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Sep 14, 2017 at 6:53 vote accept Bjørn Kjos-Hanssen
Aug 18, 2017 at 0:17 vote accept Bjørn Kjos-Hanssen
Aug 18, 2017 at 15:45
Aug 18, 2017 at 0:17 answer added Bjørn Kjos-Hanssen timeline score: 1
Apr 14, 2017 at 20:09 comment added Samuel Coskey My student Samuel Dworetzky showed that the isomorphism relation for countable models of ZFC is Borel complete (in the sense of Borel complexity theory). I don't think it affects this particular question, though.
Apr 11, 2017 at 21:36 history edited Bjørn Kjos-Hanssen CC BY-SA 3.0
forgot to mention con(ZFC)
Apr 11, 2017 at 21:22 comment added Joel David Hamkins @FrançoisG.Dorais Why not expand your comment to an answer?
Apr 11, 2017 at 20:52 comment added François G. Dorais By reading the axioms (i.e. every axiom is satisfied, as a formalized statement) it's a $\Pi^0_{\omega+1}$ set (since satisfying a first-order sentence is $\Pi^0_n$ for some $n$). I think that it's not $\Pi^0_n$ for $n<\omega$ amounts to the fact that ZFC is not finitely axiomatizable.
Apr 11, 2017 at 19:18 comment added Joel David Hamkins I think that Sam Coskey was looking at the isomorphism relation on these models under Borel reducibility, aiming to place it in the hierarchy of Borel reducibility.
Apr 11, 2017 at 19:14 history edited Bjørn Kjos-Hanssen CC BY-SA 3.0
added 59 characters in body
Apr 8, 2017 at 16:46 comment added Andreas Blass "Complete for level $\omega$" looks plausible.
Apr 8, 2017 at 14:23 comment added Andreas Blass Usually, "complete for X" means (1) in X and (2) at least as complicated as anything else in X. Presumably, by "in some sense complete for finite levels of the Borel hierarchy", you meant only (2).
Apr 8, 2017 at 7:29 history asked Bjørn Kjos-Hanssen CC BY-SA 3.0