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.

If $C \subseteq \Omega$ is a non-empty chain, then $\bigcup C$ is again a filter on $S$ and thus 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 $B \in \mathcal{F}'$. Since $C$ is a chain one can assume w.l.o.g. $\mathcal{F}\subset \mathcal{F}'$ and thus $A \in \mathcal{F}'$ in which case also $A\cap B \in \mathcal{F}'$. 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.