F-Sigma Sets form Lattice

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

Let $\mathcal F$ be the collection of all $F_\sigma$ sets of $T$.

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

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

Let $F, F'$ be $F_\sigma$ sets of $T$.

We have $F_\sigma$ Sets Closed under Union, so that $F \cup F' \in \mathcal F$.

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

Similarly, we have $F_\sigma$ Sets Closed under Intersection, and so $F \cap F' \in \mathcal F$.

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

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

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

Also see

 * $G_\delta$ Sets Form Lattice