Editing Continuous function (section)

==== Closure operator and interior operator definitions ====

In terms of the [[Interior (topology)|interior]] and [[Closure (topology)|closure]] operators, we have the following equivalences,

{{math theorem|name=Theorem|note=|style=|math_statement=Let <math>f: X \to Y</math> be a mapping between topological spaces. Then the following are equivalent.
{{ordered list|type=lower-roman
  | <math>f</math> is continuous;
  | for every subset <math>B \subseteq Y,</math> <math>f^{-1}\left(\operatorname{int}_Y B\right) \subseteq \operatorname{int}_X\left(f^{-1}(B)\right);</math>
  | for every subset <math>A \subseteq X,</math> 
<math>f\left(\operatorname{cl}_X A\right) \subseteq \operatorname{cl}_Y \left(f(A)\right).</math>
}}

}}
{{collapse top|title=Proof|left=true}}
''Proof.''{{spaces|em}}'''i ⇒ ii'''.{{spaces|en}}
Fix a subset <math>B</math> of <math>Y.</math> Since <math>\operatorname{int}_Y B</math> is open.
and <math>f</math> is continuous, <math>f^{-1}(\operatorname{int}_Y B)</math> is open in <math>X.</math>
As <math>\operatorname{int}_Y B \subseteq B,</math> we have <math>f^{-1}(\operatorname{int}_Y B) \subseteq f^{-1}(B).</math>
By the definition of the interior, <math>\operatorname{int}_X\left(f^{-1}(B)\right)</math> is the largest open set contained in <math>f^{-1}(B).</math> Hence <math>f^{-1}(\operatorname{int}_Y B) \subseteq \operatorname{int}_X\left(f^{-1}(B)\right).</math>

'''ii ⇒ iii'''.{{spaces|en}}
Fix <math>A\subseteq X</math> and let <math>x\in\operatorname{cl}_X A.</math> Suppose to the contrary that <math>f(x)\notin\operatorname{cl}_Y\left(f(A)\right),</math>
then we may find some open neighbourhood <math>V</math> of <math>f(x)</math> that is disjoint from <math>\operatorname{cl}_Y\left(f(A)\right)</math>. By '''ii''', <math>f^{-1}(V) = f^{-1}(\operatorname{int}_Y V) \subseteq \operatorname{int}_X \left(f^{-1}(V)\right),</math> hence <math>f^{-1}(V)</math> is open. Then we have found an open neighbourhood of <math>x</math> that does not intersect <math>\operatorname{cl}_X A</math>, contradicting the fact that <math>x\in\operatorname{cl}_X A.</math>
Hence <math>f\left(\operatorname{cl}_X A\right) \subseteq \operatorname{cl}_Y \left(f(A)\right).</math>

'''iii ⇒ i'''.{{spaces|en}}
Let <math>N\subseteq Y</math> be closed. Let <math>M = f^{-1}(N)</math> be the preimage of <math>N.</math>
By '''iii''', we have <math>f\left(\operatorname{cl}_X M\right) \subseteq \operatorname{cl}_Y \left(f(M)\right).</math>
Since <math>f(M) = f(f^{-1}(N)) \subseteq N,</math>
we have further that <math>f\left(\operatorname{cl}_X M\right) \subseteq \operatorname{cl}_Y N = N.</math>
Thus <math>\operatorname{cl}_X M \subseteq f^{-1}\left(f(\operatorname{cl}_X M)\right) \subseteq f^{-1}(N) = M.</math>
Hence <math>M</math> is closed and we are done.
{{collapse bottom}}

If we declare that a point <math>x</math> is {{em|close to}} a subset <math>A \subseteq X</math> if <math>x \in \operatorname{cl}_X A,</math> then this terminology allows for a [[plain English]] description of continuity: <math>f</math> is continuous if and only if for every subset <math>A \subseteq X,</math> <math>f</math> maps points that are close to <math>A</math> to points that are close to <math>f(A).</math> Similarly, <math>f</math> is continuous at a fixed given point <math>x \in X</math> if and only if whenever <math>x</math> is close to a subset <math>A \subseteq X,</math> then <math>f(x)</math> is close to <math>f(A).</math>

Instead of specifying topological spaces by their [[Open set|open subsets]], any topology on <math>X</math> can [[Equivalence of categories|alternatively be determined]] by a [[Kuratowski closure operator|closure operator]] or by an [[interior operator]].  
Specifically, the map that sends a subset <math>A</math> of a topological space <math>X</math> to its [[Closure (topology)|topological closure]] <math>\operatorname{cl}_X A</math> satisfies the [[Kuratowski closure axioms]]. Conversely, for any [[Kuratowski closure operator|closure operator]] <math>A \mapsto \operatorname{cl} A</math> there exists a unique topology <math>\tau</math> on <math>X</math> (specifically, <math>\tau := \{ X \setminus \operatorname{cl} A : A \subseteq X \}</math>) such that for every subset <math>A \subseteq X,</math> <math>\operatorname{cl} A</math> is equal to the topological closure <math>\operatorname{cl}_{(X, \tau)} A</math> of <math>A</math> in <math>(X, \tau).</math> If the sets <math>X</math> and <math>Y</math> are each associated with closure operators (both denoted by <math>\operatorname{cl}</math>) then a map <math>f : X \to Y</math> is continuous if and only if <math>f(\operatorname{cl} A) \subseteq \operatorname{cl} (f(A))</math> for every subset <math>A \subseteq X.</math>

Similarly, the map that sends a subset <math>A</math> of <math>X</math> to its [[Interior (topology)|topological interior]] <math>\operatorname{int}_X A</math> defines an [[interior operator]]. Conversely, any interior operator <math>A \mapsto \operatorname{int} A</math> induces a unique topology <math>\tau</math> on <math>X</math> (specifically, <math>\tau := \{ \operatorname{int} A : A \subseteq X \}</math>) such that for every <math>A \subseteq X,</math> <math>\operatorname{int} A</math> is equal to the topological interior <math>\operatorname{int}_{(X, \tau)} A</math> of <math>A</math> in <math>(X, \tau).</math> If the sets <math>X</math> and <math>Y</math> are each associated with interior operators (both denoted by <math>\operatorname{int}</math>) then a map <math>f : X \to Y</math> is continuous if and only if <math>f^{-1}(\operatorname{int} B) \subseteq \operatorname{int}\left(f^{-1}(B)\right)</math> for every subset <math>B \subseteq Y.</math><ref>{{cite web|title=general topology - Continuity and interior|url=https://math.stackexchange.com/q/1209229|website=Mathematics Stack Exchange}}</ref>