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Let $X$ be Hausdorff. Given any countable discrete $D \subset X$, is the closure $cl(D)$ as a subspace of $X$ Frechet-Urysohn?

Added: If I may ask more, is it must be discretely generated?

A space is called discretely generated if for every $A\subset X$ with $x \in cl(A)$ there is a discrete $D\subset A$ such that $x \in cl(D)$.

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Paul
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1 Answers1

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Not necessarily. Consider the Stone–Čech compactification $\beta \omega$ of $\omega$. We know that $\omega$ countable and discrete. However $\overline{ \omega }$ is not Fréchet, since $\omega$ is a dense subset of $\beta \omega$, and we know that $\beta \omega$ is not Fréchet (as pointed out in this answer).

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user642796
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