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Fix a natural number, $n \geq 1$. Consider the class, M, of all sets hereditarily ordinal-definable using some $\Sigma_n$ formula. Since there is a universal $\Sigma_n$ formula, M is definable. Is M necessarily a model of ZF? It seems to me that it is closed under Godel operations and almost universal for the same reasons that HOD is, and therefore a model of ZF. But I feel like I'm missing something, since I've never heard anything about this model.

If it is a model of ZF, where can I learn more about it? Has anybody done any research about it? How does it relate to HOD?

Note: I require $n \geq 1$ because the formula witnessing that HOD is almost universal is $\Sigma_1$.

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up vote 6 down vote accepted

It follows from the Reflection Principle that every ordinal definable set is ordinal definable by a $\Sigma_2$-formula in the language of set theory. Indeed, if $A = \{x : \phi(x,\bar\alpha)\}$, then $$\exists\gamma(\gamma \in \mathrm{Ord} \land \bar\alpha \in \gamma \land \forall x(x \in A \leftrightarrow x \in V_\gamma \land V_\gamma \vDash \phi(x,\bar\alpha))).$$ Therefore, there is some $\gamma \in \mathrm{Ord}$ such that $$x \in A \leftrightarrow \exists U(U = V_\gamma \land x,\bar\alpha \in U \land U \vDash \phi(x,\bar\alpha)),$$ which is $\Sigma_2$ since $U = V_\gamma$ is $\Pi_1$.

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I agree about $\Sigma_2$ but not about the reason. The quantifier $\exists\gamma$ isn't relevant, since you're allowed an ordinal parameter (like $\gamma$) anyway. I think the relevant formula defining $x\in A$ is $\exists z\,(z=V_\gamma \land z\models\phi(x,\alpha)$. The clause $z=V_\gamma$ here is $\Pi_1$, so the whole formula is $\Sigma_2$. – Andreas Blass Aug 13 '11 at 3:46
In my preceding comment, add a final parenthesis to the "relevant formula", so that it's a formula. – Andreas Blass Aug 13 '11 at 3:48
Well, of course! Thanks for the correction. – François G. Dorais Aug 13 '11 at 3:53

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