The irrationals form a residual $G_\delta$ set that is not $F_\sigma$

Refuted claim: $\mathbb{Q}$ is a $G_\delta$ subset of $\mathbb{R}$ (FALSE: $\mathbb{Q}$ is a $G_\delta$ subset of $\mathbb{R}$); equivalently, by complementation ($F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$), the irrationals are $F_\sigma$.

The witness is the set $X := \mathbb{R} \setminus \mathbb{Q}_{\mathbb{R}}$ of irrationals (The rationals embed densely in the reals). It is $G_\delta$, being $\bigcap_n (\mathbb{R} \setminus \{e(n)\})$ for any enumeration $e$ of the rationals, and it is residual, its complement being a countable union of singletons; but it is not $F_\sigma$, and that is the failure of the refuted claim. The refutation is carried out in full in $\mathbb{Q}$ is $F_\sigma$, meager and not $G_\delta$, while the irrationals are $G_\delta$, residual and not $F_\sigma$; this item records the witness and the three properties that make it the right one.

• The asymmetry is real and is not a defect of the definitions. The two classes $F_\sigma$ and $G_\delta$ are exchanged by complementation, but a particular set need not lie in both: $\mathbb{Q}_{\mathbb{R}}$ is $F_\sigma$ and not $G_\delta$, and $X$ is $G_\delta$ and not $F_\sigma$. A set lying in both classes is a genuinely stronger condition, satisfied for instance by every open set and every closed set.

• What forces it is the Baire category theorem, through the fact that $\mathbb{R}$ is not meager (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) while $\mathbb{Q}_{\mathbb{R}}$ is. Both are dense; the rationals are countable and the irrationals are uncountable. No cardinality or density argument distinguishes them in the required way; the distinction is one of category.

• $X$ is large in both senses. It is residual, so it is large in category; and it is not null. For if it were, then, $\mathbb{Q}_{\mathbb{R}}$ being null (Every at most countable subset of $\mathbb{R}$ has measure zero), one could interleave a cover of each with slack $\varepsilon \cdot 2^{-1}$ and obtain a cover of $\mathbb{Q}_{\mathbb{R}} \cup X = \mathbb{R}$ of total length at most $\varepsilon$, which A sequence of intervals covering $[a,b]$ has total length at least $b - a$, so no interval of positive length has measure zero forbids already for $[0,1]$. Interleaving two covers needs no choice principle, unlike the countably infinite case (A countable union of measure-zero sets has measure zero, by countable choice).