Suppose $X$ is a complete metric space. Is the complement of a countable union of nowhere-dense sets a dense set?

Question

I am able to prove this if we have a countable collection $\{E_n\}$ of closed nowhere-dense subsets, however I am being asked to prove it for a countable collection of nowhere-dense sets that need not be closed. I haven't been able to come up with a way to show it. Any hints would be greatly appreciated!

Answer 1

If all $E_n$ are closed and nowhere dense, $\bigcup_n E_n$ is of first category and so its complement is $\bigcap_n E_n^{\complement}$, which is the intersection of countably many open and dense subspaces, so dense by Baire's theorem.

If the $E_n$ are just nowhere dense, then by definition $F_n = \overline{E_n}$ is closed and nowhere dense and

$$(\bigcup_n E_n)^{\complement} \supseteq (\bigcup_n F_n)^{\complement} = \bigcap_n F_n^{\complement}$$ and the right hand side is already dense by the closed case hence so is the larger set $(\bigcup_n E_n)^{\complement}$ too.