Difference between revisions of "Continuous map"

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{{Refactor notice}}
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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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==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|T(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0]}}
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{{Begin Inline Theorem}}
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'''UNPROVED: '''{{Note|I suspect that this is the same as {{M|\forall\mathcal{O}\in\mathcal{K}[f(x_0)\in\mathcal{O}\implies f^{-1}(\mathcal{O})\in\mathcal{J}\wedge x_0\in f^{-1}(\mathcal{O})]}} - this is basically the same just on open sets instead}}
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{{Begin Inline Proof}}
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{{Todo|{{Note|Investigate and prove the highlighted claim}}}}
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(Leave any notes to self here)
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{{End Proof}}{{End Theorem}}
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===Claim 1===
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{{Begin Theorem}}
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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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{{Todo|Do this}}
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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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{{Todo|Fill this out}}
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{{End Proof}}
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{{End Theorem}}
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==References==
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<references/>
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{{Definition|Topology|Metric Space}}
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=Old page=
 
==First form==
 
==First form==
 
The first form:
 
The first form:

Revision as of 04:22, 22 November 2015

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

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]:

UNPROVED: I suspect that this is the same as [ilmath]\forall\mathcal{O}\in\mathcal{K}[f(x_0)\in\mathcal{O}\implies f^{-1}(\mathcal{O})\in\mathcal{J}\wedge x_0\in f^{-1}(\mathcal{O})][/ilmath] - this is basically the same just on open sets instead




TODO: Investigate and prove the highlighted claim


(Leave any notes to self here)

Claim 1

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




TODO: Do this


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




TODO: Fill this out


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