FALSE: $\mathbb{Q}$ is a $G_{\delta }$ subset of $\mathbb{R}$

False claim: $\mathbb{Q}$, that is the set ${\mathbb{Q}}_{\mathbb{R}}$ of rationals inside $\mathbb{R}$ (The rationals embed densely in the reals), is a $G_{\delta }$ set ($F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$): there is a sequence $(V_n)$ of open subsets of $\mathbb{R}$ with ${\mathbb{Q}}_{\mathbb{R}}={\bigcap }_n{V}_n$.

The claim looks plausible by symmetry. ${\mathbb{Q}}_{\mathbb{R}}$ is $F_{\sigma }$, being a countable union of singletons; the irrationals are $G_{\delta }$, being a countable intersection of complements of singletons; and the two classes are exchanged by complementation. So one expects each set to belong to both classes. It does not: the symmetry between the two classes says nothing about a single set, and the obstruction is the Baire category theorem.

• What is true about $\mathbb{Q}$. It is $F_{\sigma }$, meager, of measure zero, dense, and countable. What fails is only the $G_{\delta }$ property, and its failure is a genuine theorem about $\mathbb{R}$, resting on completeness 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. Inside $\mathbb{Q}$ itself the corresponding claim is true and trivial, $\mathbb{Q}$ being the whole space there.

• The dual false statement is not recorded separately, because it is the same statement: the irrationals fail to be $F_{\sigma }$ exactly because ${\mathbb{Q}}_{\mathbb{R}}$ fails to be $G_{\delta }$ ($F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$). The witness is The irrationals form a residual $G_{\delta }$ set that is not $F_{\sigma }$ ↗.

• Context, not a result of this library. In classical analysis the set of points at which a real function is continuous is always $G_{\delta }$, and it is the false statement above that then rules out a function continuous at every rational and at no irrational. That classical result is not proved here, and continuity is not available at this point in the reading order; the connection is recorded as orientation and nothing on this page depends on it.