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The relations $\in$ and $\subsetneq$ seem so similar in some sense. For example they are equal on ordinal numbers. So there is a natural question about their possible similar behaviors on the constructible universe or proper class of all sets for example in the case of Kunen inconsistency theorem.

Question (1): Is there a non-trivial elementary embedding $j:\langle V,\subsetneq\rangle \longrightarrow \langle V,\subsetneq\rangle$?

Question (2): Is there a non-trivial elementary embedding $j:\langle L,\subsetneq\rangle \longrightarrow \langle L,\subsetneq\rangle$?

Question (3): What is the consistency strength of existence of such embeddings relative to large cardinal axioms? Particularly what is the position of existence of a non-trivial elementary embedding from $\langle L,\subsetneq\rangle$ to itself relative to existence of $0^{\sharp}$?

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    $\begingroup$ Note that $\in$ and $\subsetneq$ are very different. The former is well-founded and the latter is not. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 7, 2013 at 21:30
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    $\begingroup$ Indeed Asaf, and the former need not be transitive while the latter always is. $\endgroup$
    – Rachid Atmai
    Commented Oct 7, 2013 at 22:44
  • $\begingroup$ @Asaf & Carlo: You are right. But $\in$ and $\subsetneq$ seem simultaneously similar and different! It just depends on the property which we want to compare them with each other. As I mentioned in the case of ordinal numbers they are in the strongest similarity which is equality! $\endgroup$
    – user36136
    Commented Oct 8, 2013 at 6:10
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    $\begingroup$ Ali, indeed there is a strong similarity in the case of ordinals, and while they do serve as a "spine" for the universe of $\sf ZFC$, they don't quite catch everything. It's the sets of ordinals which do; and for sets of ordinals $\subsetneq$ and $\in$ are two very different relations. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 8, 2013 at 6:18

1 Answer 1

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$ \newcommand\ofnoteq{\subsetneq}$

It is a very nice question!

The answer is that there are numerous definable automorphisms of $\langle V,\ofnoteq\rangle$. To see this, let $f:V\to V$ be any permutation of the universe, and define the induced function $\pi:V\to V$ by $\pi(x)=f[x]$, the image of $x$ under $f$. For example, $\pi$ maps the singleton $\{ a\}$ to the singleton $\{f(a)\}$. Using the fact that $f$ is a permutation, it is not difficult to see that $x\ofnoteq y\iff \pi(x)\ofnoteq\pi(y)$, and furthermore that $\pi$ is a bijection, and hence it is an automorphism of the universe with respect to $\ofnoteq$, and consequently an elementary embedding, which is nontrivial precisely when $f$ is.

So there is no large cardinal strength here to be found, and the lesson appears to be that $\ofnoteq$ does not capture much of the intended set-theoretic structure.

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  • $\begingroup$ Dear Joel thanks. And your answer is very very nice. $\endgroup$
    – user36136
    Commented Oct 8, 2013 at 4:35
  • $\begingroup$ And a natural project could be classifying all binary relations $E$ such that the structure $\langle V,E\rangle$ has no non-trivial elementary embedding as same as $\langle V,\in\rangle$. $\endgroup$
    – user36136
    Commented Oct 8, 2013 at 5:05
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    $\begingroup$ There goes a nice set theory final exam question. $\endgroup$
    – Andrej Bauer
    Commented Oct 8, 2013 at 7:32
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    $\begingroup$ Perhaps one might say, Welcome to the internet, Andrej! :-) But are you suggesting that we shouldn't answer interesting questions like this? I haven't seen this one before, and I'm glad to have learned about it. Do you have more such "final exam" questions? $\endgroup$
    – Joel David Hamkins
    Commented Oct 9, 2013 at 13:04
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    $\begingroup$ This answer eventually led to further work and my joint paper with Makoto Kikuchi on Set-theoretic Mereology. See jdh.hamkins.org/set-theoretic-mereology. $\endgroup$
    – Joel David Hamkins
    Commented Sep 3, 2016 at 2:40

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