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Let $M$ denote a well-founded set-sized model of ZFC. The descent value of $M$ will be defined as the value of $n$ returned by the following process.

Initialization. Let $n$ equal $0$ and $X$ equal $M$.

Step. If $L_{\omega_1^X}^X$ doesn't satisfy ZFC according to $M$, halt and output $n$. Otherwise, increment $n$, let $X$ equal $L_{\omega_1^X}^X$, and repeat.

Question. Assuming sufficiently powerful large cardinal axioms, is it true that for every natural $n\geq 0,$ there exists a well-founded model $M$ of ZFC whose descent value is $n$?

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up vote 6 down vote accepted

No, the descent value is always at most $1$ in any model of ZFC, whether it is well-founded or not. To see this, observe that if the descent value isn't $0$, then on the next step you have $X=L_{\omega_1^M}^M$, and so $X$ satisfies $V=L$, and so $\omega_1^X$ is the $\omega_1^L$ inside $M$, and $L_{\omega_1^L}$ never satisifes ZFC, as it thinks that every ordinal is countable.

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Joel, it is a good thing you're here to save me my flights of imagination, or I would probably believe some very odd things. ;-) –  goblin Jun 10 '14 at 12:20
I think your idea will have a chance to succeed if you adjoin a predicate to $L$ as you descend, rather than going all the way down to $L$. –  Joel David Hamkins Jun 10 '14 at 12:24

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