Kunen showed that Reinhardt cardinals are inconsistent in ZFC. But his proof is a bit technical for a non-set-theorist to follow. Berkeley cardinals are stronger than Reinhardt cardinals. You can refute them in ZFC by observing that every Berkeley cardinal is Reinhardt, and then appealing to Kunen's theorem.
Question 1: Is there a more direct refutation of Berkeley cardinals in ZFC, perhaps one which might be more digestible by a non-set-theorist?
Recall that a definably-Berkeley cardinal is a cardinal $\kappa$ such that whenever $M$ is a transitive set with $\kappa \in M$, there are elementary embeddings $M \to M$ with arbitrarily large critical point $\alpha < \kappa$. A Reinhardt cardinal is when this holds for $M = V_\kappa$ (EDIT: In fact the "Reinhardt cardinal is defined to be the critical point of the embedding). So perhaps there is some other $M$ one can cook up which obviously fails the Berkeley property? Maybe $M = \kappa + 1$, for example?
(A Berkeley cardinal is a cardinal $\kappa$ such that whenver $M$ is a transitive set with $\kappa \in M$ and $A \subseteq M$, there are elementary embeddings $j : (M,A) \to (M,A)$ with arbitrarily large critical point below $\kappa$. Although it seems sometimes one restricts the definition to apply only when $M = V_\alpha$? I'm not sure if that's equivalent or weaker.)
Question 2: Let $\kappa$ be a cardinal (probably uncountable, regular, limit, measurable,etc.). Can it be the case in ZFC that there exists a nontrivial elementary embedding $\kappa + 1 \to \kappa + 1$?
Best Answer
Yes, it is easier to refute Berkeleys than Reinhardts. There is a very simple refutation of Berkeleys in ZFC that is due to Woodin. It is part of the motivation for his contention that Berkeley cardinals should be inconsistent with ZF.
Towards a contradiction, assume ZFC plus $\delta$ is the least Berkeley. For each $\alpha< \delta$, choose a transitive set $M_\alpha$ containing $\delta$ such that there is no $j:M_\alpha \to M_\alpha$ with critical point less than $\alpha$. Let $\gamma$ be the supremum of the ranks of the $M_\alpha$. Since $\delta$ is Berkeley there is an elementary $i : V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $i(\vec M) = \vec M$. Notice that $i$ has no fixed points $\alpha$ between its critical point and $\delta$. Otherwise $i(M_\alpha) = M_\alpha$, but then $i\restriction M_\alpha$ is an elementary embedding from $M_\alpha$ to $M_\alpha$ with critical point less than $\alpha$.
Therefore $\delta$ has countable cofinality: in fact, $\delta = \sup_{n<\omega} \kappa_n$ where $\kappa_n = i^n(\kappa)$. But applying Berkeliness again, we can get another embedding $k: V_{\gamma+1} \to V_{\gamma+1}$ with critical point below $\delta$ such that $k(\vec N) = \vec N$ where $N_n = M_{\kappa_n}$. Clearly $k$ fixes each $N_n$ (since $k(n) = n$). But taking $n$ large enough that $\kappa_n > \text{crit}(k)$, $k\restriction N_n$ is an elementary embedding that contradicts the definition of $M_{\kappa_n}$.