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Let X be a compact metrizable space. Would you help me to prove that X has a countable basis. Thanks.

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One ingredient in the usual proof is to uwse that, instead of using ball neighborhoods $B(x,\epsilon) where $\epsilon$ is a real number, you can just use rational $\epsilon$ values. And from compactness you can choose large sets of points which uniformly pepper the space with each point epsilon close to one of the points. That's the basic idea as I recall for this. – coffeemath Nov 10 '12 at 6:24
@coffeemath I think you dropped a dollar sign there ;) – Jeremy Nov 10 '12 at 7:03
Jeremy: yes, dropped the dollar sign. Hope I don't drop an actual dollar. :) – coffeemath Nov 10 '12 at 7:09
up vote 5 down vote accepted

HINT: For each positive integer $n$ let $\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}$; this is an open cover of $X$, so it has a finite subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.

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