Proposition 4

Let $X, Y$ be two uniform spaces and let $H$ be a set of uniformly continuous mappings of $X$ into $Y$. Let $\hat{X}, \hat{Y}$ be the Hausdorff completions of $X, Y$ respectively, and let $\tilde{H}$ denote the set of mappings $\hat{u} : \hat{X} \to \hat{Y}$ as $u$ runs through $H$ (Chapter II, § 3, no. 7, Proposition 15). Then $H$ is uniformly equicontinuous if and only if $\tilde{H}$ is uniformly equicontinuous.

We recall that the diagram

is commutative, where $i$ and $j$ are the canonical mappings; moreover, the uniformity of $X$ (resp. $Y$) is the inverse image under $i$ (resp. $j$) of that of $\hat{X}$ (resp. $\hat{Y}$). Hence $H$ is uniformly equicontinuous if and only if its image under the mapping $u \to j \circ u$ is uniformly equicontinuous (Proposition 3), and we may already restrict ourselves to the case where $Y$ is Hausdorff and complete; moreover, if $\tilde{H}$ is uniformly equicontinuous then so is $H$, because it is the image of $\tilde{H}$ under the mapping $\hat{u} \to \hat{u} \circ i$; thus it remains to prove the converse when $Y = \hat{Y}$. Let $V$ be a closed entourage of $Y$; by hypothesis there is an entourage $U$ of $X$ such that the relations $(x, x') \in U$ and $u \in H$ imply that $(u(x), u(x')) \in V$. Now, if $U'$ is the image of $U$ under $i \times i$, the closure $\overline{U'}$ of $U'$ in $\hat{X} \times \hat{X}$ is an entourage of $\hat{X}$ (Chapter II, § 3, no. 7, Proposition 12); the hypothesis implies that, whenever $(z, z') \in U'$ and $u \in H$, we have $(\hat{u}(z), \hat{u}(z')) \in V$. Since $V$ is closed and $\hat{u}$ is continuous, we have also $(\hat{u}(t), \hat{u}(t')) \in V$ for all $(t, t') \in \overline{U'}$ and all $u \in H$; this completes the proof.