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A standard result in topology is that if $(K,\mathcal{T})$ is a compact space, and $f$ a continuous surjective map into another space $L$, then $L$ is compact.

I'm curious about what happens when we tweak the conditions a bit. Suppose $(S,d)$ and $(T,\delta)$ are metric spaces, with $(T,\delta)$ compact. Is there a known result of a necessary and sufficient condition for a subset $A\subset\mathcal{C}(S,T)$ to be compact also? Thanks for any proof or reference.

Here $\mathcal{C}(S,T)$ is the set of continuous functions from $S$ into $T$ with the standard metric $\rho(f,g)=\sup_{s\in S}\delta(f(s),g(s))$.

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    [Arzelà–Ascoli's theorem](http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Generalizations) seems relevant.2011-10-05
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    Maybe proper maps are relevant too. Not sure though, because I don't fully understand the question.2011-10-05
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    How would the Arzelà–Ascoli theorem apply here? The generalization there shows a necessary and sufficient condition when the domain is a compact Hausdorff space and the codomain is an arbitrary metric space, not when the domain is an arbitrary metric space, and the codomain a compact metric space.2011-10-05

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