Ultrafilter Lemma/Proof 1

Theorem

Let $S$ be a set.

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

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.

Without loss of generality, 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.

$\blacksquare$

Axiom of Choice

This proof depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Also known as

The Ultrafilter Lemma can also be referred to as:

the ultrafilter principle

the ultrafilter theorem

and abbreviated UL or UF.