# Every complete, countable metric space has a discrete, dense subset.

Given a complete, countable metric space, say $X$, I'd like to show it has a discrete, dense subset. This seems like an application of the Baire Category Theorem, but that doesn't seem to go anywhere. Any help would be appreciated.

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## 1 Answer

Consider the collection $I$ of all isolated points of $X$. (By the Baire Category Theorem $I$ is nonempty, but that is somewhat immaterial for the moment.) Note that $I$ is then a discrete subspace of $X$. If $I$ were not dense, then $U = X \setminus \overline{I}$ is a nonempty (open) set without isolated points. From here we can construct in the usual manner a Cantor set as a subset of $X$, contradicting that $X$ is countable! (The construction goes as in the linked answer, just ensuring that the $x_\sigma$ are chosen from $U$.)

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Ah, so I was heading in the right direction it seems. Thanks for your help. –  Alexander Sibelius Mar 12 '13 at 4:58