Is a Compact Hausdorff Space Metrizable if it is Locally Metrizable?

[Chris L T521]

Here's this week's problem.

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Problem: A space $X$ is locally metrizable if each point $x\in X$ has a neighborhood that is metrizable in the subspace topology. Show that a compact Hausdorff space $X$ is metrizable if and only if it is locally metrizable.

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Hint:

Spoiler

Show that $X$ is a finite union of open subspaces, each of which has a countable basis.

 

[Chris L T521]

No one answered this week's question. You can find the solution below.

Spoiler

Proof: ($\Rightarrow$) Any metrizable space is locally metrizable since a subspace of a metric space is yet again a metric space.

($\Leftarrow$) Let $X$ be compact, Hausdorff, and locally metrizable. Thus, $X$ is normal (and hence, regular) as a compact Hausdorff space. To show that $X$ has a countable basis, we note that local metrizability of $X$ means that for each $x\in X$, there is an open neighborhood $U_x$ or $x$ which is metrizable. Then the set $X\backslash U_x=U_x^c$ is closed in $X$ and is disjoint from $x$. Thus, by regularity of $X$, there are disjoint open sets $V_x\supset\{x\}$ and $W_x\supset U_x^c$. The set $\overline{V_x}$ is a closed compact subset that is contained in $U_x$. We can repeat this procedure for each $x\in X$. Since $X$ is compact, there exists a finite subcover of sets $V_{x_1},V_{x_2},\ldots,V_{x_k}$. Each space $\overline{V_{x_i}}$ is compact as a closed subspace of a compact space, and is metrizable as a subset of $U_{x_i}$. Hence, $\overline{V_{x_i}}$ is second countable with basis $\mathcal{B}_i$. The collection $\mathcal{B}=\bigcup_{i=1}^k \mathcal{B}_i$ is then a countable basis for $X$. Since $X$ has a countable basis, we now apply the Urysohn metrization theorem to see that $X$ is metrizable.$\hspace{.25in}\blacksquare$