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$X$ is called metrizable Lindelöf space if $X$ is a metrizable space and every open covering of $X$ contains a countable subcovering. Would you help me prove that $X$ has a countable basis? Thanks

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  • $\begingroup$ This question from Munkres book exercise 30 $\endgroup$ – beginner Nov 10 '12 at 6:26
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    $\begingroup$ Note that every compact space is Lindelöf, so this result implies the one in your other question. $\endgroup$ – Brian M. Scott Nov 10 '12 at 6:27
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HINT: Essentially the same hint that I gave for this question works here. 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 countable subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.

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In the metrizable space $X$, the following conditions are equivalent, cf. General Topology by Engelking Page 255:

$X$ is separable

$X$ is second countable

$X$ is Lindelöf

$X$ has countable extent

$X$ is star countable

$X$ is CCC

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  • $\begingroup$ You should have said 256 not 255 $\endgroup$ – Neil hawking Nov 7 '18 at 12:03
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I didn't want to start a new thread but wanted to see if my proof is kosher.

Note: This proof requires the assumption that every metrizable space with a countable dense subset has a countable basis.

Let $X$ be a metrizable Lindelhof space.

(Then as above)

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 countable subcover $\mathscr{B}_n$. Consider $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$.

I want to show $\mathscr{B}$ is dense in $X$.

Let $x\in X$, then let $B(x,\epsilon)$ be a basis element containing $x$. Then there exists an $n$ s.t. $x\in B(x', n)$ for some $x'$. But this implies that $x' \in B(x,\epsilon)$. So $x$ $\in \overline{\mathscr{B}}$, therefore $\overline{\mathscr{B}} = X$. Therefore $X$ has a countable basis since it contains a countable dense subset.

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