1
$\begingroup$

Consider the full Solovay model $N=M[G]$ of set theory. Let HOD be the set of hereditarily ordinal definable elements of $M[G]$. It is known that in $N$ every set of reals definable from ordinals and reals is Baire measurable.

My question is:

In HOD, is every set definable from ordinals and reals (in HOD) Baire measurable? (i.e. is the property that a definable set is Baire measurable preserved by passage from $N$ to HOD?). If we consider also HOD(HOD) does the same remain true?

Edit: I rephrase my question to be: is there a model of set theory $M$ such that in HOD$^{M[G]}$ every set of reals definable from ordinals and reals is Baire measurable? Can we find a model of set theory in which HOD, HOD(HOD) etc.. all satisfy the same property above?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

There's not enough information in the question. If $M\models V=L$, then in $M$ we have that $\mathrm{HOD}^M=L^M=M$, and therefore $\mathrm{HOD}^{\mathrm{HOD}^M}=L^M=M$ as well.

In particular we have that in both instances sets of real definable from ordinals and reals (in $M$, which is the same as $\mathrm{HOD}^M$ in this case) are the same sets of reals in $L$, and so there are sets without the Baire property.

Improve this answer
$\endgroup$
13
  • $\begingroup$ I know that HOD has sets that it thinks are not Baire measurable. My question is that in HOD, are definable sets Baire measurable? $\endgroup$
    – user38200
    Feb 11, 2014 at 19:32
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Feb 11, 2014 at 19:33
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Feb 11, 2014 at 19:37
  • $\begingroup$ I don't understand the downvote. $\endgroup$
    – Asaf Karagila ♦
    Feb 11, 2014 at 21:09
  • $\begingroup$ @AsafKaragila: There are lots of wonderful papers about what can happen by iterating HOD. $\endgroup$
    – François G. Dorais
    Feb 12, 2014 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.