Ultrafilter Lemma/Proof 1

Proof
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.