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 $(X,\mathcal D\_{p\Sigma\_{\mathcal D}})$.




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**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.



Precompact reflection in diagonal uniform spaces
You may want to look at G. Preuss, [*Foundations of topology: an approach to convenient topology*](http://books.google.ca/books?id=YY14k_ntfq0C) (Springer, 2000). 


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 $D$.


