Finding many subsets of $V_{\lambda+2}$ stable under $j:V\prec V$

Question

Working over $\mathsf{ZF}$ with an embedding $j:V\prec V$ with a critical point $\kappa$. Take $\lambda=\sup_{n<\omega}j^n(\kappa)$. (You may assume $\mathsf{DC}_\lambda$ if you need, but I am not sure it is consistent with a Reinhardt cardinal.)

My question is: is there a family $\mathcal{A}\subseteq V_{\lambda+2}$ such that

• $j(\mathcal{A})=\mathcal{A}$, $j(x)=x$ for all $x\in\mathcal{A}$, and
• $\lvert\mathcal{A}\rvert\ge\lvert V_\lambda\rvert$?

Note that if we take $\mathcal{A}=\{x\subseteq V_{\lambda+1}\mid x=j(x)\}$, then $j(\mathcal{A})\neq \mathcal{A}$ since $\alpha\in\mathcal{A}$ for all $\alpha<\kappa$. Hence $\mathcal{A}$ has to be a proper subset of $\{x\subseteq V_{\lambda+1}\mid x=j(x)\}$.

If it has a negative answer, what is the least $\alpha$ which ensures the existence of a family $\mathcal{A}\subseteq V_\alpha$ with the mentioned conditions?

Answer 1

There can be no such set $\mathcal A$. In fact, no set $\mathcal A$ with $j(\mathcal A) = \mathcal A$ and $j(x) = x$ for all $x\in \mathcal A$ can surject onto $\kappa$, and this does not require that the codomain of $j$ is $V$. The reason is that if $f :\mathcal A\to \kappa$ were a surjection, then $j(f) : \mathcal A\to j(\kappa)$ would be a surjection; but for all $x\in \mathcal A$, $j(f)(x) = j(f)(j(x)) = j(f(x)) = f(x)$, so $j(f) = f$, and in particular $\kappa\notin \text{ran}(j)$, a contradiction.