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 $\left({\mathcal F, \subseteq}\right)$ 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 Union of Subsets, conclude that $F \cup G \subseteq I$ as $F, G \subseteq I$.

From Cardinality of Set Union, $\left\vert{F \cup G}\right\vert \le \left\vert{F}\right\vert + \left\vert{G}\right\vert$, 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 $\left({\mathcal F, \subseteq}\right)$ is a directed set.

$\blacksquare$