Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 9 down vote accepted

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 $U$.)

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.