What would be the usage of free ultrafilter?

And how is it important?

BTW, I know the concept of free ultrafilter, so I only need explanation on usage and importance.


In topology, ultrafilters, including free ultrafilters, are used to construct the Čech-Stone and Wallman compactifications. They provide a nice generalization of sequences, sufficient (as sequences are not) to describe all Hausdorff topologies. This answer by Martin Sleziak to an earlier question explains one of the less familiar uses, and it and the comments appended to it have some references that you might find useful.

In model theory they are the basis of the ultraproduct construction, which among other things provides easier proofs of a number of results and a nice way to construct non-standard models of the real numbers that justify working with infinitesimals.

In set theory they provide one natural way of looking at measurable cardinals, one of the more natural and straightforward types of large cardinal.

In functional analysis they have been used to give standard proofs of results previously proved using techniques from non-standard analysis, as in John J. Buoni, Albert Klein, Brian M. Scott, & Bhushan J. Wadhwa, On Power Compactness in a Banach Space, Indiana Univ. Math. J., Vol. 32, No. 2 (1983), pp. 177-185; DOI:10.1512/iumj.1983.32.32015.

In a wide variety of areas they provide easier or nicer proofs (at least from some points of view) of results that can be proved by other means. For example, one of the nicest proofs of Arrow’s Impossibility theorem uses a free ultrafilter.

  • $\begingroup$ I think your answer is great, but I respectfully disagree on the ultrafilter proof of Arrow's theorem being one of the easier ones. See here for some very simple proofs. $\endgroup$ – Michael Greinecker May 27 '12 at 9:27
  • $\begingroup$ @Michael: I should have separated the two statements more clearly: for that one I really did mean nicest more than easiest. I agree that there are easy non-ultrafilter proofs; I’m not sure that I consider them all that much easier, but that’s a matter of taste, and I won’t argue with anyone who dies. $\endgroup$ – Brian M. Scott May 27 '12 at 9:41
  • $\begingroup$ @Martin: Thanks for adding the reference. $\endgroup$ – Brian M. Scott May 27 '12 at 9:42

[Free] Ultrafilters on $\mathbb N$ give canonical examples for sets which are neither Lebesgue measurable nor have the Baire property.

The existence of free ultrafilters follows from [a weak form of] the axiom of choice it turns out that both the assertions: "Every sets of real numbers is Lebesgue measurable" and "Every sets of real numbers have the Baire property" are inconsistent with the axiom of choice.

Indeed the Vitali sets make a far easier example of a non-measurable set, but when was the last time you ran into a counterexample of the Baire property?

To add on that, free ultrafilters are useful to determine convergence in general topological spaces (much like nets) and therefore useful to determine compactness (every ultrafilter converges).

We also have a tight connection between logic and model theory and compactness, topology and ultrafilters. First-order logic has a nice property called the compactness theorem which can be used to build models of theories. Ultrafilters can be used in a construction known as Ultraproduct to construct new models in a sorta-similar way.

It is worth noting that the connection between ultrafilters and model theory is a lot deeper. (For example Shelah's papers with Malliaris abstracts of which can be found here under numbers 996-999.)


Non-principal (free) complete ultrafilters define measurable cardinals in set theory. If the filter is principle then the ultrapower natural embedding would be trivial (identity). If, on the other hand, it is non-priciple, then there would be some ordinal (critical point) that is moved by this embedding, i.e. the embedding is non-trivial, and the situation becomes much more intersting (many problems receive proper solutions). T. Jech "Set Theory" is a good reference for measurable cardinals.

  • 3
    $\begingroup$ I think the correct term is non-principal not non-principle. $\endgroup$ – Martin Sleziak May 27 '12 at 9:07

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