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I have a few questions I have been thinking about that I could definitely use some insights on:

Question 1. Since a Woodin cardinal is a "local" notion, defined with respect to some rank-initial segment of the universe, is there a set object that witnesses that a cardinal is Woodin? In other words, can we prove $\kappa$ is Woodin if and only if there is an $S$ such that... ? (I am aware of the fact that local versions of strong cardinals have extenders which witness local strongness.)

Question 2. If there is such an $S$, does $L[S]$ or $L(S)$ satisfy that $\kappa$ is Woodin? In general, what would be the large cardinals of $L(S)$ or $L[S]$?

Thank you!

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  • $\begingroup$ (1) Yes. $\kappa$ is Woodin iff there is an $S$ such that $S$ is a $V_\alpha$ for $\alpha$ large enough that $\kappa<\alpha$ and $V_\alpha$ models $\mathsf{ZFC}$ with replacement restricted to $\Sigma_2$ formulas, and $V_\alpha$ models that $\kappa$ is Woodin. $\endgroup$ Mar 21 '14 at 2:59
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    $\begingroup$ You probably do not want that, so you may want to clarify. By the way, if $\kappa$ is Woodin, $L(V_\kappa)$ sees that $\kappa$ is Woodin, but it may fail to be a model of choice. However, one may add a well-ordering of $V_\kappa$ to this model by forcing (with initial segments), and this preserves Woodinness. This model does not see $V_\kappa^\sharp$ or any significant large cardinals above $\kappa$ (of course, it may have a class of indiscernibles). Again, probably you are searching for a different sort of $S$, and clarifying may help. $\endgroup$ Mar 21 '14 at 3:03
  • $\begingroup$ It might be worth adding to @Andres' comment that while choice may fail in $L(V_\kappa)$, this failure cannot happen in ranks below $\kappa$ (as $V$ and $L(V_\kappa)$ agree on everything below rank $\kappa$). $\endgroup$
    – Asaf Karagila
    Mar 21 '14 at 13:39
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Let's interpret your question as asking for the quantifier complexity of the assertion "$\kappa$ is a Woodin cardinal".

More specifically, one natural way to interpret your question, whether $\kappa$ is Woodin if and only if "there is $S$ such that..." is to interpret the ellipses as having only bounded quantifiers. This amounts to asking whether "$\kappa$ is Woodin" is a $\Sigma_1$ property of $\kappa$. For this, the answer is no, this level of complexity is much too simple, because $\Sigma_1$ properties are trivially preserved upwards to larger models, and we can destroy the Woodin-ness of $\kappa$ by forcing to collapse it, for example. So in this sense, the property of $\kappa$ being a Woodin cardinal cannot be witnessed by the existence of a single set $S$. (Of course, it is witnessed in one direction, as Andres explains in the comments, in that if $\kappa$ is Woodin, then it is Woodin in $L[S]$ for some set $S$, but this is not an equivalence unless one imposes requirements on $S$ that increase the complexity.)

Meanwhile, the property expressing that $\kappa$ is Woodin is $\Delta_2$. It is $\Sigma_2$, because as Andres points out in the comments, $\kappa$ is Woodin just in case there are extenders witnessing all the required $\lt\kappa$-strongness embeddings that arise in the definition of being Woodin, and so any $V_\alpha$ with $\alpha>\kappa$ can see that $\kappa$ is Woodin. Any such property is $\Sigma_2$. For the same reason, the property of not being Woodin is also visible inside any such $V_\alpha$, and so not being Woodin in also $\Sigma_2$. So the assertion "$\kappa$ is a Woodin cardinal" has complexity $\Delta_2$.

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