Is $(\sf ZF^*GC - Extensionality)$ bi-interpretable with $\sf ZF^*GC$?
Where $\sf ZF^*GC$ is $\sf ZF - Replacement + Separation + Collection + Global \ Choice$
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asked Nov 3, 2023 at 7:51
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$\begingroup$ How do you express GC in the first-order language of set theory? You have a new class relation? And then this relation is allowed in the collection scheme etc.? $\endgroup$– Joel David HamkinsCommented Nov 3, 2023 at 15:23
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$\begingroup$ @JoelDavidHamkins we add a new primitive total unary function C and add the axioms $y \in x \to C(x) \in x$ and $C(\emptyset) =\emptyset$, and of course $C $ is allowed in collection and separation $\endgroup$– Zuhair Al-JoharCommented Nov 3, 2023 at 15:44
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1 Answer
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Joel Hamkins has detected a gap in the proposed proof below. At the moment I do not know of another strategy to the answer the question, but I am leaving it below as signpost to others who might fall in the same trap that I did :)
WARNING: The proposed proof below does not work since swapping two urelements does not preserve the global choice function (even though it preserves the set-theoretical structure).
This answer uses the same idea as the one given in a comment of mine on James Hanson's answer to this MO question.
The basic idea is that one quick way to rule out the bi-interpretability between two theories is to show that some group arises as an automorphism group of a model of one of the theories, but not the other since it is well-known that bi-interpretable models have the same automorphism group.
With the above in mind, it suffices to note that by removing Extensionality from $\mathsf{ZF^*+ GC}$ (equivalently $\mathsf{ZF+GC}$) we can easily build a model of the resulting theory that has an automorphism of order 2 by swapping two urelements. But it is well-known that no model of ZFC can have an automorphism of finite order (basically because in the presence of the axiom of choice, let alone global choice, any set can be "coded" by a subset of ordinals)
answered Nov 3, 2023 at 18:39
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2$\begingroup$ Swapping the urelements a,b, will not induce an automorphism of the structure with the choice function, since the choice function C will choose one of them from {a,b}. $\endgroup$ Commented Nov 3, 2023 at 20:25
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1$\begingroup$ @JoelDavidHamkins Thanks for the prompt correction Joel. I will soon revise. $\endgroup$ Commented Nov 3, 2023 at 22:00
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$\begingroup$ Ali, I had also tried to answer the question with rigidity issues (before you had posted), but had realized the problematic issue of respecting the choice function. I wonder whether we need to make to more duplication than just urelements, but duplicate every set, in such a way that every possible choice is realized by C in a copy of the given set. This might lessen the role of C somewhat and enable a rigidity argument, but I haven't been able to push it through. I think they might be bi-interpretable. $\endgroup$ Commented Nov 4, 2023 at 0:19