Welcome to MathOverflow

Question

6

1

Is it known to be consistent with ZF that there is no non-principal ultrafilter on any infinite set? (Feel free to use your favorite interpretation of "infinite" in this context. If infinite just meant infinite ordinals, that would be fine, too. You may use all kinds of large cardinals.)

flag	asked Sep 27 2011 at 11:22 Stefan Geschke 10.6k●1●15●37

François G. Dorais · Accepted Answer · 2011-09-27 11:55:08Z UTC

10

Yes, it is consistent with ZF that every ultrafilter is principal. This is a result of Andreas Blass, A model without ultrafilters, Bull. Acad. Polon. Sci. 25 (1977), 329–331.

answered Sep 27 2011 at 11:55

François G. Dorais♦
23.1k●2●52●116

3

And it is also contained in: MR0480028 (58 #227) Pincus, David; Solovay, Robert M. Definability of measures and ultrafilters. J. Symbolic Logic 42 (1977), no. 2, 179–190. – Simon Thomas Sep 27 2011 at 12:09

Thanks for the answers. I was under the wrong impression that only the existence of ultrafilters on particular sets had been looked at – Stefan Geschke Sep 27 2011 at 12:10

Existence of non-principal ultrafilters on sets

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.