
<index-1|natural numbers|<pageref|auto-64>>
<index-1|neighborhood|<pageref|auto-197>>
<index-1|neighborhood filterbase|<pageref|auto-217>>
<index-1|norm of partition|<pageref|auto-240>>
<index-1|normal space|<pageref|auto-196>>
<index-1|normed space|<pageref|auto-158>>
<index-1|normed topology|<pageref|auto-159>>
<index-1|n-th derivative|<pageref|auto-265>>
<index-1|odd number|<pageref|auto-68>>
<index-1|open ball|<pageref|auto-149>>
<index-1|open mapping|<pageref|auto-172>>
<index-1|Open Mapping Theorem|<pageref|auto-236>>
<index-1|open neighborhood|<pageref|auto-198>>
<index-1|open set|<pageref|auto-128>>
<index-1|operator norm|<pageref|auto-184>>
<index-1|partial differential|<pageref|auto-260>>
<index-1|partial function|<pageref|auto-18>>
<index-1|partially ordered set|<pageref|auto-41>>
<index-1|partition <with|mode|math|\<cal-P\>> of
<with|mode|math|[a,b]>|<pageref|auto-238>>
<index-1|permutation|<pageref|auto-30>>
<index-1|preimage|<pageref|auto-20>>
<index-1|preorder|<pageref|auto-37>>
<index-1|product of family of sets|<pageref|auto-34>>
<index-1|product topology|<pageref|auto-141>>
<index-1|projective mapping|<pageref|auto-35>>
<index-1|pseudo metric space|<pageref|auto-145>>
<index-1|pseudo normed space|<pageref|auto-157>>
<index-1|real inner product space|<pageref|auto-165>>
<index-1|real numbers|<pageref|auto-59>>
<index-1|real part|<pageref|auto-111>>
<index-1|recursive definition|<pageref|auto-76>>
<index-1|regular space|<pageref|auto-195>>
<index-1|restricted function|<pageref|auto-21>>
<index-1|Riemann integral|<pageref|auto-245>>
<index-1|Riemann sum|<pageref|auto-244>>
<index-1|right group action|<pageref|auto-93>>
<index-1|Rolle's theorem|<pageref|auto-274>>
<index-1|<with|mode|math|<big|sum><rsub|i=1><rsup|n>s<rsub|i>>|<pageref|auto-87>>
<index-1|<with|mode|math|<big|sum><rsub|i=n><rsup|m>s<rsub|i>>|<pageref|auto-88>>
<index-1|second countable space|<pageref|auto-201>>
<index-1|segment|<pageref|auto-67>>
<index-1|semigroup|<pageref|auto-84>>
<index-1|sequence|<pageref|auto-81>>
<index-1|sequential compact set|<pageref|auto-214>>
<index-1|series|<pageref|auto-231>>
<index-1*|set>
<index-2|difference|<pageref|auto-7>>
<index-2|intersection|<pageref|auto-5>>
<index-2|pair|<pageref|auto-9>>
<index-2|powerset|<pageref|auto-8>>
<index-2|product|<pageref|auto-10>>
<index-2|relation|<pageref|auto-12>>
<index-2|subset|<pageref|auto-4>>
<index-2|union|<pageref|auto-6>>
<index-1|size of a set|<pageref|auto-74>>
<index-1|spanning set|<pageref|auto-116>>
<index-1|square root|<pageref|auto-61>>
<index-1|strict linear order relation|<pageref|auto-47>>
<index-1|subalgebra|<pageref|auto-123>>
<index-1|subbasis (topology)|<pageref|auto-139>>
<index-1|subordinate filterbase|<pageref|auto-220>>
<index-1|subspace|<pageref|auto-105>>
<index-1|subspace topology|<pageref|auto-129>>
<index-1|support|<pageref|auto-90>>
<index-1|supremum|<pageref|auto-52>>
<index-1*|surjection>
<index-2|function|<pageref|auto-23>>
<index-1|<with|mode|math|T<rsub|M>>|<pageref|auto-326>>
<index-1|<with|mode|math|T(\<varphi\>,p)>|<pageref|auto-310>>
<index-1|tag|<pageref|auto-242>>
<index-1|tagged partition|<pageref|auto-243>>
<index-1|tangent space|<pageref|auto-307>>
<index-1|tangent vector|<pageref|auto-306>>
<index-1|tangential curves|<pageref|auto-305>>
<index-1|<with|mode|math|T<rsub|p>M>|<pageref|auto-309>>
<index-1|toplinear isomorphism|<pageref|auto-205>, <pageref|auto-280>>
<index-1|topological group|<pageref|auto-207>>
<index-1|topological manifold|<pageref|auto-289>>
<index-1|topological space|<pageref|auto-127>>
<index-1|topological vector space|<pageref|auto-204>>
<index-1|topology|<pageref|auto-126>>
<index-1|totally ordered set|<pageref|auto-42>>
<index-1|transformation of vector components|<pageref|auto-315>>
<index-1|transitive left action|<pageref|auto-95>>
<index-1|transitive right action|<pageref|auto-98>>
<index-1|trivial fiber bundle|<pageref|auto-209>>
<index-1|Tychonoff's theorem|<pageref|auto-224>>
<index-1|<with|mode|math|U<rsub|x>>|<pageref|auto-251>>
<index-1|uncountable set|<pageref|auto-80>>
<index-1|uniform continuous mapping|<pageref|auto-173>>
<index-1|upper bound|<pageref|auto-50>>
<index-1|upper bound property|<pageref|auto-54>>
<index-1|upperbound|<pageref|auto-39>>
<index-1|<with|mode|math|V<rsup|\<varphi\>>>|<pageref|auto-329>>
<index-1|<with|mode|math|v<rsub|p>[f]>|<pageref|auto-312>>
<index-1|<with|mode|math|V[f]>|<pageref|auto-331>>
<index-1|vector bundle|<pageref|auto-210>>
<index-1|vector field|<pageref|auto-328>>
<index-1|vector space|<pageref|auto-104>>
<index-1|well ordered set|<pageref|auto-43>>
<index-1|Zorn's lemma|<pageref|auto-45>>
<index-1|<with|mode|math|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>|<pageref|auto-185>>
</the-index>
</body>
<\initial>
<\collection>
<associate|language|english>
<associate|page-medium|paper>
<associate|page-screen-height|1209600tmpt>
<associate|page-screen-width|1381120tmpt>
<associate|page-type|a4>
</collection>
</initial>
<\references>
<\collection>
<associate|(x,h)-\<gtr\>Df(x)(h) differentiability|<tuple|4.50|149>>
<associate|Archimedean ordering property|<tuple|1.168|39>>
<associate|Axiom of choice|<tuple|1.46|?>>
<associate|Baire category theorem|<tuple|3.281|127>>
<associate|Bolzano Weierstrass|<tuple|3.175|?>>
<associate|C-r is a local property|<tuple|4.41|146>>
<associate|C1 of real or complex functions|<tuple|4.36|145>>
<associate|Cauchy sequence|<tuple|3.56|?>>
<associate|Cr and partial derivates|<tuple|4.52|149>>
<associate|Cr between normed spaces os Cr between
manifolds|<tuple|6.40|183>>
<associate|Cr on arbitrary sets|<tuple|4.55|152>>
<associate|Differential of a billinear mapping|<tuple|4.27|143>>
<associate|Differential of composition of function and
a+tx|<tuple|6.91|199>>
<associate|Differential of constant function|<tuple|4.11|139>>
<associate|I+H has inverse if norm of H is less then 1|<tuple|5.20|163>>
<associate|I-H has inverse if norm of H is lower then 1|<tuple|5.19|161>>
<associate|Jacobi's identity|<tuple|6.105|205>>
<associate|Jacobian|<tuple|4.31|144>>
<associate|L(X1,..L(Xn,Y)..) is Banach if Y is Banach|<tuple|3.155|101>>
<associate|Lie Product|<tuple|6.101|?>>
<associate|Lie Product is a derivation|<tuple|6.103|?>>
<associate|Lie algebra|<tuple|6.104|204>>
<associate|Lie product|<tuple|6.102|204>>
<associate|Lie product of derivations is a derivation|<tuple|6.81|?>>
<associate|Lie product of derivations is a derivation on a commutative
algebra|<tuple|6.81|197>>
<associate|Open Mapping Theorem|<tuple|3.282|128>>
<associate|Re and absolute value|<tuple|2.59|65>>
<associate|Riemann Sum|<tuple|3.292|131>>
<associate|Riemann sums can be aproximated close|<tuple|3.241|?>>
<associate|T(f,p) is bijective|<tuple|6.49|185>>
<associate|TM definition|<tuple|6.67|190>>
<associate|The inversion theorem|<tuple|5.27|167>>
<associate|TpM is a vector space|<tuple|6.50|185>>
<associate|TpM is bijective|<tuple|6.31|?>>
<associate|Tychonoff|<tuple|3.237|118>>
<associate|[a,b] is compact|<tuple|3.209|111>>
<associate|[a,b] is connected|<tuple|3.306|135>>
<associate|a chart is differentiable|<tuple|6.35|181>>
<associate|a isometry between nomred vectorspaces is a
homeomorphisme|<tuple|3.118|91>>
<associate|a metric space is regular|<tuple|3.170|105>>
<associate|absolute value|<tuple|1.93|?>>
<associate|absolute value of complex numbers|<tuple|2.53|65>>
<associate|absolute value of real numbers|<tuple|1.143|35>>
<associate|accumulation and closure|<tuple|3.218|113>>
<associate|algebra|<tuple|2.80|68>>
<associate|algebra of germs|<tuple|6.63|188>>
<associate|all zeroes in sum|<tuple|2.21|58>>
<associate|alternative definition of Baire sets|<tuple|3.37|76>>
<associate|alternative definition of fnite sum|<tuple|2.14|56>>
<associate|alternative definition of norm of linear map|<tuple|3.133|95>>
<associate|apply basis vector on function|<tuple|6.56|187>>
<associate|atlas in a normed space|<tuple|6.13|174>>
<associate|atlas on open subset of manifold|<tuple|6.24|178>>
<associate|auto-1|<tuple|1|5>>
<associate|auto-10|<tuple|1.9|6>>
<associate|auto-100|<tuple|2.2|60>>
<associate|auto-101|<tuple|2.29|60>>
<associate|auto-102|<tuple|2.34|61>>
<associate|auto-103|<tuple|2.3|61>>
<associate|auto-104|<tuple|2.38|61>>
<associate|auto-105|<tuple|2.40|62>>
<associate|auto-106|<tuple|2.43|63>>
<associate|auto-107|<tuple|2.48|64>>
<associate|auto-108|<tuple|2.48|64>>
<associate|auto-109|<tuple|2.50|64>>
<associate|auto-11|<tuple|1.1.2|6>>
<associate|auto-110|<tuple|2.53|65>>
<associate|auto-111|<tuple|2.56|65>>
<associate|auto-112|<tuple|2.56|65>>
<associate|auto-113|<tuple|2.64|66>>
<associate|auto-114|<tuple|2.65|66>>
<associate|auto-115|<tuple|2.66|66>>
<associate|auto-116|<tuple|2.67|66>>
<associate|auto-117|<tuple|2.68|66>>
<associate|auto-118|<tuple|2.71|67>>
<associate|auto-119|<tuple|2.75|67>>
<associate|auto-12|<tuple|1.13|6>>
<associate|auto-120|<tuple|2.78|68>>
<associate|auto-121|<tuple|2.4|68>>
<associate|auto-122|<tuple|2.80|68>>
<associate|auto-123|<tuple|2.81|69>>
<associate|auto-124|<tuple|3|71>>
<associate|auto-125|<tuple|3.1|71>>
<associate|auto-126|<tuple|3.1|71>>
<associate|auto-127|<tuple|3.1|71>>
<associate|auto-128|<tuple|3.1|71>>
<associate|auto-129|<tuple|3.3|71>>
<associate|auto-13|<tuple|1.16|7>>
<associate|auto-130|<tuple|3.6|72>>
<associate|auto-131|<tuple|3.8|72>>
<associate|auto-132|<tuple|3.12|73>>
<associate|auto-133|<tuple|3.15|73>>
<associate|auto-134|<tuple|3.15|73>>
<associate|auto-135|<tuple|3.15|73>>
<associate|auto-136|<tuple|3.17|73>>
<associate|auto-137|<tuple|3.19|73>>
<associate|auto-138|<tuple|3.23|74>>
<associate|auto-139|<tuple|3.27|75>>
<associate|auto-14|<tuple|1.17|7>>
<associate|auto-140|<tuple|3.28|75>>
<associate|auto-141|<tuple|3.30|75>>
<associate|auto-142|<tuple|3.34|76>>
<associate|auto-143|<tuple|3.35|76>>
<associate|auto-144|<tuple|3.2|77>>
<associate|auto-145|<tuple|3.38|77>>
<associate|auto-146|<tuple|3.38|77>>
<associate|auto-147|<tuple|3.39|77>>
<associate|auto-148|<tuple|3.39|77>>
<associate|auto-149|<tuple|3.41|77>>
<associate|auto-15|<tuple|1.18|7>>
<associate|auto-150|<tuple|3.41|77>>
<associate|auto-151|<tuple|3.42|77>>
<associate|auto-152|<tuple|3.42|77>>
<associate|auto-153|<tuple|3.50|78>>
<associate|auto-154|<tuple|3.52|78>>
<associate|auto-155|<tuple|3.57|79>>
<associate|auto-156|<tuple|3.57|79>>
<associate|auto-157|<tuple|3.59|80>>
<associate|auto-158|<tuple|3.59|80>>
<associate|auto-159|<tuple|3.65|81>>
<associate|auto-16|<tuple|1.1.3|7>>
<associate|auto-160|<tuple|3.71|83>>
<associate|auto-161|<tuple|3.73|83>>
<associate|auto-162|<tuple|3.77|83>>
<associate|auto-163|<tuple|3.78|83>>
<associate|auto-164|<tuple|3.3|83>>
<associate|auto-165|<tuple|3.79|83>>
<associate|auto-166|<tuple|3.79|83>>
<associate|auto-167|<tuple|3.81|84>>
<associate|auto-168|<tuple|3.81|84>>
<associate|auto-169|<tuple|3.88|85>>
<associate|auto-17|<tuple|1.20|7>>
<associate|auto-170|<tuple|3.4|87>>
<associate|auto-171|<tuple|3.96|87>>
<associate|auto-172|<tuple|3.101|88>>
<associate|auto-173|<tuple|3.110|90>>
<associate|auto-174|<tuple|3.114|91>>
<associate|auto-175|<tuple|3.116|91>>
<associate|auto-176|<tuple|3.117|91>>
<associate|auto-177|<tuple|3.117|91>>
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<associate|derivative of product|<tuple|4.29|143>>
<associate|diameter of closed bounded set|<tuple|3.58|?>>
<associate|difference of Rieman sums|<tuple|3.293|131>>
<associate|difference of difference of sets|<tuple|1.4|5>>
<associate|differentiability and equivalent norms|<tuple|4.8|138>>
<associate|differentiability and restricted mappings|<tuple|4.9|138>>
<associate|differentiability of a multilinear mapping|<tuple|4.24|?>>
<associate|differentiability of composition of differentiable
mappings|<tuple|4.21|141>>
<associate|differentiability of functions to a field|<tuple|4.14|?>>
<associate|differentiability of map to product|<tuple|4.26|142>>
<associate|differentiability of product|<tuple|4.28|143>>
<associate|differentiability of products of metric
spaces|<tuple|4.22|142>>
<associate|differentiability of real functions and
manifolds|<tuple|6.32|180>>
<associate|differentiability of real or complex
functions|<tuple|4.16|139>>
<associate|differentiability on [a,b]|<tuple|4.57|153>>
<associate|differentiability on the real numbers|<tuple|4.18|?>>
<associate|differentiability under the integral sign|<tuple|4.55|149>>
<associate|differentiable curve|<tuple|6.41|183>>
<associate|differentiable function to a normed space|<tuple|6.37|182>>
<associate|differentiable mappings between manifolds and
composition|<tuple|6.39|182>>
<associate|differentiable mappings on a open subset of a
maifold|<tuple|6.33|181>>
<associate|differentiable vector fields property one|<tuple|6.77|196>>
<associate|differential of linear mapping|<tuple|4.12|139>>
<associate|differential of multiparameter function|<tuple|4.24|142>>
<associate|differentiation under the integral sign|<tuple|4.54|149>>
<associate|directional derivative|<tuple|6.52|186>>
<associate|directional derivative is linear|<tuple|6.53|186>>
<associate|each filterbase converges to exactly one
point|<tuple|3.222|113>>
<associate|elimination of zeroes|<tuple|2.10|?>>
<associate|equality of multiple appliance|<tuple|3.159|101>>
<associate|equivalence of multilinear and linear continuous
mappings|<tuple|3.164|103>>
<associate|equivalence of norms|<tuple|3.72|83>>
<associate|equivalence of norms on product of real
numbers|<tuple|3.242|119>>
<associate|equivalence of product|<tuple|1.89|22>>
<associate|equivalence relation|<tuple|1.17|7>>
<associate|every compact metric space is complete|<tuple|3.269|123>>
<associate|every finite subset of a hausdorf space is
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<associate|every subset of a countable set is countable|<tuple|1.219|49>>
<associate|example of a convergent serie|<tuple|3.278|125>>
<associate|example of convergent sequence|<tuple|3.250|120>>
<associate|example of derivative|<tuple|4.58|155>>
<associate|example of differentiable manifold 1|<tuple|6.20|161>>
<associate|existance of maximal filterbase|<tuple|3.233|116>>
<associate|existance of maximum and minimum in the image of a continuous
function of a compect set|<tuple|3.241|119>>
<associate|expansion of function that is infinite
differentiable|<tuple|6.93|200>>
<associate|extremum and derivate|<tuple|5.2|157>>
<associate|faithfull left action|<tuple|2.28|60>>
<associate|families and bijectives|<tuple|1.49|?>>
<associate|families and surjective mappings|<tuple|1.86|21>>
<associate|family of one set|<tuple|1.223|50>>
<associate|family of sets|<tuple|1.81|21>>
<associate|fiber bundle|<tuple|3.188|107>>
<associate|filter base has the finite intersection
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<associate|filterbase and closure|<tuple|3.226|115>>
<associate|filterbase and continuity 1|<tuple|3.228|115>>
<associate|filterbase and continuity 2|<tuple|3.229|115>>
<associate|filterbases and compactness|<tuple|3.236|117>>
<associate|filterbases and products|<tuple|3.230|115>>
<associate|finer basis|<tuple|3.26|74>>
<associate|finer metric|<tuple|3.51|78>>
<associate|finer norm|<tuple|3.67|82>>
<associate|finite dimensional space norm and inner
product|<tuple|3.93|86>>
<associate|finite dimensional spaces are isometric|<tuple|3.94|86>>
<associate|finite intersection property|<tuple|3.157|?>>
<associate|finite product of countable sets is
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<associate|finite product of sets|<tuple|1.224|51>>
<associate|finite product topology|<tuple|3.33|76>>
<associate|finite sets|<tuple|1.175|40>>
<associate|finite subset of filterbase|<tuple|3.212|112>>
<associate|finite union of finite sets|<tuple|1.191|43>>
<associate|function extension|<tuple|1.25|8>>
<associate|function of ith variable|<tuple|3.137|96>>
<associate|fundamental theorem of calculus|<tuple|5.6|126>>
<associate|fundamental theorem of calculus (1)|<tuple|5.6|158>>
<associate|fundamental theorem of calculus (2)|<tuple|5.7|159>>
<associate|fv is differentiable|<tuple|4.49|148>>
<associate|general linear group|<tuple|3.128|93>>
<associate|general product with a zero|<tuple|2.33|61>>
<associate|general sum|<tuple|2.22|58>>
<associate|general sum is finite sum in the finite case|<tuple|2.25|59>>
<associate|generalized chain rule|<tuple|4.48|148>>
<associate|generalized intervals|<tuple|1.123|30>>
<associate|generalized intervals and connectness|<tuple|3.307|136>>
<associate|generating basis in topology|<tuple|3.23|74>>
<associate|graph of a function|<tuple|1.8|?>>
<associate|group of products|<tuple|2.5|53>>
<associate|hom(X,Y)|<tuple|3.125|93>>
<associate|ibjection and projection|<tuple|1.98|?>>
<associate|image and preimage|<tuple|1.66|18>>
<associate|image by a injective function of a
intersection|<tuple|1.36|12>>
<associate|image of a restricted function|<tuple|1.32|11>>
<associate|image of inversion of injective function|<tuple|1.64|18>>
<associate|image of open set by chart is open|<tuple|6.12|174>>
<associate|image of preimage|<tuple|1.24|8>>
<associate|image of preimage of a function|<tuple|1.59|15>>
<associate|image of preimage of function|<tuple|1.52|?>>
<associate|image of projection|<tuple|1.108|26>>
<associate|image of union or intersection|<tuple|1.30|10>>
<associate|image restriction|<tuple|1.49|15>>
<associate|image restriction of a injection|<tuple|1.50|15>>
<associate|implicit function theorem|<tuple|5.35|171>>
<associate|in Banach spaces linear continuous mappings are
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<associate|induced basis on induced vector space|<tuple|2.79|68>>
<associate|induced topology|<tuple|3.116|91>>
<associate|induced vector space structure|<tuple|2.78|68>>
<associate|inductive subsets of natural numbers|<tuple|1.151|36>>
<associate|inequality and sum|<tuple|2.10|54>>
<associate|inf,sup condition|<tuple|1.117|29>>
<associate|inf\<less\>sup|<tuple|1.118|29>>
<associate|infinite subsets of natural numbers|<tuple|1.201|46>>
<associate|injection and projection|<tuple|1.98|25>>
<associate|injection extension|<tuple|1.52|16>>
<associate|injection from infinite set|<tuple|1.189|43>>
<associate|inner product norm is a norm|<tuple|3.92|86>>
<associate|inner product of neutral elements|<tuple|3.84|84>>
<associate|integral of function to linear operators|<tuple|3.300|134>>
<associate|integration of constant|<tuple|3.301|134>>
<associate|intermediate bijective function|<tuple|1.74|20>>
<associate|intersection of balls|<tuple|3.44|77>>
<associate|intersection of finite open sets is open|<tuple|3.2|71>>
<associate|introduction of the multilinear norm|<tuple|3.149|99>>
<associate|inverse function|<tuple|1.37|12>>
<associate|inverse induced topology|<tuple|3.117|91>>
<associate|inverse mapping is c-inifinite|<tuple|5.21|163>>
<associate|inverse of a local chart|<tuple|6.7|?>>
<associate|inverse of a restricted function|<tuple|1.32|?>>
<associate|inverse of bijective mapping|<tuple|1.70|19>>
<associate|inverse of continuous linear mappings between Banach spaces is
continuous|<tuple|3.283|129>>
<associate|inverse of diffeomorphisme|<tuple|5.24|166>>
<associate|inverse of isometry metric|<tuple|3.53|78>>
<associate|inverse of isomorphisme|<tuple|2.76|67>>
<associate|inverse of isomorphisme norm|<tuple|3.74|83>>
<associate|inverse of number bigger then one|<tuple|1.134|34>>
<associate|inverse of open map is continuous|<tuple|3.102|88>>
<associate|inverse of product|<tuple|1.127|33>>
<associate|inverse of product in a field|<tuple|2.31|60>>
<associate|inverse of product of many elements in a
field|<tuple|2.36|61>>
<associate|inverse of projection|<tuple|1.94|23>>
<associate|inverse of sum|<tuple|2.23|?>>
<associate|invertible linear transformation|<tuple|3.127|93>>
<associate|isometric normed spaces|<tuple|3.76|83>>
<associate|isometry and balls|<tuple|3.55|79>>
<associate|kwadrat is positive|<tuple|1.140|35>>
<associate|lagranges theorem|<tuple|5.4|158>>
<associate|left(right) group action|<tuple|2.27|60>>
<associate|limit and continuity|<tuple|3.252|120>>
<associate|limit of a sum|<tuple|3.258|121>>
<associate|limit point in metric space property|<tuple|3.186|?>>
<associate|limit point of sequences|<tuple|3.267|123>>
<associate|limit points and continuous functions|<tuple|3.99|?>>
<associate|limit preserves inequality|<tuple|3.214|?>>
<associate|limit preserves inequality (1)|<tuple|3.256|121>>
<associate|limit preserves inequality (2)|<tuple|3.257|121>>
<associate|linear combination of convergent series|<tuple|3.263|122>>
<associate|linear continuous mappings are C infinity|<tuple|4.43|146>>
<associate|linear map applied to neutral element yields the neutral
element|<tuple|3.126|93>>
<associate|linear mappings are C infinity|<tuple|4.33|?>>
<associate|linear maps between finite dimensional spaces are
continuous|<tuple|3.243|119>>
<associate|linear maps between products of the real spaces are
continuous|<tuple|3.130|94>>
<associate|linear open mappings|<tuple|3.165|104>>
<associate|linear order|<tuple|1.111|27>>
<associate|linearity of integral|<tuple|3.298|133>>
<associate|local derivation|<tuple|6.88|199>>
<associate|local derivation of constant is 0|<tuple|6.92|200>>
<associate|local vector as a function|<tuple|6.64|189>>
<associate|local vector as a function is a derivation|<tuple|6.66|189>>
<associate|localized derivation|<tuple|6.87|198>>
<associate|make inverse of natural lower then e|<tuple|1.169|39>>
<associate|manifold defined on a open subset of a
manifold|<tuple|6.23|177>>
<associate|map of finite set is finite|<tuple|1.192|44>>
<associate|mapping extension|<tuple|1.65|18>>
<associate|mapping of N into a finite set|<tuple|1.195|44>>
<associate|mapping of a filterbase|<tuple|3.227|115>>
<associate|mapping with {1,..,n}|<tuple|1.178|40>>
<associate|maximal atlas of a differential structure|<tuple|6.14|?>>
<associate|maximal atlas of a differential structure
(1)|<tuple|6.18|176>>
<associate|maximal filterbase and mapping|<tuple|3.235|117>>
<associate|maximal filterbase characterization|<tuple|3.232|116>>
<associate|maximum (minumu,) of finite sets|<tuple|1.134|?>>
<associate|maximum filterbase and convergence|<tuple|3.234|117>>
<associate|maximum minimum of finite sets|<tuple|1.193|44>>
<associate|maximum of finite set of numbers|<tuple|1.149|?>>
<associate|maximum of products|<tuple|3.68|82>>
<associate|maximum of two sets|<tuple|1.121|29>>
<associate|metric from norm|<tuple|3.63|81>>
<associate|metric space is hausdorf|<tuple|3.167|105>>
<associate|metric topology|<tuple|3.46|77>>
<associate|mulilinearity and coefficients|<tuple|3.145|97>>
<associate|multilinear mapping|<tuple|3.139|96>>
<associate|multilinear mappings and neutral element|<tuple|3.147|98>>
<associate|n multilinear mapping yields n-1 multilinear
map|<tuple|3.144|97>>
<associate|n-multilinear map becomes linear map to n-1 multilinear
map|<tuple|3.163|103>>
<associate|natural numbers are infinite|<tuple|1.187|42>>
<associate|natural numbers has a minimal separation of
1|<tuple|1.155|37>>
<associate|naturals are well ordered|<tuple|1.164|39>>
<associate|naturals are wellordered|<tuple|1.91|?>>
<associate|negative of a integer is a integer|<tuple|1.157|37>>
<associate|neighborhood filterbase|<tuple|3.213|112>>
<associate|neighborhood filterbase converts|<tuple|3.219|113>>
<associate|norm and absolute value|<tuple|3.64|81>>
<associate|norm of a finite sum|<tuple|3.60|80>>
<associate|norm of a multilinear mapping|<tuple|3.150|99>>
<associate|norm of finite product of normed spaces|<tuple|3.69|82>>
<associate|norm of muliple appliance|<tuple|3.162|103>>
<associate|norm properties of the norm of a linear map|<tuple|3.132|94>>
<associate|normed space is a manifold|<tuple|6.25|178>>
<associate|normed space properties|<tuple|3.61|80>>
<associate|normed vector space of continuous linear
mappings|<tuple|3.135|96>>
<associate|norms mappings are continuous in the product of
reals|<tuple|3.113|90>>
<associate|number between two different numbers|<tuple|1.139|34>>
<associate|odd power|<tuple|1.209|47>>
<associate|open mapping in metric spaces|<tuple|3.109|89>>
<associate|open sets in metric topology|<tuple|3.47|77>>
<associate|open subset of metric space is a differentiable
manifold|<tuple|6.26|178>>
<associate|ordering of a finite set|<tuple|1.196|44>>
<associate|othere definition of Cauchy|<tuple|3.247|120>>
<associate|partial coordinates are differentiable|<tuple|4.18|140>>
<associate|partioned sum|<tuple|2.19|57>>
<associate|patched topology|<tuple|3.19|?>>
<associate|positive integers are natural numbers|<tuple|1.154|36>>
<associate|power function is continuous|<tuple|3.122|92>>
<associate|power of a continuous linear map is linear and
continuous|<tuple|5.17|161>>
<associate|power of number bigger then one|<tuple|1.204|46>>
<associate|power of odd is odd|<tuple|1.137|?>>
<associate|power of x\<less\>1|<tuple|1.216|49>>
<associate|preimage and injective partial functions|<tuple|1.38|?>>
<associate|preimage image of union intersection of a family of
sets|<tuple|1.96|23>>
<associate|preimage of a composed|<tuple|1.37|?>>
<associate|preimage of a composed mapping|<tuple|1.60|14>>
<associate|preimage of a mapping|<tuple|1.48|?>>
<associate|preimage of composition|<tuple|1.28|9>>
<associate|preimage of difference|<tuple|1.67|18>>
<associate|preimage of difference of sets|<tuple|1.40|14>>
<associate|preimage of injective mapping|<tuple|1.38|13>>
<associate|preimage of union intersection of a family of
sets|<tuple|1.57|?>>
<associate|preimage of union or intersection|<tuple|1.29|10>>
<associate|principle of nested intervals|<tuple|3.208|110>>
<associate|principle of recursive definition|<tuple|1.200|45>>
<associate|principle of strong induction|<tuple|1.167|39>>
<associate|product of Cr functions is Cr|<tuple|4.53|149>>
<associate|product of compact subspaces|<tuple|3.238|118>>
<associate|product of continuous functions is
continuous|<tuple|3.121|92>>
<associate|product of countable sets|<tuple|1.226|51>>
<associate|product of equal real numbers|<tuple|2.42|62>>
<associate|product of finite sets is finite|<tuple|1.185|42>>
<associate|product of metric spaces|<tuple|3.58|79>>
<associate|product of naturals is countable|<tuple|1.221|50>>
<associate|product of non zero numbers|<tuple|1.141|35>>
<associate|product of powers|<tuple|1.207|47>>
<associate|product of real numbers and \<less\>=|<tuple|2.41|62>>
<associate|product of zero|<tuple|2.39|61>>
<associate|product topology|<tuple|3.30|75>>
<associate|product topology of subspace topolgies|<tuple|3.32|76>>
<associate|product with 0|<tuple|1.128|34>>
<associate|projection map is C-infinite|<tuple|4.46|147>>
<associate|projection map is continuous|<tuple|3.105|88>>
<associate|properties of two filterbases|<tuple|3.221|113>>
<associate|property of V1[V2]p]p|<tuple|6.98|?>>
<associate|property of integral|<tuple|3.299|133>>
<associate|property of max|<tuple|1.150|?>>
<associate|property of natural numbers|<tuple|1.152|36>>
<associate|proving f is Cn based on Cn-1 of Df|<tuple|4.40|146>>
<associate|real continuous mappings on a compact
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<associate|real numbers|<tuple|1.124|33>>
<associate|real smooth functions on smooth manifolds|<tuple|6.78|197>>
<associate|real space is complete|<tuple|3.270|123>>
<associate|recursive definition 1|<tuple|1.197|44>>
<associate|recursive definition 2|<tuple|1.198|45>>
<associate|recursive finite product|<tuple|1.225|51>>
<associate|reflexitivity|<tuple|1|?>>
<associate|remove zero from sum|<tuple|2.10|?>>
<associate|restricted function|<tuple|1.31|10>>
<associate|restriction and composition of functions|<tuple|1.63|17>>
<associate|restriction and preimage|<tuple|1.33|11>>
<associate|restriction of C-r mapping|<tuple|4.39|146>>
<associate|restriction of a bijection|<tuple|1.51|16>>
<associate|restriction of a bijective mapping|<tuple|1.72|19>>
<associate|restriction of a differential mapping between
manifolds|<tuple|6.36|182>>
<associate|restriction of a injection|<tuple|1.43|?>>
<associate|restriction of a injective mapping|<tuple|1.52|?>>
<associate|restriction of a mapping|<tuple|1.58|17>>
<associate|restriction of composition|<tuple|1.34|11>>
<associate|restriction of composition of a mapping|<tuple|1.62|17>>
<associate|restriction of composition of mappings|<tuple|1.48|?>>
<associate|restriction of injection|<tuple|1.48|15>>
<associate|restriction of local chart in differentiable
manifold|<tuple|6.22|177>>
<associate|restriction of surjection|<tuple|1.22|?>>
<associate|restrictionof a bijection|<tuple|1.22|?>>
<associate|rolles theorem|<tuple|5.3|157>>
<associate|scalair product and sum|<tuple|2.44|63>>
<associate|schwartz inequality|<tuple|3.90|85>>
<associate|sequence of points in a closed set of a metric
space|<tuple|3.224|?>>
<associate|sequence of points in a set of a metric
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<associate|set indexed by itself|<tuple|1.80|20>>
<associate|set of derivations on smooth functions|<tuple|6.84|198>>
<associate|set of local vectors is isomorph with set of local
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<associate|shorthand notation of function|<tuple|1.59|17>>
<associate|size of a set|<tuple|1.183|41>>
<associate|size of finite sets|<tuple|1.113|?>>
<associate|smallest element of segment|<tuple|1.163|38>>
<associate|space of linear continuous maps to a Banach space is
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<associate|split of a sum|<tuple|2.18|57>>
<associate|splitting of a integral|<tuple|3.302|134>>
<associate|sqrt of sum of squares|<tuple|1.102|?>>
<associate|squareroot is monotone|<tuple|1.146|36>>
<associate|sub basis topology|<tuple|3.27|75>>
<associate|sub differential structure|<tuple|6.20|?>>
<associate|subalgebra|<tuple|2.81|69>>
<associate|subserie property|<tuple|3.262|122>>
<associate|subset of a finite set|<tuple|1.186|42>>
<associate|subspace|<tuple|2.40|62>>
<associate|subspace topology|<tuple|3.3|71>>
<associate|subspace topology of a metric space|<tuple|3.49|78>>
<associate|subspace topology of subspace topology|<tuple|3.4|71>>
<associate|subspaces|<tuple|2.20|?>>
<associate|sum and inequalities|<tuple|2.15|?>>
<associate|sum and switch|<tuple|2.12|55>>
<associate|sum of C-r functions|<tuple|4.37|?>>
<associate|sum of C-r functions is C-r|<tuple|4.42|146>>
<associate|sum of a linear combination|<tuple|2.45|63>>
<associate|sum of basevectors 1|<tuple|2.72|67>>
<associate|sum of basevectors 2|<tuple|2.73|67>>
<associate|sum of constant to a serie|<tuple|3.240|?>>
<associate|sum of continuous functions is continuous|<tuple|3.120|92>>
<associate|sum of derivatives|<tuple|4.20|141>>
<associate|sum of differences|<tuple|2.15|56>>
<associate|sum of differentiable functions between
manifolds|<tuple|6.34|?>>
<associate|sum of differentiable functions is
differentiable|<tuple|4.19|140>>
<associate|sum of majorants|<tuple|2.9|54>>
<associate|sum of partial function|<tuple|4.17|140>>
<associate|sum of partitioned sets|<tuple|2.20|?>>
<associate|sum of partitions|<tuple|2.17|56>>
<associate|sum of positive numbers is positive|<tuple|3.85|84>>
<associate|sum of sums|<tuple|2.26|59>>
<associate|sum with zeroes|<tuple|2.10|?>>
<associate|sum(product) of differentiable functions between
manifolds|<tuple|6.38|182>>
<associate|sum, scalair product of multiple appliance|<tuple|3.160|102>>
<associate|surjection extension|<tuple|1.53|16>>
<associate|symmetry of multiple appliance|<tuple|3.158|101>>
<associate|terms of a convergent serie goes to 0|<tuple|3.261|122>>
<associate|terms of convergent sequence goes to 0|<tuple|3.238|?>>
<associate|test for countability|<tuple|1.220|49>>
<associate|test for finitness|<tuple|1.188|42>>
<associate|the Banach Fixed Point Theorem|<tuple|3.279|126>>
<associate|the canonical basis of product of reals|<tuple|2.74|67>>
<associate|the fundamental theorem of calculus (3)|<tuple|5.10|159>>
<associate|there exists a isometry between a finite dimensional normed
space and the product of reals with a induced norm|<tuple|3.119|92>>
<associate|tirangle inequality|<tuple|3.76|?>>
<associate|toplinear isomorphism|<tuple|3.184|106>>
<associate|topological vector space|<tuple|3.182|106>>
<associate|topologies and isometries|<tuple|3.56|79>>
<associate|topology and vectorspace on {p}XTpM|<tuple|6.68|190>>
<associate|topology of complex numbers|<tuple|3.55|?>>
<associate|topology of real numbers|<tuple|3.66|81>>
<associate|transformation rule of the components of a
vector|<tuple|6.58|187>>
<associate|translation is C-inifinity|<tuple|4.44|146>>
<associate|triangle inequality|<tuple|3.91|85>>
<associate|trival fiber bundle|<tuple|3.189|107>>
<associate|trivial filterbase|<tuple|3.214|112>>
<associate|union and intersection of a family|<tuple|1.97|24>>
<associate|union of disjoint sets|<tuple|1.5|5>>
<associate|union of disjoint sets and difference and
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<associate|union of families|<tuple|1.77|20>>
<associate|union of families (1)|<tuple|1.78|20>>
<associate|union of family|<tuple|1.50|?>>
<associate|union of finite sets|<tuple|1.190|43>>
<associate|unique recursive definition|<tuple|1.199|45>>
<associate|unique recursive definition 1|<tuple|1.124|?>>
<associate|unique recursive definition 2|<tuple|1.124|?>>
<associate|uniqueness of imit|<tuple|3.214|?>>
<associate|uniqueness of inverse, negative|<tuple|1.126|33>>
<associate|uniqueness of limit|<tuple|3.251|120>>
<associate|upper bound property implies lower bound
property|<tuple|1.120|29>>
<associate|vector bundle|<tuple|3.190|107>>
<associate|vector field|<tuple|6.74|194>>
<associate|vector field on a finite differential
manifold|<tuple|6.76|195>>
<associate|vector field represents two functions|<tuple|6.83|198>>
<associate|vector space of mappings|<tuple|2.43|63>>
<associate|vector space of multilinear continuous
mappings|<tuple|3.151|100>>
<associate|vector space of multilinear mappings|<tuple|3.146|97>>
<associate|zero is absorbing element in a field|<tuple|2.30|60>>
<associate|zero sum|<tuple|1.142|35>>
<associate|zeroes in general sums|<tuple|2.23|59>>
<associate|zeroes in sum|<tuple|2.11|?>>
<associate|zorn's lemma|<tuple|1.110|27>>
<associate|{1,..n}*{1,..,m} is finite|<tuple|1.184|41>>
</collection>
</references>
<\auxiliary>
<\collection>
<\associate|idx>
<tuple|<tuple|set|subset>|<pageref|auto-4>>
<tuple|<tuple|set|intersection>|<pageref|auto-5>>
<tuple|<tuple|set|union>|<pageref|auto-6>>
<tuple|<tuple|set|difference>|<pageref|auto-7>>
<tuple|<tuple|set|powerset>|<pageref|auto-8>>
<tuple|<tuple|set|pair>|<pageref|auto-9>>
<tuple|<tuple|set|product>|<pageref|auto-10>>
<tuple|<tuple|set|relation>|<pageref|auto-12>>
<tuple|<tuple|composition of relations>|<pageref|auto-13>>
<tuple|<tuple|equivalence relation>|<pageref|auto-14>>
<tuple|<tuple|equivalence class>|<pageref|auto-15>>
<tuple|<tuple|domain>|<pageref|auto-17>>
<tuple|<tuple|partial function>|<pageref|auto-18>>
<tuple|<tuple|image>|<pageref|auto-19>>
<tuple|<tuple|preimage>|<pageref|auto-20>>
<tuple|<tuple|restricted function>|<pageref|auto-21>>
<tuple|<tuple|injection|function>|<pageref|auto-22>>
<tuple|<tuple|surjection|function>|<pageref|auto-23>>
<tuple|<tuple|bijection|function>|<pageref|auto-24>>
<tuple|<tuple|<with|mode|<quote|math>|f<rsup|\|C>>>|<pageref|auto-25>>
<tuple|<tuple|mapping>|<pageref|auto-27>>
<tuple|<tuple|function>|<pageref|auto-28>>
<tuple|<tuple|bijective sets>|<pageref|auto-29>>
<tuple|<tuple|permutation>|<pageref|auto-30>>
<tuple|<tuple|family>|<pageref|auto-32>>
<tuple|<tuple|family of sets>|<pageref|auto-33>>
<tuple|<tuple|product of family of sets>|<pageref|auto-34>>
<tuple|<tuple|projective mapping>|<pageref|auto-35>>
<tuple|<tuple|preorder>|<pageref|auto-37>>
<tuple|<tuple|maximal element>|<pageref|auto-38>>
<tuple|<tuple|upperbound>|<pageref|auto-39>>
<tuple|<tuple|chain>|<pageref|auto-40>>
<tuple|<tuple|partially ordered set>|<pageref|auto-41>>
<tuple|<tuple|totally ordered set>|<pageref|auto-42>>
<tuple|<tuple|well ordered set>|<pageref|auto-43>>
<tuple|<tuple|axiom of choice>|<pageref|auto-44>>
<tuple|<tuple|Zorn's lemma>|<pageref|auto-45>>
<tuple|<tuple|strict linear order relation>|<pageref|auto-47>>
<tuple|<tuple|maximum>|<pageref|auto-48>>
<tuple|<tuple|minimum>|<pageref|auto-49>>
<tuple|<tuple|upper bound>|<pageref|auto-50>>
<tuple|<tuple|lower bound>|<pageref|auto-51>>
<tuple|<tuple|supremum>|<pageref|auto-52>>
<tuple|<tuple|infimum>|<pageref|auto-53>>
<tuple|<tuple|upper bound property>|<pageref|auto-54>>
<tuple|<tuple|lower bound property>|<pageref|auto-55>>
<tuple|<tuple|generalized interval>|<pageref|auto-56>>
<tuple|<tuple|<with|mode|<quote|math>|\<bbb-R\>>>|<pageref|auto-58>>
<tuple|<tuple|real numbers>|<pageref|auto-59>>
<tuple|<tuple|absolute value>|<pageref|auto-60>>
<tuple|<tuple|square root>|<pageref|auto-61>>
<tuple|<tuple|inductive set>|<pageref|auto-62>>
<tuple|<tuple|<with|mode|<quote|math>|\<bbb-N\><rsub|0>>>|<pageref|auto-63>>
<tuple|<tuple|natural numbers>|<pageref|auto-64>>
<tuple|<tuple|<with|mode|<quote|math>|\<bbb-Z\>>>|<pageref|auto-65>>
<tuple|<tuple|integer numbers>|<pageref|auto-66>>
<tuple|<tuple|segment>|<pageref|auto-67>>
<tuple|<tuple|odd number>|<pageref|auto-68>>
<tuple|<tuple|even number>|<pageref|auto-69>>
<tuple|<tuple|finite set>|<pageref|auto-71>>
<tuple|<tuple|infinite set>|<pageref|auto-72>>
<tuple|<tuple|countable>|<pageref|auto-73>>
<tuple|<tuple|size of a set>|<pageref|auto-74>>
<tuple|<tuple|recursive definition>|<pageref|auto-76>>
<tuple|<tuple|countable infinite set>|<pageref|auto-78>>
<tuple|<tuple|countable set>|<pageref|auto-79>>
<tuple|<tuple|uncountable set>|<pageref|auto-80>>
<tuple|<tuple|sequence>|<pageref|auto-81>>
<tuple|<tuple|semigroup>|<pageref|auto-84>>
<tuple|<tuple|group>|<pageref|auto-85>>
<tuple|<tuple|abelian group>|<pageref|auto-86>>
<tuple|<tuple|<with|mode|<quote|math>|<big|sum><rsub|i=1><rsup|n>s<rsub|i>>>|<pageref|auto-87>>
<tuple|<tuple|<with|mode|<quote|math>|<big|sum><rsub|i=n><rsup|m>s<rsub|i>>>|<pageref|auto-88>>
<tuple|<tuple|<with|mode|<quote|math>|<big|sum><rsub|a\<in\>A>a>>|<pageref|auto-89>>
<tuple|<tuple|support>|<pageref|auto-90>>
<tuple|<tuple|<with|mode|<quote|math>|<big|sum><rsub|a\<in\>A>a>>|<pageref|auto-91>>
<tuple|<tuple|left group action>|<pageref|auto-92>>
<tuple|<tuple|right group action>|<pageref|auto-93>>
<tuple|<tuple|faithful left action>|<pageref|auto-94>>
<tuple|<tuple|transitive left action>|<pageref|auto-95>>
<tuple|<tuple|free left action>|<pageref|auto-96>>
<tuple|<tuple|faithful right action>|<pageref|auto-97>>
<tuple|<tuple|transitive right action>|<pageref|auto-98>>
<tuple|<tuple|free right action>|<pageref|auto-99>>
<tuple|<tuple|field>|<pageref|auto-101>>
<tuple|<tuple|<with|mode|<quote|math>|f<rsup|n>>>|<pageref|auto-102>>
<tuple|<tuple|vector space>|<pageref|auto-104>>
<tuple|<tuple|subspace>|<pageref|auto-105>>
<tuple|<tuple|<with|mode|<quote|math>|M(X,V)>>|<pageref|auto-106>>
<tuple|<tuple|<with|mode|<quote|math>|\<bbb-C\>>>|<pageref|auto-107>>
<tuple|<tuple|complex numbers>|<pageref|auto-108>>
<tuple|<tuple|complex conjugate>|<pageref|auto-109>>
<tuple|<tuple|absolute value (complex)>|<pageref|auto-110>>
<tuple|<tuple|real part>|<pageref|auto-111>>
<tuple|<tuple|imaginary part>|<pageref|auto-112>>
<tuple|<tuple|linear combination>|<pageref|auto-113>>
<tuple|<tuple|linear independent set>|<pageref|auto-114>>
<tuple|<tuple|linear dependent set>|<pageref|auto-115>>
<tuple|<tuple|spanning set>|<pageref|auto-116>>
<tuple|<tuple|basis>|<pageref|auto-117>>
<tuple|<tuple|basis of <with|mode|<quote|math>|\<bbb-R\><rsup|n>>>|<pageref|auto-118>>
<tuple|<tuple|isomorphism>|<pageref|auto-119>>
<tuple|<tuple|induced vector space structure>|<pageref|auto-120>>
<tuple|<tuple|algebra>|<pageref|auto-122>>
<tuple|<tuple|subalgebra>|<pageref|auto-123>>
<tuple|<tuple|topology>|<pageref|auto-126>>
<tuple|<tuple|topological space>|<pageref|auto-127>>
<tuple|<tuple|open set>|<pageref|auto-128>>
<tuple|<tuple|subspace topology>|<pageref|auto-129>>
<tuple|<tuple|inner set>|<pageref|auto-130>>
<tuple|<tuple|closed set>|<pageref|auto-131>>
<tuple|<tuple|closure>|<pageref|auto-132>>
<tuple|<tuple|limit point>|<pageref|auto-133>>
<tuple|<tuple|accumulation point>|<pageref|auto-134>>
<tuple|<tuple|derived set>|<pageref|auto-135>>
<tuple|<tuple|finer topology>|<pageref|auto-136>>
<tuple|<tuple|basis (topology)>|<pageref|auto-137>>
<tuple|<tuple|generation basis>|<pageref|auto-138>>
<tuple|<tuple|subbasis (topology)>|<pageref|auto-139>>
<tuple|<tuple|box topology>|<pageref|auto-140>>
<tuple|<tuple|product topology>|<pageref|auto-141>>
<tuple|<tuple|dense subset>|<pageref|auto-142>>
<tuple|<tuple|Baire set>|<pageref|auto-143>>
<tuple|<tuple|pseudo metric space>|<pageref|auto-145>>
<tuple|<tuple|metric space>|<pageref|auto-146>>
<tuple|<tuple|limit of a function>|<pageref|auto-147>>
<tuple|<tuple|<with|mode|<quote|math>|lim<rsub|h\<rightarrow\>x>f(x)>>|<pageref|auto-148>>
<tuple|<tuple|open ball>|<pageref|auto-149>>
<tuple|<tuple|<with|mode|<quote|math>|B<rsub|d>(x,\<varepsilon\>)>>|<pageref|auto-150>>
<tuple|<tuple|closed ball>|<pageref|auto-151>>
<tuple|<tuple|<with|mode|<quote|math>|<wide|B|\<bar\>><rsub|d>(x,\<varepsilon\>)>>|<pageref|auto-152>>
<tuple|<tuple|equivalent metrics>|<pageref|auto-153>>
<tuple|<tuple|isometry>|<pageref|auto-154>>
<tuple|<tuple|bounded set>|<pageref|auto-155>>
<tuple|<tuple|diameter>|<pageref|auto-156>>
<tuple|<tuple|pseudo normed space>|<pageref|auto-157>>
<tuple|<tuple|normed space>|<pageref|auto-158>>
<tuple|<tuple|normed topology>|<pageref|auto-159>>
<tuple|<tuple|equivalent norms>|<pageref|auto-160>>
<tuple|<tuple|isometry>|<pageref|auto-161>>
<tuple|<tuple|convergent sequence>|<pageref|auto-162>>
<tuple|<tuple|Cauchy sequence>|<pageref|auto-163>>
<tuple|<tuple|real inner product space>|<pageref|auto-165>>
<tuple|<tuple|inner product space>|<pageref|auto-166>>
<tuple|<tuple|complex inner product space>|<pageref|auto-167>>
<tuple|<tuple|inner product space>|<pageref|auto-168>>
<tuple|<tuple|inner product norm>|<pageref|auto-169>>
<tuple|<tuple|continuous mapping>|<pageref|auto-171>>
<tuple|<tuple|open mapping>|<pageref|auto-172>>
<tuple|<tuple|uniform continuous mapping>|<pageref|auto-173>>
<tuple|<tuple|homeomorphism>|<pageref|auto-174>>
<tuple|<tuple|induced topology>|<pageref|auto-175>>
<tuple|<tuple|inverse induced topology>|<pageref|auto-176>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-T\><rsup|-1><rsub|f,Y>>>|<pageref|auto-177>>
<tuple|<tuple|linear mapping>|<pageref|auto-179>>
<tuple|<tuple|<with|mode|<quote|math>|Hom(X,Y)>>|<pageref|auto-180>>
<tuple|<tuple|invertible linear transformation>|<pageref|auto-181>>
<tuple|<tuple|general linear group>|<pageref|auto-182>>
<tuple|<tuple|L(X,Y)>|<pageref|auto-183>>
<tuple|<tuple|operator norm>|<pageref|auto-184>>
<tuple|<tuple|<with|mode|<quote|math>|{1,\<ldots\>,i-1,i+1,\<ldots\>,n}>>|<pageref|auto-185>>
<tuple|<tuple|multilinear mapping>|<pageref|auto-186>>
<tuple|<tuple|<with|mode|<quote|math>|Hom(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>>|<pageref|auto-187>>
<tuple|<tuple|<with|mode|<quote|math>|L(X<rsub|1>,\<ldots\>,X<rsub|n>;Y)>>|<pageref|auto-188>>
<tuple|<tuple|multilinear continuous mapping>|<pageref|auto-189>>
<tuple|<tuple|<with|mode|<quote|math>|L(X<rsub|n>,\<ldots\>,L(X<rsub|1>,Y))>>|<pageref|auto-190>>
<tuple|<tuple|<with|mode|<quote|math>|f(x<rsub|m>:\<ldots\>:x<rsub|n>)>>|<pageref|auto-191>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-P\>>>|<pageref|auto-192>>
<tuple|<tuple|Hausdorff space>|<pageref|auto-194>>
<tuple|<tuple|regular space>|<pageref|auto-195>>
<tuple|<tuple|normal space>|<pageref|auto-196>>
<tuple|<tuple|neighborhood>|<pageref|auto-197>>
<tuple|<tuple|open neighborhood>|<pageref|auto-198>>
<tuple|<tuple|fundamentally system of
neighborhoods>|<pageref|auto-199>>
<tuple|<tuple|first countable set>|<pageref|auto-200>>
<tuple|<tuple|second countable space>|<pageref|auto-201>>
<tuple|<tuple|limit of sequence>|<pageref|auto-202>>
<tuple|<tuple|topological vector space>|<pageref|auto-204>>
<tuple|<tuple|toplinear isomorphism>|<pageref|auto-205>>
<tuple|<tuple|topological group>|<pageref|auto-207>>
<tuple|<tuple|fiber bundle>|<pageref|auto-208>>
<tuple|<tuple|trivial fiber bundle>|<pageref|auto-209>>
<tuple|<tuple|vector bundle>|<pageref|auto-210>>
<tuple|<tuple|compact space>|<pageref|auto-212>>
<tuple|<tuple|compact set>|<pageref|auto-213>>
<tuple|<tuple|sequential compact set>|<pageref|auto-214>>
<tuple|<tuple|filterbase>|<pageref|auto-216>>
<tuple|<tuple|neighborhood filterbase>|<pageref|auto-217>>
<tuple|<tuple|convergent filterbase>|<pageref|auto-218>>
<tuple|<tuple|accumulate (filterbase)>|<pageref|auto-219>>
<tuple|<tuple|subordinate filterbase>|<pageref|auto-220>>
<tuple|<tuple|filterbase on a set>|<pageref|auto-221>>
<tuple|<tuple|maximal filterbase>|<pageref|auto-222>>
<tuple|<tuple|Tychonoff's theorem>|<pageref|auto-224>>
<tuple|<tuple|Heine-Borel theorem>|<pageref|auto-225>>
<tuple|<tuple|Cauchy property>|<pageref|auto-227>>
<tuple|<tuple|Convergence property>|<pageref|auto-228>>
<tuple|<tuple|complete metric space>|<pageref|auto-229>>
<tuple|<tuple|Banach space>|<pageref|auto-230>>
<tuple|<tuple|series>|<pageref|auto-231>>
<tuple|<tuple|Banach space>|<pageref|auto-232>>
<tuple|<tuple|contraction>|<pageref|auto-233>>
<tuple|<tuple|Banach fixed point theorem>|<pageref|auto-234>>
<tuple|<tuple|Baire Category Theorem>|<pageref|auto-235>>
<tuple|<tuple|Open Mapping Theorem>|<pageref|auto-236>>
<tuple|<tuple|partition <with|mode|<quote|math>|\<cal-P\>> of
<with|mode|<quote|math>|[a,b]>>|<pageref|auto-238>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-P\><rsub|1>#\<cal-P\><rsub|2>>>|<pageref|auto-239>>
<tuple|<tuple|norm of partition>|<pageref|auto-240>>
<tuple|<tuple|<with|mode|<quote|math>|\<mu\>(\<cal-P\>)>>|<pageref|auto-241>>
<tuple|<tuple|tag>|<pageref|auto-242>>
<tuple|<tuple|tagged partition>|<pageref|auto-243>>
<tuple|<tuple|Riemann sum>|<pageref|auto-244>>
<tuple|<tuple|Riemann integral>|<pageref|auto-245>>
<tuple|<tuple|connected topological space>|<pageref|auto-247>>
<tuple|<tuple|connected set>|<pageref|auto-248>>
<tuple|<tuple|<with|mode|<quote|math>|U<rsub|x>>>|<pageref|auto-251>>
<tuple|<tuple|<with|mode|<quote|math>|\<varepsilon\>-mapping>>|<pageref|auto-252>>
<tuple|<tuple|differentiable mapping at x>|<pageref|auto-253>>
<tuple|<tuple|differential>|<pageref|auto-254>>
<tuple|<tuple|differential of constant>|<pageref|auto-255>>
<tuple|<tuple|differential of linear function>|<pageref|auto-256>>
<tuple|<tuple|derivate>|<pageref|auto-257>>
<tuple|<tuple|<with|mode|<quote|math>|<frac|df(x)|dx>(y)>>|<pageref|auto-258>>
<tuple|<tuple|chain rule>|<pageref|auto-259>>
<tuple|<tuple|partial differential>|<pageref|auto-260>>
<tuple|<tuple|Jacobian matrix>|<pageref|auto-261>>
<tuple|<tuple|first derivative>|<pageref|auto-263>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|1>>>|<pageref|auto-264>>
<tuple|<tuple|n-th derivative>|<pageref|auto-265>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|n>>>|<pageref|auto-266>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|\<infty\>>>>|<pageref|auto-267>>
<tuple|<tuple|differentiation under the integral
sign>|<pageref|auto-268>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|r>> on
set>|<pageref|auto-270>>
<tuple|<tuple|local weak minimum>|<pageref|auto-273>>
<tuple|<tuple|local weak maximum>|<pageref|auto-274>>
<tuple|<tuple|Rolle's theorem>|<pageref|auto-275>>
<tuple|<tuple|Lagrange's theorem>|<pageref|auto-276>>
<tuple|<tuple|Fundamental theorem of Calculus
(classical)>|<pageref|auto-277>>
<tuple|<tuple|Fundamental theorem of calculus>|<pageref|auto-278>>
<tuple|<tuple|convex set>|<pageref|auto-279>>
<tuple|<tuple|Fundamental theorem of claculus
(generalized)>|<pageref|auto-280>>
<tuple|<tuple|toplinear isomorphism>|<pageref|auto-281>>
<tuple|<tuple|<with|mode|<quote|math>|Gl(X)>>|<pageref|auto-282>>
<tuple|<tuple|<with|mode|<quote|math>|H<rsup|n>>>|<pageref|auto-283>>
<tuple|<tuple|diffeomorphism>|<pageref|auto-284>>
<tuple|<tuple|Inverse function theorem>|<pageref|auto-286>>
<tuple|<tuple|Implicit function theorem>|<pageref|auto-287>>
<tuple|<tuple|topological manifold>|<pageref|auto-290>>
<tuple|<tuple|local chart>|<pageref|auto-291>>
<tuple|<tuple|local coordinate system>|<pageref|auto-292>>
<tuple|<tuple|atlas>|<pageref|auto-293>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|r>> compatible
charts>|<pageref|auto-294>>
<tuple|<tuple|atlas of class <with|mode|<quote|math>|C<rsup|r>>>|<pageref|auto-295>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|r>> compatable
atlasses>|<pageref|auto-296>>
<tuple|<tuple|differentiable structure>|<pageref|auto-297>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-C\><rsup|(r)>[\<cal-A\>]>>|<pageref|auto-298>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-M\>(\<cal-A\>)>>|<pageref|auto-299>>
<tuple|<tuple|differentiable manifold>|<pageref|auto-300>>
<tuple|<tuple|differentiable mapping between
manifolds>|<pageref|auto-302>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|r>>>|<pageref|auto-303>>
<tuple|<tuple|differential curve>|<pageref|auto-305>>
<tuple|<tuple|tangential curves>|<pageref|auto-306>>
<tuple|<tuple|tangent vector>|<pageref|auto-307>>
<tuple|<tuple|tangent space>|<pageref|auto-308>>
<tuple|<tuple|<with|mode|<quote|math>|\<bbb-T\>[c]>>|<pageref|auto-309>>
<tuple|<tuple|<with|mode|<quote|math>|T<rsub|p>M>>|<pageref|auto-310>>
<tuple|<tuple|<with|mode|<quote|math>|T(\<varphi\>,p)>>|<pageref|auto-311>>
<tuple|<tuple|directional derivative>|<pageref|auto-312>>
<tuple|<tuple|<with|mode|<quote|math>|v<rsub|p>[f]>>|<pageref|auto-313>>
<tuple|<tuple|<with|mode|<quote|math>|\<partial\><rsub|i>(p)>>|<pageref|auto-314>>
<tuple|<tuple|coordinate vector fields>|<pageref|auto-315>>
<tuple|<tuple|transformation of vector components>|<pageref|auto-316>>
<tuple|<tuple|<with|mode|<quote|math>|\<equiv\><rsub|p>>>|<pageref|auto-317>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-D\>(M,p)>>|<pageref|auto-318>>
<tuple|<tuple|germs>|<pageref|auto-319>>
<tuple|<tuple|<with|mode|<quote|math>|\<varepsilon\>(M,p)>>|<pageref|auto-320>>
<tuple|<tuple|algebra of germs>|<pageref|auto-321>>
<tuple|<tuple|<with|mode|<quote|math>|\<Delta\><rsub|v<rsub|p>>>>|<pageref|auto-322>>
<tuple|<tuple|derivation>|<pageref|auto-323>>
<tuple|<tuple|<with|mode|<quote|math>|\<Delta\><rsub|v<rsub|p>>>>|<pageref|auto-324>>
<tuple|<tuple|<with|mode|<quote|math>|\<Delta\><rsub|v<rsub|p>>>>|<pageref|auto-325>>
<tuple|<tuple|<with|mode|<quote|math>|T<rsub|M>>>|<pageref|auto-327>>
<tuple|<tuple|vector field>|<pageref|auto-329>>
<tuple|<tuple|<with|mode|<quote|math>|V<rsup|\<varphi\>>>>|<pageref|auto-330>>
<tuple|<tuple|directional derivative>|<pageref|auto-331>>
<tuple|<tuple|<with|mode|<quote|math>|V[f]>>|<pageref|auto-332>>
<tuple|<tuple|<with|mode|<quote|math>|C<rsup|\<infty\>>(M)>>|<pageref|auto-333>>
<tuple|<tuple|<with|mode|<quote|math>|\<cal-D\>(M)>>|<pageref|auto-334>>
<tuple|<tuple|derivation (local)>|<pageref|auto-335>>
<tuple|<tuple|local derivation>|<pageref|auto-336>>
<tuple|<tuple|<with|mode|<quote|math>|\<Omega\><rsub|p>>>|<pageref|auto-337>>
<tuple|<tuple|<with|mode|<quote|math>|\<Omega\><rsub|p>>>|<pageref|auto-338>>
<tuple|<tuple|Lie product>|<pageref|auto-339>>
<tuple|<tuple|Lie algebra>|<pageref|auto-340>>
<tuple|<tuple|Jacobi's identity>|<pageref|auto-341>>
<tuple|<tuple|Jacobi's identity>|<pageref|auto-342>>
</associate>
<\associate|toc>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|1<space|2spc>Preliminaries>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-1><vspace|0.5fn>
1.1<space|2spc>Set, relations and functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-2>
<with|par-left|<quote|1.5fn>|1.1.1<space|2spc>Sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-3>>
<with|par-left|<quote|1.5fn>|1.1.2<space|2spc>Relations
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-11>>
<with|par-left|<quote|1.5fn>|1.1.3<space|2spc>Partial functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-16>>
<with|par-left|<quote|1.5fn>|1.1.4<space|2spc>Functions
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-26>>
<with|par-left|<quote|1.5fn>|1.1.5<space|2spc>Families
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-31>>
<with|par-left|<quote|1.5fn>|1.1.6<space|2spc>Axiom of choice
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-36>>
<with|par-left|<quote|1.5fn>|1.1.7<space|2spc>Order Relations
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-46>>
<with|par-left|<quote|1.5fn>|1.1.8<space|2spc>The real numbers
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-57>>
<with|par-left|<quote|1.5fn>|1.1.9<space|2spc>Finite and infinite sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-70>>
<with|par-left|<quote|1.5fn>|1.1.10<space|2spc>Principle of recursive
definition <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-75>>
<with|par-left|<quote|1.5fn>|1.1.11<space|2spc>Countable sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-77>>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|2<space|2spc>Algebra>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-82><vspace|0.5fn>
2.1<space|2spc>Groups <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-83>
2.2<space|2spc>Fields <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-100>
2.3<space|2spc>Vector spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-103>
2.4<space|2spc>Algebras <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-121>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|3<space|2spc>Topology>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-124><vspace|0.5fn>
3.1<space|2spc>Topological spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-125>
3.2<space|2spc>Metric and normed spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-144>
3.3<space|2spc>Inner product spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-164>
3.4<space|2spc>Continuity <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-170>
<with|par-left|<quote|1.5fn>|3.4.1<space|2spc>Linear maps and
continuity <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-178>>
<with|par-left|<quote|1.5fn>|3.4.2<space|2spc>Separation
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-193>>
<with|par-left|<quote|1.5fn>|3.4.3<space|2spc>Topological Vector Spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-203>>
<with|par-left|<quote|1.5fn>|3.4.4<space|2spc>Bundles
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-206>>
3.5<space|2spc>Compact Spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-211>
<with|par-left|<quote|1.5fn>|3.5.1<space|2spc>Filter bases in Spaces
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-215>>
<with|par-left|<quote|1.5fn>|3.5.2<space|2spc>Product of compacts
subsets <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-223>>
3.6<space|2spc>Complete Spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-226>
3.7<space|2spc>Integration in Banach space
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-237>
3.8<space|2spc>Connected Sets <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-246>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|4<space|2spc>Differentiability
in normed vector spaces> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-249><vspace|0.5fn>
4.1<space|2spc>Differentiability <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-250>
4.2<space|2spc>Higher order differentiability
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-262>
4.3<space|2spc>Differentiability on general sets
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-269>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|5<space|2spc>Inverse
function theorem> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-271><vspace|0.5fn>
5.1<space|2spc>Intermediate value theorem
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-272>
5.2<space|2spc>The inverse function theorem
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-285>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|6<space|2spc>Differentiable
manifolds> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-288><vspace|0.5fn>
6.1<space|2spc>Manifolds <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-289>
6.2<space|2spc>Differentiable mappings between manifolds
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-301>
6.3<space|2spc>Tangent spaces <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-304>
6.4<space|2spc>The tangent bundle <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-326>
6.5<space|2spc>Vector fields <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-328>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|7<space|2spc>Notations>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-343><vspace|0.5fn>
<vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|Index>
<datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
<no-break><pageref|auto-344><vspace|0.5fn>
</associate>
</collection>
</auxiliary>
- [TeXmacs] GNU TeXmacs 1.0.7.5 with experimental PDF support fails to export to PDF, Marc Mertens, 07/24/2010
- Re: [TeXmacs] GNU TeXmacs 1.0.7.5 with experimental PDF support fails to export to PDF, Gubinelli Massimiliano, 07/29/2010
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