Timeline for What axioms are needed to show that the range of a finitely additive diffuse measure on $\mathbb N$ is not closed?
Current License: CC BY-SA 4.0
7 events
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| Nov 26 at 10:24 | comment | added | Elliot Glazer | It does not. Even the Hahn-Banach theorem does not imply that $\omega$ (or any set) admits a non-principal ultrafilter. See Pincus and Solovay's Definability of Measures and Ultrafilters. I wouldn't be surprised if their model of HB with no np ultrafilters in fact has that all diffuse measures have range $[0,1].$ | |
| Nov 26 at 9:34 | comment | added | Emil Jeřábek | I do not know that, but it's plausible that it doesn't. | |
| Nov 26 at 9:07 | comment | added | aduh | @EmilJeřábek Indeed, thanks for the correction. But does PM$_\omega$ imply that there exists a non-principal ultrafilter on $\omega$? Admittedly, I've been assuming it doesn't, although I don't have a reference for this off the top of my head. | |
| Nov 26 at 7:39 | comment | added | Emil Jeřábek | The mere existence of a nonprincipal ultrafilter, even specifically on $\omega$ (as you apparently mean from the context), is a quite weak principle, and certainly does not imply the Hahn-Banach theorem. You are probably confusing it with the Ultrafiler Lemma, which states that every proper filter on every set extends to an ultrafilter. | |
| Nov 26 at 0:21 | comment | added | aduh | @ElliotGlazer No, not to me. Good point! | |
| Nov 25 at 23:59 | comment | added | Elliot Glazer | Is it even clear that Hahn-Banach implies there is a diffuse measure whose range is not $[0,1]$? | |
| Nov 25 at 22:27 | history | asked | aduh | CC BY-SA 4.0 |