The image of a compact set is compact
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Flesh out with references, proof is small but easy and can wait
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Statement
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces, let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath] and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map. Then:
 if [ilmath]A[/ilmath] is compact then [ilmath]f(A)[/ilmath] is compact.
Grade: A
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I want Mendelson. Also should I make a note that considered as a subspace or compactnessasasubset are the same?
Proof
 Let [ilmath]\mathcal{U} [/ilmath] be an open cover of [ilmath]f(A)[/ilmath] and then showing it has a finite subcover
 The proof depends on [ilmath]\{f^{1}(U)\ \vert\ U\in\mathcal{U} \} [/ilmath] being an open cover of [ilmath]A[/ilmath] by sets open in [ilmath](X,\mathcal{ J })[/ilmath]
 The open part comes from the [ilmath]U[/ilmath] be open sets, then by the definition of continuity [ilmath]f^{1}(U)[/ilmath] is open in [ilmath]X[/ilmath] for each [ilmath]U\in\mathcal{U} [/ilmath]
 We need to show [math]A\subseteq\bigcup_{U\in\mathcal{U} }f^{1}(U)[/math], by the impliessubset relation and then the definition of union we see:
 [math]\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][/math] [math]\iff\forall x\in A\exists U\in\mathcal{U}[x\in f^{1}(U)][/math]
 In this case [ilmath]\{f^{1}(U)\ \vert\ U\in\mathcal{U} \} [/ilmath] is an open cover of [ilmath]A[/ilmath]
 Then we see, by compactness of [ilmath]A[/ilmath]:
 there is some [ilmath]n\in\mathbb{N} [/ilmath] such that [ilmath]\{f^{1}(U_1),\ldots,f^{1}(U_n)\} [/ilmath]
 The proof depends on [ilmath]\{f^{1}(U)\ \vert\ U\in\mathcal{U} \} [/ilmath] being an open cover of [ilmath]A[/ilmath] by sets open in [ilmath](X,\mathcal{ J })[/ilmath]
Grade: C
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References
