Finite Subsets form Directed Set

Theorem
Let $I$ be a set.

Denote with $\mathcal F$ the set of finite subsets of $I$.

Let $\subseteq$ be the subset relation on $\mathcal F$.

Then $\struct {\mathcal F, \subseteq}$ is a directed set.

Proof
From Subset Relation is Ordering, we know that $\subseteq$ is an ordering, hence also a preordering.

Now let $F, G \in \mathcal F$.

From Set Union Preserves Subsets, conclude that $F \cup G \subseteq I$ as $F, G \subseteq I$.

From Cardinality of Set Union, $\card {F \cup G} \le \card F + \card G$, implying that $F \cup G$ is a finite set.

Hence $F \cup G \in \mathcal F$.

Furthermore, $F \subseteq F \cup G$ and $G \subseteq F \cup G$.

It follows that $\struct {\mathcal F, \subseteq}$ is a directed set.