Timeline for For which theories does ZFC without global choice prove the existence of a proper class monster model?
Current License: CC BY-SA 4.0
26 events
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| Jul 12, 2018 at 22:21 | answer | added | James E Hanson | timeline score: 6 | |
| Jul 7, 2018 at 21:19 | comment | added | James E Hanson | Oh for some reason I thought we were talking about Robinson arithmetic. I'm really not sure why. By extensions I meant completions of the theory. | |
| Jul 7, 2018 at 12:18 | comment | added | Emil Jeřábek | I'm not quite sure what you mean by extensions (the theory is complete, while extensions in a larger language can interpret anything). I think Presburger arithmetic cannot interpret much, as the definable sets are very simple. FWIW, Presburger arithmetic $Th(\mathbb Z,+,<)$ is NIP, and $Th(\mathbb Z,+)$ is superstable. | |
| Jul 6, 2018 at 17:28 | comment | added | James E Hanson | Ah thank you, I realized I'm not familiar enough with models of Presburger arithmetic. Which theories can extensions of Presburger arithmetic interpret? | |
| Jul 6, 2018 at 17:16 | comment | added | Emil Jeřábek | (Expanding it to some class model of Presburger arithmetic is trivial: e.g., fix an order on $\hat{\mathbb Z}$, and take the lexicographic product. This won't be saturated.) | |
| Jul 6, 2018 at 17:14 | comment | added | Emil Jeřábek | $\hat{\mathbb Z}$ is the profinite integers. A monster model of Presburger arithmetic has to look like this: it's $No\times\hat{\mathbb Z}$ as a group, with the induced order on No being the usual order, and on $\hat{\mathbb Z}$ some order that makes $\mathbb Z$ a convex subgroup (this choice shouldn't even matter). What remains to be determined is how the orders on No and $\hat{\mathbb Z}$ sit next to each other. I thought that this could be described definably from a set of parameters, considering that $\hat{\mathbb Z}$ is a set, but now I'm no longer sure it's so easy. | |
| Jul 6, 2018 at 16:45 | comment | added | James E Hanson | Also what is $\hat{\mathbb{Z}}$? | |
| Jul 6, 2018 at 16:42 | comment | added | James E Hanson | That seems like a really good case to think about, but why do you think it can be extended to a saturated model of Presburger arithmetic rather than just some proper class sized model? | |
| Jul 6, 2018 at 16:40 | comment | added | Emil Jeřábek | Nevertheless, $\mathrm{No}\times\hat{\mathbb Z}$ is a monster model of $Th(\mathbb Z,+)$, and I suspect this can be ordered in a suitable way (using the standard order on No) to make it a monster model of full Presburger arithmetic. | |
| Jul 6, 2018 at 16:25 | comment | added | Emil Jeřábek | Yes, I figured meanwhile that already the additive reduct (Oz,+) is not even 1-saturated, as the only congruence types realized are those of ordinary integers. | |
| Jul 6, 2018 at 16:17 | comment | added | nombre | @EmilJeřábek: The ring $\mathbf{Oz}$ of Omnific integers is not saturated since $\mathbb{Z}$ is definable in it by the sentence $\varphi[n]$ saying that no relation $x^2=2y^2$ may hold for non zero $x,y$ between $-|n|$ and $|n|$ (where the order is definable using the fact that the quotient field is real closed). | |
| Jul 6, 2018 at 15:55 | comment | added | James E Hanson | My gut instinct is that that's way too good to be true, but I have no idea how to approach it. | |
| Jul 6, 2018 at 15:53 | history | edited | James E Hanson | CC BY-SA 4.0 |
Some thoughts about DLO and the random graph
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| Jul 6, 2018 at 6:31 | comment | added | Emil Jeřábek | Speaking of surreal numbers, the ring of omnific integers isn't saturated, by any chance? This would have huge interpretability strength. | |
| Jul 5, 2018 at 17:36 | comment | added | Joel David Hamkins | I'm fine with ultrapowers, of course. I just didn't know exactly what someone means by "EM functor" as a set-theoretic construction, since the issues here seem to be about set-theoretic implementations of model theoretic ideas. | |
| Jul 5, 2018 at 16:14 | history | edited | James E Hanson | CC BY-SA 4.0 |
added 906 characters in body
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| Jul 5, 2018 at 15:51 | comment | added | James E Hanson | I've added a more explicit description of the EM model with $Ord$ for a spine. Did you also want an elaboration on the ultrapower of a class model thing? I thought that was sort of common in set theory. | |
| Jul 5, 2018 at 15:45 | history | edited | James E Hanson | CC BY-SA 4.0 |
Details of EM model with Ord for a spine construction
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| Jul 5, 2018 at 12:25 | comment | added | Joel David Hamkins | Could you elaborate on the model-construction methods you mention in the case of the uncountably categorical theories and the other cases? I don't quite see how the model-constructions can be done uniformly. | |
| Jul 5, 2018 at 12:15 | comment | added | Joel David Hamkins | I wonder if the iterated ultrapower of a model $M\models T$ by an ultrafilter $\mu$ on $\omega$, but iterated along the surreal line (using equivalence classes of finite-support functions from the surreals to $M$) has a chance to be saturated? Basically, do we get the extra (uncountable) saturation by using a saturated linear order for the iteration? | |
| Jul 5, 2018 at 11:39 | comment | added | Joel David Hamkins | Regarding the surreal line, it seems to me that it is saturated (as an order, and I think this also implies saturation as a field) in ZFC, even if without global choice we cannot seem to undertake the back-and-forth argument necessary for universality. So there is no proxy happening there. | |
| Jul 5, 2018 at 11:34 | comment | added | Joel David Hamkins | A comment on the question (which I like very much): in general, you cannot formulate the concept of "saturated" for class-sized models in mere ZFC, because you will need truth predicates in order to do so, but ZFC does not prove the existence of truth predicates. Basically, for a class model to be a saturated is not a first-order property in set theory. But of course, sometimes we can define a truth predicate, for example, if the model arises from an elementary chain. So I guess we should interpret your request for a saturated model to require the truth predicate that certifies it. | |
| Jul 5, 2018 at 11:29 | comment | added | Joel David Hamkins | See also my question here mathoverflow.net/q/227849/1946, which asks whether the universality (proxy for saturation?) of the surreal line is a weak global choice principle. | |
| Jul 5, 2018 at 7:40 | comment | added | Emil Jeřábek | Ah, never mind. It’s here: mathoverflow.net/q/229094 , but it does not give any saturation. | |
| Jul 5, 2018 at 7:35 | comment | added | Emil Jeřábek | I seem to remember that some time ago, there was an answer (by Joel Hamkins?) saying that all consistent theories have such models in ZFC. However, I can’t find it now. | |
| Jul 5, 2018 at 4:43 | history | asked | James E Hanson | CC BY-SA 4.0 |