# 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 $\mathcal F = \left\{ {T \in \mathcal P \left({S}\right) : \exists B \in \mathcal I: B \subseteq T}\right\}$.

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

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

Because $\mathcal A$ has the finite intersection property:

$\varnothing \notin \mathcal I$

Because each element of $\mathcal F$ is a superset of some element of $\mathcal I$:

$\varnothing \notin \mathcal F$

Let $P \in \mathcal F$ and $P \subseteq Q \subseteq S$.

Then:

$\exists B \in \mathcal I: B \subseteq P$

and so:

$B \subseteq Q \subseteq S$

Thus:

$Q \in \mathcal F$

Let $P \in \mathcal F$ and $Q \in \mathcal F$.

Then:

$\exists B, C \in \mathcal I: B \subseteq P$ 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, by definition, $\mathcal F$ is a filter on $S$.

By the Ultrafilter Lemma, there exists an ultrafilter $\mathcal U$ on $S$ such that:

$\mathcal F \subseteq \mathcal U$

Because $\mathcal A \subseteq \mathcal F$:

$\mathcal A \subseteq \mathcal U$

$\blacksquare$