Show that our proof of Ascoli’s theorem goes through if \(\mathbb{R}^n\) is replaced by any metric space in which all closed bounded subspaces are compact.
Solution
In the proof of Ascoli’s theorem, the only property of the target \(\mathbb R^n\) used beyond metrizability is that for each \(x\in X\), pointwise boundedness places the set \[ \{f(x):f\in\mathcal F\} \] in a compact set, and likewise finitely many bounded target sets arising in the finite-net argument have compact closure. If every closed bounded subset of the metric target \(Y\) is compact, these steps remain valid verbatim.
More explicitly, equicontinuity reduces uniform control on a compact domain to finitely many evaluation points. Pointwise boundedness makes the possible values at each of these finitely many points lie in bounded subsets of \(Y\); their closures are compact by hypothesis, hence totally bounded. Finite \(\varepsilon\)-nets in these compact closures then give a finite \(\varepsilon\)-net for \(\mathcal F\) in the uniform metric. Thus \(\mathcal F\) is totally bounded. The closure of an equicontinuous family is equicontinuous, and pointwise limits remain in the compact value closures; so the closure of \(\mathcal F\) is complete and totally bounded, hence compact, as in the Euclidean proof. Hence Ascoli’s theorem holds for every proper metric target (every closed bounded subset compact).