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Suppose $\alpha$ is admissible and $\beta<\alpha$. We know that $L_\alpha$ is an admissible set (by definition), but adding subsets of $\beta$ to $L_\alpha$ might break admissibility: while set forcing preserves admissibility, we might have e.g. $\alpha$ countable, $\beta=\omega$, and $X\subseteq\omega$ codes a well-ordering of length $>\alpha$ - then adding $X$ to $L_\alpha$ breaks admissibility.

My question is about when we can be certain this won't happen. Specifically:

Suppose $\alpha$ is admissible, $\beta<\alpha$. Under what conditions is there an admissible set $A$ containing $V_\beta$ with height $\alpha$?

(NOTE: I assume all admissible sets are transitive. I think that's standard, but I don't have Barwise' book on hand, so I just want to clarify.)

This is a really broad question, since there are of course three parameters kicking around:

  • What's $\alpha$?

  • What's $\beta$?

  • What's $V$? In particular, exactly how far is $V$ from $L$?

I'm especially interested in the following more specific twist on this question:

Suppose $\kappa$ is a cardinal in $V$ and $\alpha$ is the least admissible $>\kappa$. What sort of large cardinal properties can $\kappa$ satisfy and still have an admissible set containing $V_\kappa$ of height $\alpha$?

For instance, can $\kappa$ be measurable?


EDIT: A related question (which I meant to include originally): for any $\kappa$ whatsoever, of course there is an $\alpha$ which is the least height of an admissible set containing $V_\kappa$; and this $\alpha$ is admissible. Turning this around yields a potentially interesting function: say that the robustness of an admissible ordinal $\alpha$, $rob(\alpha)$, is the supremum of the $\beta<\alpha$ such that there is an admissible set of height $\alpha$ containing $V_\beta$. Then it's reasonable to ask what we can say about this function. That is:

How does $rob$ depend on the large cardinal structure of $V$?

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  • $\begingroup$ If $0^\#$ exists, for any ordinal $\kappa \geq \omega$, the least $\kappa$-admissible is strictly smaller than the least $(\kappa,0^\#)$-admissible. Do you want to look at large cardinals compatible with $L$? $\endgroup$ Apr 6, 2016 at 7:41

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