The Mitchell order, unclear step in the proof of well-foundedness

Question

Here in the first answer, I do not understand the step that there exist $\beta_\eta$ with $j(\beta_\eta)=\alpha_\eta$ for $\eta<\theta$ for some fixed $\theta<\kappa$ and for all $a_\eta\in M$.

I think that this is not true in general, but I don't know why we can be happy with this in that particular proof that $M^\kappa\subset M$ . This last condition is what I would like to prove.

Answer 1

It's a requirement that Noah poses on the sequence $(\alpha_\eta \mid \eta < \theta)$ - he doesn't claim that this is true for all sequences of length $< \kappa$.

Here is the relevant quote - the emphasize is mine.

Suppose $a_\eta\in M$ with $a_\eta=j(b_\eta)$, for $\eta<\theta$ for some fixed $\theta<\kappa$. Then we have $j(\langle b_\eta\rangle_{\eta<\theta})=\langle a_\eta\rangle_{\eta<\theta}$: [..]

As I've pointed out in an earlier comment, this is in fact not always true. Consider for example the sequence $(\alpha_0) = (\kappa)$ of length $1$. Since $\kappa = \mathrm{crit}(j)$, there is no $\beta_0$ such that $j(\beta_0) = \alpha_0$. Hence we cannot apply Noah's argument to this sequence.