# 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?

• If $X$ is Baire, then a comeagre set is dense. Complete metric spaces are Baire. – Daniel Fischer Nov 3 '15 at 15:50
• And if $X$ is not Baire, a comeagre set need not be dense and can in fact be empty: every subset of $\Bbb Q$ is comeagre. – Brian M. Scott Nov 3 '15 at 17:40

## 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.

• It seems like the meat of this is showing that complete metric spaced are Baire, and everything else is pretty easy symbol manipulation. Is it easy to show that complete metric spaces are Baire? – Richard Rast Nov 3 '15 at 22:54
• It's not very hard. But I assumed you knew that theorem already. It's in any decent topology book that covers metric spaces. – Henno Brandsma Nov 4 '15 at 5:18