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?

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.
• 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. – Asaf Karagila Feb 11 '14 at 19:33
• 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. – Asaf Karagila Feb 11 '14 at 19:37