Difference between revisions of "Continuous map"

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{{Refactor notice|grade=A}}
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:: '''Note: ''' there are a few different conditions for continuity, there's also continuity at a point. This diagram is supposed to show how they relate to each other.
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{| class="wikitable" border="1"
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|-
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| style="font-size:1.2em;" |
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{{MM|1=\begin{xy}\xymatrix{
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\text{Continuous} \ar@2{<->}[d]_-{\text{claim }1} \ar@2{<.>}[drr] &  & \\
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{\begin{array}{lr}\text{Continuous at }x_0\\ \text{(neighbourhood)}\end{array} } \ar@2{<->}[rr]_{\text{claim }2} & & {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(sequential)}\end{array} }
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}\end{xy} }}
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| '''Note that: '''
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* All arrow denote logical [[implies]], or "if and only if"
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* Dotted arrows show immediate results of the claims on this page
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|-
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! Overview
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! Key
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|}
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__TOC__
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==[[/Refactoring tasks]]==
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{{/Refactoring tasks}}
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==Definition==
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Given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}} we say that a [[map]], {{M|f:X\rightarrow Y}} is continuous if<ref name="KMAPI">Krzysztof Maurin - Analysis - Part 1: Elements</ref>:
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* {{M|\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}]}}
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That is to say:
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* The [[pre-image]] of every set open in {{M|Y}} under {{M|f}} is open in {{M|X}}
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==Continuous at a point==
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Again, given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}}, and a point {{M|x_0\in X}}, we say the [[map]] {{M|f:X\rightarrow Y}} is ''continuous at {{M|x_0}}'' if<ref name="KMAPI"/>:
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* {{M|\forall N\subseteq Y}}[[neighbourhood|{{M|\text{ neighbourhood to } }}]]{{M|f(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0]}}
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===Claim 1===
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{{Begin Theorem}}
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[[Continuous map/Claim: continuous iff continuous at every point|Claim]]: The [[mapping]] {{M|f}} is continuous {{M|\iff}} it is continuous at every point
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{{Begin Proof}}
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{{:Continuous map/Claim: continuous iff continuous at every point}}
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{{End Proof}}{{End Theorem}}
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==Sequentially continuous at a point==
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Given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}}, and a point {{M|x_0\in X}}, a [[function]] {{M|f:X\rightarrow Y}} is said to be ''continuous at {{M|x_0}}'' if<ref name="KMAPI"/>:
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* {{MM|1=\forall (x_n)_{n=1}^\infty\left[\lim_{n\rightarrow\infty}(x_n)=x\implies\lim_{n\rightarrow\infty}(f(x_n))=f(x)\right]}} (Recall that {{M|1=(x_n)_{n=1}^\infty}} denotes a [[sequence]], see [[Limit (sequence)]] for information on limits)
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===Claim 2===
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{{Begin Theorem}}
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Claim: {{M|f}} is continuous at {{M|x_0}} using the neighbourhood definition {{M|\iff}} it is continuous at {{M|x_0}} using the sequential definition
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{{Begin Proof}}
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'''Proof: ''' neighbourhood {{M|\implies}} sequential:
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: Let {{M|1=(x_n)_{n=1}^\infty}} be given, and that {{M|(x_n)\rightarrow x}} - we wish to show that {{M|1=\lim_{n\rightarrow\infty}(f(x_n))=f(x)}}
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:: Let {{M|\epsilon > 0}} be given.
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::: By hypothesis, as {{M|B_\epsilon(f(x))}} is a neighbourhood of {{M|f(x)}} then {{M|f^{-1}(B_\epsilon(f(x)))}} is a neighbourhood to {{M|x}}
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:::: So {{M|\exists\delta>0[B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))}}
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::::* By the [[implies-subset relation]] if {{M|B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))}} then {{M|a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))}}
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::: Choose {{M|N\in\mathbb{N} }} such that {{M|n>N\implies d_1(x_n,x)<\delta}} (which we can do because {{M|(x_n)\rightarrow x}})
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:::* Note that {{M|d_1(x_n,x)<\delta\implies x_n\in B_\delta(x)}}
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::: So {{M|x_n\in B_\delta(x)\implies x_n\in f^{-1}(B_\epsilon(f(x)))}}
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:::* But if {{M|a\in f^{-1}(B)}} then {{M|f(a)\in B}} as {{M|f^{-1}(B)}} contains exactly the things which map to an element of {{M|B}} under {{M|f}}
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::: So {{M|f(x_n)\in B_\epsilon(f(x))}}
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:: But {{M|f(x_n)\in B_\epsilon(f(x))\iff d_2(f(x_n),f(x))<\epsilon}}
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: This completes the proof
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:* we have shown that given an {{M|\epsilon > 0}} that there exists an {{M|N}} such that for {{M|n>N}} we have {{M|d_2(f(x_n),f(x))<\epsilon}} which is exactly the definition of {{M|(f(x_n))\rightarrow f(x)}}
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'''Proof: ''' sequential {{M|\implies}} neighbourhood:
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{{Todo|This - See Analysis I - Maurin page 44 if stuck}}
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[[Category:First-year friendly]]
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{{End Proof}}
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{{End Theorem}}
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==References==
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<references/>
 
{{Definition|Topology|Metric Space}}
 
{{Definition|Topology|Metric Space}}
  
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=Old page=
 
==First form==
 
==First form==
 
The first form:
 
The first form:
  
 
<math>f:A\rightarrow B</math> is continuous at <math>a</math> if:<br />
 
<math>f:A\rightarrow B</math> is continuous at <math>a</math> if:<br />
<math>\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon</math> (note the [[Implict qualifier|implicit <math>\forall x\in A</math>]])
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<math>\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon</math> (note the [[Implicit qualifier|implicit <math>\forall x\in A</math>]])
  
 
==Second form==
 
==Second form==
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<math>\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}</math> - that is the pre-image of all open sets in <math>(A,\mathcal{J})</math> is open.
 
<math>\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}</math> - that is the pre-image of all open sets in <math>(A,\mathcal{J})</math> is open.
  
{{Todo|The most important Theorem, that these two continuity definitions are the same}}
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==Equivalence of definitions==
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[[Continuity definitions are equivalent]]
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{{Definition|Topology|Metric Space}}

Latest revision as of 01:44, 14 October 2016

Grade: A
This page is currently being refactored (along with many others)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
Note: there are a few different conditions for continuity, there's also continuity at a point. This diagram is supposed to show how they relate to each other.

[math]\begin{xy}\xymatrix{ \text{Continuous} \ar@2{<->}[d]_-{\text{claim }1} \ar@2{<.>}[drr] & & \\ {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(neighbourhood)}\end{array} } \ar@2{<->}[rr]_{\text{claim }2} & & {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(sequential)}\end{array} } }\end{xy}[/math]

Note that:
  • All arrow denote logical implies, or "if and only if"
  • Dotted arrows show immediate results of the claims on this page
Overview Key

/Refactoring tasks

Add the following:

  1. A map is continuous if and only if the pre-image of every closed set is closed
  2. A map is continuous if and only if each point in the domain has an open neighbourhood for which the restriction of the map is continuous on

Definition

Given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] we say that a map, [ilmath]f:X\rightarrow Y[/ilmath] is continuous if[1]:

  • [ilmath]\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}][/ilmath]

That is to say:

  • The pre-image of every set open in [ilmath]Y[/ilmath] under [ilmath]f[/ilmath] is open in [ilmath]X[/ilmath]

Continuous at a point

Again, given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath], and a point [ilmath]x_0\in X[/ilmath], we say the map [ilmath]f:X\rightarrow Y[/ilmath] is continuous at [ilmath]x_0[/ilmath] if[1]:

Claim 1

Claim: The mapping [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] it is continuous at every point


IMG 20151122 220401.jpg

  • A quick proof I did on some scrap - click for full version

This requires one or more proofs to be written up neatly and is on a to-do list for having them written up. This does not mean the results cannot be trusted, it means the proof has been completed, just not written up here yet. It may be in a notebook, some notes about reproducing it may be left in its place, perhaps a picture of it, so forth. The message provided is:
See image


Sequentially continuous at a point

Given two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath], and a point [ilmath]x_0\in X[/ilmath], a function [ilmath]f:X\rightarrow Y[/ilmath] is said to be continuous at [ilmath]x_0[/ilmath] if[1]:

  • [math]\forall (x_n)_{n=1}^\infty\left[\lim_{n\rightarrow\infty}(x_n)=x\implies\lim_{n\rightarrow\infty}(f(x_n))=f(x)\right][/math] (Recall that [ilmath](x_n)_{n=1}^\infty[/ilmath] denotes a sequence, see Limit (sequence) for information on limits)

Claim 2

Claim: [ilmath]f[/ilmath] is continuous at [ilmath]x_0[/ilmath] using the neighbourhood definition [ilmath]\iff[/ilmath] it is continuous at [ilmath]x_0[/ilmath] using the sequential definition


Proof: neighbourhood [ilmath]\implies[/ilmath] sequential:

Let [ilmath](x_n)_{n=1}^\infty[/ilmath] be given, and that [ilmath](x_n)\rightarrow x[/ilmath] - we wish to show that [ilmath]\lim_{n\rightarrow\infty}(f(x_n))=f(x)[/ilmath]
Let [ilmath]\epsilon > 0[/ilmath] be given.
By hypothesis, as [ilmath]B_\epsilon(f(x))[/ilmath] is a neighbourhood of [ilmath]f(x)[/ilmath] then [ilmath]f^{-1}(B_\epsilon(f(x)))[/ilmath] is a neighbourhood to [ilmath]x[/ilmath]
So [ilmath]\exists\delta>0[B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))[/ilmath]
  • By the implies-subset relation if [ilmath]B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))[/ilmath] then [ilmath]a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))[/ilmath]
Choose [ilmath]N\in\mathbb{N} [/ilmath] such that [ilmath]n>N\implies d_1(x_n,x)<\delta[/ilmath] (which we can do because [ilmath](x_n)\rightarrow x[/ilmath])
  • Note that [ilmath]d_1(x_n,x)<\delta\implies x_n\in B_\delta(x)[/ilmath]
So [ilmath]x_n\in B_\delta(x)\implies x_n\in f^{-1}(B_\epsilon(f(x)))[/ilmath]
  • But if [ilmath]a\in f^{-1}(B)[/ilmath] then [ilmath]f(a)\in B[/ilmath] as [ilmath]f^{-1}(B)[/ilmath] contains exactly the things which map to an element of [ilmath]B[/ilmath] under [ilmath]f[/ilmath]
So [ilmath]f(x_n)\in B_\epsilon(f(x))[/ilmath]
But [ilmath]f(x_n)\in B_\epsilon(f(x))\iff d_2(f(x_n),f(x))<\epsilon[/ilmath]
This completes the proof
  • we have shown that given an [ilmath]\epsilon > 0[/ilmath] that there exists an [ilmath]N[/ilmath] such that for [ilmath]n>N[/ilmath] we have [ilmath]d_2(f(x_n),f(x))<\epsilon[/ilmath] which is exactly the definition of [ilmath](f(x_n))\rightarrow f(x)[/ilmath]

Proof: sequential [ilmath]\implies[/ilmath] neighbourhood:


TODO: This - See Analysis I - Maurin page 44 if stuck



References

  1. 1.0 1.1 1.2 Krzysztof Maurin - Analysis - Part 1: Elements

Old page

First form

The first form:

[math]f:A\rightarrow B[/math] is continuous at [math]a[/math] if:
[math]\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon[/math] (note the implicit [math]\forall x\in A[/math])

Second form

Armed with the knowledge of what a metric space is (the notion of distance), you can extend this to the more general:

[math]f:(A,d)\rightarrow(B,d')[/math] is continuous at [math]a[/math] if:
[math]\forall\epsilon>0\exists\delta>0:d(x,a)<\delta\implies d'(f(x),f(a))<\epsilon[/math]
[math]\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))[/math]

In both cases the implicit [math]\forall x[/math] is present. Basic type inference (the [math]B_\epsilon(f(a))[/math] is a ball about [math]f(a)\in B[/math] thus it is a ball in [math]B[/math] using the metric [math]d'[/math])

Third form

The most general form, continuity between topologies

[math]f:(A,\mathcal{J})\rightarrow(B,\mathcal{K})[/math] is continuous if
[math]\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}[/math] - that is the pre-image of all open sets in [math](A,\mathcal{J})[/math] is open.

Equivalence of definitions

Continuity definitions are equivalent