Editing Boolean prime ideal theorem (section)

==The ultrafilter lemma==
{{Main|Ultrafilter lemma}}
{{See also|Set-theoretic topology}}

A filter on a set {{mvar|X}} is a nonempty collection of nonempty subsets of {{mvar|X}} that is closed under finite intersection and under superset. An ultrafilter is a maximal filter.
The ultrafilter lemma states that every filter on a set {{mvar|X}} is a subset of some [[ultrafilter]] on {{mvar|X}}.<ref>{{citation
 | last = Halpern | first = James D.
 | year = 1966
 | title = Bases in Vector Spaces and the Axiom of Choice
 | journal = [[Proceedings of the American Mathematical Society]]
 | publisher = American Mathematical Society
 | jstor = 2035388
 | doi = 10.1090/S0002-9939-1966-0194340-1 | doi-access = free
 | volume = 17
 | issue = 3
 | pages = 670–673
 }}</ref>
An ultrafilter that does not contain finite sets is called "non-principal". The ultrafilter lemma, and in particular the existence of non-principal ultrafilters (consider the filter of all sets with finite complements), can be proven from [[Zorn's lemma]].

The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice. The idea behind the proof is that the subsets of any set form a Boolean algebra partially ordered by inclusion, and any Boolean algebra is representable as an algebra of sets by [[Stone's representation theorem]].

If the set {{mvar|X}} is finite then the ultrafilter lemma can be proven from the axioms ZF. This is no longer true for infinite sets; an additional axiom {{em|must}} be assumed. [[Zorn's lemma]], the [[axiom of choice]], and [[Tychonoff's theorem]] can all be used to prove the ultrafilter lemma. The ultrafilter lemma is strictly weaker than the axiom of choice.

The ultrafilter lemma has many [[Filters in topology|applications in topology]]. The ultrafilter lemma can be used to prove the [[Hahn-Banach theorem]] and the [[Alexander subbase theorem]].