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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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Noah, now you can't say the author of what you're reading doesn't care. ;) –  J. M. May 12 '11 at 2:11
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I found the article @ arxiv.org/PS_cache/math/pdf/9909/9909029v1.pdf. Is this the version you've got? (It helps, when referring to a publication (and available on-line) to include a link to it.) –  amWhy May 12 '11 at 2:13
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@Amy: A quibble: it would be better in general to link to the abstract page rather than to the PS_Cache, since the links to the abstract page are guaranteed to be persistent while the cache files aren't (moreover, not everyone prefers pdf). –  t.b. May 12 '11 at 2:20
    
@Theo: Sorry about that. Thanks for the information; I'll be sure to follow your advice in the future. –  amWhy May 12 '11 at 2:23
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Even though the question was answered, $\theta$ can be a very general term, and it is usually rude to just send us looking for the paper. Please try and include the definition of $\theta$ the next time you use it. –  Asaf Karagila May 12 '11 at 5:22
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up vote 7 down vote accepted

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. –  Noah S May 12 '11 at 2:59
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