Locally compact Hausdorff and Second countable space and nested compact sequences

Question

I have seen a theorem in my lecture notes without proof. It says:

If $X$ is a locally compact Hausdorff and second countable space, then it can be written as a nested sequence of compact subsets of $X$.

I could write $X$ as a union of compact sets because for every $x\in X$, there is a basis element and a compact set name $B_{n}$ and $C$ such that $x\in B_{n} \subseteq C_{n}$. Moreover, $X$ can be written as a union of disjoint compact sets by taking all finite intersections of them. How it can be written as a nested sequence of compact sets?

score 4 · Accepted Answer · answered Jun 09 '16 at 10:19

4

Take $K_n =\bigcup_{\ell=1}^n B_\ell $. This is an increasing (nested) sequence of compact sets.

answered Jun 09 '16 at 10:19

PhoemueX

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why $B_{l}$ s are compact? you mean their closures? @PhoemueX – heaven-of-intensity Jun 09 '16 at 10:22
@KNP: If I interpreted what you wrote correctly, then the $B_\ell $ are compact. But finite unions of compact sets are compact. – PhoemueX Jun 09 '16 at 10:28
$B_{n}$ are the basis elements and they are just open but I got the solution. If I consider the closures of $B_{n}$ then they are compact and their finite union is the answer. Thank you! – heaven-of-intensity Jun 09 '16 at 10:40

Locally compact Hausdorff and Second countable space and nested compact sequences

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