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Joel David Hamkins
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This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements.

The question of whether a set of reals $A$ is measurable or not is something that can be checked in a comparatively small rank-initial segment of the universe, in $V_{\omega+3}$ or so, and for this reason, it is a local property, which therefore has complexity at worst $\Delta_2$.

This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements.

This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements.

The question of whether a set of reals $A$ is measurable or not is something that can be checked in a comparatively small rank-initial segment of the universe, in $V_{\omega+3}$ or so, and for this reason, it is a local property, which therefore has complexity at worst $\Delta_2$.

Source Link
Joel David Hamkins
  • 236.3k
  • 44
  • 777
  • 1.4k

This is impossible; there is no model of ZFC like that. The reason is that the set of non-measurable sets of reals (or non-Baire sets, respectively) is definable, and moreover $\Sigma_2$ definable; so under the first part of your conditions, it would have a definable member, which would violate the second part of your requirements.