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Is there a model of set theory in which:

  1. Every projectively definable family of sets of reals has an OD or projectively definable member;

  2. Every OD or projectively definable set of reals has the property of Baire.

PS: By a projectively definable family of sets of reals I mean:

There exists a formula $\varphi(\Gamma)$ where $\Gamma$ is supposed to be a set of (tuples) of reals and $\mathcal{F}$ is the family of sets defined by: $$\Gamma \in \mathcal{F} \;{\rm iff} \; \varphi(\Gamma,a)$$ but $\varphi(\Gamma,a) : Q_1 x_1 \dots Q_n x_n Q'_1 z_1 \dots Q'_m z_m \psi(x,z,a,\Gamma)$ with $x$ and $z$ reals and integers respectively, and the $Q_i, Q'_j$ are quantifiers $\in \{ \forall, \exists\}$, $a$ are real parameters. $\psi$ is a $\Delta_0$ formula in the variables $x_i,z_n$ and $\Gamma$.

$\varphi$ and $\psi$ are formulas in the language of arithmetic.

Note that $\Gamma \subset \mathbb{R}^2$ to be precise. An example of $\varphi$ would be: $$\forall x \exists y (x,y) \in \Gamma \wedge [\forall z \forall t (((x,z) \in \Gamma) \wedge ((x,t) \in \Gamma)) \rightarrow z=t]$$ ($\Gamma$ is a function, here $x,y,z,t$ are assumed to be real numbers).

Note that $\mathcal{F}$ may have cardinality $2^{2^{\aleph_0}}$ so we cannot parametrize it by reals.

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  • $\begingroup$ I would understand your phrase "projectively definable family of sets of reals" to refer to a set of sets of reals consisting of the sections of a projective subset of $\mathbb{R}^2$. But on this interpretation, property 1 is trivial. So could you say more precisely what you mean? $\endgroup$ – Joel David Hamkins Nov 4 '13 at 17:18
  • $\begingroup$ @JDH: I edited accordingly, thanks. $\endgroup$ – user38200 Nov 4 '13 at 17:36
  • $\begingroup$ Could you clarify the formal language more precisely? What kind of assertions are allowed with $\psi$? If you mean that $\psi$ has no further quantifiers, then it seems that you can't say anything much about $\Gamma$ inside $\psi$, or about how it relates to the $x_i$ and $z_j$, since the elements of $\Gamma$ are sets of reals, but none of the other variables are sets of reals. $\endgroup$ – Joel David Hamkins Nov 4 '13 at 17:56
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    $\begingroup$ (And please, do not downvote this.) $\endgroup$ – Andrés E. Caicedo Nov 4 '13 at 19:02
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    $\begingroup$ @AndresCaicedo, the user does not intend what you said (and what I had thought at first with my first comment), a projective subset of the plane. Rather, his notion amounts to a family defined by a projective property considered in second-order logic, as picking out a family of sets. $\endgroup$ – Joel David Hamkins Nov 5 '13 at 4:35
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Let me ignore the OD issue for a moment, and just prove that there is always a projectively definable family of sets of reals, with no projective member. Indeed, there is such a family consisting of a single set.

Namely, let $S$ be the full satisfaction relation on the reals for projective truth. Thus, $S$ consists of all the pairs $(\varphi,x)$, where $\varphi$ is a projective statement and $\varphi(x)$ holds. The set $S$ is not itself projective, for otherwise we could diagonalize against the projective sets by forming the set $\{ n\mid \neg\varphi(n,n)\}$, which would have to be projective, but cannot be defined by any projective formula.

But now the point is that the singleton family ${\cal F}=\{S\}$ is definable by a formula of your type, and is therefore a projectively definable family of sets of reals in the sense of your question. The definition of this family simply notes that $S$ is the only set of pairs that obeys the Tarskian definition of truth, so that it gets the right answer on the atomic formulas, and also does the right thing on Boolean combinations and on quantifiers. That is, the set $S$ exhibits certain internal features — the recursive Tarskian definition of truth — and these features are expressible by quantifying only over integers and reals, provided that we can also refer to membership in $S$ in the expressions.

This example rules out property 1, if one should ask for a projective member of the family. But since projective truth is ordinal definable, it doesn't rule out statement 1 as it is stated, and for this I have to think a bit more.

In any case, if $V=HOD$ then every set is OD, and so any model of $V=HOD$ will satisfy statement 1 in the ordinal-definable version.

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Probably "model of set theory" was intended to mean "model of $\mathsf{ZFC}$" but I thought it might be worth mentioning that if we consider models of $\mathsf{ZF}$ then the answer is yes—provided that for (1) we are satisfied with a member that is $\text{OD}_a$ for some real $a$ instead of a $\text{OD}$ member, which is clearly too much to hope for.

Let $M$ be a model of $\mathsf{ZF} + \mathsf{AD}^+$, for example $M = L(\mathbb{R})$ under the assumption that there are infinitely many Woodin cardinals with a measurable cardinal above them. Letting $a$ be a real parameter, Woodin's basis theorem for $\Sigma^2_1(a)$ says that every nonempty $\Sigma^2_1(a)$ collection of sets of reals has a $\Delta^2_1(a)$ member.

This is even stronger than the desired result, because on one hand every projective property can be expressed in a $\Sigma^2_1(a)$ way for some real $a$, and on the other hand every $\Delta^2_1(a)$ set of reals is definable from $a$.

Moreover (2) holds in this model $M$ because the Axiom of Determinacy implies that every set has the property of Baire.

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  • $\begingroup$ I should mention that for $L(\mathbb{R})$ the basis theorem is due to Solovay, and also that (if we allow real parameters) the model $L(\mathbb{R})$ is a sort of trivial example of (1) because it satisfies "every set is ordinal definable from a real." $\endgroup$ – Trevor Wilson Nov 7 '13 at 18:20
  • $\begingroup$ Do you have a reference for Woodin's basis theorem? $\endgroup$ – user38200 Nov 8 '13 at 18:55
  • $\begingroup$ @user38200 See John Steel's paper on the derived model theorem: math.berkeley.edu/~steel/papers/dm.ps, Section 7 $\endgroup$ – Trevor Wilson Nov 8 '13 at 20:29
  • $\begingroup$ Can't we force the axiom of choice over a model of $ZF+AD^+$ as in a paper by Steel ("Two consequences of determinacy consistent with choice")? $\endgroup$ – user38200 Nov 10 '13 at 14:08
  • $\begingroup$ @user38200 That's right, or if you don't mind CH then you can just force with $\text{Col}(\omega_1,\mathbb{R})$. I'm not sure what the complexity of the family of non-ground-model sets of reals would be in such models. $\endgroup$ – Trevor Wilson Nov 10 '13 at 20:05

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