Timeline for Does ZF+AD have any unusual arithmetic consequences?
Current License: CC BY-SA 4.0
9 events
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| Jun 22, 2019 at 18:27 | comment | added | Timothy Chow | @NoahSchweber : Depending on one's taste, one might regard your final highlighted statement as comforting rather than annoying. It could be interpreted as reassuring people who really only care about arithmetic consequences that they need not worry unduly that those crazy mathematicians who assume wild axioms about sets will go off on the wrong track. | |
| Jun 21, 2019 at 20:31 | comment | added | Noah Schweber | @TimCampion Whoops, looks like I spoke to soon! Interesting. | |
| Jun 21, 2019 at 19:48 | comment | added | Tim Campion | I see, thanks! If ZF+Reinhardt is known to contain inner models of ZFC+I0, that would answer another old question of mine. | |
| Jun 21, 2019 at 19:44 | vote | accept | Tim Campion | ||
| Jun 21, 2019 at 19:44 | comment | added | Noah Schweber | Sorry, I thought you were still comparing with AD. For Reinhardt vs. I0 for example, you'd want to show that any model of ZF + Reinhardt has an inner model of ZFC + I0. I believe this is known to be true but I'm not sure. (Certainly i think almost everyone believes that if they're each consistent then they're arithmetically compatible.) | |
| Jun 21, 2019 at 19:34 | comment | added | Tim Campion | I'm confused -- granted that ZF+Reinhardt proves determinacy holds in $L(\mathbb R)$, we see that ZF+Reinhardt is stronger than ZF+AD in arithmetic consequences. But it's the other way around that we'd like -- we'd like to know that, say, the arithmetic consequences of ZF+Reinhardt and ZFC+I0 are consistent with each other, and I don't see how the argument gets us there... | |
| Jun 21, 2019 at 19:31 | comment | added | Noah Schweber | @TimCampion Nope, those aren't any different either: they also prove that determinacy holds in $L(\mathbb{R})$. The mystery about them is whether they're consistent, but it is known that even if they are consistent they're extremely strong. Again, no set-theoretic axioms are (assuming their consistency) known to provide contradictory arithmetic facts, ignoring ones which are themselves arithmetic statements specifically chosen to do so (e.g. inconsistency statements). | |
| Jun 21, 2019 at 19:29 | comment | added | Tim Campion | Thanks! I suppose this line of argument would be circumvented by something like ZF+ Reinhardt or ZF+Berkeley. I gather from the lack of discussion on my earlier linked question that not much is known about these theories -- so perhaps they might yield "alternative arithmetic facts"... (Of course, because so little is known it's certainly questionable whether such theories have any kind of "inner coherence" to them.) | |
| Jun 21, 2019 at 19:20 | history | answered | Noah Schweber | CC BY-SA 4.0 |