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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?

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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.

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  • $\begingroup$ 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
    Commented 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 ♦
    Commented 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 ♦
    Commented Feb 11, 2014 at 19:37
  • $\begingroup$ I don't understand the downvote. $\endgroup$
    – Asaf Karagila ♦
    Commented 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
    Commented Feb 12, 2014 at 0:44

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