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 $\AA \subseteq \FF$:
- $\AA \subseteq \UU$
$\blacksquare$