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This may be a silly question, but I keep coming back to it.

Let $j:V\prec M$ be a non-trivial elementary embedding with $M$ a transitive class and $\kappa$ the critical point of $j$. Define the critical sequence for $j$ as usual, setting $\kappa_0=\kappa$ and $\kappa_n=j^n(\kappa)$ (i.e., the $n^{th}$ iterate of $j$ evaluated at $\kappa$). Finally, let $\lambda=\sup_{n<\omega}(\kappa_n)$. Then $j(\lambda)=\lambda$ and $\lambda$ is a strong limit of cofinality $\omega$.

The trouble I'm having is the claim that $j(\lambda^+)=\lambda^+$. The claim follows from the string of inequalities: $\lambda^+\leq j(\lambda^+)=(\lambda^+)^M\leq\lambda^+$ However, I don't understand why the final inequality is true. I feel as if I'm just overlooking something obvious.

For reference, the string of inequlaities is from proof 2 (due to Woodin) of Kunen's Inconsistency on page 320 in Kanamori's The Higher Infinite.

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I think you might just be thinking too hard about it. First of all, $\lambda^+ \in M$ and trivially $\lambda < \lambda^+$, and $M$ also knows this fact. Therefore $M$'s version of the cardinal successor of $\lambda$ cannot exceed $\lambda^+$. –  Arthur Fischer Jul 25 '12 at 21:06
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1 Answer

up vote 2 down vote accepted

We have indeed that $j(\lambda)=\lambda$. Since $j$ is elementary $j(\lambda^+)$ is the successor of $j(\lambda)=\lambda$ in $M$, so indeed $j(\lambda^+)=(\lambda^+)^M$.

However if if $M\subseteq V$ is an inner model then $(\alpha^+)^M\leq(\alpha^+)^V$ for trivial reasons.

Therefore we have the equality $\lambda^+=j(\lambda^+)$.

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I knew this question was silly. Thanks for grounding me back in reality. –  Everett Piper Jul 25 '12 at 21:19
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