Why is the Mitchell order well-founded?

Question

Here, on page 358, I do not understand why we reach a contradiction once we show $U_1>U_2>...>U_n>...$ is a descending sequence in $M$? What does it contradict that this order holds in $M$?

Answer 1

We start with the least $\kappa$ on which the Mitchell order is ill-founded. Now pick a Mitchell-descending sequence $U_0>U_1>...$ of measures on $\kappa$. Let $j:V\rightarrow M$ be the elementary embedding associated to $U_0$. Since $j$ is elementary, we have that $M\models$"$j(\kappa)$ is the least cardinal where the Mitchell sequence is ill-founded."

Since $\kappa<j(\kappa)$, this means that $M$ must not see the sequence $U_1>U_2>...$ - either because it doesn't contain $\{U_1, U_2, ...\}$ or because it doesn't believe $U_i>U_{i+1}$ for some $i$. So by showing that this sequence exists and is Mitchell-descending in $M$, we'll have a contradiction.

Since $U_i<U_0$ for $i>0$, we have $U_i\in M$ for all $i>0$. This means that the whole sequence is in $M$ (why?). So now *all we have to do is show $M\models U_i>U_{i+1}$ for $i>0$, and we'll be done. And this is what Jech does in the remainder of the proof.