# Is a compact Hausdorff uniform space fine?

Let $\mathcal D_1$ and $\mathcal D_2$ be two uniformities on $X$ which produce the same topologies on $X$ (say $\mathcal T= \mathcal T _{\mathcal D_1}=\mathcal T _{\mathcal D_2}$).

If $(X,\mathcal T)$ is compact and Hausdorff, is $\mathcal D_1$ the same as $\mathcal D_2$?

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Yes, compact Hausdorff spaces have a unique uniformity compatible with their topology.

I assume you already know that the system $\mathcal{D}_0$ of all neighborhoods of the diagonal $\Delta = \{(x,x) \in X \times X \mid x \in X\}$ of $X \times X$ is a uniformity on a compact Hausdorff space $X$. This is not trivial, however, it is not very hard to prove.

Suppose $\mathcal{D}$ is another uniformity on $(X,\mathcal{T})$ inducing the topology. Then $\mathcal{D} \subseteq \mathcal{D}_0$. Let us show the reverse inclusion.

Since $X$ is Hausdorff, the diagonal $\Delta = \{(x,x) \in X \times X \mid x \in X\}$ is equal to the intersection of all closures of entourages of the diagonal: $$\Delta = \bigcap_{U \in \mathcal{D}} \overline{U}.$$ Let $W \supseteq \Delta$ be an open neighborhood of the diagonal, so $W \in \mathcal{D}_0$. Then $$X \times X = W \cup \Delta^c = W \cup \bigcup_{U \in \mathcal{D}} \overline{U}^c$$ and by compactness of $X \times X$ there are $U_1,\dots,U_n$ such that $$X \times X = W \cup \overline{U}_{1}^c \cup \dots \cup \overline{U}_{n}^c,$$ so $W \supseteq \overline{U}_{1} \cap \dots \cap \overline{U}_{n} \supseteq U_1 \cap \dots \cap U_n$ and hence $W \in \mathcal{D}$.

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I'm trying to prove it more elementarily.

Suppose $\mathcal D$ and $\mathcal E$ are twoHaussdorff and compact uniformities on $X$ and $\mathcal T_{\mathcal D}=\mathcal T_{\mathcal E}$ and $\mathcal D\subsetneqq \mathcal E$. I'm trying to reach a contradiction.

There is some $E\in \mathcal E\setminus\mathcal D$. So $$(\forall D\in \mathcal D)(D\nsubseteq E)$$ $$\Rightarrow (\forall D\in \mathcal D)(\exists (x,y)\in D)((x,y)\notin E)$$ So there a net $n:(\mathcal D,\subseteq^{-1})\to X^2$ such that $$(\forall D\in \mathcal D)(n(D)\in D)$$ and $$(\forall D\in \mathcal D)(n(D)\notin E)$$

Also there's a symmetric $E'\in \mathcal E$ such that

$$E'oE'\subseteq E$$

$X^2$ is compact and so $n$ has a convergent subnet. that is, there's some increasing cofinal function: $$\theta:(P,\le)\to (\mathcal D,\subseteq^{-1})$$ and some $(a,b)\in X^2$ such that $$no\theta \to(a,b)$$ For each $D\in \mathcal D$ there's some symmetric $D'\in \mathcal D$ such that: $$D'oD'oD'\subseteq D$$ On the other hand, since $\theta$ is cofinal and increasing, there's some $p\in P$ such that: $$(\forall x \ge p)(\theta (x)\subseteq D')$$ $$\Rightarrow (\forall x \ge p)\left(n(\theta (x))\in \theta (x)\subseteq D'\right)$$ $$\Rightarrow(a,b)\in \overline {D'} \subseteq D'oD'oD'\subseteq D$$ $D$ is arbitrary: $$(a,b)\in \bigcap_{D\in \mathcal D} D=\Delta$$ $$\Rightarrow a=b$$ $$\Rightarrow no\theta \to (a,a)$$

So there' some $p\in P$ such that $$n(\theta(p)) \in E'[a]\times E'[a]\subseteq \bigcup_{x\in X} E'[x]\times E'[x] =E'oE'\subseteq E$$ $$\Rightarrow (\exists D\in \mathcal D)(n(D)\in E)$$ a contradiction.

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