Timeline for Do Measurable Cardinals Exist? (assuming ZFC)
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Jul 8, 2013 at 22:14 | comment | added | François G. Dorais | @Tom: The first issue is that measurable cardinals are defined in terms of two-valued probability measures. Also, the way I read Dudley's definition, every cardinal from the first real-valued measurable on is "Dudley measurable" since there is no restriction that says that the measure couldn't concentrate on a smaller set. Finally, the phrase "measure 0 cardinal" is hopelessly confusing. | |
| Jul 8, 2013 at 22:14 | comment | added | Andrés E. Caicedo | (A bit too peculiar in its notational conventions, though.) | |
| Jul 8, 2013 at 22:11 | comment | added | Asaf Karagila♦ | Fremlin's book is a wonderful (and free!) reference. | |
| Jul 8, 2013 at 22:08 | comment | added | Andrés E. Caicedo | A more careful discussion of measurable cardinals from an analyst point of view is in Federer's Geometric measure theory. And for a more modern treatment, Fremlin's treatise on Measure theory. (Set theoretic issues are mostly in volume 5). | |
| Jul 8, 2013 at 22:06 | comment | added | Andrés E. Caicedo | I see now (after reading François's comment) that Dudley is also using "measurable" in a nonstandard fashion, meaning what we more commonly call (atomlessly) real-valued measurable. It is consistent that the continuum is real valued measurable, that is, there is a probability measure on $\mathcal P(\mathbb R)$ that is zero on singletons. This is equivalent to the existence of a (necessarily, not translation invariant) extension of Lebesgue measure to all subsets of $[0,1]$. ("Consistent:" It is equiconsistent with the existence of measurables, relative to $\mathsf{ZFC}$.) | |
| Jul 8, 2013 at 22:05 | comment | added | François G. Dorais | @Andres: The definition that Dudley gives is closer to that of a real-valued measurable; the continuum could be real-valued measurable. | |
| Jul 8, 2013 at 22:03 | comment | added | Andrés E. Caicedo | The continuum hypothesis is irrelevant here. The continuum is never measurable. The point is that if $\kappa$ is measurable, then it is uncountable, and if $\lambda<\kappa$, then also $2^\lambda<\kappa$. | |
| Jul 8, 2013 at 22:02 | comment | added | Asaf Karagila♦ | @Tom: I have to admit that the quoted text doesn't make a lot of sense to me either, and i mainly addressed the problem that seemed to baffle you: do measurable cardinals exist in $\sf ZFC$. I'm glad to see that it helped you. | |
| Jul 8, 2013 at 22:00 | comment | added | Tom LaGatta | @François, could you describe more? If you claim it's incorrect, a critique would be appreciated. | |
| Jul 8, 2013 at 21:54 | comment | added | François G. Dorais | The formulation by Dudley is confusingly imprecise and incorrect in all interpretations I can think of. | |
| Jul 8, 2013 at 21:54 | history | edited | Asaf Karagila♦ |
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| Jul 8, 2013 at 21:49 | vote | accept | Tom LaGatta | ||
| Jul 8, 2013 at 21:36 | comment | added | Andrés E. Caicedo | You cannot prove in first-order logic that the existence of measurables is consistent with $\mathsf{ZFC}$. I would not consider this an open problem, just as I do not consider open the consistency of arithmetic. (There is nothing to edit about open problems in the Wikipedia article, and $\mathsf{AD}$ is incompatible with choice, so it is really the wrong thing to focus on in this case.) | |
| Jul 8, 2013 at 21:36 | answer | added | Asaf Karagila♦ | timeline score: 18 | |
| Jul 8, 2013 at 21:32 | history | asked | Tom LaGatta | CC BY-SA 3.0 |