Ultrafilter lemma

Remarks

Proof idea: consider the proper filters that contain $F$, ordered by inclusion. This family is nonempty, since it contains $F$. Every nonempty chain of them has an upper bound, its union, which is still a proper filter. By Zorn's lemma there is a maximal one, and it is an ultrafilter. Applied to the power set of a set $X$, the lemma says that every filter on $X$ lies in an ultrafilter on $X$. For an infinite $X$ and the filter of cofinite sets, this gives non-principal ultrafilters. Applied to the filter generated by an element $a \ne 0$, it gives an ultrafilter containing $a$, which is the key step of Stone's representation theorem. The lemma is not provable in ZF alone, but it is strictly weaker than the axiom of choice over ZF; it is equivalent to the Boolean prime ideal theorem.

Going deeper

For a student

The ultrafilter lemma is the form of choice that much of algebra and topology needs. It proves the compactness theorem of propositional logic, Tychonoff's theorem for compact Hausdorff spaces, and the representation of Boolean algebras as fields of sets. For a countable Boolean algebra no choice is needed: one lists the elements and decides them one at a time.