# A Compact Hausdorff space which is locally metrizable is metrizable.

This is exercise 7 from section 34 in Munkres. The hint given is to show that the space is a union of finitely many subspaces which are second countable. This question has been asked before A compact Hausdorff space is metrizable if it is locally metrizable.

However, I'm still confused. The idea appears to be: let $U_x$ be a metrizable neighborhood of $x$ for each $x\in X$. Since $X$ is compact, there are $x_1,\ldots,x_n$ for which $\{U_{x_k}\}$ form a finite subcover. If we knew that each $U_{n_k}$ was second countable, we could conclude that $X$ was metrizable. This is because $X$ is regular and second countable.

I can't find any reason why the $U_{n_k}$ are second countable though.

• @Mirko Why do we know that each $U_{x_k}$ is second countable? – Tim Raczkowski Dec 12 '15 at 23:43
• It's also true for paracompact Hausdorff spaces, though the proof is a bit harder. – Henno Brandsma Dec 13 '15 at 7:14

Choose $U_x$ compact ( with interiors covering $X$) and metrizable. Then they are second countable. ( a compact metric space is second countable).

• The $U_x$ are open. – Tim Raczkowski Dec 12 '15 at 23:44
• @Tim Raczkowski: Inside every open metrizable open neighborhood of $x$ choose a compact subset containing $x$ in the interior. That one will have a countable basis. Finitely many of these will cover $x$. The interior of these sets will have a countable basis. So $X$ will have a countable basis. – Orest Bucicovschi Dec 12 '15 at 23:50
• @Tim Raczkowski: The fact used is that for every point $x$ in an open subset $U$ of a compact topological space there exists $K$ compact so that $x \in \overset{\circ}{K }$ and $K \subset U$. – Orest Bucicovschi Dec 12 '15 at 23:53
• Right. Let's say that for each $U_{x_k}$ we have a $K_{x_k}$. It's not clear to me that that the $K_{x_k}$ cover $X$. Most likely $K_{n_k}\subsetneq U_{n_k}$. – Tim Raczkowski Dec 12 '15 at 23:56
• @TimRaczkowski : Interior of a set – Orest Bucicovschi Dec 13 '15 at 0:05

For each $p\in X$ let $U_p$ be a ndhd of $p$ such that $U_p$ is metrizable. Since $X$ is a compact Hausdorff space, it is a regular space, so there is an open $V_p$ such that $p\in \overline {V_p} \subset U_p.$ Since $X$ is a compact Hausdorff space, $\overline { V_p}$ is compact. And $\overline { V_p}$ is metrizable because it is a subset of $U_p.$ Now $\cup \{V_p :p\in X\}=X$, so $\cup \{V_p :p\in F\}=X$ for some finite $F$ because $X$ is compact. Each $\overline {V_p}$ is second -countable because it is metrizable and compact. Any subspace of a second-countable space is second-countable. So each $V_p$ is second countable, and is open, and $F$ is finite, so $X=\cup \{V_p:p\in F\}$ is second-countable.

2. In a regular space $$X$$, for every point $$x\in X$$ and every open set $$U$$ containing it, there exists another open set $$V$$ such that $$x\in V\subseteq\overline{V}\subseteq U$$
Using the first three claims, we can prove that for every point $$x$$ in an open subset $$U$$ of a compact topological space, there exists a compact set $$K$$ so that $$x\in \mathring{K}$$ (interior of $$K$$) and $$K\subseteq U$$. This compact set $$K$$ will be exactly $$\overline{V}$$ from the second claim mentioned above.
Then, if we take an open cover of $$X$$ using the interiors of compact sets $$K_x$$, it has a finite sub-cover, and therefore we can build a finite cover of $$X$$ using compact metrizable sets $$K_x$$. According to the 4th claim, each of these sets has a countable basis, and we can show that the union of these bases is a countable basis for all of $$X$$. Therefore, $$X$$ is second countable, and from there we can continue as you mentioned in your question.