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My analysis prof constructed a non-measurable subset of $\mathbb{R}$ today. A student noticed he used axiom of choice and asked for a construction that doesn't use it. Then, another student responded that there exist models of ZF where every subset of $\mathbb{R}$ is measurable--which blows my mind. Can anybody show me a proof of this? Also, ZF is just a set of axioms, what did my classmate mean by "models of ZF"?

  • https://mathoverflow.net/a/42220/158937 – Rushabh Mehta Nov 10 '21 at 01:21
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    If you don't know what models are, I'd suggest learning some logic before diving into this question. – Rushabh Mehta Nov 10 '21 at 01:21
  • See the Solovay Model But I second the comment that you should first look up what models are. – lulu Nov 10 '21 at 01:22
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    Without $AC$ you can't construct a non-measurable set. But you can't prove that they don't exist, either. So your title is not quite right. – TonyK Nov 10 '21 at 01:26
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    @Ian "ZF-C is equiconsistent with ZF-C+axiom of determinacy which does get you that" That's not true: $\mathsf{ZF+AD}$ proves the consistency of $\mathsf{ZF}$ (this is nontrivial, but basically $\mathsf{ZF+AD}\vdash$ "$L_{\omega_1}\models\mathsf{ZFC}$"), and so has strictly greater consistency strength. (In fact it has much greater consistency strength.) – Noah Schweber Nov 10 '21 at 01:42
  • @NoahSchweber Thanks for the correction, been a while since I reviewed this topic. – Ian Nov 10 '21 at 01:46
  • Without Alexandra Ocasio-Cortez everything is possible. – markvs Nov 10 '21 at 02:06

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