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

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

I tried to use the fact of compact space. But I do not know if the opens are compact subspaces.

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  • $\begingroup$ I tried to prove the hint: show that $X$ is a finite union of open subspaces, each of which has a countable basis. $\endgroup$ – Jarbas Dantas Silva Nov 3 '13 at 21:14
  • $\begingroup$ I tried to use the fact of compact space. But i do not know if the opens are compacts subspaces. $\endgroup$ – Jarbas Dantas Silva Nov 3 '13 at 21:15
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Of course, if $X$ is metrizable then it is also locally metrizable.

For the other direction, you could really follow the hint. Supposed that $X$ is locally metrizable, there exists an open cover $\bigcup_{x\in X}U_x$ of $X$ where $U_x$ is a metrizable neighborhood of $x$. Because of compactness, it has a finite subcover: $\bigcup_{i<n} U_i=X$ where $U_i:=U_{x_i}$ for some $x_i$.

Because of metrizability, each $U_i$ has countable base $(V_{i,j})_j$ of open subsets, so that all finite intersections of these are still countable, and they give a base of $X$.

The space also has to satisfy the separation axiom $T_3$, but this holds as $X$ is Hausdorff and compact.

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  • $\begingroup$ The double implication was my misreading. Gone now. $\endgroup$ – dfeuer Nov 3 '13 at 22:40
  • $\begingroup$ Not every metrizable space has a countable base. Must all the compact ones? $\endgroup$ – dfeuer Nov 3 '13 at 22:47
  • $\begingroup$ @dfeuer: Yes.$\,$ $\endgroup$ – Brian M. Scott Nov 4 '13 at 7:59
  • $\begingroup$ @BrianM.Scott, a clue? $\endgroup$ – dfeuer Nov 4 '13 at 8:01
  • $\begingroup$ @dfeuer: It’s Lindelöf, and in metrizable spaces Lindelöf, separable, and second countable are equivalent. You can get separability by taking for each $n\in\Bbb N$ the cover by open $2^{-n}$-balls, extracting a finite (or even just countable) subcover, and letting $D_n$ be the set of centres of the members of the subcover; $D=\bigcup_{n\in\Bbb N}D_n$ is a countable dense subset of $X$. Then $\{B(x,2^{-n}):x\in D\text{ and }n\in\Bbb N\}$ is a countable base for $X$. $\endgroup$ – Brian M. Scott Nov 4 '13 at 8:05
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For every $x\in X$, there exists a neighborhood $U_x$ which is metrizable. These neighborhoods cover $X$, i.e., $X=\bigcup_x U_x$. Now use the definition of compactness to reduce this to a finite union, $X=U_1\cup\ldots\cup U_n$. Each of these sets is metrizable, so pick metrics which are defined locally on each $U_i$. Lastly, use a partition of unity to patch together the local metrics into a global one.

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  • $\begingroup$ Can you show construction of metric on $X\times X$ using Partition of unity. Means how metrices on $U_i$ can be patched up $\endgroup$ – Sushil Mar 8 '18 at 16:31
  • $\begingroup$ @Sushil I don't think I understand your question. Maybe you could elaborate. Is your question: "Given a metrizable space $X$, show that $X\times X$ is metrizable." If so, this can be done without a partition of unity. For example, on can define the metric on the product as $d_{X\times X}((x,y),(x',y'))=d_X(x,x')+d_X(y,y')$. I hope this helps. $\endgroup$ – D Wiggles Mar 13 '18 at 3:07
  • $\begingroup$ Actually I don't get how partition of unity will help in getting metric. $\endgroup$ – Sushil Mar 13 '18 at 11:16
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[Hint: Using normality/regularity]

The space $X$ is compact Hausdorff, so it is normal/regular.

Now for each $x\in X$, there is a metrizable open neighborhood $U_{x}$ of $x$, for this $U_{x}$, by regularity, there is an open neighborhood $V_{x}$ of $x$ such that $x\in V_x \subset V_{x}^{-} \subset U_{x}$. Notice that the closure $V_x^{-}$ of $V_x$ is a closed subset of a compact space and a subspace of a metrizable space, therefore it is a compact metrizable space and has a countable basis, so $V_x$ has a countable basis.

Since $X$ is compact, and the family $\{V_x\}_{x\in X}$ of opens covers $X$, there is a finite subcover in which every member is open and has a countable basis. Hence $X$ has a countable basis.

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