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I apologize for asking the same question twice, since my last question was not really understood and there seems to be a problem preventing me from comment of editing the question.

Is there a model of ZFC such that:

  1. Every OD set of reals is measurable.

  2. Every OD ${\hspace{.03 in}\it family}$ of sets of reals contains at least one OD definable set.

Thank you

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  • $\begingroup$ What does "OD definable" mean? $\;$ $\endgroup$ – user5810 Oct 21 '13 at 2:56
  • $\begingroup$ I presume the OP means "every OD family of sets of reals contains at least one OD set of reals;" is this correct? $\endgroup$ – Noah Schweber Oct 21 '13 at 2:56
  • $\begingroup$ It should be "nonempty" family to avoid trivial contradiction. $\endgroup$ – Joel David Hamkins Oct 21 '13 at 3:02
  • $\begingroup$ The other question is here: mathoverflow.net/questions/145392/measurable-and-definable-sets $\endgroup$ – Joel David Hamkins Oct 21 '13 at 3:48
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The answer to this question also is negative. (I take statement 2 to be only about non-empty families.)

First, your statement 2 implies that every real is ordinal definable, since the set of singletons of non-OD reals is definable, but can't have any OD member.

In particular, your statement 2 implies that there is a definable well-ordering of the reals, without assuming AC, using the HOD order.

Second, your statement 2 implies that every set of reals is ordinal definable, since the collection of non-ordinal definable sets of reals is a definable family, but has no ordinal-definable members.

But this contradicts statement 1, since we can define a non-measurable set using the Vitali argument.

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  • $\begingroup$ This doesn't necessarily hold if "OD definable set" means something other than "OD set". $\hspace{1.2 in}$ $\endgroup$ – user5810 Oct 21 '13 at 3:07
  • $\begingroup$ Ricky, yes, by "OD definable set" I took the OP to mean "OD set" or equivalently "Ord-definable set". It may simply be a typo... $\endgroup$ – Joel David Hamkins Oct 21 '13 at 3:16
  • $\begingroup$ If we restrict to defiable families of sets of cardinalities continuum, do we have a positive answer? $\endgroup$ – user38200 Oct 21 '13 at 7:20
  • $\begingroup$ @user38200, that doesn't affect things, since we can easily replace each real with a continuum of of reals, equidefinable from it. For example, consider the family of sets $A_x$, where $x$ is not OD, consisting of all reals whose even digits are all the digits of $x$. This now has size continuum, and the family of all such $A_x$ is OD, but there can be no OD member, since from $A_x$ we can recover $x$. So once again, every real is OD, and so we have an OD well-order, from which we can define an OD non-meausurable set. $\endgroup$ – Joel David Hamkins Oct 25 '13 at 13:40

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