$\mathcal{E}(\lambda)\neq\emptyset$ if and only if there is a nontrivial elementary embedding $j:V_\lambda\rightarrow V_\lambda$.
My question is why this can even be consistent. A function $j:A\rightarrow B$ for $A,B\subseteq V_\lambda$ is an elementary embedding in $V_\lambda$ if and only if $j$ is a function in $V_\lambda$ (which is true), and for every first-order $\varphi$:
- $V_\lambda\models\forall a_0\in A((A\models\varphi(a_0))\Leftrightarrow(\exists b_0\in B((a_0,b_0)\in j\land B\models\varphi(b_0)))$
- $V_\lambda\models\forall a_0,a_1\in A((A\models\varphi(a_0,a_1))\Leftrightarrow(\exists b_0,b_1\in B((a_0,b_0)\in j\land (a_1,b_1)\in j\land B\models\varphi(b_0,b_1)))$
- ...
However, it seems like this is true if and only if $j:A\rightarrow B$ is an actual elementary embedding.
First question: Why isn't the rank of every I3 cardinal a model of the existence of a Reinhardt cardinal?
Second question: Given a sufficiently large cardinal $\kappa$ (inaccessible, perhaps), when does $V_\kappa$ satisfy that $j$ is a nontrivial elementary embedding for some $j\subseteq V_\kappa$?
Replacement fails in $V_\lambda$, since $f(n)=j^n(\kappa)$ is definable and with domain $\omega$, but it isn't a set in $V_\lambda$.
So yes, $V_\lambda$ satisfies "some set theory + choice + there is a Reinhardt cardinal". But it doesn't satisfy enough set theory to prove Kunen's theorem. Specifically, there isn't enough Replacement going on in $V_\lambda$ for the proof to work.