I have a question about proposition 6.4. of I0 and rank-into-rank axioms by Dimonte 2017. The question is about the direct limit of a directed system of rank-into-rank elementary embeddings defined as follows:
Definition 6.3. Let $j : M_0 \prec M_1$. Then we say that $(M_\alpha , j^\alpha)$ is the $\alpha$-th iterate if:
- $j^0=j$;
- $j^{\alpha+1} = j^\alpha (j^\alpha)$;
- $M_{\alpha+1} = j^\alpha (M_\alpha)$ is well-founded;
- $j_{\gamma, \alpha+1} = j^\alpha \circ j^{\alpha+1} \circ ... \circ j^\gamma$. So $j_{\alpha, \alpha+1} = j^\alpha$
- if $\alpha$ is a limit, then $(M_\alpha, j^\alpha, j_{n, \alpha})$ is the direct limit of the system $(M_\beta, j_{\beta, \gamma})$ with $\beta, \gamma \lt \alpha$
where $M_0$, and $M_n$ for every finite $n$, is $V_\lambda$, where $\lambda$ is the supremum of the critical sequence of $j$.
The proof of proposition 6.4. begins as follows:
Suppose $j : V_\lambda \prec V_\lambda$ is $\omega$-iterable, let $\langle \kappa_n : n \in \omega \rangle$ be its critical sequence. Then $M_\omega$ is the set of equivalence classes of $(n, a)$ such that $a \in V_\lambda$, where if $n \lt m$ then $(n, a)$ is equivalent to $(m, b)$ iff $j_{n,m} (a) =b$. Also, $j_{n, \omega} (a)$ is the class of $[(n, a)]$.
We prove by induction that $j_{n, \omega} (\alpha) = \alpha$ for $\alpha \in \kappa_n$. Note that if $\beta \lt \kappa_n$, then for every $m \gt n$, $j_{n,m} (\beta) = \beta$, therefore $[(n, \beta)] = [(m, \beta)]$. So for any $m \in \omega$, $\beta \lt \lambda$, $[(m, \beta)] \lt [(n, \alpha)]$ iff $[(n, \beta)] \lt [(n, \alpha)]$: If $m \le n$, then $j_{m,n} (\beta) \lt \alpha \lt \kappa_n = j_{m,n} (\kappa_m)$ therefore by elementarity $\beta \lt \kappa_m$ and $[(m, \beta)] = [(n, \beta)]$. If $m \gt n$, then $\beta \lt j_{n,m} (\alpha) = \alpha$, therefore also $[(n, \beta)] = [(n, \beta)]$. So $j_{n, \omega} (\alpha) = [(n, a)] = \{[(n, \beta)] : \beta \lt \alpha \}$, that is by induction $\alpha$
In the following paragraph I have corrected two assumed typos, please tell me if the original makes sense.
We want to calculate now $j_{0, \omega} (\kappa_0)$. If $\alpha \lt j_{0, \omega} (\kappa_0)$, then there exist $n \in \omega$ such that $\alpha = j_{n, \omega} (\beta)$ for some $\beta \lt \lambda$. But then $j_{n, \omega} (\beta) \lt j_{0, \omega} (\kappa_0) = j_{n, \omega} (j_{0, n} (\kappa_0)) = j_{n, \omega} (\kappa_n)$, therefore $\alpha = j_{n, \omega} (\beta) = \beta \lt \lambda$. On the other hand, if $\alpha \lt \lambda$ then there exists $n \in \omega$ such that $\alpha \lt \kappa_n$, so $j_{n, \omega} (\alpha) = \alpha \lt j_{n, \omega} (\kappa_n) = j_{0, \omega} (\kappa_0)$. Therefore $j_{0, \omega} (\kappa_0) = \lambda$.
Why doesn't this proof go through without the assumption that $M_\omega$ is well-founded? If it did, then it would prove that rank-into-rank embeddings are inconsistent, for proposition 6.4 implies that the least rank-into-rank embedding $j$ has ill-founded $j_{0, \omega} (\kappa_0)$