Let $x$ and $y$ be any two points of $A \cup B$, and let $Z \in A \cap B$. Then $(x, z) \in V$ and $(z, y) \in V$, so that $(x, y) \in \overline{V}$.

If $x$ is any point of $X$ and $V$ is any symmetric entourage of $X$, then the neighbourhood $V(x)$ of $x$ is $\overline{V}$-small. If $\mathfrak{F}$ is a filter which converges to $x$, there is a set of $\mathfrak{F}$ contained in $V(x)$, and therefore $V$-small.

Clearly every filter which is finer than a Cauchy filter is a Cauchy filter.

Let $g = f \times f$. If $V'$ is an entourage of $X'$, then $\overline{g^{-1}(V')}$ is an entourage of $X$, and the image under $f$ of a $\overline{g^{-1}(V')}$-small set is $V'$-small; hence the result.

It follows in particular that if the uniformity of a uniform space $X$ is replaced by a coarser uniformity, then every Cauchy filter with respect to the original uniformity remains a Cauchy filter with respect to the new uniformity.

This fact can be easily remembered in the following form: the finer the uniformity, the fewer Cauchy filters there are.

The condition is necessary by Proposition 3. Conversely, suppose that it is satisfied, and let $U(V_{i_1}, \ldots, V_{i_n})$ be an entourage of the uniformity $\mathfrak{U}$ [§ 2, no. 3, formula (1)]. By hypothesis, for each index $k$ there is a set $M_k \in \mathfrak{B}$ such that $f_{i_k}(M_k)$ is $V_{i_k}$-small ($1 \leq k \leq n$). Let $M$ be a set of $\mathfrak{B}$ contained in $M_k$ for $1 \leq k \leq n$; then for each pair of points $x, x'$ of $M$ we have $[f_{i_k}(x), f_{i_k}(x')] \in V_{i_k}$ for $1 \leq k \leq n$, so that

If $M, M'$ are in $\mathfrak{B}$ and $V, V'$ are in $\mathfrak{G}$, then there is a set $M'' \in \mathfrak{B}$ (resp. $V'' \in \mathfrak{G}$) such that $M'' \subset M \cap M'$ (resp. $V'' \subset V \cap V'$); hence $V''(M'') \subset V(M) \cap V'(M')$ and therefore the sets $V(M)$ ($M \in \mathfrak{B}, V \in \mathfrak{G}$) indeed form a base of a filter $\mathfrak{F}_0$ on $X$. Further, if $M$ is $V$-small, then $V(M)$ is $V$-small; hence $\mathfrak{F}_0$ is a Cauchy filter and is clearly coarser than $\mathfrak{F}$. To complete the proof it is enough to show that if $\mathfrak{G}$ is a Cauchy filter coarser than $\mathfrak{F}$, then $\mathfrak{G}$ is finer than $\mathfrak{F}_0$. For each $M \in \mathfrak{B}$ and each $V \in \mathfrak{G}$ there is a set $N \in \mathfrak{G}$ which is $V$-small; since $N \in \mathfrak{F}$, $N$ meets $M$; hence $N \subset V(M)$ and so $V(M) \in \mathfrak{G}$.

Take $\mathfrak{F}$ in Proposition 5 to be the filter of all subsets of $X$ which contain $x$, and take $\mathfrak{B}$ to consist of the single element $\{x\}$.

There is a filter $\mathfrak{G}$ which is finer than both $\mathfrak{F}$ and $\mathfrak{B}(x)$ (Chapter I, § 7, no. 2, Proposition 4); since $\mathfrak{F}$ is a Cauchy filter, so is $\mathfrak{G}$. If $\mathfrak{F}_0$ is the unique minimal Cauchy filter coarser than $\mathfrak{F}$, then both $\mathfrak{F}_0$ and $\mathfrak{B}(x)$ are minimal Cauchy filters coarser than $\mathfrak{G}$. Hence $\mathfrak{F}_0 = \mathfrak{B}(x)$, which shows that $\mathfrak{F}$ converges to $x$.

This is a consequence of Corollary 2.

Let $V$ be any entourage of $X$; then there is an open entourage $U \subset V$ ($\S$ 1, no. 2, Corollary 2 to Proposition 2). For each subset $M$ of $X$, $U(M)$ is open and contained in $V(M)$; hence the result, in view of Proposition 5.

This criterion shows the importance of complete spaces in all questions involving the notion of limit: if a function takes its values in a complete space we can prove the existence of a limit without knowing in advance the value of the limit; this would be impossible if the definition of limit were the only criterion of convergence at our disposal.

A uniformity which is finer than the uniformity of a complete space need not be a uniformity of a complete space (Exercise 2). However, we have the following proposition:

The conditions are clearly necessary, because $\mathfrak{T}_2$ is coarser than $\mathfrak{T}_1$. Conversely, suppose that the conditions are satisfied, and let $x$ be a limit point of $\mathfrak{F}$ with respect to $\mathfrak{T}_2$; we shall show that $x$ is a limit of $\mathfrak{F}$ with respect to $\mathfrak{T}_1$. Let $V$ be a symmetric entourage of $\mathfrak{U}_1$ which is closed in the topology $\mathfrak{T}_2 \times \mathfrak{T}_2$. By hypothesis, $\mathfrak{F}$ contains a set $M$ which is V-small; hence if $x' \in M$ we have $M \subset V(x')$. But $V(x')$ is closed in the topology $\mathfrak{T}_2$; hence $x$, which lies in the closure of $M$ with respect to $\mathfrak{T}_2$, must belong to $V(x')$. It follows that $M \subset \hat{V}(x)$, and the proposition is proved.

For every Cauchy filter with respect to $U_1$ is then a Cauchy filter with respect to $U_2$ and therefore converges in the topology $\mathfrak{T}_2$.

Note that the hypotheses of the Corollary to Proposition 7 are satisfied when $\mathfrak{T}_1 = \mathfrak{T}_2$ (§ 1, no. 2, Corollary 2 to Proposition 2).

Let $X$ be a complete space and let $A$ be a closed subspace of $X$. If $\mathfrak{F}$ is a Cauchy filter on $A$, then it is a Cauchy filter base on $X$ (no. 1, Proposition 3) and therefore converges to a point $x \in X$; but since $A$ is closed we have $x \in A$, and therefore $\mathfrak{F}$ converges in the subspace $A$.

Now let $\overline{A}$ be a non-closed subset of a Hausdorff uniform space $X$, and let $b \in \overline{A} - A$. The trace $\mathfrak{B}_A$ on $A$ of the neighbourhood filter $\mathfrak{B}$ of $b$ in $X$ is a Cauchy filter on $A$; but it cannot converge to a point $c \in A$, otherwise $c$ would be a limit point of $\mathfrak{B}$ (no. 2, Proposition 5, Corollary 3) which is absurd since $b \neq c$ and $X$ is Hausdorff.

It is enough to show that every minimal Cauchy filter $\mathfrak{F}$ on $X$ is convergent. Since $A$ is dense and since every set of $\mathfrak{F}$ has a non-empty interior (no. 2, Corollary 4 to Proposition 5), the trace $\mathfrak{F}_A$ of $\mathfrak{F}$ on $A$ is a Cauchy filter on $A$, hence converges to a point $x_0 \in X$. Since $\mathfrak{F}$ is coarser than the filter on $X$ generated by $\mathfrak{F}_A$, it follows that $\mathfrak{F}$ converges to $x_0$ (no. 2, Corollary 3 to Proposition 5).

The first assertion is a consequence of the characterization of Cauchy filters and convergent filters on a product space (no. 1, Corollary 2 to Proposition 4 and Chapter I, § 7, no. 6, Corollary 1 to Proposition 10). Conversely, suppose

is complete (the $X_i$ being non-empty) and let $\mathfrak{F}_x$ be a Cauchy filter on $X_x$. For each $i \neq x$ let $\mathfrak{F}_i$ be a Cauchy filter on $X_i$, and consider the product filter (Chapter I, § 6, no. 7)

$\mathfrak{F}$ is a Cauchy filter (no. 1, Corollary 2 to Proposition 4), hence is convergent, and therefore so is each $\operatorname{pr}_x \mathfrak{F} = \mathfrak{F}_x$ (Chapter I, § 7, no. 6, Corollary 1 to Proposition 10).

For $X$ is Hausdorff and can be identified with a closed subspace of $\prod_{i \in I} X_\alpha$ (Chapter I, § 8, no. 2, Corollary 2 to Proposition 7); the corollary therefore follows from Proposition 10 and Proposition 8 (no. 4).

An inverse limit of complete Hausdorff uniform spaces $X_\alpha$ can be empty, even if all the $X_\alpha$ are non-empty and all the $f_{\alpha\beta}$ are surjective, as is shown by the case of discrete spaces (Set Theory, Chapter III, § 1, Exercise 32). However, we have the following theorem:

(α_n) is cofinal in I. We shall write f_{mn} in place of f_{α_mα_n} for m ≤ n, and we shall put f_{n, n+1}(X_{α_{n+1}}) = Y_n. Then, if m ≤ n, f_{mn}(Y_n) is dense in Y_m; for by definition f_{m, n+1}(X_{α_{n+1}}) is dense in

f_{m, m+1}(X_{α_{m+1}}) = Y_m, and f_{m, n+1}(X_{α_{n+1}}) = f_{mn}[f_{n, n+1}(X_{α_{n+1}})] = f_{mn}(Y_n).

By induction on n and k we can define a fundamental system (V_{kn})_{k∈\mathbf{N}} of closed symmetric entourages of X_{α_n} for each n such that

(2) $\overset{2}{V}_{k+1, n} \subset V_{kn}$,

(3) $(f_{n, n+1} \times f_{n, n+1})(V_{k, n+1}) \subset V_{kn}$.

In effect, let (U_{kn})_{k∈\mathbf{N}} be a fundamental system of entourages of X_{α_n}. If we suppose that the V_{kn} have been defined for a given n and for all k ∈ \mathbf{N}, then since f_{n, n+1} is uniformly continuous we can define the entourage V_{k, n+1} by induction on k so that (3) is satisfied and

The assertion follows.

Now let x_0 ∈ Y_0. We shall show that for each integer k > 0 there is a point z ∈ X such that [x_0, f_{α_0}(z)] ∈ V_{k-1, 0}; this will prove the theorem. Since f_{n, n+1} (Y_{n+1}) is dense in Y_n, we can define by induction a sequence of points x_n ∈ Y_n such that

(4)

By reason of (3) it follows that if m ≤ n then

(5)

From this we conclude that for fixed m the sequence (f_{mn}(x_n))_{n≥m} is a Cauchy sequence in X_{α_m} and therefore converges to a point z_m; for by induction it follows from (5) that, for each pair of integers p ≥ m, q > 0, we have

(6)

and by virtue of (2) it is clear that the right-hand side of (6) is contained in V_{k+p-1, m}. Let q increase indefinitely; we infer in particular that, for m = p = 0, we have (x_0, z_0) ∈ V_{k-1, 0}, since V_{k-1, 0} is closed. On the other hand, from the relations z_m = \lim_{n→∞} f_{mn}(x_n) and from the continuity of f_{m, m+1}, we deduce that f_{m, m+1}(z_{m+1}) = z_m for each m ≥ 0. For each γ ∈ I there is at least one integer n such that α_n ≥ γ; putting $z_\gamma = f_{\gamma, \alpha_n}(z_n)$ we verify immediately that $z_\gamma$ does not depend on the value of $n$ such that $\alpha_n \geq \gamma$, and that the family $(z_\alpha)_{\alpha \in I}$ so defined is a point $z$ of $X = \varprojlim X_\alpha$. Since $f_{\alpha_0}(z) = z_0$, the proof is complete.

Let each $X_\alpha$ carry the discrete uniformity, and apply Theorem 1.

which satisfies the following condition: for each $\alpha \in I$ there is an index $\beta \geq \alpha$ such that for all $\gamma \geq \beta$ we have $f_{\alpha \gamma}(\overline{u}_\gamma^{-1}(x_\gamma)) = f_{\alpha \beta}(\overline{u}_\beta^{-1}(x'_\beta))$. Then there is an element $x \in X$ such that $u(x) = x'$.

Apply Theorem 1 to the inverse system of sets $\overline{u}_\alpha^{-1}(x'_\alpha)$, each carrying the discrete uniformity.

This follows from the theorem of extension by continuity (loc. cit.) because $X'$ is regular (§ 1, no. 2, Proposition 3) and because on $X'$ convergent filters are the same as Cauchy filters.

When $X$ is also a uniform space, there is the following theorem:

The existence of $\overline{f}$ is an immediate consequence of Propositions 3 and 11 of no. 1. Let us show that $\overline{f}$ is uniformly continuous. Let $V'$ be a closed symmetric entourage of $X'$, and let $V$ be an entourage of $X$ such that, when $x$ and $y$ are in $A$ and are $V$-close, then $f(x)$ and $f(y)$ are $V'$-close. We may assume that $V$ is the closure in $X \times X$ of an entourage $W$ of $A$ ($\S$ 2, no. 4, Proposition 6). We have $[ \bar{f}(x), \bar{f}(y) ] \in V'$ when $(x, y) \in W$; since $\bar{f} \times \bar{f}$ is continuous in $X \times X$ (Chapter I, $\S$ 4, no. 1, Proposition 1) we have also $[ \bar{f}(x), \bar{f}(y) ] \in V'$ when $(x, y) \in V = \overline{W}$, since $V'$ is closed (Chapter I, $\S$ 2, no. 1, Theorem 1).

Q.E.D.

$f$ is uniformly continuous in $Y_1$, hence (Theorem 2) extends to a uniformly continuous mapping $\bar{f} : X_1 \to X_2$. Likewise the inverse $g$ of $f$ extends to a uniformly continuous mapping $\bar{g} : X_2 \to X_1$. The function $\bar{g} \circ \bar{f}$ is therefore a continuous mapping of $X_1$ into itself whose restriction to $Y_1$ is the identity mapping; by the principle of extension of identities (Chapter I, $\S$ 8, no. 1, Corollary 1 to Proposition 2) $\bar{g} \circ \bar{f}$ is therefore the identity mapping of $X_1$; similarly $\bar{f} \circ \bar{g}$ is the identity map of $X_2$. Consequently (Set Theory, R, $\S$ 2, no. 12) $\bar{f}^{-1}$ and $\bar{g}$ are bijections and are inverses of each other; they are also uniformly continuous and are therefore isomorphisms ($\S$ 2, no. 1, Proposition 2).

It should be remarked that if $f$ is a bijective uniformly continuous mapping of $Y_1$ onto $Y_2$, its extension by continuity need be neither injective nor surjective (Exercise 3).

(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 $\mathfrak{X}$.