Difference between revisions of "Continuous map"
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Claim: The [[mapping]] {{M|f}} is continuous {{M|\iff}} it is continuous at every point | Claim: The [[mapping]] {{M|f}} is continuous {{M|\iff}} it is continuous at every point | ||
{{Begin Proof}} | {{Begin Proof}} | ||
− | {{Todo| | + | [[File:IMG 20151122 220401.jpg|200px]] |
+ | * A quick proof I did on some scrap - click for full version | ||
+ | {{Todo|Write up proof}} | ||
+ | [[Category:Writeup]] | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
+ | |||
==Sequentially continuous at a point== | ==Sequentially continuous at a point== | ||
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"/>: | 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"/>: |
Revision as of 22:33, 22 November 2015
- 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:
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Overview | Key |
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Contents
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]:
- [ilmath]\forall N\subseteq Y[/ilmath][ilmath]\text{ neighbourhood to } [/ilmath][ilmath]T(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0][/ilmath]
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
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
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.