This question arises from an issue arising in user38200's recent question concerning models of set theory in which every definable set has a definable element. In my answer to that question, with François's help, it turned out that $V=\text{HOD}$ is equivalent to the assertion that every $(\Sigma_2\wedge\Pi_2)$-definable set has an ordinal-definable element.

The question here is whether this level of complexity can be reduced to $\Sigma_2$ or not.

Question 1. Is there a model of $\text{ZFC}+V\neq\text{HOD}$ in which every $\Sigma_2$-definable set has an ordinal definable element?

In the equivalence to $V=\text{HOD}$, the $\Sigma_2\wedge\Pi_2$ level of complexity arises because in any model of $\text{ZFC}+V\neq\text{HOD}$, the set of all minimal-rank non-OD sets is defined by a formula of complexity $\Sigma_2\wedge\Pi_2$, but has no ordinal-definable elements. I know that this set cannot in general be defined provably by a $\Sigma_2$ formula, since we could preserve the truth of any $\Sigma_2$ formula by forcing up high so as to code sets into the GCH pattern (or some other desired method of coding), and thereby change which sets down low were in OD without affecting that given $\Sigma_2$ assertion. But it isn't clear to me how that observation answers the question here, since one would seem to need to construct a counterexample model of $V\neq\text{HOD}$, where all $\Sigma_2$-definable sets had OD members, and forcing to change that one set doesn't seem to do this.

I am also interested in the case of $\Pi_2$-definability.

Question 2. Is there a model of $\text{ZFC}+V\neq\text{HOD}$ in which every $\Pi_2$-definable set has an ordinal-definable element?

  • By a $\Sigma_2$-definable set, you mean a unique set satisfying a $\Sigma_2$-formula $\phi(x)$, or a set of the form $\{x:\phi(x)\}$ for a $\Sigma_2$-formula $\phi(x)$? – Emil Jeřábek Sep 14 '14 at 12:40
  • The former. If $V\neq\text{HOD}$, then the answer at the other question shows that the set of minimal-rank non-OD sets is characterized as the unique set with a certain $\Sigma_2\wedge\Pi_2$ property. – Joel David Hamkins Sep 14 '14 at 12:59
  • The relevant $\Sigma_2$ properties in your answer (i.e., the last two conditions) say that a bunch of sets are in OD. As far as I understand it, a witness to the outer existential quantifier here is an ordinal and a defining formula (or something to that effect), and these are explicitly well ordered. Maybe you could use this to $\Pi_2$-define the pair $\langle x,y\rangle$, where $x$ is the original set, and $y$ is the least witness to the existential quantifier? – Emil Jeřábek Sep 14 '14 at 13:49
  • Every OD set $x$ does indeed have a minimal ordinal $\theta$ such that $x$ is definable in $V_\theta$. So you are proposing to define the set of such pairs $\langle x,\theta\rangle$? – Joel David Hamkins Sep 14 '14 at 13:52
  • 1
    I guess I rather want to define $\langle A,V_\theta\rangle$, where $A$ is the set of minimal-rank non-OD sets, and $\theta$ is minimal such that every $x\in A$ of rank less than the rank of $A$ is definable in $V_{\theta'}$ for some $\theta'<\theta$. Or maybe with the sequence $\langle V_{\theta'}:\theta'<\theta\rangle$ in place of $V_\theta$ if it makes the formula simpler. – Emil Jeřábek Sep 14 '14 at 14:04
up vote 7 down vote accepted

Update. (June, 2017) François Dorais and I have completed a paper growing out of this answer and our others on related posts.

F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? (arχiv:1706.07285)

Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.

Read more at the blog post.

$\newcommand\ZFC{\text{ZFC}} \newcommand\HOD{\text{HOD}} \newcommand\P{\mathbb{P}} \newcommand\Q{\mathbb{Q}}$

Original answer. It turns out that question 1 has a positive answer, while question 2 has a negative answer.

The reader may find it useful to know of the characterization of the $\Sigma_2$ properties as the semi-local properties, those which are equivalent to an assertion of the form $\exists\theta\ V_\theta\models\psi$, where $\psi$ can have any complexity. In particular, whenever a $\Sigma_2$ property $\varphi(A)$ is true of a set $A$, it is because there is some $V_\theta$ satisfying something about $A$. We may therefore preserve that $\Sigma_2$ fact about $A$, while forcing over $V$, provided that we only force up high and preserve $V_\theta$, the rank-initial-segment of the universe up to $\theta$.

First, let's consider the positive answer to question 1.

Theorem 1. Every model of $\ZFC$ has a forcing extension satisfying $V\neq\HOD$, in which every $\Sigma_2$-definable set has an ordinal-definable element.

Proof idea: perform a forcing iteration, considering each $\Sigma_2$ formula in turn, where we try to freeze the set defined by that formula and then code one of its elements (if any) into the GCH pattern high above the witness to that $\Sigma_2$ property. In the end, every nonempty $\Sigma_2$-definable set will contain an ordinal-definable element.

Proof. Start with $V$ as a ground model. Without loss, by forcing if necessary, we may assume that $V$ satisfies $V=\HOD$, so that there is a definable well-ordering of the universe. Enumerate the $\Sigma_2$ formulas $\varphi_0,\varphi_1,\ldots$, and so on. Note that we may refer to $\Sigma_2$-truth since there is a universal truth predicate for truth of bounded complexity (so there will be no issues with Tarski's theorem on the non-definability of truth). We define a full-support forcing iteration $\P$ of length $\omega$, where the forcing at each stage will become progressively more highly closed. At the first stage, we consider the formula $\varphi_0$, and ask: is there a forcing extension $V[g_0]$ in which $\varphi_0$ holds of a nonempty set $A_0$? If so, we perform such a forcing (choose the least poset forcing this), and let $\lambda_0$ be the smallest $\beth$-fixed point above the size of that forcing so that also $\varphi_0$ is witnessed in $V_{\lambda_0}^{V[g_0]}$. Next, perform additional $\leq\lambda_0$-closed forcing over $V[g_0]$ to an extension $V[g_0][h_0]$, where $h_0$ forces to code one of the elements of $A_0$ into the GCH pattern above $\lambda_0$. This preserves the definition of $A_0$ by $\varphi_0$, while ensuring that $A_0$ has an ordinal definable element. Now, let $\theta_1$ be well above this coding, and continue.

At stage $n$, we have the partial extension $V^{(n)}=V[g_0][h_0]\cdots[g_{n-1}][h_{n-1}]$, which performed forcing below the cardinal $\theta_n$. We ask whether we can perform $\leq\theta_n$-forcing so that $\varphi_n$ holds of a nonempty set $A_n$ in the extension. If so, we do that forcing, let $\lambda_n$ be large enough to witness the $\Sigma_2$ property for $\varphi_n$, and then perform GCH coding above that so as to make an element of $A_n$ ordinal-definable, and let $\theta_{n+1}$ larger than all that. (Otherwise, we ignore $\varphi_n$ and let $\theta_{n+1}=\theta_n$.)

Consider the corresponding extension $V[G]$, where $G\subset\P$ is $V$-generic. Finally, we force to add a Cohen subset $H\subset\delta$, where $\delta$ is a regular cardinal above $\sup_n\theta_n$, since this will force $V\neq\HOD$ in $V[G][H]$. The final desired model is $V[G][H]$. Because we used full support, it follows that the tail forcing in $\P$ after stage $n$ is $\leq\theta_n$-closed, as is the forcing to add $H$, and so preserves sets of rank below $\lambda_n$. Thus, if $\varphi_n$ defines a nonempty set in $V[G][H]$, then at stage $n$ we would have observed that it was possible to force it to hold of a nonempty set (with forcing that was sufficiently closed), and so we would have treated it at stage $n$. That is, we would have forced to code one of its elements into the GCH pattern, afterwards always preserving that definition and this coding. So in the case that $\varphi_n$ does define a nonempty set in $V[G][H]$, then the stage $n$ forcing exactly ensured that one of the elements of this set was coded into the GCH pattern of $V[G][H]$ and was therefore ordinal-definable there. The later stages of forcing were arranged so as to preserve all these definitions.

Thus, $V[G][H]$ is a model of $V\neq\HOD$, as the forcing to add $H$ is weakly homogeneous, such that every $\Sigma_2$-definable nonempty set has an ordinal-definable element. QED

The argument reminds me of the forcing iteration proof of the maximality principle, where one forces a given statement to be true, if it is possible to force it in such a way that it remains true in all further forcing extensions. The end result is a model where any statement that could become necessarily true by forcing, is already true, and this is precisely what the maximality principle asserts.

Meanwhile, Emil's insightful suggestion in the comments leads to a negative answer to question 2.

Theorem 2. If $V\neq\HOD$, then there is a nonempty $\Pi_2$-definable set (with no parameters) containing no ordinal-definable element.

Proof. Let $A$ be the set of minimal-rank non-OD sets. That is, $A$ consists of all non-OD sets of rank $\alpha$, where $\alpha$ is minimal such that there is any non-OD set of rank $\alpha$. In my answer to this related question, I had proved that $A$ is characterized by a $\Sigma_2\wedge\Pi_2$ property.

Emil's idea was to consider not $A$, but a related set, namely the set $U=A\times V_\theta$, where $\theta$ is the smallest ordinal such that $A\in V_\theta$ and $V_\theta\models A$ is the set of minimal-rank non-OD sets.

The set $U$ is defined by the following property: $U$ consists of the cartesian product $U=A\times B$ of two sets $A$ and $B$ such that the set $B$ has the form $B=V_\theta$ for some ordinal $\theta$ such that $A\in V_\theta$ and $V_\theta\models "A$ is the set of minimal-rank OD sets and there is no $\theta'<\theta$ for which $V_{\theta'}\models A$ is the set of minimal-rank non-OD sets"; and finally, the elements of $A$ really are not in OD.

This property altogether has complexity $\Pi_2$, due mainly to the last clause. The first part, requiring that $U$ has the form $A\times B$, is $\Delta_0$. The next part, asserting that $B$ has the form $B=V_\theta$ for some ordinal $\theta$ has complexity $\Pi_1$, essentially because one need only assert that $B$ is transitive and satisfies some minimal set theory such that it thinks it is a $V_\theta$, and such that $B$ contains all subsets of any of its elements, so that it is using the true power set operation. The properties asserting that $V_\theta$, that is, $B$, satisfies certain complication assertions has complexity $\Delta_0$, since all quantifiers are bounded by $B$ and hence ultimately by $U$. And finally, asserting that the elements of $A$ are really not ordinal-definable has complexity $\Pi_2$, since "$x\in\text{OD}$'' has complexity $\Sigma_2$, as any instance of ordinal-definability reflects to some $V_\theta$ and hence is locally verifiable; thus, the assertion $\forall x\in A\ x\notin\text{OD}$ has complexity $\Pi_2$.

So altogether, the set $U=A\times V_\theta$ is $\Pi_2$-definable, but it can have no ordinal-definable elements, since every element of $U$ has the form $(a,b)$ for some $a\in A, b\in V_\theta$, and if the pair $(a,b)$ were ordinal-definable, then $a$ would be ordinal-definable, contradicting $a\in A$ and the fact that every member of $A$ is not ordinal-definable. QED

Note that the proof is completely uniform, in that the definition of the set does not depend on the model in any way. Rather, we have a $\Pi_2$ definition that $\ZFC+V\neq\HOD$ proves is a nonempty set disjoint from OD.

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.