# On the Compact Uniformization Theorem

I just read through a proof of the Compact Uniformization Theorem, and I follow it up to the very last line. The proof is:

Compact Uniformization Threorem. If $X$ is a compact regular space, then the neighborhood filter $\mathcal{N}_\Delta$ of the diagonal $\Delta\subset X\times X$ is a uniform structure on $X$.

Suppose that $\mathcal{N}_\Delta$ does not have the property that for any entourage $E$, there exists an entourage $D$ such that $D\circ D\subseteq E$. Then there exists an open neighborhood $E\in\mathcal{N}_\Delta$ such that

$$(D\circ D)\setminus E\neq\emptyset\qquad (D\in\mathcal{N}_\Delta)$$

Thus, the family

$$\mathcal{C}=\{(D\circ D)\setminus E\ | \ D\in\mathcal{N}_\Delta\}$$

consists of nonempty subsets and is a filter-base on $X\times X\setminus E$.

The latter begin a closed subset of the product of two compact spaces is compact by Tychonoff's Theorem. It follows that $\mathcal{C}$ has a cluster point $(p,q)\in X\times X\setminus E$.

Note the neighborhood $E(q)$ of $q$ does not contain $p$, and thus the neighborhood $E^{-1}(p)$ of $p$ does not contain $q$.

So by a previous Lemma, there exists an open neighborhood $D$ of $\Delta$ such that $D\circ D$ is disjoint with some neighborhood of $(p,q)$. It follows that $(p,q)$ cannot be a cluster point of $\mathcal{C}$, a contradiction.

I just don't see why it immediately follows that $(p,q)$ is not a cluster point of $\mathcal{C}$, simply because there is some open neighborhood of the diagonal that is disjoint from one of the neighborhoods of $(p,q)$. Can someone please explain?

The previous lemma is:

Let $p$ and $q$ be a pair of points in a regular topological space $X$ such that $\mathcal{N}_p\neq\mathcal{N}_q$. Then there exists an open cover $\mathcal{U}$ of $X$ and a neighborhood $W$ of $(p,q)$ such that $$W\cap(\Delta_\mathcal{U}\circ\Delta_\mathcal{U})=\emptyset.$$

If it's not clear, $$\Delta_\mathcal{U}=\bigcup_{U\in\mathcal{U}}U\times U.$$

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Please re-check your definition for a cluster point of a filter-base. The usual definition says that $x$ is a cluster point of filter base $B$ if $\forall B_0\in B$ and $\forall U$ nbhd of $x$, $B_0\cap U\neq \emptyset$. So yeah, that last statement follows. – Willie Wong Dec 10 '10 at 15:01
@Cromarty: sorry, I removed my first comment after I realized that I didn't read the antecedant that $\mathcal{C}$ is a filter-base. – Willie Wong Dec 10 '10 at 15:03
To see if I'm getting this straight, let $W$ be a neighborhood $(p,q)$, and then $D\circ D\cap W=\emptyset$. But how is $D\circ D$ in the filter base $\mathcal{C}$, in order to have the contradiction? – yunone Dec 10 '10 at 15:05
@Willie, no problem, I figure it's good to have the lemma there anyway. – yunone Dec 10 '10 at 15:06
Ah, but since $D\circ D \setminus E \subset D\circ D$, you have that $(D\circ D \setminus E) \cap W = \emptyset$ either. That is, if you make a set smaller, you cannot make it intersect more points. – Willie Wong Dec 10 '10 at 15:25