# Ultrafilter Lemma/Corollary

## Theorem

Let $S$ be a non-empty set.

Let $\AA$ be a set of subsets of $S$.

Suppose that $\AA$ has the finite intersection property.

Then there is an ultrafilter $\UU$ on $S$ such that $\AA \subseteq \UU$.

## Proof

Let $\II$ be the set of intersections of non-empty finite subsets of $\AA$.

Let $\FF = \set {T \in \powerset S: \exists B \in \II: B \subseteq T}$.

Note that $\AA \subseteq \II \subseteq \FF$.

$\FF$ is a filter on $S$:

Because $\AA$ has the finite intersection property:

$\O \notin \II$

Because each element of $\FF$ is a superset of some element of $\II$:

$\O \notin \FF$

Let $P \in \FF$ and $P \subseteq Q \subseteq S$.

Then:

$\exists B \in \II: B \subseteq P$

and so:

$B \subseteq Q \subseteq S$

Thus:

$Q \in \FF$

Let $P \in \FF$ and $Q \in \FF$.

Then:

$\exists B, C \in \II: B \subseteq P$ and $C \subseteq Q$

Then:

$B \cap C \in \II$

and:

$B \cap C \subseteq P \cap Q \subseteq S$

Thus:

$P \cap Q \in \FF$

Thus, by definition, $\FF$ is a filter on $S$.

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

$\FF \subseteq \UU$

Because $\AA \subseteq \FF$:

$\AA \subseteq \UU$

$\blacksquare$