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Suppose V is a model of Godel-Berney's set theory with the axiom of choice. A well-known result of Kunen says that there can be no elementary embedding $V$ to itself. This result further implies that there can be embedding from $V$ to $V$ which is $\Sigma_1$-elementary.

Is it consistent (with say Godel-Berney's set theory with the axiom of choice) that there is a $\Delta_1$-elementary embedding from $V$ to $V$? How about a $\Delta_0$-elementary embedding?

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The answer is yes, the existence of such embeddings is equiconsistent over ZFC with a measurable cardinal.

If $\kappa$ is measurable, then there is a fully elementary embedding $j:V\to M$ into a transitive class $M$ with critical point $\kappa$. By composing this map with the inclusion $M\subset V$, we may view $j:V\to V$. The observation to make is that this remains $\Delta_0$-elementary, since every transitive class is $\Delta_0$-elementary in the universe $M\prec_{\Sigma_0}V$, and so we may view $j:V\to V$ as a composition of the original elementary embedding with this $\Delta_0$-elementary inclusion.

In fact, the embedding $j:V\to V$ is $\Delta_1$-elementary, since $M$ is actually $\Delta_1$-elementary in $V$, as $\Sigma_1$ assertions go up and $\Pi_1$ assertions go down. Since $M$ and $V$ are elementarily equivalent, they agree on the equivalence or non-equivalence of any two formulas.

(Note that there is a subtle issue about what exactly one means by $\Delta_1$-elementary, since there is no such thing as a $\Delta_1$-formula, unlike $\Delta_0$. For example, in the argument, it was important that $M$ and $V$ agreed on the equivalence of the $\Sigma_1$ nad $\Pi_1$ formulas in question. It can happen, however, that two models of ZFC do not necessarily agree like this, and there is a difference between $\Delta_1$ and provably-$\Delta_1$.)

Conversely, let us show that measurable cardinals are required. If $j:V\to V$ is a nontrivial $\Delta_0$-elementary embedding, then I claim, first, that there is a critical point, that is, a first ordinal that is moved. Notice that $j$ must take ordinals to ordinals. Assume toward contradiction that $j$ fixes all ordinals. Let $a$ be an $\in$-minimal set with $j(a)\neq a$. So $a\subset j(a)$ since $j(x)=x$ for all $x\in a$. By the axiom of choice, we may well-order $a$ with a relation $\lhd$ in some order-type $\gamma=|a|$. It follows that $j(\lhd)$ well-orders $j(a)$ in order-type $j(\gamma)=\gamma$. Furthermore, the $\beta^{th}$ element of $a$ gets mapped to the $j(\beta)=\beta^{th}$ element of $j(a)$ with respect to $j(\lhd)$. So $j$ is onto $j(a)$, and therefore $j(a)=j"a=a$, a contradiction. So we have shown that there must be a critical point $\kappa$, the least ordinal with $\kappa<j(\kappa)$. Secondly, now, let $\mu$ be the set of $X\subseteq\kappa$ with $\kappa\in j(X)$. It follows now that $\mu$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$. This is easily seen to be a nonprincipal filter, and it is $\kappa$-complete, since if $X_\alpha\in \mu$ for all $\alpha<\gamma$, some $\gamma<\kappa$, then $j(\langle X_\alpha\mid\alpha<\gamma\rangle=\langle j(X_\alpha)\mid\alpha<\gamma\rangle$, and $\kappa$ is in every element of the right hand side and hence in the intersection of the right-hand side. So $\cap_\alpha X_\alpha\in\mu$.

Lastly, let me point out that if one weakens much more, below $\Delta_0$, then one gets the embeddings in ZFC. This idea grew out of my work in:

Hamkins, Joel David, Every countable model of set theory embeds into its own constructible universe, J. Math. Log. 13, No. 2, Article ID 1350006, 27 p. (2013). DOI, ZBL1326.03046, blog.

In that article, I consider the $\in$-embeddings, which are embeddings $j$ for which merely $$x\in y\iff j(x)\in j(y).$$ This is a weakening of $\Delta_0$-elementarity. These are precisely the self-embeddings of the structure $\langle V,\in\rangle$ in the model-theorists sense of the term for a relational structure.

Theorem. ZFC proves that the map defined by the following recursion is an $\in$-embedding $j:V\to V$. $$j(x)=\{j(y)\mid y\in x\}\cup\{\{\emptyset,x\}\}$$

See theorem 19 and lemma 18 in the paper. Basically, it is clear that $y\in x\to j(y)\in j(x)$ by definition. And conversely, $j$ is injective, since we can recover $x$ from the last part of $j(x)$, and if $j(y)\in j(x)$, then it must have come from $y\in x$, since $j(y)\neq\{\emptyset,x\}$ as $j(z)\neq\emptyset$ and $\emptyset\notin j(z)$ for any $z$.

Let me also mention my question, Can there be an embedding $j:V\to L$?, which asks whether one can have in GBC an $\in$-embedding from $V$ to $L$, in cases where $V\neq L$. The full question remains open, although we now know some things about partial answers, for work-in-progress.

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  • $\begingroup$ What about fully $\Sigma_1$-elementary embeddings or even $\Delta_2$-elementary embeddings? $\endgroup$ – Keith Millar Oct 19 '18 at 20:59

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