Timeline for What consistency results follow the assumption: $\forall\alpha(\aleph_{\alpha+1}\nleq2^{\aleph_\alpha})$?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2019 at 18:25 | comment | added | Asaf Karagila♦ | Thanks @Trevor, once again we must refer to The Unpublished Works of W.H. Woodin, Esquire, ed. A.R.D. Mathias... I wish I could I'm surprised. :-) | |
| Nov 1, 2019 at 18:23 | comment | added | Trevor Wilson | @Asaf I think it's still unpublished and I don't know the proof, but you can see the theorem stated more specifically on p. 2 of arxiv.org/pdf/1904.01815.pdf. Probably Nam could tell you something about the proof. | |
| Oct 31, 2019 at 17:14 | comment | added | Asaf Karagila♦ | @Trevor: Where can one find a proof of Woodin's theorem that you mention in the comment to Yair about the supercompactness of $\omega_1$? (I am sure that you told me before on this site, but I can't seem to find that sort of information...) | |
| Nov 1, 2015 at 17:55 | comment | added | Asaf Karagila♦ | It's interesting to note that if the HOD conjecture is true, then this result is inconsistent with the existence of an extendible cardinal, as it implies $V$ is far far away from $\sf HOD$. | |
| May 10, 2015 at 6:59 | comment | added | Asaf Karagila♦ | That's an interesting question. I think that the issue here is similar to supercompactness. While "useful" is something that seem to imply a strong connection to aleph cardinals (e.g. $\mathcal P_\kappa(\lambda)$ rather than $P_\kappa(X)$ for arbitrary $X\geq\kappa$); these are often notably weaker than their choice-y counterparts. | |
| May 10, 2015 at 6:37 | comment | added | Trevor Wilson | @Asaf I wonder if there is a useful choiceless version of subcompactness. If so, then perhaps the natural order of things could be restored. | |
| May 10, 2015 at 6:13 | comment | added | Asaf Karagila♦ | If it's weaker than a subcompact, then it cannot be supercompact! :-) | |
| May 10, 2015 at 0:25 | comment | added | Trevor Wilson | @Asaf Sure, I meant for the right definition :-) Anyway, the point is that the form of supercompactness of $\omega_1$ that I'm referring to is weaker than a subcompact by the aforementioned theorem of Woodin, yet still implies that square fails everywhere (and in fact that every uncountable cardinal is threadable.) | |
| May 9, 2015 at 23:40 | comment | added | Asaf Karagila♦ | Trevor, I think that depends on the definition of "supercompact". | |
| May 9, 2015 at 17:51 | comment | added | Trevor Wilson | @YairHayut I think it is weaker without $\mathsf{AC}$. Woodin has proved that $\mathsf{AD} + \mathsf{DC} + {}$"$\omega_1$ is supercompact" is consistent relative to a proper class of Woodin limits of Woodins. | |
| May 9, 2015 at 17:40 | comment | added | Yair Hayut | The failure of square at a successor to singular is conjectured to be equiconsistent with a subcompact cardinal. | |
| May 9, 2015 at 7:38 | comment | added | Trevor Wilson | Yes, that's probably what I should have said. Although the difference is small compared with how inaccurate my guess probably was to begin with :-) | |
| May 9, 2015 at 7:32 | comment | added | Asaf Karagila♦ | Proper class of those? | |
| May 9, 2015 at 7:31 | comment | added | Trevor Wilson | I would guess that it's significantly stronger, say a Woodin limit of Woodin cardinals. | |
| May 9, 2015 at 7:29 | comment | added | Asaf Karagila♦ | Thanks, that's interesting. Do you have some educated guess to make (mine is that $(\bullet)$ is roughly equiconsistent with a proper class of Woodin cardinals). | |
| May 9, 2015 at 7:18 | comment | added | Trevor Wilson | ($\bullet$) is known in certain contexts (typically below $\Theta$) as the boldface $\mathsf{GCH}$. But I don't know if anyone has looked at its consistency strength specifically. So what I said was just based on its consequences for failure of covering. | |
| May 9, 2015 at 7:09 | comment | added | Asaf Karagila♦ | Awesome. Is $(\bullet)$ somewhat a known axiom, or can you just say these things because of some obvious consequences like failure of the covering lemma for all sort of inner models? | |
| May 8, 2015 at 23:27 | history | answered | Trevor Wilson | CC BY-SA 3.0 |