EDIT: it appears that my original question has some confusion between auto-morphisms and elementary embeddings as it is obvious from the answer below, therefore I'll clarify here what I exactly want.

Can we have the following?

A transitive model $M$ of ZF-Regularity such that we have an external injective function $j$ from $P(M)$ to $P(M)$ such that $j$ is an ismomorphism on $\in$ between $P(M)$ and $range(j)$. Now we demand that $M$ is a subset of $range(j)$, and that every definable subset of $M$ [from parameters in $M$] would be sent by $j$ to an element of $M$.

Where "$\kappa $ is a definable subset of $M$ from parameters $w_1,..,w_n \in M$", is defined as: $$ \forall y (y \in \kappa \leftrightarrow y \in M \wedge \phi(y,w_1,..,w_n))$$ for some formula $\phi(y,w_1,..,w_n)$ that is written in the language of set theory [with the only exception of allowing $j $ to be appear on parameters, so $j(w_i), j^{-1}(w_i)$ are allowed].

We know that $range(j) \neq M$, because the parity of $M$ is different from $P(M)$.

The question is if there is a clear inconsistency with that?

Is the following known to be consistent with some extension of $\text{ZF}$?

There is a transitive model $M$ of $\text{ZF}$ such that there is an external non-trivial elementary embedding $j$ from $P(M)$ to $P(M)$ (where $P$ is the known Power operator) such that $M \subset range(j)$ and such that every definable subset $\kappa$ of $M$ (parameter free or from parameters in $M$) in the language of set theory (i.e. doesn't use the symbol $j$) is sent by $j$ to an element of $M$[i.e. $j(\kappa) \in M$] and such that the symbol $j$ can be used freely in the instances of Replacement and Separation.

To clarify what is meant by "$\kappa $ is definable subset of $M$ from parameters $w_1,..,w_n \in M$", is to mean that: $$ \forall y (y \in \kappa \leftrightarrow y \in M \wedge \phi(y,w_1,..,w_n))$$ for some formula $\phi(y,w_1,..,w_n)$ that is written in the language of set theory.

In particular: is the above consistent with $\text{ZF + Reinhardt cardinal}$?

  • $\begingroup$ "and such that the symbol $j$ can be used freely in the instances of Replacement and Separation." What do you mean by this? What structure is supposed to satisfy these strengthenings of Replacement and Separation? Certainly not $M$ or $P(M)$ themselves ... $\endgroup$ – Noah Schweber Jul 12 '18 at 23:21
  • $\begingroup$ I don't understand your question, but certain ideas in it seem to have a family resemblance with ideas in the wholeness axiom, introduced by Paul Corazza. See cantorsattic.info/Wholeness_axioms. Corazza was interested in the idea of adding $j$ explicitly to the language, plus the assertion that it is an elementary embedding of the universe to itself, and then looking at how much set theory one might have in the language with $j$. This seems to be in part what you also are doing. $\endgroup$ – Joel David Hamkins Jul 12 '18 at 23:28
  • $\begingroup$ @JoelDavidHamkins, yes part of it is like that, but there are additional requirements here, that of the definable subsets of $M$ as specified in the head post being able to be moved internally in $M$ by automorphism $j$ $\endgroup$ – Zuhair Al-Johar Jul 13 '18 at 4:51

EDIT: Based on the comments, I think it's worth clarifying a bit of the nature of the Boffa-Jensen construction of a model of NFU (which appears to be part of the motivation for this question).

In this construction, we do not have an elementary embedding; rather, we begin with a model $M$ of ZFC and an automorphism - unfortunately for our context denoted "$j$" in Holmes' article, but which I will call "$a$" - which moves a rank, that is, which is not the identity on the ordinals. I think the conflation of automorphism and elementary embedding is generating some confusion here.

Incidentally, a fun exercise is to show that any nontrivial automorphism of a model of ZFC must move an ordinal, while this is not true of ZF! I believe this was first observed by Harvey Friedman. A cute result of Cohen is the construction of a (necessarily ill-founded) model of ZF with a nontrivial automorphism of finite order, and it was in Cohen's paper that I first saw this fact quoted.

Now we come to a key point: the automorphism $a$ is completely external, and the mere existence of a nontrivial automorphism immediately implies that $M$ is illfounded. Put better: any well-founded model of ZFC, or ZF, or even much weaker, has no nontrivial automorphisms. (Think about the least rank of an element moved by a hypothetical nontrivial automorphism …) Indeed, when we speak of automorphisms of models of set theory these are always understood as being completely external to the models in question; note that this is in contrast with the situation for elementary embeddings, where there are various flavors of definability. Similarly, the connection with ill-foundedness breaks down for elementary embeddings: assuming $0^\sharp$ exists, there is a nontrivial elementary embedding of $L$ into itself, but $L$ is of course well-founded. Automorphisms and elementary embeddings are really very different objects.

Anyways, in the Boffa-Jensen construction we take an $M$ and $a$ as above, and fix some $M$-ordinal $\alpha$ with $a(\alpha)\not=\alpha$. WLOG we have $a(\alpha)<\alpha$ (otherwise replace $a$ with $a^{-1}$. The structure $\mathcal{B}$ associated to the data $(M, a, \alpha)$ has domain $(V_\alpha)^M$, and elementhood relation $\varepsilon$ given by $$x\varepsilon y\iff a(x)\in^My\wedge y\in^MV_{a(\alpha)+1}.$$ Everything in $V_\alpha^M\setminus V_{a(\alpha)+1}^M$ winds up being an urelement in the sense of $\mathcal{B}$: if $y\in V_\alpha^M\setminus V_{a(\alpha)+1}^M$, then we never have $x\varepsilon y$ for any $x\in V_\alpha^M$.

(Why is $V_\alpha^M\setminus V_{a(\alpha)+1}^M$ nonempty? Well, one way to guarantee this would be to make sure that $M$ is an $\omega$-model, that is, the "finite ordinals" in the sense of $M$ are in fact all finite; however, even this assumption is unnecessary, since $a(\alpha)$ and $\alpha$ must have the same "$M$-parity" and no ordinal has the same parity as its successor.)

However, we're seeing something very specific to automorphisms here. An elementary embedding $M\preccurlyeq M$ simply wouldn't help us do the same thing: it is crucial that $a$ moves ranks downwards. Having an "upward-moving" automorphism is still fine, since we can invert it, but elementary embeddings aren't surjective in general so no similar trick will work there. Indeed, I know of no way to get a model of NFU (or similar) from, for example, $0^\sharp$ in a natural way (remember that $0^\sharp$ is equivalent in a sense to a nontrivial elementary embedding from $L$ into itself; this isn't a contradiction with Kunen, since "$0^\sharp$ exists" implies $V\not=L$).

It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. (If not, what do you mean?) Under this interpretation the answer to your question is no. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an element of our "power structure" $\mathfrak{S}$, not as a subset of it).

But $M$ is a definable element of $\mathfrak{S}$: $M$ is the unique $z\in\mathfrak{S}$ such that for all $a\in\mathfrak{S}$ we have $\mathfrak{A}\models\exists x(a\in x)\iff a\in z$. So $M$ can't be moved by an elementary embedding.

As far as I can tell, this argument (or a version of it) works for any reasonable envisioning of $P(M)$ as a structure. For example, another possible interpretation of your question eschews "$\in$" but has the same result. We look at a structure $\mathfrak{S}$ whose elements correspond to subsets of $M$ (if you want, the domain of $\mathfrak{S}$ is exactly $P(M)$). The language of $\mathfrak{S}$ is just $\{\subseteq\}$, interpreted in the obvious way.

The domain of $\mathfrak{S}$ has a special subset $Ground$: namely, the set of all $a$ with $\{x\in M: x\in a\}=\{x\in M: M\models x\in b\}$ for some $b\in M$. These are the "internal sets" of $M$.

The first two requirements you put on $j$ are elementarity and $j(a)\in D$ whenever $a$ is a definable subset of $M$. The latter means in particular that we must have $j(M)\in D$ (note that $M\in \mathfrak{S}$). But then $j$ certainly can't be elementary, since $M$ is the unique $a\in\mathfrak{S}$ satisfying $\forall x(x\subseteq M)$ (and elementary maps can't move definable elements.

  • $\begingroup$ first: what is $\mathfrak{A}$? Now: $M$ is the domain of a model of $ZF$, $M$ is both a subset and an element of $P(M)$, $j$ is an automorphism from $P(M)$ to $P(M)$ that moves subsets of $M$ that are definable in the language of set theory (with parameters in $M$ or without any parameters) in $M$. You say that this is not possible, but by then how I am to understand Boffa models that uses automorphism that moves a rank internally? those are used in the proof of $NFU$, in those models you do have $j(M)$ being an element of $M$, see math.boisestate.edu/~holmes/preserves3.pdf (page 5). $\endgroup$ – Zuhair Al-Johar Jul 13 '18 at 4:48
  • 1
    $\begingroup$ @ZuhairAl-Johar $M$ is not a subset of $P(M)$ unless $M$ is transitive ... $\endgroup$ – Noah Schweber Jul 13 '18 at 4:52
  • $\begingroup$ Yes, $M$ is transitive. $\endgroup$ – Zuhair Al-Johar Jul 13 '18 at 4:53
  • $\begingroup$ @ZuhairAl-Johar And where precisely did you state that in your question? $\endgroup$ – Noah Schweber Jul 13 '18 at 4:54
  • $\begingroup$ OK guilty as charged. I'll write that. But anyhow your argument doesn't seem to be affected by that, isn't it? $\endgroup$ – Zuhair Al-Johar Jul 13 '18 at 4:55

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