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Kunen showed that there is no nontrivial $j: V \rightarrow_e V$. One might wonder what happens in $\mathsf{ZFC}$ with atoms.

Let's denote the universe by $U$. We aren't assuming that the atoms form a set, or even that there are no more atoms than pure sets. Of course, if there at least two atoms, there will be nontrivial automorphisms on $U$, so that isn't very interesting. The following seems like a better way of translating Kunen's question into this framework: let's call an elementary embedding atrivial if it is nontrivial but it is the identity mapping on atoms. Can there be an atrivial $j: U \rightarrow_e U$?

Two comments: 1) I've tried to prove that there can be no such $j$ by, in essence, quotienting out distinctions among Urelemente (similar to a Fraenkel-Mostowski model) and building a $j'$ that inherits nontriviality, but without success; 2) note that every object has at most set-many atoms in its transitive closure, so one can find a set of atoms $A$ such that $j$ is nontrivial on the restricted hierarchy $U(A)$ built over those atoms alone; unfortunately, there's no guarantee that $j``U(A)$ is a class built up only over set-many atoms.

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  • $\begingroup$ The non-trivial clause should be about ordinals moved. Not about identity. $\endgroup$ – Asaf Karagila Dec 31 '17 at 10:09
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    $\begingroup$ @Asaf, it seems fine to me to have it be about identity. It follows that there must be ordinals moved. For example, if you well-order a set and all ordinals are fixed, then $j(x)$ would have to be the same as $j"x$, and so there could be no $\in$-minimal set that is moved. $\endgroup$ – Joel David Hamkins Dec 31 '17 at 12:38
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    $\begingroup$ But actually, one can also argue without AC in ZF with atoms that every non-identity $j:V\to V$ that fixes all atoms must move ordinals. If $j$ fixes all ordinals, then it fixes $V_\alpha(A)$ for any set of atoms $A$. Now argue by induction on rank that $j$ fixes every element. $\endgroup$ – Joel David Hamkins Dec 31 '17 at 13:55
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Theorem. The Kunen inconsistency works over ZFC with atoms. That is, in this theory, there is no non-identity elementary embedding $j:V\to V$ that fixes every atom.

Proof. Suppose that $j:V\to V$ is an elementary embedding fixing every atom. If $j$ is not the identity embedding, then I claim that $j$ must move an ordinal. To see this, let $u$ be any $\in$-minimal element that is moved by $j$. Let us well-order $u$ in some order type $\beta$. Since ordinals are fixed by $j$, it follows that the $\alpha^{th}$ element of $u$ is carried by $j$ to the $j(\alpha)^{th}=\alpha^{th}$ element of $j(u)$, but since those elements are fixed (by minimality of $u$) and the total length of the well-order is fixed, it follows that $j(u)$ is simply $u$ itself, contradicting the choice of $u$. (In fact, by an argument on ranks, one can eliminate the use of choice in this part of the argument; but you still need it for the other part.)

So $j$ is not the identity on ordinals. We may now restrict $j$ to the pure part of the universe. Let $W$ be the class of sets having no atoms in their transitive closures. This is a model of ZFC, without atoms. Since this is definable, it follows that $j\upharpoonright W$ is a nontrivial elementary embedding from $W$ to $W$. This contradicts the usual Kunen inconsistency. For example, one gets that $j\upharpoonright V_{\lambda+2}^W$ is a set in $W$, where $\lambda$ is the supremum of the critical sequence, and this violates the usual ZFC version of the Kunen inconsistency in $W$. $\Box$

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  • $\begingroup$ Thank you. Out of curiosity, is the use of $V$ to denote the entire universe when Urelemente are in play standard? I find it a bit ungainly to have it sometimes denote the entire universe and sometimes just the pure sets, but if it's an established convention, I'll conform. (I realize that people with pluralist or anti-Platonist tendencies may not view what's going on here in quite this way.) $\endgroup$ – Beau Madison Mount Dec 31 '17 at 17:50
  • $\begingroup$ Oh, I always use $V$ that way, and I see now that you had used a different notation in your question; I apologize for the confusion. I'm not sure what the standard notation is when using this theory in the literature. $\endgroup$ – Joel David Hamkins Dec 31 '17 at 18:16

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