Timeline for Absoluteness, reflection to ctms, and choice in outer models
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2017 at 19:17 | comment | added | Joel David Hamkins | My point is that without DC, we nevertheless have that ZF proves: for every $\alpha$ and every countable subset $\{a_0,a_1,\ldots\}\subset V_\alpha$, there is a countable transitive set $M$ that has the same theory as $V_\alpha$ and which has objects $b_n$ realizing the same types in $M$ as the $a_n$'s realize in $V_\alpha$. This is very like having a countable elementary submodel of $V_\alpha$, but short enough of it to be provable in ZF. | |
| May 14, 2017 at 19:11 | comment | added | Asaf Karagila♦ | @Joel: Yes, but if you could name a countable subtree, you would be done already. | |
| May 14, 2017 at 14:02 | comment | added | Joel David Hamkins | You can improve elementarily equivalent to allow for countably many named constants, so this is closer to being elementary. Those constants will have the same type in the countable model as they have in $V_\alpha$. | |
| May 14, 2017 at 14:00 | comment | added | Asaf Karagila♦ | @Joel: Probably. I guess you can get to something like "there is an elementary equivalent transitive model", but not as far as elementary submodel from the LS side of things. For the DC side of things, I don't think there's much to prove there. If you can find a countable subtree, you're fine, if not then you're probably not fine. | |
| May 14, 2017 at 13:56 | comment | added | Joel David Hamkins | I wonder if one can combine that observation with my answer below to see that in ZF one can prove certain definable instances of DC? | |
| May 14, 2017 at 13:54 | comment | added | Joel David Hamkins | Ah, of course, because if you have a countable model with the relation then you can obviously find a path. | |
| May 14, 2017 at 13:49 | vote | accept | Elliot Glazer | ||
| May 14, 2017 at 13:46 | comment | added | Asaf Karagila♦ | @Joel: In general, yes. DC is equivalent to stating that every $V_\alpha$ has a countable elementary submodel. (For "sufficiently large $\alpha$" is enough here, of course.) | |
| May 14, 2017 at 13:38 | comment | added | Joel David Hamkins | Meanwhile, my answer shows how we can solve the problem without necessarily obtaining countable elementry substructures of $V_\alpha$. So a natural follow-up question would be whether one needs DC for these instances of the downward LS theorem. | |
| May 14, 2017 at 12:51 | answer | added | Joel David Hamkins | timeline score: 8 | |
| May 14, 2017 at 12:35 | comment | added | Joel David Hamkins | Since ZF proves the reflection theorem, we have that $\sigma$ is true in some $V_\alpha$. The question is whether we can find a sufficiently elementary countable substructure of $V_\alpha$. We use DC to construct such a countable elementary substructure, and the way I think about the question, it is asking essentially whether one can prove sufficient instances here of the downward Löwenheim-Skolem theorem without DC. | |
| May 14, 2017 at 1:00 | history | edited | Elliot Glazer | CC BY-SA 3.0 |
Correcting a minor error in my argument
|
| May 13, 2017 at 19:45 | history | edited | Elliot Glazer | CC BY-SA 3.0 |
Made a clarification in response to Joel's comment.
|
| May 13, 2017 at 19:43 | comment | added | Elliot Glazer | It is a theorem that if $V \models ZF+DC,$ then $\Sigma_1$ sentences are downward absolute to $L.$ This is because $ZF+DC \vdash HC \prec_1 V,$ so any $\Sigma_1$ sentence is equivalent to a $\Sigma_1^{HC}$ sentence, which is equivalent to a $\Sigma_2^1$ sentence, which is downward absolute to $L$ by Shoenfield absoluteness. The point of this question is determining whether we can get rid of the use of DC in these arguments. | |
| May 13, 2017 at 17:36 | comment | added | Asaf Karagila♦ | Well. $\Pi_1$ statements are downwards absolute. If $M\models\lnot\exists x\varphi(x,y)$ for some $y\in L$, then $L$ ought to satisfy the same. | |
| May 13, 2017 at 17:20 | comment | added | Elliot Glazer | Asaf, the point is to prove downwards absoluteness. Joel, the latter. Each of these three claims are all relativized to $M,$ under the hopes that they can be proven from ZF alone. | |
| May 13, 2017 at 17:15 | comment | added | Joel David Hamkins | You seem to be missing the hypothesis that $M$ is transitive, which I believe you have in mind. Or do you mean instead to assert merely that $M$ thinks $M_0$ is transitive? | |
| May 13, 2017 at 17:08 | comment | added | Asaf Karagila♦ | The second claim is easy to verify: $\Delta_0$ statement are absolute, so $\Sigma_1$ is upwards absolute. | |
| May 13, 2017 at 16:57 | history | asked | Elliot Glazer | CC BY-SA 3.0 |