# Consistency hierarchy between models with many urelements

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

• 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? Sep 19, 2015 at 7:48
• @AsafKaragila I didn't yet, but I will (I guess you mean the "Axiom of Choice" book). Sep 19, 2015 at 8:02
• Either that or Set Theory. In the latest edition it appears in Chapter 15, if I recall correctly. Sep 19, 2015 at 8:25
• @AsafKaragila you were quite correct. As I have summarized it all in a short answer, I think we should delete our comments now Sep 20, 2015 at 8:01

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.