Must a comeager set be dense?

Say, on a complete separable metric space. Separable probably doesn't matter.

It's easy to see the opposite; if $D$ is a dense $G_\delta$ set, it's a countable intersection of open sets which contain $D$, so must also be dense. Consequently the complements of those sets are closed sets with no interior, so must be nowhere dense. And so their union (the complement of $D$) must be meager, so $D$ is comeager. Great.

It's not hard to extend this logic a bit more; if $C$ is any comeager set, then there is a $C'\subset C$ where $C'$ is comeager and $G_\delta$. But must $C'$ be dense? Must $C$ be dense?

I suppose the real question then is this: if $C\subset X$ is comeager, must $C$ be dense? (the converse definitely does not hold, see $\mathbb Q\subset\mathbb R$). By the reduction in the previous paragraph, it's enough to consider $C$ to be $G_\delta$, but that doesn't obviously lead to density. Does it?

1 Answer

Let $A \subseteq X$ be comeagre. So $X \setminus A$ is meagre, so $X \setminus A = \cup_n B_n$, where all $B_n$ are nowhere dense.

Note that $X \setminus \overline{B_n} \subseteq X \setminus B_n$, and the left hand side is open and dense (because $\overline{B_n}$ is closed and nowhere dense). So when $X$ is Baire (in particular when $X$ is completely metrisable), $\cap_n (X \setminus B_n) = X \setminus (\cup_n B_n) = A$ is dense.

When $X$ is not Baire, this fails: $X = \mathbb{Q}$, then every subset of $X$ is comeagre (all subsets are meagre, being countable), but many are non-dense.