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.

  • $\begingroup$ without paracompactness, $\omega_1$ would be a counterexample, so that property is essential. $\endgroup$ – Henno Brandsma Jun 26 '18 at 8:38
  • $\begingroup$ 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 $D\subseteq E$? $\endgroup$ – user479859 Jun 26 '18 at 8:42
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    $\begingroup$ Why do you want to know? Is there some application that needs this? $\endgroup$ – Henno Brandsma Jun 26 '18 at 9:10
  • $\begingroup$ And do you want to know this for any $\mathcal{U}$ that is compatible or just one of them? $\endgroup$ – Henno Brandsma Jun 26 '18 at 9:11
  • $\begingroup$ 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 $D\subseteq E$. $\endgroup$ – user479859 Jun 26 '18 at 9:23

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.

  • $\begingroup$ 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? $\endgroup$ – user479859 Jun 26 '18 at 9:35
  • $\begingroup$ @user479859 why would an infinite union of compacts be compact? My answer says no for non-compact $X$. You cannot always find such a $D$. $\endgroup$ – Henno Brandsma Jun 26 '18 at 9:52

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