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Jan 14, 2023 at 12:56 history edited Zuhair Al-Johar CC BY-SA 4.0
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Jan 14, 2023 at 2:31 comment added Gabe Goldberg @HanulJeon I think $\text{AC}_{\lambda^+}$ suffices to split $S^{\lambda^+}_\omega$ into $\lambda^+$ disjoint stationary sets, contrary to Woodin's proof. (The measurability of $\lambda^+$ is just a slight elaboration on Woodin's proof in the $\text{DC}_{\lambda^+}$ context; it's Woodin's proof plus Ulam splitting.)
Jan 14, 2023 at 2:08 comment added Hanul Jeon Could we argue that $\mathsf{DC}_{\lambda^+}$ does not hold because we can carry over Woodin's proof of Kunen inconsistency with $\mathsf{DC}_{\lambda^+}$? Does the measurability of $\lambda^+$ give a stronger fact that $\mathsf{AC}_{\lambda^+}$ is incompatible with the Reinhardtness?
Jan 13, 2023 at 17:46 history edited Gabe Goldberg CC BY-SA 4.0
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Jan 12, 2023 at 21:05 vote accept Zuhair Al-Johar
Jan 12, 2023 at 20:38 history edited Gabe Goldberg CC BY-SA 4.0
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Jan 12, 2023 at 20:33 history answered Gabe Goldberg CC BY-SA 4.0