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It is known that the following is true in the Solovay model (SM) for ZFC: any countable OD (ordinal-definable) set $X$ of reals necessarily consists of OD elements. What about countable OD sets of higher rank?

Theorem. In SM, if $\,A\ne\emptyset$ is a countable OD set of sets of reals then any its element (a set of reals) is OD itself.

Sketch (only works provided sets in $A$ are pairwise disjoint). As being OD is an OD property, it suffices to prove that $A$ contains at least one OD element. Suppose that this is not the case. There is a countable ordinal $\xi$ and a $Coll(\omega,\xi)$-name $t\in V$ ($V$ being the ground model of ZFC whose SM-extension we are talking about) such that the union $U$ of all sets in $A$ (an OD set of reals) contains a real of the form $t[f]$, where $f\in\xi^\omega$ is $Coll(\omega,\xi)$-generic over $V$. Now define an equivalence relation $E$ on $\xi^\omega$ such that $f\,E\,g$ iff either the reals $t[f],t[g]$ belong to the same set in $A$ or both $t[f],t[g]$ do not belong to $U$. Obviously $E$ is OD and has at most countably many equivalence classes. As the domain of $E$ is the same as reals, it suffices to conclude that all $E$-classes are OD, therefore some sets in A are OD too, a contradiction.

Question. Is it true in SM that any countable OD set (no restrictions to the nature of its elements) has an OD element ?

(After some attempts to solve it.) Still the problem persists, and it seems to be focused in the following conjecture.

Conjecture. In the Solovay model for ZFC (= the Levy collapse extension of L), let $X$ be a countable or finite non-empty OD (ordinal-definable) set of sets of reals. Then $X$ contains a proper OD subset $\emptyset\ne X'\subsetneqq X$.

The conjecture is equally open for many other generic models not especially designed to make something definable - like for instance a simple Cohen or random-Solovay extension $L[a]$ of $L$.

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  • $\begingroup$ Sorry, it just really bugged me, I had to edit it. :-) $\endgroup$ – Asaf Karagila Dec 8 '14 at 20:44

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