1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ why $B_{l}$ s are compact? you mean their closures? @PhoemueX $\endgroup$ – Kiarash Jun 9 '16 at 10:22
  • $\begingroup$ @KNP: If I interpreted what you wrote correctly, then the $B_\ell $ are compact. But finite unions of compact sets are compact. $\endgroup$ – PhoemueX Jun 9 '16 at 10:28
  • $\begingroup$ $B_{n}$ 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! $\endgroup$ – Kiarash Jun 9 '16 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.