View source for 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

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: '''Note: ''' this might be called the ''local criterion for continuity''
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==Statement==
Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]], and let {{M|f:X\rightarrow Y}} be a [[map]] between them, then{{rITTMJML}}:
* {{M|f:X\rightarrow Y}} is [[continuous]] {{iff}} for each point {{M|x\in X}} there exists an [[open neighbourhood]] of {{M|x}}, say {{M|U}}, such that {{M|f\big\vert_{U}:U\rightarrow Y}}<ref group="Note">This denotes the {{link|restriction|function}} of {{M|f}} to {{M|U}}, so {{M|f\big\vert_U:U\rightarrow Y}} by {{M|1=f\big\vert_U:u\mapsto f(u)}}</ref> is [[continuous]]
** Symbolically, the right hand side can be written:
*** {{M|1=\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})]}}<ref group="Note">Do I want {{M|\implies}} rather than {{M|\wedge}}?
{{Todo|Explore}}</ref>
** And the full statement
{{Todo|Symbolical representation of entire statement, including continuity!}}
==Proof==
{{Requires proof|review=true|msg=Needs review, not all that happy with it}}
===Continuous {{M|\implies}} Statement ===
Suppose {{M|f:X\rightarrow Y}} is continuous. We wish to show {{M|1=\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})]}}
* Let {{M|x\in X}} be given.
** Then take {{M|X\in\mathcal{J} }} as our [[open neighbourhood]] of {{M|x}}. Automatically {{M|x\in X}} and {{M|f\big\vert_X:X\rightarrow Y}} is just {{M|f:X\rightarrow Y}} and is continuous, we're done.
===Statement {{M|\implies}} continuous===
Suppose {{M|1=\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})]}}, we wish to show {{M|f:X\rightarrow Y}} is continuous.
* Let {{M|U\in\mathcal{K} }} be given (so {{M|U}} is [[open set|open]] in {{Top.|Y|K}}), we must show {{M|f^{-1}(U)}} is open in {{Top.|X|J}}.
** For any/all {{M|x\in f^{-1}(U)}} there exists an [[open neighbourhood]] {{M|V_x\in\mathcal{J} }} of {{M|x}} such that {{M|f\big\vert_{V_x}:V_x\rightarrow Y}} is [[continuous]], specifically this means:
*** {{M|(f\big\vert_{V_x})^{-1}(U)}} is [[relatively open]] in {{M|V_x}} (as we must consider {{M|V_x}} with the [[subspace topology]] it inherits from {{Top.|X|J}})
**** But as {{M|V_x}} is itself open, and [[the relatively open sets of an open subspace are open]]<ref group="Note">This is easy to see even without the theorem. If say {{M|U}} is open in {{M|V_x}} then there exists an [[open set]], {{M|W}}, open in {{M|X}} such that {{M|1=U=V_x\cap W}}, but both {{M|V_x}} and {{M|W}} are open in {{M|X}}! So by the properties of being a [[topology]], their intersection is open too.</ref> we see {{M|(f\big\vert_{V_x})^{-1}(U)}} is open in {{Top.|X|J}} too.
*** We see that {{M|1=(f\big\vert_{V_x})^{-1}(U):=\{x\in V_x\ \vert\ f\big\vert_{V_x}(x)\in U\} }} {{M|1==\{x\in V_x\ \vert\ f(x)\in U\}=f^{-1}(U)\cap V_x}}
*** So {{M|1=(f\big\vert_{V_x})^{-1}(U)=f^{-1}(U)\cap V_x}} is an [[open neighbourhood]] to {{M|x\in f^{-1}(U)}}
** Since {{M|x\in f^{-1}(U)}} was arbitrary, we see this is true for all {{M|x\in f^{-1}(U)}}
** We notice that for every point, {{M|x\in f^{-1}(U)}}, there is an [[open neighbourhood]] to {{M|x}} contained in {{M|f^{-1}(U)}}. 
** Recall [[a set is open if and only if every point in the set has an open neighbourhood contained within the set]]
{{Todo|Check this. Also there may be an alternative union construction we could use. Union of the form {{M|1=\bigcup_{x\in f^{-1}(U)}(f\big\vert_{V_x})^{-1}(U)}}}}
==See also==
* [[Pasting lemma]]
==Notes==
<references group="Note"/>
==References==
<references/>
{{Theorem Of|Topology}}