Let $S$ be a set.
Let $S$ be a non-empty set.
Let $\mathcal A$ be a set of subsets of $S$.
Suppose that $\mathcal A$ has the finite intersection property.
Then there is an ultrafilter $\mathcal U$ on $S$ such that $\mathcal A \subseteq \mathcal U$.
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 $S$.
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 $\mathcal F, \mathcal F' \in C$ with $A \in \mathcal F$ and $B \in \mathcal F'$.
We have that $C$ is a chain.
Without loss of generality, let $\mathcal F \subset \mathcal F'$.
Thus $A \in \mathcal F'$.
- $A \cap B \in \mathcal F'$
- $A, B \in \bigcup C$
- $\mathcal F \subseteq \mathcal F'$
The maximality of $\mathcal F'$ is in this context equivalent to $\mathcal F'$ being an ultrafilter.
Axiom of Choice
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.
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.
Boolean Prime Ideal Theorem
This theorem depends on the Boolean Prime Ideal Theorem (BPI).
As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.