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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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True, you can establish set theory as the foundation for most mathematics, even set theory. But there is no need to use both tags in questions which are not really set theoretical in their content. In fact, it is best to use neither in questions like this and the previous one (which I have also retagged earlier). –  Asaf Karagila Jan 21 '13 at 8:26

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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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