Timeline for Question about HOD
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2014 at 4:19 | comment | added | Asaf Karagila♦ | @Andres: Thank you for this information (which I somehow missed on my inbox). In either case, I think that the OP's question is ill-posed. (Except the last edit) | |
| Feb 12, 2014 at 19:05 | comment | added | Andrés E. Caicedo | @AsafKaragila And the project of axiomatizing the Solovay model is not quite solved. Woodin has results indicating sufficient conditions for $L(\mathbb R)$ to be a Solovay model, but this is not the same. (Some of my work with Richard Ketchersid on natural model of determinacy grew out of the desire to understand what such an axiomatization would look like.) | |
| Feb 12, 2014 at 19:03 | comment | added | Andrés E. Caicedo | @AsafKaragila Well, not really. In the context of large cardinals, $L(\mathbb R)$ of $V$ is a Solovay model. The standard meaning here is that the $L(\mathbb R)$ of $V$ is (perhaps in some forcing extension) elementary equivalent to the $L(\mathbb R)$ of $M[G]$ for some $M$ model of $\mathsf{ZFC}$, and some $G$ that is $M$-generic for the appropriate Levy collapse. In fact, what we have is an elementary embedding from such $L(\mathbb R)$ into the $L(\mathbb R)$ of $V$. Clearly, none of this is taking place in the neighborhood of $L$. | |
| Feb 12, 2014 at 5:19 | comment | added | Asaf Karagila♦ | @François: My point was (and I apologize for the many pings) that the OP wrote "the full Solovay model" which is often takes place as $L[G]$ and its $\mathrm{HOD}$ is in fact $L$. So iterating $\mathrm{HOD}$s has no value. The question is now edited to reflect this, and the answer to that I do not know off hand. | |
| Feb 12, 2014 at 0:57 | comment | added | Asaf Karagila♦ | Recall that we treat "Solovay models" as $L(\Bbb R)$ which satisfy certain axioms (Woodin has a nice axiomatization, if I recall correctly). | |
| Feb 12, 2014 at 0:53 | comment | added | Asaf Karagila♦ | @François: And therefore further information is needed to properly answer this question. Especially since often in the context of Solovay's model, $M\models V=L$ | |
| Feb 12, 2014 at 0:51 | comment | added | François G. Dorais | @Asaf: True, but that's one of the possibilities. | |
| Feb 12, 2014 at 0:48 | comment | added | Asaf Karagila♦ | @François: I know. None of which assume $\mathrm{HOD}=L$, though. | |
| Feb 12, 2014 at 0:44 | comment | added | François G. Dorais | @AsafKaragila: There are lots of wonderful papers about what can happen by iterating HOD. | |
| Feb 11, 2014 at 22:00 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
Clarified the answer.
|
| Feb 11, 2014 at 21:09 | comment | added | Asaf Karagila♦ | I don't understand the downvote. | |
| Feb 11, 2014 at 19:37 | comment | added | Asaf Karagila♦ | Every inner model of $L$ is $L$. If $\mathrm{HOD}=L$, then you have $\mathrm{HOD(HOD)}=L$, and $\mathrm{HOD(HOD(HOD))}=L$ and $\mathrm{HOD(HOD(HOD(HOD)))}=L$, and so on and so forth. So nothing is different, and the answer is still negative to your question. | |
| Feb 11, 2014 at 19:33 | comment | added | Asaf Karagila♦ | And my point is that the question depends on more information. If $M\models V=L$ then $\mathrm{HOD}^M=M=L^M$. In that case, no. They don't have to have the Baire property. | |
| Feb 11, 2014 at 19:32 | comment | added | user38200 | I know that HOD has sets that it thinks are not Baire measurable. My question is that in HOD, are definable sets Baire measurable? | |
| Feb 11, 2014 at 19:30 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |