Timeline for The First Failure of GCH in Large Cardinals Smaller than Measurables
Current License: CC BY-SA 3.0
11 events
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| Nov 10, 2013 at 14:39 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 10, 2013 at 14:09 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 2, 2013 at 18:54 | comment | added | Joel David Hamkins | Yes, Carlo, I agree with that. But being $\Sigma_2$-extendible is much weaker than being $\Sigma_2$-correct, which are the cardinals you get by reflection. Indeed, every $\Sigma_2$-correct cardinal is a limit of $\Sigma_2$-extendible cardinals. | |
| Nov 2, 2013 at 18:51 | comment | added | Rachid Atmai | Actually one can prove that $\Sigma_2$-extendible cardinals exist just using the reflection theorem. There is cub-many such cardinals, provided they're not inaccessible (just as you've said). | |
| Nov 2, 2013 at 17:56 | comment | added | Joel David Hamkins | That result about strongly Ramsey cardinals (as well as the definition of these cardinals) is due to Victoria Gitman. | |
| Nov 2, 2013 at 16:12 | comment | added | Joel David Hamkins | @MohammadGolshani, no, Ramsey cardinals are not necessarily $\Sigma_2$-reflecting, and one can make the GCH fail first at them. This is easier to see with the strongly Ramsey cardinals, which can easily be made indestructible by $\text{Add}(\kappa,1)$. | |
| Nov 2, 2013 at 16:11 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 2, 2013 at 12:58 | vote | accept | CommunityBot | ||
| Nov 2, 2013 at 12:58 | comment | added | user42090 | Interesting and a bit surprising! Thanks Joel. | |
| Nov 2, 2013 at 12:54 | comment | added | Mohammad Golshani | Dear Prof. Hamkins, your answer is really interesting. Are Ramsey cardinals $\Sigma_2$ reflecting, and if not, then can they be the first cardinal violating the GCH | |
| Nov 2, 2013 at 12:39 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |