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 $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ contains 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.

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.

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 $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ contains 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: $$p\mathcal D\subseteq \lbrace D\in \mathcal D \mid (\exists F\subset X: F\text{ is finite})(D[F]=X) \rbrace$$

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

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 $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ contains 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.

Each diagonal uniform space $(X,\mathcal D)$ can be derived from a 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 $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ contains 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 uniformity).

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

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

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 $(X,\Sigma)$ is denoted by $(X,p\Sigma)$ where $p\Sigma$ contains 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: $$p\mathcal D\subseteq \lbrace D\in \mathcal D \mid (\exists F\subset X: F\text{ is finite})(D[F]=X) \rbrace$$

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