Ultrafilter on a set
and that is maximal with respect to inclusion, in the sense that there does not exist a strictly larger collection of subsets of
This article deals specifically with ultrafilters on a set and does not cover the more general notion.
On the other hand, there exists models of ZF where every ultrafilter on a set is principal.
[1]: 186 Usually, only free ultrafilters lead to non-trivial constructions.
Some authors do not include non-degeneracy (which is property (1) above) in their definition of "filter".
However, the definition of "ultrafilter" (and also of "prefilter" and "filter subbase") always includes non-degeneracy as a defining condition.
is the set A prefilter or filter base is a non-empty and proper (i.e.
and any of the following equivalent conditions are satisfied:[2][3] A filter subbase that is ultra is necessarily a prefilter.
The next theorem shows that every ultrafilter falls into one of two categories: either it is free or else it is a principal filter generated by a single point.
Consequently, free ultrafilters can only exist on an infinite set.
[12] This result is not necessarily true for an infinite family of filters.
However, the preimage of an ultrafilter is not necessarily ultra, not even if the map is surjective.
The elementary filter induced by an infinite sequence, all of whose points are distinct, is not an ultrafilter.
Every proper filter is equal to the intersection of all ultrafilters containing it.
can be extended to a free ultrafilter if and only if the intersection of any finite family of elements of
That is, there exist models in which the axioms of ZF hold but the ultrafilter lemma does not.
There also exist models of ZF in which every ultrafilter is necessarily principal.
Every filter that contains a singleton set is necessarily an ultrafilter and given
is finite then the ultrafilter lemma can be proven from the axioms ZF.
The existence of free ultrafilter on infinite sets can be proven if the axiom of choice is assumed.
Under ZF, the axiom of choice is, in particular, equivalent to (a) Zorn's lemma, (b) Tychonoff's theorem, (c) the weak form of the vector basis theorem (which states that every vector space has a basis), (d) the strong form of the vector basis theorem, and other statements.
However, the ultrafilter lemma is strictly weaker than the axiom of choice.
[17] Other authors attribute this discovery to Bedřich Pospíšil (following a combinatorial argument from Fichtenholz, and Kantorovitch, improved by Hausdorff).
[18][19] Under ZF, the axiom of choice can be used to prove both the ultrafilter lemma and the Krein–Milman theorem; conversely, under ZF, the ultrafilter lemma together with the Krein–Milman theorem can prove the axiom of choice.
[citation needed] The Rudin–Keisler ordering (named after Mary Ellen Rudin and Howard Jerome Keisler) is a preorder on the class of powerset ultrafilters defined as follows: if
extends the natural numbers, may possess, which prove useful in various areas of set theory and topology.
Walter Rudin proved that the continuum hypothesis implies the existence of Ramsey ultrafilters.
[32] In fact, many hypotheses imply the existence of Ramsey ultrafilters, including Martin's axiom.
Saharon Shelah later showed that it is consistent that there are no P-point ultrafilters.
is Ramsey if and only if it is minimal in the Rudin–Keisler ordering of non-principal powerset ultrafilters.
