Ultrafilter Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set.

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


Corollary

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 $S$.

From Subset Relation is Ordering, the subset relation "$\subseteq$" makes $\Omega$ 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 $\mathcal F, \mathcal F' \in C$ 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 $A \in \mathcal F'$.

Hence:

$A \cap B \in \mathcal F'$

In particular:

$A, B \in \bigcup C$

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.

$\blacksquare$


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.


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.

$\blacksquare$


Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI).

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.