Theorem 1

On a compact space $X$ there is exactly one uniformity compatible with the topology of $X$; the entourages of this uniformity are all the neighbourhoods of the diagonal $\Delta$ in $X \times X$. Furthermore, $X$ endowed with this uniformity is a complete uniform space.

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.