Working in a suitable extension of $\sf ZF-Reg.$, can we have a transitive model $M$ of $\sf ZF-Reg.$ such that for every set in $M$ there is a partition on it in $M$ all compartments of which are non-singleton finite sets. And such that any subset $X$ of $M$ that is a family of pairwise disjoint larger than singleton finite sets then $X \in M$.
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asked Aug 10, 2022 at 18:18
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No, this is not possible: the ordinals of $M$ will always provide a counterexample.
Even if regularity fails, the class $\mathsf{Ord}$ of ordinals cannot be a set. But we can partition $\mathsf{Ord}$ into (say) two-element sets via the equivalence relation $$\alpha\sim\beta\quad\iff\quad \sup\{\lambda: 2\cdot\lambda<\alpha\}=\sup\{\lambda: 2\cdot\lambda<\beta\}.$$ The classes of this relation are $\{0,1\},\{2,3\},...,\{\omega,\omega+1\}, ...$ etc.
(Perhaps more elegantly, let $\approx$ be any equivalence relation on $\omega$ all of whose classes are finite and have at least two elements. Then consider the relation $\approx'$ defined by $\alpha\approx'\beta$ iff there is a limit ordinal $\lambda$ and finite ordinals $m,n$ such that $\lambda+m=\alpha$, $\lambda+n=\beta$, and $m\approx n$.)
answered Aug 10, 2022 at 19:26
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$\begingroup$ Yes! I had the feel that well ordered sets can violate what I'm saying, I think we need to require $X$ being non-well orderable, for this to be possible. So, the idea is that all non-well orderable classes of pairwise disjoint larger than singleton finite sets is a set. $\endgroup$ Commented Aug 10, 2022 at 20:38
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$\begingroup$ I'll post that as a new question. $\endgroup$ Commented Aug 10, 2022 at 20:52
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$\begingroup$ Is the last condition supposed to be $m\approx n$? $\endgroup$ Commented Aug 10, 2022 at 23:39
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