Ultrafilter Lemma

Theorem
Let $$X$$ be a set.

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

Proof
Let $$\Omega$$ be the set of filters on $$X$$.

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 $$X$$ and thus an upper bound of $$C$$.

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.

Comment
Note that this result, dependent as it is upon Zorn's Lemma, is a consequence of the Axiom of Choice.