Timeline for Reference Request: Independence of the ultrafilter lemma from ZF
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30, 2017 at 18:03 | comment | added | Joel David Hamkins | Do we have a model of ZF in which every set has a nonprincipal ultrafilter, but some filter does not extend to an ultrafilter? | |
| Apr 30, 2017 at 14:59 | comment | added | François G. Dorais | @JoelDavidHamkins: $P(X)/I$ is not necessarily isomorphic to the Boolean algebra of subsets of a set. For example, $P(\omega)/fin$ is not atomic. | |
| Apr 30, 2017 at 12:05 | comment | added | Joel David Hamkins | @FrançoisG.Dorais Is that actually stronger than saying every set has an ultrafilter? Supppose $F$ is a nonprincipal filter on $X$. I can form the quotient $P(X)/I$ for the dual ideal (let's also add the finite sets to the ideal), and then if I put an ultrafilter $F^*$ on the quotient, don't a get an ultrafilter on $X$ extending $F$? | |
| Mar 25, 2011 at 9:14 | vote | accept | Greg Graviton | ||
| Mar 22, 2011 at 21:02 | vote | accept | Greg Graviton | ||
| Mar 25, 2011 at 9:14 | |||||
| Mar 22, 2011 at 20:44 | answer | added | Andreas Blass | timeline score: 21 | |
| Mar 22, 2011 at 16:40 | answer | added | arsmath | timeline score: 9 | |
| Mar 22, 2011 at 11:33 | comment | added | François G. Dorais | @Zsbán: Neither. The Ultrafilter Lemma says that any filter on a set can be extended to an ultrafilter on that set. | |
| Mar 22, 2011 at 11:04 | comment | added | Zsbán Ambrus | Is the ultrafilter lemma that there exists a non-principal ultrafilter over any infinite set, or only that there exists a non-principal ultrafilter over some set? | |
| Mar 22, 2011 at 10:18 | comment | added | Ed Dean | Jech's book The Axiom of Choice covers this material. | |
| Mar 22, 2011 at 10:14 | answer | added | Michael Greinecker | timeline score: 7 | |
| Mar 22, 2011 at 9:31 | history | asked | Greg Graviton | CC BY-SA 2.5 |