# Precompact reflection in diagonal uniform spaces

Each diagonal uniform space $(X,\mathcal D)$ can be derived from the covering uniform space $(X,\Sigma_{\mathcal D})$ and each covering uniform space $(X,\Sigma)$ can be derived from the diagonal uniform space $(X,\mathcal D_{\Sigma})$.

The precompact reflection of the covering uniform space $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ consists of any covering for $X$ with a finite refinement in $\Sigma$.

Now we can define the precompact reflection of the diagonal uniform space $(X,\mathcal D)$ (denoted by $(X,p\mathcal D)$) as $(X,\mathcal D_{p\Sigma_{\mathcal D}})$.

My question is:

How can we define the precompact reflection of $(X,\mathcal D)$ directly? (without converting it to covering uniform space).

One can easily prove: $$p\mathcal D\subseteq \lbrace D\in \mathcal D \mid (\exists F\subseteq X: F\text{ is finite})(D[F]=X) \rbrace$$

but it seems $\supseteq$ is not always true. I could not find a counterexample for it.

In $\S$4.3.2.19, he proves your assertion that $(X,p\Sigma)$ is in fact a precompact uniform space. Moreover, his $\S$4.3.2.15 makes it clear that in terms of diagonal entourages $$p\mathcal{D} = \{ D\in \mathcal{D}\mid \exists (D'\in \mathcal{D}, \text{ finite } F\subseteq X) : D'\circ D'\subseteq D, D'[F] = X \}.$$ The inclusion from left to right is easy to see, recalling that the existence of such a $D'_\sigma$ for a diagonal entourage $D_\sigma$ formed from a finitely refinable cover $\sigma$ follows from the fact that it has a finitely refinable star-refinement $\sigma'$ (otherwise $p\Sigma$ is not a uniformity). The other inclusion follows from the refinement sequence $\sigma_{D'}\prec \{D(x) \mid x\in F\}\prec \sigma_D$, for covers generated by the diagonal entourages $D'$ and $D$.
• How is $\sigma_D$ defined? And do you have the counterexample in last part? – user31967 Jul 31 '13 at 23:06
• In the usual way, $\sigma_D = \{ D(x) \mid x\in X \}$. And sorry, I don't have a counterexample. I only looked at the general argument. – Igor Khavkine Aug 1 '13 at 14:36
• So by $D[F]$ in last line you don't mean $D[F]$ you mean $\{D[a] \mid a\in F\}$! – user31967 Aug 1 '13 at 14:55