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Let $X$ be a (Tychonoff) separable topological space (whose weight does not exceed $2^\omega$) and $M$ be a metrizable subspace of $X$. Is true that $M$ is separable?

user34870
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  • Maybe this can be useful: http://math.stackexchange.com/questions/516886/prove-that-a-subset-of-a-separable-set-is-itself-separable – Kolmin Sep 17 '15 at 15:39

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No. Let $X=[0;1]^{\frak c}$ be a Tychonoff cube. Then $X$ is separable, $w(X)=\frak c$, but it contains a discrete space $M=\{e_\alpha:\alpha<\frak c\}$ (where each $e_\alpha=(e_\alpha)_{\beta, \beta<\frak c}$, $(e_\alpha)_{\alpha}=1$, and $(e_\alpha)_{\beta}=0$ provided $\beta\ne\alpha$) of size $\frak c$.

Alex Ravsky
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    Indeed, any completely regular space of weight $\le 2^\omega$ embeds as a subspace into $X$ (universal space theorem), among which there are many non-separable ones. – Henno Brandsma Sep 18 '15 at 14:17
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Let $S$ be the Sorgenfrey line, and let $X=S\times S$, the Sorgenfrey plane. $S$ is separable, since $\Bbb Q$ is dense in $S$, so $X$ is separable. $X$ is first countable and of cardinality $2^\omega$, so its weight is clearly no more than $2^\omega$ (and in fact it is $2^\omega$. However, the set $D=\{\langle x,-x\rangle:x\in S\}$ is a closed, discrete subset of cardinality $2^\omega$. Being discrete, it is metrizable, and it’s clearly not separable.

Brian M. Scott
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