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 $\mathcal D_2$?
 A: 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 $W \in \mathcal{D}$.
A: 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.
