G-delta set

Remarks

Every open set is $G_\delta$: take $U_n = A$ for all $n$. In a metric space with metric $d$, every closed set $F$ is $G_\delta$, since $F = \bigcap_n \set{x : d(x, F) < 1/n}$.

Example: the set of irrational numbers is $\bigcap_{q \in \Q} (\R \setminus \set{q})$. Each $\R \setminus \set{q}$ is open and $\Q$ is countable, so the irrationals form a $G_\delta$ set.

Non-example: $\Q$ is not a $G_\delta$ subset of $\R$. If $\Q = \bigcap_n U_n$ with $U_n$ open, each $U_n$ is open and dense, and so are the sets $\R \setminus \set{q}$ for $q \in \Q$. These countably many dense open sets have empty intersection, which contradicts the Baire category theorem.

A set is $G_\delta$ if and only if its complement is an F-sigma set.