Ultrafilter Lemma

Theorem
Let $S$ be a set.

Every filter on $S$ is contained in an ultrafilter on $S$.

Proof from the Axiom of Choice
Let $\Omega$ be the set of filters on $S$.

From Subset Relation is Ordering, the subset relation makes $\struct {\Omega, \subseteq}$ a partially ordered set.

Let $C \subseteq \Omega$ be a non-empty chain.

Then $\bigcup C$ is again a filter on $S$.

Thus $\bigcup C$ is an upper bound of $C$.

Indeed, if $A, B \in \bigcup C$ then there there are filters $\FF, \FF' \in C$ with $A \in \FF$ and $B \in \FF'$.

We have that $C$ is a chain.

, let $\FF \subset \FF'$.

Thus $A \in \FF'$.

Hence:
 * $A \cap B \in \FF'$

In particular:
 * $A, B \in \bigcup C$

For any $\FF \in \Omega$ there is therefore by Zorn's Lemma a maximal element $\FF'$ such that:
 * $\FF \subseteq \FF'$

The maximality of $\FF'$ is in this context equivalent to $\FF'$ being an ultrafilter.

Proof from the Boolean Prime Ideal Theorem
Order the subsets of $S$ by reverse inclusion.

Then the result follows trivially from the Boolean Prime Ideal Theorem.

Also known as
This axiom may be called the ultrafilter principle or the ultrafilter theorem, and may be abbreviated UL or UF.