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In their paper "The Role of the Foundation Axiom in the Kunen Inconsistency" (arXiv:1311.0814 [Math.LO]), Daghighi, Golshani, Hamkins, and Jerabek show that the patterns of possibility for the existence of nontrivial automorphisms and nontrivial elementary embeddings of the universe in models of set theory without Foundation take the following form:

{$id_{V}$}$\subseteq$$Aut(V)$$\subseteq$$Eem(V)$

where $id_{V}$ is just the identity mapping from $V$ to $V$, $Aut(V)$ are the automorphisms from $V$ to $V$,and $Eem(V)$ are the elementary embeddings from $V$ to $V$.

In fact, they prove that there are models of $ZFC^{-f}$ that realize each of these four separating refinements of {$id_{V}$}$\subseteq$$Aut(V)$$\subseteq$$Eem(V)$:

i). {$id_{V}$}=$Aut(V)$=$Eem(V)$

ii). {$id_{V}$}$\subsetneq$$Aut(V)$=$Eem(V)$

iii).{$id_{V}$}=$Aut(V)$$\subsetneq$$Eem(V)$

iv). {$id_{V}$}$\subsetneq$$Aut(V)$$\subsetneq$$Eem(V)$

My question is simply this:

Are there models of $NGB+{\lnot}AC$ that realize each of the four separating refinements?

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  • $\begingroup$ Why the downvote? Why is this question deemed 'unhelpful'? What, if anything, is wrong with this question? $\endgroup$ – Thomas Benjamin Apr 12 '15 at 8:11
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    $\begingroup$ Did you intentionally drop the superscript, so that you are requiring foundation? Then (ii) and (iv) are impossible, the straightforward proof by $\in$-induction that every automorphism is the identity does not need any choice. (i) is easy (e.g., take a symmetric extension violating choice of a model of V=L with no inaccessible cardinals), and (iii) is a restatement of the open problem whether the existence of Reinhardt cardinals is consistent. $\endgroup$ – Emil Jeřábek Apr 12 '15 at 9:57
  • $\begingroup$ @EmilJeřábek: Yes, I did intentionally drop the superscript. My purpose in doing so was to see if $NGB+{\lnot}AC+{\lnot}RR$ was Consistent. If you are correct in determining that (ii) and (iv) are impossible, then what of Prof. Hamkins' contention that his and Palumbo's model (see Thm. 5) that shows that $ZF+{\lnot}RR$ is consistent if $ZF$ is translates into a model of $NGB+{\lnot}AC+{\lnot}RR$? Such a model would, by definition of ${\lnot}RR$, seemingly have to realize either (ii) or (iv)--so which does it satisfy (please enlighten me)? Also, Yair's answer to my previous question $\endgroup$ – Thomas Benjamin Apr 12 '15 at 12:16
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    $\begingroup$ If you extend RR to classes, it only makes $\neg RR$ weaker. It says that there exists a class $A$ such that for every relation $R$, $(A,R)$ is not rigid. This does not in any way contradict the fact that there is a different class, namely $V$, which does carry a rigid relation. Yair’s answer says what it says: if it is consistent with NBG that there is a nontrivial elementary embedding $V\to V$, then the existence of such an elementary embedding is also consistent with there being a set that does not carry a rigid relation. $\endgroup$ – Emil Jeřábek Apr 12 '15 at 14:55
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    $\begingroup$ I have no access to the paper, however, as far as I can gather from Google books preview, he is talking about external automorphisms of the model for which replacement needn’t hold. This is a whole different game. Every first-order model has elementary extensions with lots of such automorphisms. $\endgroup$ – Emil Jeřábek Apr 24 '15 at 12:12
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I think there's some confusion over automorphisms/embeddings which are "internal" vs. those which are "external." Some comments which I hope clear things up:

  • No model of ZF has nontrivial definable (even from parameters) automorphisms. This is a proof by transfinite induction: let $\alpha$ be least such that some element of $V_\alpha$ is moved by the automorphism $A$. This $\alpha$ exists by Foundation and the definability of $A$ (this is where the argument breaks down if $A$ isn't definable - but note that it also suffices for our model to be well-founded!). Clearly $\alpha>0$, but then let $x\in V_\alpha$ be moved by $A$. Since $\alpha>0$, $x\not=\emptyset$, but for each $y\in x$ we have $A(y)=y$. So $A(x)\not=\{A(y): y\in x\}$, which means $A$ is not an automorphism.

  • On the other hand, there are models of ZF with automorphisms. This is just a fact of logic: if $T$ is any first-order theory with an infinite model, then $T$ has a model whose automorphism group is nontrivial. The proof is a nice combination of compactness and Ramsey's theorem for pairs; see https://en.wikipedia.org/wiki/Ehrenfeucht%E2%80%93Mostowski_theorem.

  • This means: as long as we talk about automorphisms of models of ZF, we mean external automorphisms. The same goes for similar set theories. Note that the crucial axiom is Foundation, which means that theories without foundation - e.g., $ZF^{-f}$ or $NF$ - may have definable automorphisms.

  • However, this sort of argument doesn't let us build automorphisms with prescribed properties. For instance, if we want a model of a theory $T$ which has a nontrivial automorphism of finite order, we might be out of luck! In particular, if any model of $T$ has a definable linear ordering, then no model of $T$ has a nontrivial automorphism of finite order! This is a good exercise.

  • A harder exercise is to show that no model of ZFC (well-founded or not) has a nontrivial automorphism of finite order; this is due to Friedman, and is quoted at the beginning of Cohen's paper. Note that this means that if we can build a model of ZF with a nontrivial automorphism of finite order, as Cohen does, that model must satisfy $\neg AC$.


To answer your question: if by automorphism/embedding we mean definable automorphism/embedding, then: with the axiom of foundation in hand, options 2 and 4 are immediately ruled out; and option 1 is achieved in e.g. any symmetric submodel of a set-generic extension of a model of $V=L$. Meanwhile, any instance of 3 would contain a Reinhardt cardinal, so it's generally believed that 3 can't happen.

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