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.)
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asked Sep 27, 2011 at 11:22
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1 Answer
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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
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5$\begingroup$ 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. $\endgroup$ Commented Sep 27, 2011 at 12:09
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$\begingroup$ Thanks for the answers. I was under the wrong impression that only the existence of ultrafilters on particular sets had been looked at $\endgroup$ Commented Sep 27, 2011 at 12:10