Let X be a compact metrizable space. Would you help me to prove that X has a countable basis. Thanks.
Compact metrizable space has a countable basis (Munkres Topology)
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$\begingroup$
general-topology
metric-spaces
compactness
second-countable
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2Jeremy: yes, dropped the dollar sign. Hope I don't drop an actual dollar. :) – 2012-11-10
1 Answers
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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 $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.