The image of a compact set is compact

Stub grade: A

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Flesh out with references, proof is small but easy and can wait

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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The message provided is:

I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?

Proof

Grade: C

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The message provided is:

Proof isn't that important as it is easy and routine.