Is there a model of set theory such that:
AC holds,
Every ordinal definable set is measurable,
Every ordinal definable set of sets of $\mathbb{R}^2$ whose projection on the first (or second axis) is ordinal definable contains a definable member.
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Sign up to join this communityIs there a model of set theory such that:
AC holds,
Every ordinal definable set is measurable,
Every ordinal definable set of sets of $\mathbb{R}^2$ whose projection on the first (or second axis) is ordinal definable contains a definable member.
The answer is no, there is no such model. And we don't even need AC.
To see this, note first that statement 3 is equivalent to the assertion that every real number is ordinal definable. The reason is that the set of reals that are ordinal definable is itself definable, and so the complement is also definable, but contains no OD member.
But from this, we can use the HOD order, which is definable, to find an ordinal-definable well-ordering of the reals. And from any such ordering, we can define a non-measurable set by the Vitali argument, which would be a definable violation of statement 2.