$X$ local compact metric space contains countable dense subset $\Rightarrow X=\cup_{j=1}^{\infty} K_j$, $K_1\subset K_2\subset…$ compact

Question

Let $X$ be locally compact metric space, that is every point is surrounded by a compact neighbourhood. Show that

If $X$ contains a countable dense subset $T$, then there are compact subsets $K_1\subset K_2\subset…$ so that $X=\cup_{j=1}^{\infty} K_j$

Let $t_i$ be elements of $T$, since $X$ is locally compact there is $\epsilon>0$ so that $\overline{B_{\epsilon}(t_i)}$ is compact and $X=\cup \{\overline{B_{\epsilon}(t_i)}|\epsilon\in \mathbb{Q}\}$. But how can I create an increasing sequences of compact sets?

@PaulFrost The rational sequence topology (or Mrówka $\Psi$) are Tychonoff separable locally compact and not $\sigma$-compact. — Henno Brandsma, Nov 03 '21 at 06:39
There need not be one $\varepsilon$ that works for all $t_i$ at the same time..And if we have a different $r_i$ per $t_i$ the claim that the union is $X$ is possibly false. — Henno Brandsma, Nov 03 '21 at 06:41

Henno Brandsma · Answer 1 · 2021-11-03T12:25:01.177

1

If we're in a metric space, a countable dense set means there is a countable base.

This means there is also a countable base $\{B_n\}$ so that $\overline{B_n}$ are all compact (follows from this).

Then apply the fact that finite unions of compact sets are compact to easily modify this to an increasing sequence of compacta.

edited Nov 03 '21 at 12:25

answered Nov 03 '21 at 06:45

Henno Brandsma

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I take $K_j=\cup{n=0}^j \overline{ B_n}$, why is then $X=\cup_{j=0}^{\infty} K_j$? – user30523 Nov 03 '21 at 08:34
@user30523 if $K_j = \bigcup_{i=0}^j \overline{B_i}$ then the union of the $K_j$ the same as those all those closures hence equal to $X$. – Henno Brandsma Nov 03 '21 at 08:38
${B_n}$ countable base of $X$ means that $X=\cup_n B_n$? – user30523 Nov 03 '21 at 08:41
@user30523 so certainly the closures too. – Henno Brandsma Nov 03 '21 at 08:41
why can I find a base so that the closures are compact? – user30523 Nov 03 '21 at 08:44
@user30523 that’s a standard fact. – Henno Brandsma Nov 03 '21 at 09:16

$X$ local compact metric space contains countable dense subset $\Rightarrow X=\cup_{j=1}^{\infty} K_j$, $K_1\subset K_2\subset…$ compact

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