Ultrafilter Lemma/Corollary

Theorem
Let $S$ be a non-empty set.

Let $\mathcal A$ be a set of subsets of $S$.

Suppose that $\mathcal A$ has the finite intersection property.

Then there is an ultrafilter $\mathcal U$ on $S$ such that $\mathcal A \subseteq \mathcal U$.

Proof
Let $\mathcal I$ be the set of intersections of non-empty finite subsets of $\mathcal A$.

Let $\mathcal F = \{ T \in \mathcal P(S) \mid \exists B \in \mathcal I: B \subseteq T \}$.

Note that $\mathcal A \subseteq \mathcal I \subseteq \mathcal F$.

$\mathcal F$ is a filter on $S$:

Since $\mathcal A$ has the finite intersection property, $\varnothing \notin \mathcal I$.

Since each element of $\mathcal F$ is a superset of some element of $\mathcal I$, $\varnothing \notin \mathcal F$.

If $P in \mathcal F$ and $P \subseteq Q \subseteq S$, then for some $B \in \mathcal I$, $B \subseteq P$, so $B \subseteq Q \subseteq S$. Thus $Q \in \mathcal F$.

If $P \in \mathcal F$ and $Q \in \mathcal F$ then for some $B, C \in \mathcal I$, $B \subseteq P$ and $C \subseteq Q$.

Then $B \cap C \in \mathcal I$ and $B \cap C \subseteq P \cap Q \subseteq S$.

Thus $P \cap Q \in \mathcal F$.

Thus $\mathcal F$ is a filter on $S$.

By the Ultrafilter Lemma, there is an ultrafilter $\mathcal U$ on $S$ such that $\mathcal F \subseteq \mathcal U$.

Since $\mathcal A \subseteq \mathcal F$, $\mathcal A \subseteq \mathcal U$.