Show that the general version of Ascoli’s theorem implies the classical version (Theorem 45.4) when \(X\) is Hausdorff.
Solution
The classical version (Theorem 45.4) states: for \(X\) compact and \(\mathcal C(X, \RR^n)\) in the uniform topology, a subset \(\mathcal F\) has compact closure if and only if \(\mathcal F\) is equicontinuous and pointwise bounded.
The general version states: for a space \(X\), a metric space \((Y, d)\), and \(\mathcal C(X, Y)\) in the topology of compact convergence, if \(\mathcal F \subseteq \mathcal C(X, Y)\) is equicontinuous and each set \(\mathcal F_a = \{f(a) : f \in \mathcal F\}\) has compact closure, then \(\mathcal F\) is contained in a compact subspace of \(\mathcal C(X, Y)\); the converse holds if \(X\) is locally compact Hausdorff.
Let \(X\) be compact Hausdorff and \(Y = \RR^n\).
1. On \(\mathcal C(X, \RR^n)\), the topology of compact convergence equals the uniform topology, and it is Hausdorff.
\(X\) is compact, so the two topologies coincide by E-2AWO8. The uniform topology is induced by a metric.
2. For \(a \in X\), \(\mathcal F_a\) has compact closure in \(\RR^n\) if and only if \(\mathcal F_a\) is bounded.
The closure of a bounded subset of \(\RR^n\) is closed and bounded, hence compact by the Heine–Borel theorem; a set with compact closure is bounded.
3. If \(\mathcal F\) is equicontinuous and pointwise bounded, then \(\overline{\mathcal F}\) is compact.
By step 2 each \(\mathcal F_a\) has compact closure, so the general version puts \(\mathcal F\) inside a compact subspace \(K\). By step 1, \(K\) is closed in the Hausdorff space \(\mathcal C(X, \RR^n)\), so \(\overline{\mathcal F} \subseteq K\) is a closed subset of a compact space, hence compact.
4. If \(\overline{\mathcal F}\) is compact, then \(\mathcal F\) is equicontinuous and pointwise bounded.
\(\mathcal F\) is contained in the compact subspace \(\overline{\mathcal F}\). A compact Hausdorff space is locally compact Hausdorff, so the converse in the general version shows that \(\mathcal F\) is equicontinuous and each \(\mathcal F_a\) has compact closure; by step 2, each \(\mathcal F_a\) is bounded.
5. Q.E.D.
Steps 3 and 4 are the two implications of the classical version.