Recall that an $F_{\sigma}$ set is, by definition, a countable union of closed sets. We see that $\mathbb{Q}$ is an $F_{\sigma}$ set: $\mathbb{Q} = \cup_{q \in \mathbb{Q}} \{q\}$. Is $\mathbb{R} \backslash \mathbb{Q}$ an $F_{\sigma}$ set? Note: $\mathbb{Q}$ and anything in $\mathbb{R} \backslash \mathbb{Q}$ are hollow.
I have no idea how to do this. I know the goal is to use Baire Caegory Theorem. I think that since anything in $\mathbb{R} \backslash \mathbb{Q}$ is hollow, we have that $\mathbb{R} \backslash \mathbb{Q}$ is a collection of hollow sets, $\{F_n\}$ in $\mathbb{R}$. I don't know how to show the sets are closed or the collection is countable. Then by Baire Category Theorem, we know that $\cup_{n=1}^{\infty}$ is hollow, but $\mathbb{Q}$ being hollow means $\mathbb{R} \backslash \mathbb{Q}$ is dense
$\mathbb{Q}$ is hollow, that tells us that $\mathbb{R} \backslash \mathbb{Q}$ is dense (which seems to contradict the note, but maybe I'm just not reading it right).
If this isn't the best way to approach the problem, please let me know the better alternative.
EDIT: I think it might be best to prove it as such: Suppose $\mathbb{R} \backslash \mathbb{Q}$ is a $F_{\sigma}$ set. Then $\mathbb{R} \backslash \mathbb{Q} \cup \mathbb{Q}$ is a countable collection of closed collow sets so by Baire Categoty Theorem, the union of this collection of sets is hollow, but $\mathbb{R} \backslash \mathbb{Q} \cup \mathbb{Q}$ is just $mathbb{R}$, which we know is dense, so we've reached a contradiction
