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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Martin Sleziak
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Bumblebee
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2The only metric spaces with finite bases are the finite discrete metric spaces. – Brian M. Scott Nov 16 '16 at 19:49
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Thank you. So, what is the wrong with my above verification? – Bumblebee Nov 16 '16 at 19:53
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3How do you prove that $\mathscr B_0$ is a subbasis for the topology? – bof Nov 16 '16 at 19:53
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2$\mathscr{B}_0$ is not a subbase unless $M$ is finite. – Brian M. Scott Nov 16 '16 at 19:54
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What's the point of using $\mathscr U$ instead of $\mathscr V={B(x,1):x\in M}?$ After all, $\mathscr V$ is a subcover of $\mathscr U$; for all you know, your finite subcover $\mathscr B$ is extracted from $\mathscr V.$ – bof Nov 16 '16 at 19:57
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@BrianM.Scott: I am trying to show that every compact metric space is second countable. At the middle of my proof,I came up with the above argument. How can I fix this? – Bumblebee Nov 16 '16 at 20:05
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2@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 $M$. – Brian M. Scott Nov 16 '16 at 20:12
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Thank you for you excellent suggestion. I helps me lot. Have a good day. – Bumblebee Nov 16 '16 at 20:24
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@Nil: You’re welcome! – Brian M. Scott Nov 16 '16 at 20:38
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@Brian M. Scott: would you like to answer this question. otherwise this remains as an unanswered question. I am glad to accept your answer. – Bumblebee Dec 05 '16 at 04:53
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@Nil: Done. $,$ – Brian M. Scott Dec 05 '16 at 07:06
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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.
Brian M. Scott
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