Prove the following.
Theorem (Arzela’s theorem, general version). Let \(X\) be a Hausdorff space that is \(\sigma\)-compact; let \(f_n\) be a sequence of functions \(f_n: X \to \mathbb{R}^k\). If the collection \((f_n)\) is pointwise bounded and equicontinuous, then the sequence \(f_n\) has a subsequence that converges, in the topology of compact convergence, to a continuous function.
[Hint: Show that \(\mathcal{C}(X, \mathbb{R}^k)\) is first-countable.]
Solution
Choose compact sets \(A_1\subset A_2\subset\cdots\) whose interiors cover the \(\sigma\)-compact Hausdorff space \(X\). On \(C(X,\mathbb R^k)\) the compact-convergence topology is metrizable, for example by \[ \rho(f,g)=\sum_{j\ge1}2^{-j}\min\{1,\sup_{A_j}|f-g|\}, \] so it is first-countable.
For each compact \(A_j\), the restricted family \(\{f_n|_{A_j}\}\) is pointwise bounded and equicontinuous. By Ascoli on the compact domain \(A_j\), its closure in the uniform topology is compact; hence every sequence has a uniformly convergent subsequence on \(A_j\). Starting with the original sequence, choose successively subsequences converging uniformly on \(A_1,A_2,\dots\). The diagonal subsequence converges uniformly on every \(A_j\), hence on every compact subset of \(X\), to a function \(f\). The function \(f\) is continuous: its restriction to each \(A_j\) is a uniform limit of continuous functions, and the interiors of the \(A_j\) form an open cover of \(X\). Thus the diagonal subsequence converges to \(f\) in compact convergence.