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Oct 23, 2022 at 22:42 vote accept Jesse Elliott
Apr 13, 2017 at 12:58 history edited CommunityBot
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Dec 13, 2014 at 15:53 comment added Yair Hayut Another way to cheat and get slightly stronger large cardinals: it is consistent that there is an ordinal $\alpha$, $V=L$ and in some generic extension (by a set forcing notion) there is an $\alpha$-Erdos cardinal, even when $\alpha$ is not countable in $L$.
Dec 13, 2014 at 4:29 answer added Mohammad Golshani timeline score: 13
Dec 13, 2014 at 1:23 comment added Andrés E. Caicedo Your suggestion is essentially it. The modern presentation of $0^\sharp$ is as what Schimmerling calls a baby mouse, this is a structure $J_\tau$ with an external measure that is iterable (and some technical definability requirements). Asking that for each $\alpha<\omega_1$ there is such a structure, but only $\alpha$-iterable, seems as strong as possible. But this is in essence the same as considering $\alpha$-Erdős cardinals.
Dec 13, 2014 at 1:12 history asked Jesse Elliott CC BY-SA 3.0