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.


You may want to look at G. Preuss, Foundations of topology: an approach to convenient topology (Springer, 2000).

In $\S$, he proves your assertion that $(X,p\Sigma)$ is in fact a precompact uniform space. Moreover, his $\S$ 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 $D$.

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  • $\begingroup$ How is $\sigma_D$ defined? And do you have the counterexample in last part? $\endgroup$ – user31967 Jul 31 '13 at 23:06
  • $\begingroup$ 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. $\endgroup$ – Igor Khavkine Aug 1 '13 at 14:36
  • $\begingroup$ So by $D[F]$ in last line you don't mean $D[F]$ you mean $\{D[a] \mid a\in F\}$! $\endgroup$ – user31967 Aug 1 '13 at 14:55
  • $\begingroup$ You're right. I was a bit sloppy. I've edited to correct that. $\endgroup$ – Igor Khavkine Aug 1 '13 at 18:15

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