View source for Continuous map

{{Refactor notice|grade=A}}
:: '''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.

{| class="wikitable" border="1"
|-
| style="font-size:1.2em;" | 
{{MM|1=\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} }}
| '''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
|}

__TOC__
==[[/Refactoring tasks]]==
{{/Refactoring tasks}}
==Definition==
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>:
* {{M|\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}]}} 
That is to say:
* The [[pre-image]] of every set open in {{M|Y}} under {{M|f}} is open in {{M|X}}
==Continuous at a point==
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"/>:
* {{M|\forall N\subseteq Y}}[[neighbourhood|{{M|\text{ neighbourhood to } }}]]{{M|f(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0]}}
===Claim 1===
{{Begin Theorem}}
[[Continuous map/Claim: continuous iff continuous at every point|Claim]]: The [[mapping]] {{M|f}} is continuous {{M|\iff}} it is continuous at every point
{{Begin Proof}}
{{:Continuous map/Claim: continuous iff continuous at every point}}
{{End Proof}}{{End Theorem}}

==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"/>:
* {{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)
===Claim 2===
{{Begin Theorem}}
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
{{Begin Proof}}
'''Proof: ''' neighbourhood {{M|\implies}} sequential:
: 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)}}
:: Let {{M|\epsilon > 0}} be given.
::: 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}}
:::: So {{M|\exists\delta>0[B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))}}
::::* 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)))}}
::: 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}})
:::* Note that {{M|d_1(x_n,x)<\delta\implies x_n\in B_\delta(x)}}
::: So {{M|x_n\in B_\delta(x)\implies x_n\in f^{-1}(B_\epsilon(f(x)))}}
:::* 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}}
::: So {{M|f(x_n)\in B_\epsilon(f(x))}}
:: But {{M|f(x_n)\in B_\epsilon(f(x))\iff d_2(f(x_n),f(x))<\epsilon}}
: This completes the proof
:* 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)}}

'''Proof: ''' sequential {{M|\implies}} neighbourhood:
{{Todo|This - See Analysis I - Maurin page 44 if stuck}}
[[Category:First-year friendly]]
{{End Proof}}
{{End Theorem}}

==References==
<references/>
{{Definition|Topology|Metric Space}}

=Old page=
==First form==
The first form:

<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 [[Implicit qualifier|implicit <math>\forall x\in A</math>]])

==Second form==
Armed with the knowledge of what a [[Metric space|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:<br />
<math>\forall\epsilon>0\exists\delta>0:d(x,a)<\delta\implies d'(f(x),f(a))<\epsilon</math><br />
<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 qualifier|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 [[Topological space|topologies]]

<math>f:(A,\mathcal{J})\rightarrow(B,\mathcal{K})</math> is continuous if<br />
<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]]

{{Definition|Topology|Metric Space}}