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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 $X$?

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  • $\begingroup$ @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] $\endgroup$ – Andreas Blass Dec 18 '13 at 16:32
  • $\begingroup$ Also, it shouldn't matter that the OP refers to complete separable metric spaces $X$ where I just said "reals". $\endgroup$ – Andreas Blass Dec 18 '13 at 16:32
  • $\begingroup$ An, I completely misunderstood the question, I thought (incorrectly) we were after something else. @user38200, sorry for the confusion. $\endgroup$ – Andrés E. Caicedo Dec 18 '13 at 16:43
  • $\begingroup$ @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? $\endgroup$ – user38200 Jan 29 '14 at 12:05
  • $\begingroup$ 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. $\endgroup$ – Andreas Blass Jan 29 '14 at 13:49

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