Timeline for The (non-)absoluteness of second-order elementary equivalence
Current License: CC BY-SA 4.0
11 events
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| May 13, 2022 at 4:58 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added a Wayback Machine link for the dead link.
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 16, 2014 at 5:45 | comment | added | Noah Schweber | I'm still interested in the remaining questions, but this is certainly enough for now! | |
| Apr 16, 2014 at 5:44 | vote | accept | Noah Schweber | ||
| Mar 29, 2014 at 19:35 | history | edited | Ali Enayat | CC BY-SA 3.0 |
edited body
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| Mar 29, 2014 at 14:17 | comment | added | Ali Enayat | Noah, I revamped my argument since contrary to my initial claim, Harrington's theorem has a limitation on $M$ (which makes perfect sense in light of the considerations of projective absoluteness discussed in Joel's answer). At the moment I do not know if the countable structures can be arranged to be ordinals. I also do not know the answer to your "quick question". | |
| Mar 29, 2014 at 14:13 | history | edited | Ali Enayat | CC BY-SA 3.0 |
added 1845 characters in body
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| Mar 29, 2014 at 8:37 | comment | added | Noah Schweber | This is a really beautiful argument. One quick question: in the resulting $M[G]$, can the two now-non-equivalent ordinals be made second-order equivalent once again? (Really, that should be "for some $M$ . . .") | |
| Mar 28, 2014 at 18:39 | comment | added | Noah Schweber | This is really interesting - thank you very much! | |
| Mar 28, 2014 at 10:27 | history | answered | Ali Enayat | CC BY-SA 3.0 |