Timeline for A weak (?) form of Shelah cardinals
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 17, 2019 at 21:28 | vote | accept | Trevor Wilson | ||
| Aug 17, 2019 at 21:28 | comment | added | Trevor Wilson | I see, thanks! On further thought, it seems like $\kappa$ Woodin and $\Sigma_3$-reflecting implies $\kappa$ weakly Shelah: given $f:\kappa\to\kappa$, use Woodinness to get some $\alpha < \kappa$ that is $\mathord{<}\kappa$-$f$-strong. Then the desired conclusion (on existence of $j$) holds for cofinally many $\bar{\kappa} < \kappa$ in place of $\kappa$. Formulating this in terms of extenders, $\Sigma_3$-reflection implies the desired conclusion holds for cofinally many $\bar{\kappa} \in \operatorname{Ord}$ in place of $\kappa$, and therefore for $\kappa$ itself (since $f$ is increasing.) | |
| Aug 17, 2019 at 18:36 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
added 8 characters in body
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| Aug 17, 2019 at 18:36 | comment | added | Asaf Karagila♦ | Oh. Shame. It would have been nice to see all very large cardinals rendered inconsistent on MathOverflow... :-P | |
| Aug 17, 2019 at 18:35 | comment | added | Gabe Goldberg | You're right, it was a typo, I meant limits of weakly Shelahs | |
| Aug 17, 2019 at 18:31 | comment | added | Asaf Karagila♦ | Wait, how is this possible? I thought Shelah cardinals are measurable. No? That means that every Shelah cardinal is a measurable weakly Shelah, so it is the limit of Shelah cardinals. Therefore the ordinals are not well-founded... | |
| Aug 17, 2019 at 18:25 | history | answered | Gabe Goldberg | CC BY-SA 4.0 |