The last part of the theorem is straightforward; for every Cauchy filter on $X$ has a cluster point [axiom (C)] and is therefore convergent ($§ 3$, no. 2, Proposition 5, Corollary 2).

Let us show next that if there is a uniformity on $X$ compatible with the topology of $X$, then the set $U$ of entourages of this uniformity is the set of all neighbourhoods of $\Delta$. We know already that every entourage is a neighbourhood of $\Delta$ ($§ 1$, no. 2, Proposition 2), hence we have to show that conversely every neighbourhood of $\Delta$ belongs to $U$. Suppose that there is a neighbourhood $U$ of $\Delta$ which is not in $U$; then the sets $V \cap C_U$, as $V$ runs through $U$, form a base of a filter $G$ on the compact space $X \times X$; consequently $G$ has a cluster point $(a, b)$ not belonging to $\Delta$. Since $U$ is a filter coarser than $G$, $(a, b)$ is also a cluster point of $U$. But the uniformity defined by $U$ is Hausdorff by hypothesis; hence the intersection of the closures of the sets of $U$ is $\Delta$ ($§ 1$, no. 2, Corollary 2 to Proposition 2 and Proposition 3); thus we arrive at a contradiction.

It remains therefore to show that the set $B$ of neighbourhoods of $\Delta$ in $X \times X$ is the set of entourages of a uniformity compatible with the topology of $X$. For this it is enough to show that $B$ is the set of entourages of a Hausdorff uniformity on $X$; for then the topology induced by this uniformity will be coarser than the topology of $X$ (Chapter I, $§ 2$, no. 2, Proposition 3) and must therefore coincide with the latter (Chapter I, $§ 9$, no. 4, Theorem 2, Corollary 3).

$B$ clearly satisfies axioms $(F_I)$ and $(F_{II})$. Let us show that axioms $(U_{II})$ and $(U_{III})$ are also satisfied and that $\Delta$ is the intersection of the sets of $B$. Take the last point first: every set consisting of a single point $(x, y) \in X \times X$ is closed, since $X$ is Hausdorff; hence if $x \neq y$, the complement of $(x, y)$ in $X \times X$ is a neighbourhood of $\Delta$. Since the symmetry $(x, y) \to (y, x)$ is a homeomorphism of $X \times X$ onto itself, it follows that $V \in \mathfrak{B}$ implies $\overline{V}^1 \in \mathfrak{B}$, whence $(\mathbf{U}_{\mathrm{II}})$. Finally, suppose that $(\mathbf{U}_{\mathrm{III}})$ is not satisfied; then there is a set $V \in \mathfrak{B}$ such that for all $W \in \mathfrak{B}$ the set $\hat{W} \cap \mathscr{C}V$ is not empty; the sets $\hat{W} \cap \mathscr{C}V$, as $W$ runs through $\mathfrak{B}$, therefore form a filter base on $X \times X$, and this filter base has a cluster point $(x, y)$ not in $\Delta$. Now, since $X$ is regular (Chapter I, , no. 2, Corollary to Proposition 1) there are disjoint open neighbourhoods $U_1$ of $x$ and $U_2$ of $y$, and there are closed neighbourhoods $V_1 \subset U_1$, $V_2 \subset U_2$ of $x$ and $y$ respectively. Let $U_3 = \mathscr{C}(V_1 \cup V_2)$, and consider the neighbourhood

It follows immediately from these definitions that if $(u, v) \in W$ and $u \in V_1$ (resp. $u \in U_1$), then we must have $v \in U_1$ (resp. $v \in U_1 \cup U_3 = \mathscr{C}V_2$); hence the neighbourhood $V_1 \times V_2$ of $(x, y)$ in $X \times X$ does not meet $\hat{W}$, and we have a contradiction. This completes the proof.

For by Alexandroff’s theorem (Chapter I, § 9, no. 8, Theorem 4) a locally compact space is homeomorphic to a subspace of a compact space.

Let $g = f \times f$, then $g$ is continuous on $X \times X$ (Chapter I, § 4, no. 1, Proposition 1, Corollary 1); hence for each open entourage $V'$ of $X'$, $\overline{g^{-1}}(V')$ is an open subset of $X \times X$, which evidently contains the diagonal. The theorem thus follows from Theorem 1, since the open entourages of $X'$ form a fundamental system of entourages (§ 1, no. 2, Proposition 2, Corollary 2).

Under the hypotheses of Theorem 2, the restriction of $f$ to any subspace $A$ of $X$ is uniformly continuous; hence (§ 3, no. 6, Theorem 2):

We may express this condition more intuitively by saying that $X$ can be covered by a finite number of sets which are as small as we please.

Let $i : X \to \hat{X}$ be the canonical mapping, then the entourages of $X$ are the inverse images under $i \times i$ of the entourages of $\hat{X}$ (§ 3, no. 7, Proposition 12). Suppose $X$ is precompact, and let $U$ be any entourage of $\hat{X}$; then there is a symmetric entourage $U'$ of $\hat{X}$ such that $\hat{U}' \subset U$. Since $\hat{X}$ is compact, there exist a finite number of points $x_j \in \hat{X}$ such that the $U'(x_j)$ (which are $U$-small) cover $\hat{X}$. If $V$ is the inverse image of $U$ by $i \times i$, the sets $i^{-1}(U'(x_j))$ are $V$-small and cover $X$. Conversely, suppose that for each entourage $V$ of $X$ there is a finite covering of $X$ by $V$-small sets. We have to show that every ultrafilter $\mathfrak{F}$ on $\hat{X}$ is convergent; since $\hat{X}$ is complete, it is enough to show that $\mathfrak{F}$ is a Cauchy filter, i.e. that for each closed entourage $U$ of $\hat{X}$ there is a $U$-small set in $\mathfrak{F}$ (§ 1, no. 2, Proposition 2, Corollary 2). Let $V$ be the inverse image of $U$ under $i \times i$, and let $(B_j)$ be a finite covering of $X$ by $V$-small sets; then the sets $C_j = i(B_j)$ are $U$-small and cover $i(X)$, so that

On the other hand, since $C_j \times C_j \subset U$, and $U$ is closed in $\hat{X} \times \hat{X}$, we have $\overline{C}_j \times \overline{C}_j \subset U$, so that the $\overline{C}_j$ are also $U$-small. Since $\mathfrak{F}$ is an ultrafilter, one of the $\overline{C}_j$ belongs to $\mathfrak{F}$ (Chapter I, § 6, no. 4, Corollary to Proposition 5).

Q.E.D.

This follows from Theorem 3 and Theorem 1 of no. 1.

The first two assertions are immediate consequences of Theorem 3. Let $X$ be a uniform space, $A$ a precompact subset of $X$, and let $i : X \to \hat{X}$ be the canonical mapping. $i(\overline{A})$ is contained in the closure of $i(A)$ in $\hat{X}$ (Chapter I, § 2, no. 1, Theorem 1), hence the closure of $i(\overline{A})$ in $\hat{X}$ is contained in a compact set and is therefore compact.

For if $i : X \to \hat{X}$ and $j : Y \to \hat{Y}$ are the canonical mappings, we have $j(f(A)) = \hat{f}(i(A))$ (§ 3, no. 7, Proposition 15) and hence $j(f(A))$ is relatively compact in $\hat{Y}$ (Chapter I, § 9, no. 4, Theorem 2, Corollary 1).

The condition is necessary by virtue of Proposition 2. Sufficiency follows from the characterization of the Hausdorff completion of $X$ given in § 3, no. 9, Proposition 18, and Tychonoff’s theorem (Chapter I, § 9, Theorem 3, Corollary).

If the proposition were false, then none of the sets $A \cap \overline{V}(B)$, where $V$ runs through the set of symmetric entourages of $X$, would be empty; hence these sets would form a filter base on $A$, which would have a cluster point $x_0 \in A$. Hence for each symmetric entourage $V$ of $X$, $\overline{V}(x_0)$ would meet $B$ and therefore, as $B$ is closed, we should have $x_0 \in B$, contrary to hypothesis.

Let $U$ be any open neighbourhood of $A$, then $B = \complement U$ is closed and does not meet $A$; hence, by Proposition 4, there is an entourage $V$ such that $V(A) \cap V(B) = \varnothing$, and therefore $V(A) \subset U$.

For each $x \in X$, let $A_x$ denote the intersection of the sets $A_{x, v}$ as $V$ runs through the set of symmetric entourages of $X$; $A_x$ is the equivalence class of $x$ for the equivalence relation "for all symmetric entourages $V$, there is a $V$-chain joining $x$ and $y$".

It is enough to show that $A_x$ is connected: for in any uniform space $X$ the component of $x$ is contained in the intersection of the neighbourhoods of $x$ which are both open and closed, and this intersection is contained in $A_x$ by Proposition 5.

Suppose $A_x$ is not connected. Since $A_x$ is closed, there are two non-empty disjoint closed sets $B$ and $C$ such that $B \cup C = A_x$. By Proposition 4 of § 3 there is an entourage $U$ of $X$ such that $U(B) \cap U(C)$ is empty.

Let $W$ be an open entourage such that $\hat{W} \subset U$, and let $H$ be the closed set which is the complement of $W(B) \cup W(C)$ in $X$. Suppose for example that $x \in B$, and let $y$ be a point of $C$. Then for each symmetric entourage $V \subset W$ one sees immediately, by induction on $i$, that every $V$-chain $(x_i)_{0 \leq i \leq n}$ joining $x$ and $y$ in $X$ must have a point in $H$, by the choice of $W$. Since by hypothesis $x$ and $y$ can be joined by a $V$-chain for each symmetric entourage $V$, we see that $H \cap A_{x, v}$ is not empty if $V \subset W$. On the other hand, if $V' \subset V$ then clearly $A_{x, v'} \subset A_{x, v}$; thus it follows that, as $V$ runs through the set of symmetric entourages of $X$, the sets $H \cap A_{x, v}$ form a filter base of closed sets in the compact space $H$. Hence all these sets have a common point and therefore $H$ meets $A_x$; but this contradicts the definition of $H$.

Let V be a relatively compact open neighbourhood of K in X (Chapter I, § 9, no. 7, Proposition 10) and let F be its frontier. Let U ⊂ \overline{V} be a set which is both open and closed with respect to \overline{V}. Then U is closed in X, and if in addition U does not meet F, then U is open in X (for then U ⊂ V and U is open in V). Hence it is enough to show that there is a subset of \overline{V} which contains K, does not meet F and is both open and closed with respect to \overline{V}.

Suppose this is not the case: then the intersections of F with the subsets of \overline{V} which contain K and are open and closed in \overline{V} form a filter base of closed sets in F; F is compact, so these sets have a common point y ∈ F. But this is absurd; since \overline{V} is compact, K is a component of \overline{V}, and by virtue of Proposition 6, the intersection of the sets which contain K and are both open and closed in \overline{V} is just K. The Corollary is thus proved.

We know from Chapter I, § II, no. 5, Proposition 9 that X/R is totally disconnected; hence we have to show that X/R is Hausdorff (Chapter I, § 10, no. 4, Proposition 8.) Let A and B be two distinct components of X. By Proposition 6 there is a symmetric entourage U of X such that no point of A can be joined to any point of B by a U-chain. The set V (resp. W) of points of X which can be joined to a point x ∈ A (resp. y ∈ B) by a U-chain is both open and closed in X (Proposition 5) and contains A (resp. B); these sets are therefore open neighbourhoods of A and B respectively, are saturated with respect to R and do not intersect. This completes the proof.