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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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    $\begingroup$ Arzelà–Ascoli's theorem seems relevant. $\endgroup$ – Rasmus Oct 5 '11 at 19:44
  • $\begingroup$ Maybe proper maps are relevant too. Not sure though, because I don't fully understand the question. $\endgroup$ – shaye Oct 5 '11 at 21:18
  • $\begingroup$ 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. $\endgroup$ – Gotye Oct 5 '11 at 22:10
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Perhaps this is a useful example: suppose $S$ is not compact and $T = [0,1]$ with the usual metric. Take $A$ to be the set of functions $f_y(x) = d(x,y)/(1 + d(x,y))$ from $S$ to $T$. Then $A$ is not compact.
In fact, take any sequence $\{y_n\}$ in $S$ with no convergent subsequence. Then the sequence $f_{y_n}$ has no subsequence converging to a member of $A$, because $|f_{y_n}(y) - f_y(y)| = d(y_n,y)/(1 + d(y_n,y))$ has no subsequence converging to 0.

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