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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$\begingroup$ In statement 3, you say "set of sets of $\mathbb{R}^2$, but I took this to be just about a "subset of $\mathbb{R}^2$". Also, it should be stated that the set is non-empty to avoid a trivial contradiction. $\endgroup$– Joel David HamkinsCommented Oct 21, 2013 at 1:34
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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.
answered Oct 20, 2013 at 23:52
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$\begingroup$ (There is the caveat that we want some choice so we can actually define Lebesgue measure and check its basic properties as usual.) $\endgroup$ Commented Oct 20, 2013 at 23:59
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$\begingroup$ @Andres, I agree. But from ZF plus statement 3 we get a well-ordering of the reals, and perhaps this enough choice to define Lebesgue measure and prove the basic properties? And further, we have $\mathbb{R}\subset\text{HOD}$, which would give rise to the $\text{HOD}$ Lebesgue measure, as ZFC holds there. $\endgroup$ Commented Oct 21, 2013 at 1:05
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$\begingroup$ That is enough choice to get its basic properties, since it gives CC($\mathbb{R}$). $\;$ $\endgroup$– user5810Commented Oct 21, 2013 at 1:30