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

Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces, and let [ilmath]f:X\rightarrow Y[/ilmath] be a map between them, then:

• [ilmath]f:X\rightarrow Y[/ilmath] is continuous if and only if for each point [ilmath]x\in X[/ilmath] there exists an open neighbourhood of [ilmath]x[/ilmath], say [ilmath]U[/ilmath], such that [ilmath]f\big\vert_{U}:U\rightarrow Y[/ilmath] is continuous
  • Symbolically, the right hand side can be written:
    • [ilmath]\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})][/ilmath]
  • And the full statement

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TODO: Symbolical representation of entire statement, including continuity!

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Proof

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Continuous [ilmath]\implies[/ilmath] Statement

Suppose [ilmath]f:X\rightarrow Y[/ilmath] is continuous. We wish to show [ilmath]\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})][/ilmath]

• Let [ilmath]x\in X[/ilmath] be given.
  • Then take [ilmath]X\in\mathcal{J} [/ilmath] as our open neighbourhood of [ilmath]x[/ilmath]. Automatically [ilmath]x\in X[/ilmath] and [ilmath]f\big\vert_X:X\rightarrow Y[/ilmath] is just [ilmath]f:X\rightarrow Y[/ilmath] and is continuous, we're done.

Statement [ilmath]\implies[/ilmath] continuous

Suppose [ilmath]\forall x\in X\exists U\in\mathcal{J}[x\in U\wedge( f\big\vert_U:U\rightarrow Y\text{ is continuous})][/ilmath], we wish to show [ilmath]f:X\rightarrow Y[/ilmath] is continuous.

• Let [ilmath]U\in\mathcal{K} [/ilmath] be given (so [ilmath]U[/ilmath] is open in [ilmath](Y,\mathcal{ K })[/ilmath]), we must show [ilmath]f^{-1}(U)[/ilmath] is open in [ilmath](X,\mathcal{ J })[/ilmath].
  • For any/all [ilmath]x\in f^{-1}(U)[/ilmath] there exists an open neighbourhood [ilmath]V_x\in\mathcal{J} [/ilmath] of [ilmath]x[/ilmath] such that [ilmath]f\big\vert_{V_x}:V_x\rightarrow Y[/ilmath] is continuous, specifically this means:
    • [ilmath](f\big\vert_{V_x})^{-1}(U)[/ilmath] is relatively open in [ilmath]V_x[/ilmath] (as we must consider [ilmath]V_x[/ilmath] with the subspace topology it inherits from [ilmath](X,\mathcal{ J })[/ilmath])
      • But as [ilmath]V_x[/ilmath] is itself open, and the relatively open sets of an open subspace are open we see [ilmath](f\big\vert_{V_x})^{-1}(U)[/ilmath] is open in [ilmath](X,\mathcal{ J })[/ilmath] too.
    • We see that [ilmath](f\big\vert_{V_x})^{-1}(U):=\{x\in V_x\ \vert\ f\big\vert_{V_x}(x)\in U\}[/ilmath] [ilmath]=\{x\in V_x\ \vert\ f(x)\in U\}=f^{-1}(U)\cap V_x[/ilmath]
    • So [ilmath](f\big\vert_{V_x})^{-1}(U)=f^{-1}(U)\cap V_x[/ilmath] is an open neighbourhood to [ilmath]x\in f^{-1}(U)[/ilmath]
  • Since [ilmath]x\in f^{-1}(U)[/ilmath] was arbitrary, we see this is true for all [ilmath]x\in f^{-1}(U)[/ilmath]
  • We notice that for every point, [ilmath]x\in f^{-1}(U)[/ilmath], there is an open neighbourhood to [ilmath]x[/ilmath] contained in [ilmath]f^{-1}(U)[/ilmath].
  • Recall a set is open if and only if every point in the set has an open neighbourhood contained within the set

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TODO: Check this. Also there may be an alternative union construction we could use. Union of the form [ilmath]\bigcup_{x\in f^{-1}(U)}(f\big\vert_{V_x})^{-1}(U)[/ilmath]

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See also

• Pasting lemma