$\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q\mapsto \widehat{q}$ (The rationals embed densely in the reals), the set usually written $\mathbb{Q}$ once the identification is made, and put $X:=\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ for the irrationals. Then:

1. ${\mathbb{Q}}_{\mathbb{R}}$ is an $F_{\sigma }$ set ($F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$) and is meager (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$);
2. $X$ is a $G_{\delta }$ set and is residual;
3. ${\mathbb{Q}}_{\mathbb{R}}$ is not a $G_{\delta }$ set, and $X$ is not an $F_{\sigma }$ set.

Claims 1 and 2 are bookkeeping. Claim 3 is the substance and is exactly where Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets is spent: no argument from the algebra of open and closed sets alone can reach it, since ${\mathbb{Q}}_{\mathbb{R}}$ and $X$ are interchanged by complementation while $F_{\sigma }$ and $G_{\delta }$ are, so any such argument would prove the same thing about both sets and about neither.

• Where the two halves of the argument part company. Claim 1 is a listing argument: step 1.2 lists ${\mathbb{Q}}_{\mathbb{R}}$, and step 1.1 shows that each real singleton is nowhere dense; claim 3 uses the completeness of $\mathbb{R}$ through A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$, inside Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets. Indeed $\mathbb{Q}$ is a $G_{\delta }$ subset of itself, being the whole space, so no argument that ignores the ambient completeness can possibly give claim 3.

• The irrationals are also uncountable (The irrationals are uncountable), by a different and much cheaper argument that needs only the countability of $\mathbb{Q}$ and the uncountability of $\mathbb{R}$. Uncountability and being residual are independent properties: $\mathbb{Q}$ is meager and countable, the Cantor set is meager and uncountable (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), and $X$ is residual and uncountable.

• The named witness for claim 3 is The irrationals form a residual $G_{\delta }$ set that is not $F_{\sigma }$ ↗, and the false statement it refutes is FALSE: $\mathbb{Q}$ is a $G_{\delta }$ subset of $\mathbb{R}$; the refutation is carried out here.

• Meagre and measure zero are not the same smallness. ${\mathbb{Q}}_{\mathbb{R}}$ is both, but the two notions diverge as soon as one leaves the countable case: $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗ writes $\mathbb{R}$ as a meager set together with a set of measure zero, and the set of measure zero there is residual. So being residual, which is what claim 2 gives for $X$, carries no information at all about size in measure.