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S Nov 5, 2020 at 17:54 history edited Glorfindel CC BY-SA 4.0
Reformatted it to make it cleaner
S Nov 5, 2020 at 17:54 history suggested Master CC BY-SA 4.0
Reformatted it to make it cleaner
Nov 5, 2020 at 17:25 review Suggested edits
S Nov 5, 2020 at 17:54
Aug 26, 2020 at 18:29 comment added Gabe Goldberg Every Berkeley cardinal is hyper-Berkeley. Clearly Berkeley cardinals of uncountable cofinality are hyper-Berkeley. Suppose $\kappa$ is hyper-Berkeley of countable cofinality and $s\subseteq \kappa$ is cofinal and ordertype $\omega$. Fix a transitive set $M$ containing $\kappa$. Bagaria-Koellner-Woodin tricks yield a $j : M\to M$ such that $j(\alpha) = \alpha$ for all $\alpha\in s$ and $\text{crit}(j) < \kappa$. The first fixed point of $j$ above $\text{crit}(j)$ is strictly below $\kappa$ since $j$ fixes every ordinal in the cofinal set $s$.
Aug 26, 2020 at 12:27 comment added Asaf Karagila (Also, try to give your questions a descriptive title.)
Aug 26, 2020 at 12:25 comment added Asaf Karagila Asking equiconsistency in the "above ZFC" portion of large cardinals is hard, since we don't actually have a lot of techniques there. Specifically, we have "more or less" direct implications (e.g. Berkeley cardinal implies there is a set model with a Reinhardt; or super-Reinhardt is Reinhardt), and forcing (e.g. start with a Berkeley cardinal of cofinality $\omega_1$, and collapse $\omega_1$ to be countable, then all the embeddings lift). And even then, the forcing results are still quite rudimentary and insufficient to provide "simple exercise" answers in most cases.
Aug 26, 2020 at 12:22 history edited Asaf Karagila CC BY-SA 4.0
edited tags
S Aug 26, 2020 at 12:19 history suggested Hanul Jeon CC BY-SA 4.0
Fix some typos
Aug 25, 2020 at 23:55 review Suggested edits
S Aug 26, 2020 at 12:19
Aug 25, 2020 at 21:43 review First posts
Aug 25, 2020 at 23:51
Aug 25, 2020 at 21:36 history asked Alex O. CC BY-SA 4.0