Timeline for Ordinal notations within non-standard models of arithmetic
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2019 at 20:12 | answer | added | andrey bovykin | timeline score: 4 | |
| Jun 14, 2019 at 19:34 | vote | accept | Fedor Pakhomov | ||
| Jun 11, 2019 at 15:11 | answer | added | Fedor Pakhomov | timeline score: 6 | |
| Jun 11, 2019 at 8:29 | comment | added | Emil Jeřábek | Feel free to write an answer yourself, there are too many things here that I don’t understand. For example, is there a more explicit description of countable recursively saturated models of $\mathrm{Th}(\alpha,{<})$? Also, for the existence of a truth definition in PA, it is not enough that the theory is decidable, we need that the proof of decidability of the theory can be formalized in PA. Now, I can well believe this is possible, but since I don’t know how the proof goes, I’m not comfortable with asserting it as a fact in an answer without justification. | |
| Jun 10, 2019 at 18:40 | comment | added | Fedor Pakhomov | @EmilJeřábek Do you want to write a proper answer about it? If not, I could write about this myself. By the way, if I am not missing anything, this observation about importance of recursively-saturated order types, enables classification all the order types of countable non-standard models of ordinal arithmetic $(\varepsilon_{\alpha},+,\cdot,\mathsf{exp})$: those are the order types that I mentioned in my answer to that question, and all the countable recursively saturated orders $A$ elementary equivalent to $\omega^{\omega}$. | |
| Jun 10, 2019 at 17:49 | comment | added | Fedor Pakhomov | @EmilJeřábek Indeed, then this answers my question (at least for the case of countable $\mathfrak{A}$). | |
| Jun 10, 2019 at 17:34 | comment | added | Emil Jeřábek | The truth definition is needed to prove that any $(\alpha)^{\mathfrak A}$ is recursively saturated. The converse implication only needs that $\mathrm{Th}(\alpha,{<})$ is (co)interpretable in PA: it follows from the resplendency of countable recursively saturated models. | |
| Jun 10, 2019 at 16:55 | comment | added | Fedor Pakhomov | @EmilJeřábek But is it enough to claim that we could obtain any recursively saturated model of $\mathsf{Th}(\alpha,<)$ as $(\alpha)^{\mathfrak{A}}$? Or just that any $(\alpha)^{\mathfrak{A}}$ is recursively saturated? | |
| Jun 10, 2019 at 16:53 | comment | added | Emil Jeřábek | Yes, that is exactly what I need. | |
| Jun 10, 2019 at 16:52 | comment | added | Fedor Pakhomov | @EmilJeřábek However if all you needed to use for your observation was that over $\mathsf{PA}$ there is a truth definition for $(\alpha,<)$, then it is the case. It is due to the fact that all $\mathsf{Th}(\alpha,<)$ are decidable. | |
| Jun 10, 2019 at 16:42 | comment | added | Fedor Pakhomov | @Emil Theories $\mathsf{Th}(\alpha,<)$ for $\alpha\ge \omega^{\omega}$ do not enjoy this kind of quantifier elimination. This is due to the fact that $(\omega^n,<)\equiv_{\Pi_n}(\omega^n(1+A),<)$, for any linear order $A$. And that there are first-order formulas distinguishing $(\omega^n,<)$ and $(\omega^{\omega},<)$. | |
| Jun 10, 2019 at 16:35 | comment | added | Emil Jeřábek | I don’t know that much about the theories of the structures $(\alpha,{<})$, but I assume they do have quantifier elimination down to some class of formulas of bounded quantifier alternation, right? If so, then models of the form $(\alpha)^{\mathfrak A}$ for countable $\mathfrak A\models\mathrm{PA}$ are exactly the recursively saturated countable models of $\mathrm{Th}(\alpha,{<})$. | |
| Jun 10, 2019 at 15:31 | history | asked | Fedor Pakhomov | CC BY-SA 4.0 |