Ultrafilter Lemma/Proof 2

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Theorem

Let $S$ be a set.

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


Proof

Let $\FF$ be a filter on $S$.

Let the power set of $S$ be ordered by inclusion.

Then, by Filter on Set is Proper Filter, $\FF$ is a proper filter on $\powerset S$.

By Filter is Ideal in Dual Ordered Set, $\FF$ is an ideal in the dual of $\powerset S$.


By Singleton of Bottom is Ideal, $\set \empty$ is an ideal in $\powerset S$.

By Ideal is Filter in Dual Ordered Set, $\set \empty$ is a filter on the dual of $\powerset S$.

By the definition of Filter on Set, $\empty \notin \FF$, so $\FF$ and $\set \empty$ are disjoint.

Therefore, by the Boolean Prime Ideal Theorem, there is a prime ideal $\GG \supseteq \FF$ that is disjoint from $\set \empty$.


By Prime Ideal is Prime Filter in Dual Lattice, $\GG$ is a prime filter on the dual of the dual of $\powerset S$.

Thus, $\GG$ is a prime filter on $\powerset S$ by Dual of Dual Ordering.

But $\GG$ is disjoint from $\set \empty$, which implies that $\empty \notin \GG$.

Therefore, $\GG$ is a proper filter.

Thus, by Proper and Prime iff Ultrafilter in Boolean Lattice, $\GG$ is an ultrafilter on $\powerset S$.

But by comparing the definitions, $\GG$ is thus an ultrafilter on $S$.

$\blacksquare$


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.


Also known as

The Ultrafilter Lemma can also be referred to as:

the ultrafilter principle
the ultrafilter theorem

and abbreviated UL or UF.