In Hrbacek and Jech (1999, p.205), they point out that "it is known that the theorem [the extension of any filter to an ultrafilter] cannot be proved in Zermelo-Fraenkel set theory alone." And in Jech (2000, p.81), he mentioned that "[i]t is known that the theorem [the Prime Ideal Theorem] cannot be proved without using the Axiom of Choice. However, it is also known that the Prime Ideal Theorem is weaker than the Axiom of Choice."

I am having a hard time finding a reference for the above claims. Can someone please give me some pointers (references) to, for instance, $\mathbf{ZF}\not\vdash \{\text{existence of non-principal ultrafilters}\}$? Thanks!

  • Hrbacek, K. and Jech, T. J. (1999). Introduction to Set Theory. Marcel Dekker, New York, third edition.
  • Jech, T. J. (2003). Set Theory. Springer-Verlag, Berlin, Heidelberg, New York, 3rd millennium ed, rev. and expanded edition.

2 Answers 2


In addition to what Noah wrote, Jech "The Axiom of Choice" has a proofs, partial proofs, or problems with hints for the following:

  1. In the first Cohen model, the axiom of [countable] choice fails; but the Boolean Prime Ideal theorem holds. Therefore every filter can be extended to an ultrafilter there.

  2. There is a model of $\sf ZF$ in which there are no free ultrafilters on $\omega$. In the same model the Hahn-Banach theorem also fails (although Hahn-Banach is strictly weaker than the ultrafilter lemma).

    You can find the full proof in Halbeisen's "Combinatorial Set Theory" (whereas Jech gives this as an exercise with a hint).

These contain better exposition from a modern point of view, compared to papers from the 1960s.

Blass proved, as Noah pointed out, that it is consistent that every ultrafilter is principal. The proof is constructed from two parts:

  1. Constructing a forcing extension where every filter on $\omega$ is principal, start from $L$.

    Proving that if there is no inner model with a measurable cardinal (e.g. if you start from $L$), then if there is no free ultrafilter on $\omega$, then there are no free ultrafilters on any well-orderable set.

  2. Prove that if $W$ is the smallest class containing all the singletons and closed under well-ordered unions, and all ultrafilters on ordinals are principal, then all ultrafilters on sets in $W$ are principal.

    And that the model constructed in the very first step, is internally equal to $W$.

The paper itself is surprisingly short, too.


See this mathoverflow question https://mathoverflow.net/questions/59157/reference-request-independence-of-the-ultrafilter-lemma-from-zf, especially the answer by Andreas Blass.

Sol Feferman proved that $ZF$ does not prove that there is a nonprincipal ultrafilter on $\omega$, in "Some applications of the notions of forcing and generic sets" http://matwbn.icm.edu.pl/ksiazki/fm/fm56/fm56129.pdf.

The stronger statement "$ZF$ does not prove that any infinite set carries a nonprincipal ultrafilter" was proved by Andreas Blass in "A model without ultrafilters."

The ultrafilter lemma was proved strictly weaker than full $AC$ by Halpern and Levy, in ""The Boolean prime ideal theorem does not imply the axiom of choice."

(Sadly, I can't find Blass or Halpern-Levy online.)

In general, the book "Consequences of the axiom of choice" by Rubin and Rubin and the accompanying website http://consequences.emich.edu/conseq.htm are invaluable for this sort of question.

  • $\begingroup$ This is very helpful. Let me look into it. Thanks! $\endgroup$
    – Logica
    Jun 11, 2015 at 14:26
  • $\begingroup$ Here is a link to Halpern-Levy. $\endgroup$
    – Logica
    Jun 11, 2015 at 14:59
  • $\begingroup$ Uhh... no. Cohen's first model satisfies $\sf BPI$ therefore there are many nonprincipal ultrafilters on $\omega$. Remember that at the day, generic extensions were often called "Cohen models". $\endgroup$
    – Asaf Karagila
    Jun 11, 2015 at 15:37
  • $\begingroup$ @AsafKaragila I've removed that sentence. I'm surprised, though - how do you get a hereditarily symmetric name for an ultrafilter? $\endgroup$ Jun 11, 2015 at 15:51
  • $\begingroup$ No. That is the heart and soul of the very difficult Halpern-Levy paper. If you trace their proof, you should find out how to define such name more explicitly. $\endgroup$
    – Asaf Karagila
    Jun 11, 2015 at 15:53

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