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Here is a theorem which said:

Every locally metrizable linearly lindelöf space $X$ is a separable metrizable space (hence, lindelöf).

Somebody said, it is enough to show that $X$ is hereditarily lindelöf. I cannot follow him. Could somebody help me?

If I may ask more, he also said, if $X$ is not hereditarily, then $X$ contains a subspace $Y$ such that $|Y|=\omega_1$ and each uncountable subspace $Z$ of $Y$ is not lindelöf. How to show such subspace $Y$ exists? Thanks ahead:)

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1 Answer 1

up vote 2 down vote accepted

Suppose that $X$ is hereditarily Lindelöf and locally metrizable. Then $X$ is $T_3$ and Lindelöf, hence paracompact, and a paracompact locally metrizable Hausdorff space is metrizable by the Smirnov metrization theorem, so $X$ is metrizable. Let $\mathscr{B}=\bigcup_{n\in\omega}\mathscr{B}_n$ be a base for $X$ such that each $\mathscr{B}_n$ is discrete. $X$ is hereditarily Lindelöf, so $\bigcup\mathscr{B}_n$ is Lindelöf for each $n\in\omega$, and therefore each $\mathscr{B}_n$ is countable, and $X$ is second countable.

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