G-Delta Sets form Lattice

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $\mathcal G$ be the collection of all $G_\delta$ sets of $T$.

Then $\left({\mathcal G, \subseteq}\right)$ is a lattice, where $\subseteq$ denotes the subset relation.

Proof
From Subset Relation is Ordering, $\subseteq$ is an ordering on $\mathcal G$.

Let $G, G'$ be $G_\delta$ sets of $T$.

We have $G_\delta$ Sets Closed under Union, so that $G \cup G' \in \mathcal G$.

From Union is Smallest Superset and Subset of Union, it follows that $G \cup G'$ is the supremum of $G$ and $G'$.

Similarly, we have $G_\delta$ Sets Closed under Intersection, and so $G \cap G' \in \mathcal G$.

From Intersection is Largest Subset and Intersection is Subset, it follows that $G \cap G'$ is the infimum of $G$ and $G'$.

Thus any two elements of $\mathcal G$ are seen to have both a supremum and an infimum in $\mathcal G$.

Hence $\left({\mathcal G, \subseteq}\right)$ is a lattice.

Also see

 * $F_\sigma$ Sets form Lattice