Theorem 3

Let $X$ be a uniform space. Then there exists a complete Hausdorff uniform space $\hat{X}$ and a uniformly continuous mapping $i : X \to \hat{X}$ having the following property:

(P) Given any uniformly continuous mapping $f$ of $X$ into a complete Hausdorff uniform space $Y$, there is a unique uniformly continuous mapping $g : \hat{X} \to Y$ such that $f = g \circ i$.

If $(i_1, X_1)$ is another pair consisting of a complete Hausdorff uniform space $X_1$ and a uniformly continuous mapping $i_1 : X \to X_1$ having the property (P), then there is a unique isomorphism $\varphi : \hat{X} \to X_1$ such that $i_1 = \varphi \circ i$.

The first statement of the theorem signifies that the pair $(i, \hat{X})$ is the solution of the universal mapping problem (Set Theory, Chapter IV, , no. 1) in which the $\Sigma$-sets are complete Hausdorff uniform spaces, the $\sigma$-morphisms are uniformly continuous mappings and the $\alpha$-mappings are uniformly continuous mappings of $X$ into a complete Hausdorff uniform space. The uniqueness of the pair $(i, \hat{X})$ up to a unique isomorphism therefore follows from the general properties of solutions of universal mapping problems (loc. cit.). It remains to prove the existence of the pair $(i, \hat{X})$

1) Definition of $\hat{X}$. Let $\hat{X}$ be the set of minimal Cauchy filters (no. 2) on $X$. We shall define a uniform structure on $\hat{X}$. For this purpose, if $V$ is any symmetric entourage of $X$, let $\tilde{V}$ denote the set of all pairs $(\mathscr{K}, \mathfrak{Y})$ of minimal Cauchy filters which have in common a $V$-small set. We shall show that the sets $\tilde{V}$ form a fundamental system of entourages of a uniform structure on $\hat{X}$:

(i) Since each $\mathscr{K} \in \hat{X}$ is a Cauchy filter, we have by definition $(\mathscr{K}, \mathscr{K}) \in \tilde{V}$ for every symmetric entourage $V$ of $X$; hence axiom $(U'_1)$ is satisfied.

(ii) If $V$ and $V'$ are two symmetric entourages of $X$, then $W = V \cap V'$ is a symmetric entourage, and every set which is $W$-small is also $V$-small and $V'$-small; hence $\tilde{W} \subset \tilde{V} \cap \tilde{V}'$, which prove $(B_1)$.

(iii) The sets $\tilde{V}$ are symmetric by definition, hence $(U''_1)$ is satisfied.

(iv) Given a symmetric entourage $V$ of $X$, let $W$ be a symmetric entourage such that $\tilde{V} \subset V$. Consider three minimal Cauchy filters $\mathscr{K}, \mathfrak{Y}, \mathfrak{B}$ such that $(\mathscr{K}, \mathfrak{Y}) \in \tilde{W}$ and $(\mathfrak{Y}, \mathfrak{B}) \in \tilde{W}$; then there are two $W$-small sets $M, N$ such that $M \in \mathscr{K} \cap \mathfrak{Y}$ and $N \in \mathfrak{Y} \cap \mathfrak{B}$. Since $M$ and $N$ belong to $\mathfrak{Y}$, $M \cap N$ is not empty and therefore (no. 1, Proposition 1) $M \cup N$ is $\tilde{W}$-small and hence $V$-small; since $M \cup N$ belongs to $\mathscr{K}$ and to $\mathfrak{B}$ we have $\tilde{W} \subset \tilde{V}$; hence $(U'''_1)$ is satisfied.

We show next that the uniform space $\hat{X}$ is Hausdorff. Let $\mathscr{K}, \mathfrak{Y}$ be two minimal Cauchy filters on $X$ such that $(\mathscr{K}, \mathfrak{Y}) \in \hat{X}$ for all symmetric entourages $V$ of $X$. It follows immediately that the sets $M \cup N$, where $M \in \mathscr{K}$ and $N \in \mathfrak{Y}$, form a base of a filter $\mathfrak{B}$ coarser than $\mathscr{K}$ and $\mathfrak{Y}$. Now $\mathfrak{B}$ is a Cauchy filter, since for every symmetric entourage $V$ of $X$ there is by hypothesis a $V$-small set $P$ belonging to both $\mathscr{K}$ and $\mathfrak{Y}$ and therefore belonging to $\mathfrak{B}$. By the definition of minimal Cauchy filters, we have $\mathscr{K} = \mathfrak{B} = \mathfrak{Y}$, and this shows that $\hat{X}$ is Hausdorff.

2) Definition of $i$; the uniform structure of $X$ is the inverse image under $i$ of that of $\hat{X}$. We know that for each $x \in X$ the neighbourhood filter $\mathfrak{B}(x)$ of $x$ in $X$ is a minimal Cauchy filter (no. 2, Proposition 5, Corollary 1). So we define $i(x) = \mathfrak{B}(x)$. Let $f = i \times i$; we shall show that for each symmetric entourage $V$ of $X$ we have $j^{-1}(\tilde{V}) \subset V \cup j^{-1}[(\tilde{V})^3]$, and this will prove our assertion (§ 2, no. 4). Now, if $[i(x), i(y)] \in \tilde{V}$, there is a $V$-small set $M$ which is a neighbourhood of each of $x$ and $y$, hence $(x, y) \in V$. Conversely, if $(x, y) \in V$, it is immediately seen that the set $V(x) \cup V(y)$ is $V$-small and is a neighbourhood of each of $x$ and $y$.

3) $\hat{X}$ is complete and $i(X)$ is dense in $\hat{X}$. The trace on $i(X)$ of a neighbourhood $\tilde{V}(\mathfrak{X})$ of a point $\mathfrak{X} \in X$ is the set of all $i(x)$ such that

This relation means that there is a $V$-small neighbourhood of $x$ in $X$ which belongs to $\mathfrak{X}$, i.e. that $x$ is an interior point of a $V$-small set of $\mathfrak{X}$. Let $M$ be the union of the interiors of all $V$-small sets of $\mathfrak{X}$; then $M$ belongs to $\mathfrak{X}$ (no. 2, Proposition 5, Corollary 4) and from what has been said it follows that $\tilde{V}(\mathfrak{X}) \cap i(X) = i(M)$. We conclude that:

(i) $\tilde{V}(\mathfrak{X}) \cap i(X)$ is not empty, hence $i(X)$ is dense in $\hat{X}$. (ii) The trace of $\tilde{V}(\mathfrak{X})$ on $i(X)$ belongs to the filter base $i(\mathfrak{X})$ on $X$; hence this filter base converges in $\hat{X}$ to the point $\mathfrak{X}$.

Now let $\mathfrak{F}$ be a Cauchy filter on $i(X)$; then from 2) above and Proposition 4 of no. 1, $\overline{i^1}(\mathfrak{F})$ is a base of a Cauchy filter $\mathfrak{G}$ on $X$. Let $\mathfrak{X}$ be a minimal Cauchy filter coarser than $\mathfrak{G}$ (no. 2, Proposition 5); then $i(\mathfrak{X})$ is a Cauchy filter base on $i(X)$ (no. 1, Proposition 3), and $\mathfrak{F} = i[\overline{i^1}(\mathfrak{F})]$ is finer than the filter whose base is $i(\mathfrak{X})$. Since the latter converges in $\hat{X}$, so does $\mathfrak{F}$, and Proposition 9 of no. 4 therefore shows that $\hat{X}$ is complete.

4) Verification of the property (P). Let $f$ be a uniformly continuous mapping of $X$ into a complete Hausdorff uniform space $Y$. Let us first show that there is a unique uniformly continuous mapping $g_0 : i(X) \to Y$ such that $f = g_0 \circ i$. Since $f$ is continuous, we have

hence if we put $g_0(i(x)) = \lim f(\mathfrak{B}(x))$, we have $f = g_0 \circ i$; so it remains to show that $g_0$ is uniformly continuous in $i(X)$. Let $U$ be an entourage of $Y$ and let $V$ be a symmetric entourage of $X$ such that the relation $(x, x') \in V$ implies $(f(x), f(x')) \in U$; we have seen in 2) that the relation $(i(x), i(x')) \in \tilde{V}$ implies $(x, x') \in V$, hence also is implied

which proves our assertion.

Let $g$ be the extension of $g_0$ by continuity to $\hat{X}$ (no. 6, Theorem 2); then $f = g \circ i$, and it is clear that $g$ is the unique continuous mapping of $\hat{X}$ into $Y$ satisfying this relation, since $i(X)$ is dense in $\hat{X}$ (Chapter I, § 8, no. 1, Proposition 2, Corollary 1).

Q.E.D.