Proposition 18

Let $X$ be a set, let $(Y_\lambda)_{\lambda \in L}$ be a family of uniform spaces, and for each $\lambda \in L$ let $f_\lambda$ be a mapping of $X$ into $Y_\lambda$. Let $X$ carry the coarsest uniformity $U$ which makes all the $f_\lambda$ uniformly continuous. Then the uniformity of the Hausdorff completion $\hat{X}$ of $X$ is the coarsest for which all the mappings $\hat{f}_\lambda : \hat{X} \to \hat{Y}_\lambda$ ($\lambda \in L$) (no. 7, Proposition 15) are uniformly continuous. Furthermore, if $j_\lambda$ is the canonical mapping of $Y_\lambda$ into $\hat{Y}_\lambda$, and if $g_\lambda = j_\lambda \circ f_\lambda$, then $\hat{X}$ may be identified with the closure in $\prod_{\lambda \in L} \hat{Y}_\lambda$ of the image of $X$ under the mapping $x \to (g_\lambda(x))$.

Let $X'$ (resp. $Y'_\lambda$) be the Hausdorff uniform space associated with $X$ (resp. $Y_\lambda$), and let $f'_\lambda : X' \to Y'_\lambda$ be the uniformly continuous mapping which makes the diagram

commutative (i being the canonical mapping).

The transitivity of initial uniformities (§ 2, no. 3, Proposition 5) shows on the one hand that $\mathfrak{U}$ is the coarsest uniformity for which the mappings $j_\lambda \circ f_\lambda : X \to Y'_\lambda$ are uniformly continuous, and on the other hand that $\mathfrak{U}$ is also the inverse image under $i$ of the coarsest uniformity $\mathfrak{U}'$ on the set $X'$ for which the $f'_\lambda$ are uniformly continuous. Now $\mathfrak{U}'$ is Hausdorff, for if $x_1, x_2$ are two points of $X$ such that $j_\lambda(f_\lambda(x_1)) = j_\lambda(f_\lambda(x_2))$ for each $\lambda \in L$, then $(x_1, x_2)$ belongs to all the entourages of $\mathfrak{U}$ and hence $i(x_1) = i(x_2)$. Proposition 17 of no. 8 therefore shows that $\mathfrak{U}'$ is the uniformity of the Hausdorff space $X'$ associated with $X$.

This being so, the bijection $x' \to (f'_\lambda(x'))$ identifies $X$ with a uniform subspace of the product $\prod_\lambda Y'_\lambda$ (§ 2, no. 6, Proposition 8). Since the $Y'_\lambda$ are Hausdorff, each $Y'_\lambda$ can be identified with a dense subspace of its completion $\hat{Y}_\lambda$, and hence $\prod_\lambda Y'_\lambda$ can be identified with a dense subspace of $\prod_\lambda \hat{Y}_\lambda$ (Chapter I, § 4, no. 3, Proposition 7). But $\prod_\lambda \hat{Y}_\lambda$ is Hausdorff and complete (no. 5, Proposition 10); the closure $\overline{X'}$ of $X'$ in $\prod_\lambda \hat{Y}_\lambda$ is therefore a complete Hausdorff subspace (no. 4, Proposition 8) which can be identified with the Hausdorff completion $\hat{X}$ of $X$; under this identification the mappings $\hat{f}_\lambda$ become the projections onto the factors $\hat{Y}_\lambda$, and the proposition is proved.