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I'm currently reading J. D. Hamkins' paper "Unfoldable cardinals and the GCH," and I've run across a comment that I think I ought to find trivial, but I don't. On page 1187, he says that $V_\theta^L\subseteq L_{\aleph_\theta}$ - why is this the case? In particular, why isn't it the case that $V_\theta^L=L_\theta$?

Thank you very much in advance.

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    @ Amy: you're quite right, I will do so in the future. @ Asaf: you're also quite right. I thought I'd defined $\theta$, but clearly I hadn't.2011-05-12

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Well, how funny that I happen to be right here as you ask the question, although I don't have my article with me.

But note that the elements of the powerset of $L_\kappa$ in $L$ appear at stages before (and unbounded in) $L_{\kappa^+}$, by the famous argument of Gödel showing the GCH in $L$. Thus, one application of powerset corresponds to the next larger cardinal in the constructibility hierarchy. And so one can prove the inclusion by induction on $\theta$.

Meanwhile, note that $V_\theta^L$ is usually not equal to $L_\theta$, since they have different cardinalities for numerous $\theta$. It is true, however, that $V_\theta^L=L_\theta$ whenever $\theta$ is a beth fixed point in $L$.

The point for the article is that every unfoldable cardinal in $L$ is strongly unfoldable there, since having the target model contain $V_\theta^L$ amounts to having it include a large $L_\alpha$, since $V_\theta^L\subset L_\alpha$ for some sufficiently large $\alpha$, and the above observation says exactly which $\alpha$ suffices. This observation is due originally to Andres Villaveces, the inventor of unfoldable cardinals, and shows that unfoldable cardinals and strongly unfoldable cardinals have the same consistency strength.

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    Thank you very much! This is perfect. I definitely should have been able to figure this one out.2011-05-12