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Oct 21, 2017 at 23:24 comment added Keith Millar In terms of ZF, Reinhardt cardinals are quite contradictory with $V=HOD$.
Dec 14, 2013 at 13:30 vote accept CommunityBot
Dec 14, 2013 at 2:39 comment added Andrés E. Caicedo (A boring tree, being linear...)
Dec 13, 2013 at 23:50 answer added Joel David Hamkins timeline score: 16
Dec 13, 2013 at 22:37 comment added user43940 @AndreasBlass: Your question produced a question for me! Is it unknown that each (discovered and undiscovered) large cardinal axiom with strictly weaker consistency strength than $0^{\sharp}$ exists is consistent with $V=L$?
Dec 13, 2013 at 21:56 comment added Andreas Blass See mathoverflow.net/questions/95406 .
Dec 13, 2013 at 21:54 comment added Andreas Blass Is the statement "all large cardinal assumptions below [$0^\sharp$] are consistent with $V=L$" intended just as a statement about the widely studied large cardinal axioms, or is it intended as a general principle (presumably based on some definition of what a "large cardinal axiom" is)?
Dec 13, 2013 at 21:15 history edited user43940 CC BY-SA 3.0
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Dec 13, 2013 at 21:06 history asked user43940 CC BY-SA 3.0