Timeline for Can a nonstandard model of $\mathsf{PA}$ be "$\Delta^1_1$-well-ordered?"
Current License: CC BY-SA 4.0
12 events
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| Mar 3, 2021 at 3:25 | comment | added | Noah Schweber | :P Just add a couple sentences - e.g. saying that in fact every completion of PA shows up, and the key model-theoretic notions are resplendence and $\omega$-saturation. That should make it long enough. Or heck, just copy-paste this! | |
| Mar 3, 2021 at 3:24 | comment | added | James E Hanson | Apparently if your answer is too short it gets converted into a comment automatically now. | |
| Mar 3, 2021 at 3:17 | vote | accept | Noah Schweber | ||
| Mar 3, 2021 at 3:16 | comment | added | Noah Schweber | Incidentally, if you want some easy rep you can post a link to this at my MSE question - that has a nice bounty on it. | |
| Mar 3, 2021 at 3:16 | comment | added | Noah Schweber | I'm pretty sure that we can outright prove in $\mathsf{ZFC}$ that every model has an elementary extension which is resplendent of the same cardinality. (The Kossak article above claims this without reference - top of the right column on the second page.) | |
| Mar 3, 2021 at 3:13 | comment | added | James E Hanson | There's a pretty direct argument that some model has no $\Delta_1^1$ cuts, but setting it up precisely is probably more work than verifying that computably saturated models work. You can build a chain of elementary extensions of a countable model of $\mathsf{PA}$ and force that any given pair of $\Sigma_1^1$ formulas either overlaps or fails to cover the whole model. In order to make sure this stays true, you expand the language at each step with new unary predicates, and then you build elementary extensions in that new bigger language. You deal with parameters by catching your tail. | |
| Mar 3, 2021 at 3:10 | comment | added | James E Hanson | You probably figured this out by now, but specialness is a kind of almost saturation that's good enough for many applications but doesn't require any set theoretic assumptions. | |
| Mar 3, 2021 at 3:09 | comment | added | James E Hanson | Yes, really we're only using resplendence and $\omega$-saturation, but full saturation or specialness are actually the only ways I know to get full resplendence. This is why a computably saturated countable model is probably good enough (countable computably saturated models are resplendent for expansions to computable theories). | |
| Mar 3, 2021 at 2:47 | comment | added | Noah Schweber | Incidentally, for those (like me) less familiar with resplendence, I recommend this quick survey by Kossak as a fun starting point. Basically, resplendent = elementary extensions preserve $\Sigma^1_1$ formulas. | |
| Mar 3, 2021 at 2:46 | comment | added | Noah Schweber | Nice! A couple questions: what is a "special model," and what does "resplendent for finite extensions" mean? Also, it seems at a glance that all you need is resplendence, since the only things about $\mathcal{M}$ you actually use are its resplendence and $\omega$-saturatedness, and the former implies the latter. Or am I missing something? | |
| Mar 2, 2021 at 2:44 | history | edited | James E Hanson | CC BY-SA 4.0 |
minor typos
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| Mar 2, 2021 at 2:06 | history | answered | James E Hanson | CC BY-SA 4.0 |