Corollary 1

Let $X$ be a topological (resp. uniform) space, let $Y$ be a Hausdorff uniform space and let $H$ be an equicontinuous (resp. uniformly equicontinuous) subset of $C(X; Y)$. Suppose that $H(x)$ is relatively compact in $Y$ for each $x \in X$. Then $H$ is relatively compact in $C(X; Y)$ with respect to the topology of compact (resp. precompact) convergence.

Let $\overline{H}$ be the closure of $H$ in $\mathscr{F}_s(X; Y)$. $\overline{H}$ is equicontinuous (resp. uniformly equicontinuous) (no. 3, Proposition 6). Moreover, we have $\overline{H}(x) \subset \overline{H(x)}$ (§ 1, no. 2, Remark 6) and therefore $\overline{H}(x)$ is also relatively compact; hence Theorem 2 shows that $\overline{H}$ is precompact with respect to $\mathscr{S}$-convergence, where $\mathscr{S}$ denotes the set of all compact (resp. precompact) subsets of $X$. Moreover, since $\overline{H(x)}$ is compact, and therefore complete, $\overline{H}$ is complete with respect to the uniformity of pointwise convergence (Chapter II, § 3, no. 5, Proposition 10 and no. 4, Proposition 8) and therefore also with respect to the uniformity of $\mathscr{S}$-convergence (§ 1, no. 5, Proposition 5, Corollary 2); $\overline{H}$ is therefore compact, since it is precompact, complete and Hausdorff (§ 1, no. 2, Proposition 1).