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I'm trying to understand the proof of this lemma. My problem lies in the existence of $V_{n(x)}$. As far as I know (from Munkres' topology), there should be a neighborhood $U_x$ of $x$ such that $\overline{U_x}$ is compact and $\overline{U_p}\subseteq V_{n(x)}$. Can anyone show me how? Thanks.

cqfd
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Boar
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    The weight $w(X)$ of a space $X$ is the least infinite cardinal $k$ such that $X$ has a base $B$ with cardinal $|B|\le k.$ (We can say " with $|B|=k$" if $X$ does not have a finite base). Theorem: If $C$ is a base for a space $X$ then $X$ has a base $B\subset C$ with $|B|\le w(X)$.... So for your Q it suffices that the family $C$ of all open subsets with compact closures is a base for $X.$ – DanielWainfleet Jul 05 '19 at 05:56
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    related question: https://math.stackexchange.com/questions/4030200/does-a-second-countable-locally-compact-hausdorff-space-admit-a-countable-basi?rq=1 – Chill2Macht Feb 06 '22 at 04:01

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That there is an open set $U_x$ containing $x$ with $\overline {U_x}$ compact is a theorem on locally compact Hausdorff spaces. There there exists some $V_n=V_{n(x)}$ contained in $U_x$ and containing $x$ because $(V_n)$ is a base. Since $V_{n(x)} \subset U_x$ it follows that the closure of $V_{n(x)} $ is compact. Since $(V_{n(x)})$ is a sub-collection of $(V_n)$ is is countable. That finishes the proof.

Edit: as pointed out by Steve it is not clear that $(V_{n(x)})$ is a base. So the proof has to me modified as follows: let $U$ be any open set and $x \in U$. Pick $U_x$ as above. Then there exists $V_{n(x)}$ such that $x \in V_{n(x)} \subset U \cap U_x$. Rest is now clear.

Kavi Rama Murthy
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  • Sir, how come does $V_{n(x)}$ become a basis? Thank you. – Boar Jul 04 '19 at 14:18
  • Let $U$ be an open set and let $q\in U$. If $\exists V\in{V_{n(x)}}$ s.t. $q\in V\subseteq U$, then ${V_{n(x)}}$ will be a basis. To find $V$, we may well consider $V_{n(q)}$ since $q$ is already in $V_{n(q)}$. But how about $V_{n(q)}\subseteq U$? I don't how to fix it. – Boar Jul 04 '19 at 14:33
  • correction: How could ${V_{n(x)}}$ be a basis? – Boar Jul 04 '19 at 14:37
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    @Steve I have edited the proof. – Kavi Rama Murthy Jul 04 '19 at 23:23
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$\{V_n\}$ is a basis for the topology. Then by definition, given any open set $U$ and a point $x\in U$, we can find a basis element $V_{n(x)}$ such that $x\in V_{n(x)}\subset U$.

cqfd
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My problem lies in the existence of $V_{n(x)}$.

Given any open set $U$ at all around $x$, the phrase "$\{V_n\}$ is a basis of the topology" means that one of the $V_n$'s can get inside of $U$ and around $x$ too. This explains the existence of $V_{n(x)}$.

rschwieb
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