Ultrafilter Lemma

Theorem
Let $S$ be a set.

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

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

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

From Subset Relation is Ordering, the subset relation "$\subseteq$" makes $\Omega$ 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 $\mathcal F, \mathcal F' \in C$ with $A \in \mathcal F$ and $B \in \mathcal F'$.

We have that $C$ is a chain.

, let $\mathcal F \subset \mathcal F'$.

Thus $A \in \mathcal F'$.

Hence:
 * $A \cap B \in \mathcal F'$

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

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

The maximality of $\mathcal F'$ is in this context equivalent to $\mathcal F'$ 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.