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I am trying to show Every compact metrizable space is second countable

My Attempt:

Let $(X,\mathfrak{T})$ be a compact metrizable space. We wish to show that it has a countable basis.

Then given $C \subset \mathcal{P}(X)$ an open cover of $X$, it has a finite subcover $\{C_n| n \in F, F \text{ is a finite set in } X\}$.

Since $X$ is metrizable, therefore $\forall x \in C_n, \exists B_n \in \mathfrak{B}$ such that $x \in B_n \subset C_n$, where $B_n$ is a metric ball, then $\{B_{n_x}| x \in C_n\}$ is an open cover of $C_n$.

Since the closure $\overline C_n$ is closed, a closed set in a compact space is compact, therefore $\overline C_n$ is compact and covered by finite many $B_{n_x}$. Since $C_n \subset \overline C_n$, so $C_n$ is covered by finite many $B_{n_x}$ as well.

Then $\bigcup\limits_{x \in C_n, n \in F} \{B_{n_x}\}$ is a finite union of finitely many metric balls that covers $X$. Therefore $\{B_{n_x}| x \in C_n, n \in F\}$ is a countable basis.

Can someone check my work? I feel I overkilled it with the closure. Is there a much easier way?

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    $\begingroup$ I don't see how your construction results in a basis. Say I have a point $x$ and an open ball of radius $0.00000000000000001$ around it. Which element of your basis contains $x$ and is contained in this ball? $\endgroup$ – Matt Samuel Jul 31 '16 at 22:36
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    $\begingroup$ HINT: For each $n\in\Bbb N$, consider the open covering by balls of radius $1/n$. $\endgroup$ – Ted Shifrin Jul 31 '16 at 22:41
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    $\begingroup$ There's no need to downvote a question of "is this correct" when it's not correct. That's the whole point of posting the question. $\endgroup$ – Matt Samuel Jul 31 '16 at 22:53
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This is not correct. You have produced a countable collection of metric balls that cover $X$, but such a collection need not be a basis. For instance, $X=[0,1]$ is covered by the single metric ball $B_2(0)$, but $\{B_2(0)\}$ is certainly not a basis (the topology it generates is the indiscrete topology!).

A sure-fire sign that what you're doing isn't right is that you haven't actually used any specific open cover of $X$. So for instance, you might just have $C=\{X\}$. In that case, compactness doesn't tell you anything (this cover has a finite subcover for any space), so your argument would work without compactness if it worked at all.

So to find a correct proof, you will need to choose some specific special open covers of $X$ to take finite subcovers of. And to get a basis, you probably want your sets to be "arbitrarily small". This suggests you might try looking at open covers by sets of small diameter.

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  • $\begingroup$ I think that many starting mathematicians fall into the fallacy of "How could such a specific construction address such a general question"? To this effect, just saying "any open cover" feels more correct, at first. +1. $\endgroup$ – Omnomnomnom Jul 31 '16 at 22:54
  • $\begingroup$ This is one of those proof from heaven type of proof. $\endgroup$ – Olórin Aug 1 '16 at 2:43

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