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Let $M$ be a compact metric space. Then the collection $\mathscr{U}=\{B(x,\frac1n):x\in M, n\in\Bbb{N}\}$ forms an open cover of $M.$ Let $\mathscr{B}_0$ be a finite sub-cover extract from $\mathscr{U}$. We can easily prove that $\mathscr{B}_0$ is a (finite) sub-basis for the topology on $M$.
Does $\mathscr{B}_0$ provide a finite basis for the metric topology on $M$?

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    $\begingroup$ The only metric spaces with finite bases are the finite discrete metric spaces. $\endgroup$ – Brian M. Scott Nov 16 '16 at 19:49
  • $\begingroup$ Thank you. So, what is the wrong with my above verification? $\endgroup$ – Bumblebee Nov 16 '16 at 19:53
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    $\begingroup$ How do you prove that $\mathscr B_0$ is a subbasis for the topology? $\endgroup$ – bof Nov 16 '16 at 19:53
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    $\begingroup$ $\mathscr{B}_0$ is not a subbase unless $M$ is finite. $\endgroup$ – Brian M. Scott Nov 16 '16 at 19:54
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    $\begingroup$ @Nil: For each $n\in\Bbb Z^+$ let $\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}$, and get a finite subcover $\mathscr{B}_n$ of $\mathscr{U}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$, and show that $\mathscr{B}$ is a countable base for $M$. $\endgroup$ – Brian M. Scott Nov 16 '16 at 20:12
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From the comments it appears that what you really want to show is that a compact metric space $X$ is second countable. To do that, for each $n\in\Bbb Z^+$ let $\mathscr{U}_n=\left\{B\left(x,\frac{1}n\right):x\in X\right\}$; $X$ is compact, so $\mathscr{U}_n$ has a finite subcover $\mathscr{B}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$; then show that $\mathscr{B}$ is a base for $X$. This takes a little work with the triangle inequality but isn't too hard.

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