# Let $(X, \mathcal{U})$ be a uniform space and $C\in\mathcal{U}$ be given. Is there a compact set $D\in\mathcal{U}$ such that $D\subseteq C$?

Let $X$ be first countable, locally compact, paracompact, Hausdorff space. We know that $X$ has a uniformity $\mathcal{U}$. Thus $(X, \mathcal{U})$ is a uniform space.

Is it true that :

For every $E\in\mathcal{U}$, there is compact set $D\in\mathcal{U}$ such that $D\subseteq E$.

Please help me to know it.

• without paracompactness, $\omega_1$ would be a counterexample, so that property is essential. – Henno Brandsma Jun 26 '18 at 8:38
• But in my research, $X$ is papracompact, Can we say that for every $E\in\mathcal{U}$, there is compact set $D\in \mathcal{U}$ with $D\subseteq E$? – user479859 Jun 26 '18 at 8:42
• Why do you want to know? Is there some application that needs this? – Henno Brandsma Jun 26 '18 at 9:10
• And do you want to know this for any $\mathcal{U}$ that is compatible or just one of them? – Henno Brandsma Jun 26 '18 at 9:11
• In my research, we can work with every $\mathcal{U}$ that is compatible. It is important for me that for every $E\in\mathcal{U}$, there is compact set $D\in\mathcal{U}$ with $D\subseteq E$. – user479859 Jun 26 '18 at 9:23

## 1 Answer

If such a $D$ existed, $\Delta_X$ would be a closed subset of it, and so $X$ would be compact as $\Delta_X \simeq X$. So this only happens in the trivial case that $X$ is compact Hausdorff.

• Since $X$ is locally compact, each $x \in X$ has an open, relatively compact neighborhood $U^. x$. Since $X$ is paracompact, the open cover $\{U^.x\}$ has a closed (and hence compact) locally finite refinement $\{V^{.}_{\alpha} \}$ . Is it true that for $C\in\mathcal{U}$, $A = C\bigcap(\bigcup_{\alpha}V^{.}_{\alpha}\times V^{.}_{\alpha})$ is compact? – user479859 Jun 26 '18 at 9:35
• @user479859 why would an infinite union of compacts be compact? My answer says no for non-compact $X$. You cannot always find such a $D$. – Henno Brandsma Jun 26 '18 at 9:52