View source for The image of a compact set is compact

{{Stub page|grade=A|msg=Flesh out with references, proof is small but easy and can wait}}
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==Statement==
Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]], let {{M|A\in\mathcal{P}(X)}} be an arbitrary [[subset of]] {{M|X}} and let {{M|f:X\rightarrow Y}} be a ''[[continuous]]'' [[map]]. Then:
* if {{M|A}} is [[compact]] then {{M|f(A)}} is compact.
{{Requires references|grade=A|I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?}}
==Proof==
* Let {{M|\mathcal{U} }} be an [[open cover]] of {{M|f(A)}} and then showing it has a finite subcover
** The proof depends on {{M|\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} }} being an [[open cover]] of {{M|A}} by sets [[open set|open]] in {{Top.|X|J}}
**# The open part comes from the {{M|U}} be [[open sets]], then by the definition of [[continuity]] {{M|f^{-1}(U)}} is open in {{M|X}} for each {{M|U\in\mathcal{U} }}
**# We need to show {{MM|A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)}}, by the [[implies-subset relation]] and then the definition of [[union]] we see:
**#* {{MM|\left(A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right)\iff\forall x\in A\left[x\in\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right]}} {{MM|\iff\forall x\in A\exists U\in\mathcal{U}[x\in f^{-1}(U)]}} 
**#* In this case {{M|\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} }} is an [[open cover]] of {{M|A}}
** Then we see, by [[compactness]] of {{M|A}}:
*** there is some {{M|n\in\mathbb{N} }} such that {{M|\{f^{-1}(U_1),\ldots,f^{-1}(U_n)\} }}
{{Requires proof|grade=C|msg=Proof isn't that important as it is easy and routine.
* Done a chunk, but not finished. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 17:20, 18 December 2016 (UTC)}}
==References==
<references/>
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{{Theorem Of|Topology}}