Does the following result hold in the full Solovay model?

In Solovay paper, "A model of set theory in which every set of reals is Lebesgue measurable", he cites in the end a theorem 4.1 (generalization of results), which says, roughly that in $\mathfrak{N}_1$ (the Solovay model), for $X$ a complete separable metric space, then every set $A \subset X$ has the property of Baire and other properties.

Is it possible to deduce a theorem which would be like: in $\mathfrak{N}=\mathfrak{M}[G]$(the full Solovay model) every set $A \subset X$ definable from a countable sequence of ordinals has the property of Baire?

Edit: This question remains unanswered, but I would like to add one detail to it: $X$ the complete separable metric space may not have a "definable" topology, so the basic open sets may not be definable from a countable sequence of ordinals. Does the same remark in the comments below still hold though?

Further Edit: What happens in the case where the topology and the metric of $X$ are definable from ordinals and $X$, in the sense that the metric is definable by a formula whose parameters are ordinals and $X$?

• @AndresCaicedo I'm confused by your comment, partly because the answer you linked to was about what can be proved in ZFC whereas the present question is (if I correctly understand it) about the model produced by Lévy-collapsing below an inaccessible, and partly because, as far as I can see, it's true in that model that any set $A$ of reals definable from a countable sequence of ordinals has the Baire property. Such an $A$ is in the Solovay model, so it has the Baire property there, and that's preserved upward to the Lévy model. [continued in next comment] – Andreas Blass Dec 18 '13 at 16:32
• Also, it shouldn't matter that the OP refers to complete separable metric spaces $X$ where I just said "reals". – Andreas Blass Dec 18 '13 at 16:32
• An, I completely misunderstood the question, I thought (incorrectly) we were after something else. @user38200, sorry for the confusion. – Andrés E. Caicedo Dec 18 '13 at 16:43
• @AndreasBlass: Does the result that you mentioned in the comment still hold when the metric and topology of $X$ are not definable from a countable sequence of ordinals? – user38200 Jan 29 '14 at 12:05
• I think the result in my previous comment can fail if the topology is not definable from a countable sequence of ordinals. Start with some wild (undefinable, lacking the Baire property) set $Z$ of reals of the cardinality of the continuum, say a Bernstein set; then apply a permutation of the reals that takes $Z$ to a definable set, say the unit interval $[0,1]$ but (therefore ) takes the standard topology of the reals to an undefinable topology $T$. The reals with topology $T$ are still a nice (Polish) space, but now the unit interval lacks the Baire property. – Andreas Blass Jan 29 '14 at 13:49