Without any choice axioms, are there free ultrafilters on the natural numbers? If not, can we prove the existence of ANY free ultrafilters, on any set?
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$\begingroup$ No. If you does not assume AC, then you cannot prove the existence of free ultrafilers on the set of natural numbers. $\endgroup$– Hanul JeonMar 10, 2014 at 8:32
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$\begingroup$ After years of wondering and days of looking, I find the answer said in-passig on another tab I had opened, within five minutes of finally asking. System works! (A free ultrafilter on the naturals implies non-measurable sets of reals.) $\endgroup$– leewzMar 10, 2014 at 8:33
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$\begingroup$ @tetori What about any other set? Are there free ultrafilters that don't require choice? I'd expect not. $\endgroup$– leewzMar 10, 2014 at 8:34
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$\begingroup$ I don't know the case of another set... $\endgroup$– Hanul JeonMar 10, 2014 at 8:38
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No. To both questions.
It is consistent that the axiom of choice fails and there are no free ultrafilters on the natural numbers; and it is also consistent that the axiom of choice fails and there are no free ultrafilters on any set.
I'm not sure who proved the first result. The second result is due to Andreas Blass.
answered Mar 10, 2014 at 9:08