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The axiom of extensionality can be weakened, asking only for non-empty sets with the same elements to be equal. Then there can be many "different empty sets" called urelements. I call this theory ZFU (I'm not sure if this coincides exactly with what is denoted by ZFU in the literature).

If we have a model ${\mathfrak A}=(M,\varepsilon)$ of ZFU where $M$ is a set and $\varepsilon$ is a binary relation on $M$, and the set of $\varepsilon$-urelements of $M$ has cardinality $\alpha$, I say that $\mathfrak A$ is an $\alpha$-model of ZFU.

Thus an $1$-model is the same thing as a model of ZFC. Also, if we have an $\alpha$-model of ZFU with $\alpha\neq 0$, then for any urelement $u$, the set of all elements of $M$ whose $\varepsilon$-transitive closure only contains $u$ as an urelement is easily seen to be a model of ZFC.

Thus, the existence of an $\alpha$-model implies the consistency of ZFC. Is the implication strict when $\alpha >1$ ? Also, is the existence of a $\beta$-model a strictly stronger claim than the existence of an $\alpha$-model, when $\alpha<\beta$.

I asked a closely related question here

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Ewan Delanoy
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  • In the usual formulation, ZFU includes non-set urelements. It's not hard to see this is more or less the same thing. Have you looked in Jech to see that ZF and ZFU are equiconsistent? – Asaf Karagila Sep 19 '15 at 07:48
  • @AsafKaragila I didn't yet, but I will (I guess you mean the "Axiom of Choice" book). – Ewan Delanoy Sep 19 '15 at 08:02
  • Either that or Set Theory. In the latest edition it appears in Chapter 15, if I recall correctly. – Asaf Karagila Sep 19 '15 at 08:25
  • @AsafKaragila you were quite correct. As I have summarized it all in a short answer, I think we should delete our comments now – Ewan Delanoy Sep 20 '15 at 08:01

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As shown in Lemma 15.47 of Jech's "Set Theory" (freely available online here ; thanks to Asaf Karagila for pointing this reference), if ZFC is consistent then there are $\alpha$-models for any $\alpha$. The idea is that the urelements should be all of equal rank, we can take e.g. $U=V_{\alpha+1}\setminus V_{\alpha}$, then build the cumulative hierarchy from there.

Ewan Delanoy
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