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Sep 11, 2023 at 22:54 history edited LSpice CC BY-SA 4.0
Lowenheim -> Löwenheim, while this is on the front page
S Sep 11, 2023 at 22:27 history suggested C7X CC BY-SA 4.0
MathJaxify, typos
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S Sep 11, 2023 at 22:27
Apr 30, 2021 at 18:04 comment added Joel David Hamkins @FrançoisG.Dorais I believe that it was Zermelo who proved that the models of second-order ZFC2 are precisely the $V_\kappa$ for inaccessible cardinals $\kappa$. This is his famous quasi-categoricity result.
Feb 16, 2010 at 3:31 history edited Joel David Hamkins CC BY-SA 2.5
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Feb 16, 2010 at 3:18 comment added Joel David Hamkins Well, if by second order you just mean GBC=Goedel-Bernays set theory (with choice), then that is not quite right either, since every ZFC model can be given a second order part satisfying GBC. If you mean KM = Kelly Morse, then you have to go a bit further, but you still don't need an inaccessible. But if by "second-order" you mean that you have V_kappa with the full V_{kappa+1} as the second order part, then this does indeed imply that kappa is inaccessible.
Feb 16, 2010 at 3:18 comment added François G. Dorais @Mike: There might be no such V_a's, though you are correct that there are plenty if there is also an inaccessible. What you say about ZFC2 is true, this is an old theorem of Shepherdson (from one of his trilogy of papers titled Inner models for set theory).
Feb 16, 2010 at 3:15 history edited Joel David Hamkins CC BY-SA 2.5
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Feb 16, 2010 at 3:11 comment added Mike Shulman Hear hear. It is much too infrequently mentioned that there are many non-inaccessible a for which Va is a model of ZFC. I think it is true, though, that k must be inaccessible as soon as Vk is a model for "second-order" ZFC, right?
Feb 16, 2010 at 3:05 history edited Joel David Hamkins CC BY-SA 2.5
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Feb 16, 2010 at 2:56 history answered Joel David Hamkins CC BY-SA 2.5