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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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