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Is it consistent with $\sf ZFC$ to have mutual elementary embeddability between distinct transitive sets?

Formally, is the following theory consistent? $${\sf ZFC} + \exists M \exists N: M,N \text { are transitive } \land M \neq N \land M \prec N \land N \prec M$$, and if consistent, then what's its consistency level?

Note: $M \prec N$ refers to existence of an elementary embedding (in the language of set theory) from $M$ to $N$.

Note: The same question has been posted as a revised question to an older one at MathStackExchange.

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    $\begingroup$ If you are going to cross-post a question from math.SE, at the very least you're expected to make it easy for people to know (on both ends!) where the other question is, so they can see what people have been saying so far. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 2, 2023 at 20:33
  • $\begingroup$ @AsafKaragila, OK, I'll do that! $\endgroup$
    – Zuhair Al-Johar
    Commented Jan 2, 2023 at 21:00

1 Answer 1

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There are examples of this where $M,N$ are also models of ZFC, in the following paper, which is joint with Monroe Eskew, Sy Friedman and Yair Hayut: https://arxiv.org/abs/2108.12355

Thus, you certainly get such a situation if $0^\sharp$ exists. (The examples constructed in the paper from $0^\sharp$ are with proper class models, but you can find some set sized restrictions of those examples.)

It's also easy to see that if there are $M,N$ proper class models of ZF/ZFC and elementary $j:M\to N$ and $k:N\to M$, then $0^\sharp$ exists.

However, your question asks about sets $M,N$, and only requires they be transitive (not be models of any particular theory). The set-sized version (even if one demanded also some theory) does not imply $0^\sharp$ exists, because if it holds in $V$ then it holds in $L$, with $M,N$ countable in $L$, by $\Sigma^1_2$-absoluteness. If $M,N$ are transitive sets modelling ZFC, and there are elementary $j:M\to N$ and $k:N\to M$, then letting $\alpha=\mathrm{Ord}\cap M=\mathrm{Ord}\cap N$, we get an elementary $j\circ k:M\to M$, and this restricts to give an elementary $\ell:L_\alpha\to L_\alpha$. Conversely, if there is an ordinal $\alpha$ and an elementary $\ell:L_\alpha\to L_\alpha$, then the construction of the paper cited above produces transitive $M,N$ and elementary $j:M\to N$ and $k:N\to M$, with $M\neq N$. As I recall, Dmytro Taranovsky had an answer on MO regarding the consistency strength of an elementary $\ell:L_\alpha\to L_\alpha$, but I can't find that at present. However, for your actual question, $M,N$ are arbitrary transitive sets, and there I don't see immediately whether one even gets an elementary $\ell:\beta\to\beta$ for some ordinal $\beta$ (for example, $\beta=$ the rank of $M$?)

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    $\begingroup$ Great. But this leaves open the question of whether you can get an example in ZFC (without assuming $0^\sharp$) if we don't require $M$ and $N$ to be models of ZFC. $\endgroup$
    – Alex Kruckman
    Commented Jan 2, 2023 at 21:44
  • $\begingroup$ @AlexKruckman, right. $\endgroup$
    – Farmer S
    Commented Jan 2, 2023 at 22:31
  • $\begingroup$ Re: your last sentence, aren't the ordinals in $M$ always definable in $M$ as "the hereditarily transitive sets"? As long as $M$ is transitive, $M$ will compute h.t.-ness correctly, unless I'm missing something. So I think that if there is an elementary embedding of $M$ into $N$ we get $M\cap Ord= M\cap Ord$ and moreover the elementary embedding restricts to a self-embedding of this ordinal. $\endgroup$
    – Noah Schweber
    Commented Jan 2, 2023 at 22:32
  • $\begingroup$ Yes, but what if e.g. $\mathrm{Ord}\cap M<\mathrm{rank}(M)$, and maybe the embeddings fix all ordinals in $\mathrm{Ord}\cap M$. Or even if $\mathrm{Ord}\cap M=\mathrm{rank}(M)$, but $M$ doesn't model enough to be able to define the function $x\mapsto \mathrm{rank}(x)$... I think the usual proof shows that if $\ell:M\to M$ is elementary and not the identity, then there has to be a set $x\in M$ such that $\mathrm{rank}(\ell(x))>\mathrm{rank}(x)$, but if $M$ doesn't understand what the ranks of sets are, maybe still $\ell(\mathrm{rank}(x))=\mathrm{rank}(x)$? $\endgroup$
    – Farmer S
    Commented Jan 2, 2023 at 22:37
  • $\begingroup$ @FarmerS Oh, I see - by "elementary" you also wanted $\mathscr{l}$ to be not the identity. $\endgroup$
    – Noah Schweber
    Commented Jan 2, 2023 at 23:29

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