Timeline for Is choice needed to establish the existence of idempotent ultrafilters?
Current License: CC BY-SA 3.0
7 events
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| May 18, 2018 at 3:16 | comment | added | David Fernandez-Breton | I just found out a very recent paper which partially answers Juris's question above: ams.org/journals/proc/2018-146-01/S0002-9939-2017-13719-8/… Apparently it suffices to assume that every filter on $\mathbb R$ can be extended to an ultrafilter, and this gives us idempotent ultrafilters on $\mathbb N$ | |
| Mar 13, 2013 at 12:48 | history | edited | Andreas Blass | CC BY-SA 3.0 |
corrected one word
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| Apr 20, 2012 at 14:16 | comment | added | Benjamin Steinberg | Yes, the question of Juris is what I had in mind. I see I was naive to think idempotents in compact left topological semigroups implies choice. Notice that since $\beta N^+$ is a free left compact left semitopological semigroup on one generator the existence of an idempotent in it implies the general case. | |
| Apr 20, 2012 at 13:14 | comment | added | Andreas Blass | Juris: That's a good question. I don't know, but my guess is negative. Start with Solovay's Lebesgue measure model (or $L(\mathbb R)$ in the presence of large cardinals), where there are no ultrafilters on $\omega$, and adjoin an ultrafilter by forcing with $P(\omega)/$fin. I'd be surprised to see an idempotent ultrafilter in the resulting model. I'd expect that the only ultrafilters would be the selective one you adjoined, ultrafilters obtained from it by iterated summation, and isomorphic copies thereof; and I'd expect none of these to be idempotent. | |
| Apr 20, 2012 at 10:57 | comment | added | Juris Steprans | But does the existence of an ultrafilter imply the existence of an idempotent? This might have been the motivation of the original question. | |
| Apr 20, 2012 at 4:08 | vote | accept | Benjamin Steinberg | ||
| Apr 20, 2012 at 0:32 | history | answered | Andreas Blass | CC BY-SA 3.0 |