If X is any uniformizable topological space, then we can show that there exists a finest covering uniformity on X by taking the union of all covering uniformities compatible with the topology of X. We know that this finest uniformity is a normal family. How to show that this finest uniformity is compatible with the topology of X?

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    $\begingroup$ It is enough to have that $U[x]$ is open for any $U$ in the finest uniformity. If $U$ is in the finest uniformity then there is a uniformity $\Phi$ compatible with the topology such that $U\in\Phi$. But then $U[x]$ is open. $\endgroup$ – OR. Jun 30 '17 at 11:29
  • $\begingroup$ @MariePierredeLeTetou it's a bit more complicated (and you work with entourage uniformities instead of covering uniformities). $\endgroup$ – Henno Brandsma Jul 4 '17 at 9:29

Suppose we have a uniformisable space $(X,\mathcal{T}_X)$. Let $\mathcal{C}(X)$ be the set of all possible covering uniformities on the set $X$. For a covering uniformity $\mu$ let $\mathcal{T}_\mu$ denote its induced topology on $X$.

Define $$\mathcal{M} = \{\mu \in \mathcal{C}(X): \mathcal{T}_\mu = \mathcal{T}_X\}$$

Then $\mathcal{M}$ is non-empty (as $\mathcal{T}_X$ is uniformisable), and is a normal family as a union of normal families. It is not necessarily a uniformity, but it is, like all normal families, a so-called subbase for a uniformity, the so-called "fine uniformity", $\mu_f$. This means that

$$\mathcal{U} \in \mu_f \Leftrightarrow \exists n \in \mathbb{N}: \forall i \le n : \exists \mu_i \in \mathcal{M} \exists \mathcal{U}_i \in \mu_i: \bigwedge_{i=1}^n \mathcal{U_i} \prec \mathcal{U}$$

In words: all covers that are refined by a meet of finitely many covers from $\mathcal{M}$. This is a uniformity (quite easy to check) that contains all of $\mathcal{M}$ and it is the smallest such uniformity.

The claim is that $\mathcal{T}_{\mu_f} = \mathcal{T}_X$. The interesting inclusion is from left to right:

Suppose $O$ is open in the topology induced by $\mu_f$. Let $x \in O$. Then by definition there is some $\mathcal{U} \in \mu_f$ such that $St(x,\mathcal{U}) \subseteq O$.
Find the $\mu_i \in \mathcal{M}$ and $\mathcal{U}_i \in \mu_i$, $i=1\ldots n$, as stated above.

But then

$$\bigcap_{i=1}^n St(x, \mathcal{U}_i) = St(x, \bigwedge_{i=1}^n \mathcal{U}_i) \subseteq St(x, \mathcal{U}) \subseteq O $$

And as all sets $St(x, \mathcal{U}_i)$ are in a generating set for $\mathcal{T}_X$ (from $\mu_i \in \mathcal{M}$) this shows that $x$ is an interior point for $O$ in $\mathcal{T}_X$. As this holds for all $x \in O$, $O \in \mathcal{T}_X$ as required. Right to left is trivial: Find a uniformity $\mu$ such that $\mathcal{T}_\mu = \mathcal{T}_X$. So $\mu \in \mathcal{M}$ so all sets $St(x, \mathcal{U})$ for $\mathcal{U} \in \mu$ are also in the generating set for $\mu_f$, as $\mu \subset \mu_f$, so $\mathcal{T}_X = \mathcal{T}_\mu \subseteq \mathcal{T}_{\mu_f}$, as required.


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