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Consider a ctm $\mathfrak{M}$ of $ZF+AD^+$. Is it possible to force over $\mathfrak{M}$ to get a model of ZFC which satisfies further the following:

  1. Every projectively definable family of sets of reals has an $OD_a$ member;

  2. Every $OD_a$ set of reals has the property of Baire; Where $a$ is any real parameter.

Here statement 1 is as in my previous question (Projectively definable family of sets of reals).

PS: I am asking this question because there is a paper by Steel & Van Wesep "Two consequences of determinacy consistent with choice" which uses a similar technique.

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  • $\begingroup$ What is $a$ in your notation $\text{OD}_a$? $\endgroup$ – Joel David Hamkins Nov 10 '13 at 14:54
  • $\begingroup$ $OD_a$ means definable from ordinals and reals. $\endgroup$ – user38200 Nov 10 '13 at 14:57
  • $\begingroup$ So I guess you mean that $a$ is an (arbitrary) real parameter. $\endgroup$ – Joel David Hamkins Nov 10 '13 at 15:04
  • $\begingroup$ Yes exactly. I already edited it in the body. $\endgroup$ – user38200 Nov 10 '13 at 15:07
  • $\begingroup$ One thing to keep in mind is that one can force over any model of ZFC so as to make any given set definable without parameters (and hence put it into OD), by forcing that adds no reals and no sets of reals and doesn't change which families are projectively definable. But this doesn't seem to answer your question. $\endgroup$ – Joel David Hamkins Nov 10 '13 at 15:14
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$\newcommand\R{\mathbb{R}}\newcommand\OD{\text{OD}}$

The answer is no.

First, notice that the family consisting of all subsets $R\subset\R^2$ that are a well-ordering of the reals is projectively definable in your sense, since we can say that $R$ is a well-ordering by quantifying only over countable objects: $R$ is reflexive, transitive, anti-symmetric and linear, every real is mentioned, and there is no infinite descending sequence (and we can code such $R$ canonically with a subset of $\R$, to avoid the need for $\R^2$ here). Thus, if statement 1 holds, we get an $\OD_a$ well-ordering of $\R$ for some real parameter $a$. But from any such well-ordering, we can define a set of reals, such as the Vitali set, that does not have the property of Baire, thereby violating property 2.

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  • 1
    $\begingroup$ An infinite sequence $\langle s_k\mid k\lt\omega\rangle$ of binary sequences $s_k$ can be coded as a single binary sequence $s(\langle i,j\rangle)=s_i(j)$ using a pairing function $\langle\cdot,\cdot\rangle$. $\endgroup$ – Joel David Hamkins Nov 10 '13 at 18:28

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