I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"):

  1. The ultrafilter lemma is independent of ZF.
  2. ZF + the ultrafilter lemma does not imply the Axiom of Choice.

I would prefer an overview article / book that links to the original papers instead of the original papers themselves. That's because I don't rely on these facts; I only want to give some context for my readers.

  • 4
    $\begingroup$ Jech's book The Axiom of Choice covers this material. $\endgroup$ – Ed Dean Mar 22 '11 at 10:18
  • $\begingroup$ Is the ultrafilter lemma that there exists a non-principal ultrafilter over any infinite set, or only that there exists a non-principal ultrafilter over some set? $\endgroup$ – Zsbán Ambrus Mar 22 '11 at 11:04
  • 3
    $\begingroup$ @Zsbán: Neither. The Ultrafilter Lemma says that any filter on a set can be extended to an ultrafilter on that set. $\endgroup$ – François G. Dorais Mar 22 '11 at 11:33
  • $\begingroup$ @FrançoisG.Dorais Is that actually stronger than saying every set has an ultrafilter? Supppose $F$ is a nonprincipal filter on $X$. I can form the quotient $P(X)/I$ for the dual ideal (let's also add the finite sets to the ideal), and then if I put an ultrafilter $F^*$ on the quotient, don't a get an ultrafilter on $X$ extending $F$? $\endgroup$ – Joel David Hamkins Apr 30 '17 at 12:05
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    $\begingroup$ @JoelDavidHamkins: $P(X)/I$ is not necessarily isomorphic to the Boolean algebra of subsets of a set. For example, $P(\omega)/fin$ is not atomic. $\endgroup$ – François G. Dorais Apr 30 '17 at 14:59

In any of the formulations mentioned (so far) in the comments, the ultrafilter lemma is independent of ZF but weaker than AC. That the strongest form (all filters can be extended to ultrafilters) doesn't imply AC is a theorem of J.D. Halpern and A. Lévy ["The Boolean prime ideal theorem does not imply the axiom of choice" in Axiomatic Set Theory, Proc. Symp. Pure Math. XIII part 1, pp. 83-134]. That ZF doesn't prove even the weakest form (there exists a nonprincipal ultrafilter on some set) is a theorem of mine ["A model without ultrafilters," Bull. Acad. Polon. Sci. 25 (1977) pp. 329-331], building on S. Feferman's construction of a model with no non-principal ultrafilters on the set of natural numbers ["Some applications of the notions of forcing and generic sets," Fundamenta Mathematicae 55 (1965) pp. 325-345].


If you're just interested in context, rather than proofs, Eric Schechter's book Handbook of Analysis and Its Foundations talks about the relation between existence of ultrafilters, various weak notions of choice, and standard results in analysis.

  • $\begingroup$ Wow, that's a great book recommendation! I'll cite this one but accept the other answer since the latter is more specific. $\endgroup$ – Greg Graviton Mar 22 '11 at 21:02

The axiom "There exists some nonprincipalultrafilter." is discussed in Horst Herrlich, "The Axiom of Choice" as WUF(?) and in Rubin and Rubin "Equivalents of the Axiom of Choice, II." under [206]. The latter is kind of the standard ressource for these kinds of questions.

Here are papers providing proofs:

For 1:

MR0480028 (58 #227) Pincus, David ; Solovay, Robert M. Definability of measures and ultrafilters. J. Symbolic Logic 42 (1977), no. 2, 179--190. JSTOR, doi: 10.2307/2272118

For 2:

MR0480027 (58 #226) Pincus, David . Adding dependent choice to the prime ideal theorem. Logic Colloquium 76 (Oxford, 1976), pp. 547--565. Studies in Logic and Found. Math., Vol. 87, North-Holland, Amsterdam, 1977. doi: 10.1016/S0049-237X(09)70445-8


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