locally compact, Hausdorff, second-countable $\Rightarrow$ paracompact

Question

I'm digging into a proof given by Frank W. Warner:

I wonder why Warner claims that $\overline{G_i}$ is compact. One way to show this is to find a compact set that contains $G_i$. Hence, $\overline{G_i}$ will be contained in this compact set. The result then follows. But how do we find such compact set? Thanks.

Are you asking about the display "$\overline{G_i}$ is compact" or the argument that each of the $G_i$ that is subsequently constructed has a compact closure? — Eric Towers, Jul 05 '19 at 15:49

freakish · Accepted Answer · 2019-07-05T16:02:41.637

2

I wonder why Warner claims that $\overline{G_i}$ is compact.

By the definition each $G_i$ is a union of some finite subcollection of $\{U_k\}$. And each $U_k$ is relatively compact, again by the definition.

And so the conclusion follows from two facts:

edited Jul 05 '19 at 16:02

answered Jul 05 '19 at 15:56

freakish

42,851

$U_k$ cannot be compact since in a Hausdorff space, a compact set must be closed. – Boar Jul 05 '19 at 23:49
1

@Steve $U$ is relatively compact, meaning it has compact closure. It wasn’t said $U$ is compact. – Henno Brandsma Jul 06 '19 at 05:16

score 1 · Answer 2 · answered Jul 05 '19 at 15:59

A basic fact (pun intended) is that a locally compact Hausdorff space $X$ has a base $\mathcal{B}$ such that all $B \in \mathcal{B}$ have compact closures.

Then $X$ being second countable implies that every base for $X$ can be thinned out to (has a subcollection that is) a countable base.

These two facts imply the first line of the proof. The rest just uses that closures commute with finite unions and finite unions of compact sets are compact.

locally compact, Hausdorff, second-countable $\Rightarrow$ paracompact

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