# Is the inclusion version of Kunen inconsistency theorem true?

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 $0^{\sharp}$?

• Note that $\in$ and $\subsetneq$ are very different. The former is well-founded and the latter is not. – Asaf Karagila Oct 7 '13 at 21:30
• Indeed Asaf, and the former need not be transitive while the latter always is. – 16278263789 Oct 7 '13 at 22:44
• @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! – user36136 Oct 8 '13 at 6:10
• 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. – Asaf Karagila Oct 8 '13 at 6:18

$\newcommand\ofnoteq{\subsetneq}$

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.

• Dear Joel thanks. And your answer is very very nice. – user36136 Oct 8 '13 at 4:35
• 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 $\langle V,\in\rangle$. – user36136 Oct 8 '13 at 5:05
• Perhaps one can find some "$\in$-like" binary relation $E$ in this direction which is a good candidate to replace with $\in$ for rebuilding the foundation of mathematics in the language of set theory but by a different axiomatization in the sense of $E$. Such relation will be interesting if $X~E~Y$ be well defined for proper classes $X$ and $Y$ as same as sets. This fact can give us an opportunity to work with sets and proper classes uniformly. And go further in the tree of large cardinals (which have a new definition using $E$). – user36136 Oct 8 '13 at 5:36
• There goes a nice set theory final exam question. – Andrej Bauer Oct 8 '13 at 7:32
• 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? – Joel David Hamkins Oct 9 '13 at 13:04