Proposition 6

Let $X$ be a topological (resp. uniform) space, let $Y$ be a uniform space and let $H$ be a subset of $\mathscr{F}(X; Y)$. Then $H$ is equicontinuous at a point $x_0 \in X$ (resp. uniformly equicontinuous) if and only if the closure $\overline{H}$ of $H$ in $\mathscr{F}_s(X; Y)$ is equicontinuous at $x_0$ (resp. uniformly equicontinuous).

The condition is sufficient, trivially. To show that it is necessary, consider an entourage $V$ of $Y$ which is closed in $Y \times Y$; by hypothesis, there is a neighbourhood $U$ of $x_0$ in $X$ (resp. an entourage $M$ of $X$) such that the relation $x \in U$ (resp. $(x', x'') \in M$) implies $(h(x_0), h(x)) \in V$ [resp. $(h(x'), h(x'')) \in V$] for all $h \in H$. Since $V$ is closed, the mappings $h \in \mathscr{F}(X; Y)$ which satisfy the relation $(h(x_0), h(x)) \in V$ for all $x \in U$ [resp. the relation $(h(x'), h(x'')) \in V$ for all $(x', x'') \in M$] form a closed subset of $\mathscr{F}_s(X; Y)$ (§ 1, no. 2, Remark 6); since this closed subset contains $H$, it contains $\overline{H}$. Hence the result, since the closed entourages of $Y$ form a fundamental system of entourages (Chapter II, § 1, no. 2, Proposition 2, Corollary 2).