Does anyone know where I can find a proof for the Condensation Lemma for the L[U] hierarchy in Set Theory?
Thanks a lot.
Does anyone know where I can find a proof for the Condensation Lemma for the L[U] hierarchy in Set Theory? Thanks a lot. 


Eran, there is some ambiguity in your question, as there are several possible interpretations of what is meant by "the $L[U]$ hierarchy". The fine details of the argument depend on the version you choose. For the nonfine structural version, Kanamori's "The higher infinite" has a very general argument, see the proof of Lemma 20.2. This argument refers to Theorem 3.3.(b), which is not proved there, but references are provided, to Devlin's "Constructibility" and Moschovakis "Descriptive Set Theory". For a fine structural version, using the old fine structure of JensenDodd, see "The Core Model", by A.J. Dodd. For a fine structural version, using the current (Mitchell style) fine structure, see "Fine structure and iteration trees" by Mitchell and Steel, and Steel's article in the Handbook of Set theory. [Edit: Added Oct. 10, 2010.] Let me add a remark about the relevance of distinguishing between fine structural or 'coarse' approaches. In the usual, nonfine structural setting, we use a predicate $U$ to build $L[U]$ and $U$ is, in $L[U]$, a normal, fine measure on some cardinal $\kappa$. (There seems to be an issue with LaTeX, so let me on occasion write $s(A,\gamma)$ for $A_\gamma$.) Note that $s(L[U],\kappa)=L_\kappa$. (Because, as one can easily verify by induction, $U\cap s(L[U],\gamma)=\emptyset$ for all $\gamma\le\kappa$. Note that this does not happen when we form $L[A]$, for $A$ a set of ordinals. But $U$ is not a set of ordinals, but rather a set of sets of ordinals.) However, as soon as we see enough of the measure, we are actually able to define new subsets not just of $\kappa$ but even of $\omega$ (for example, $0^\sharp$). Consider a countable $X\prec s(L[U],\lambda)$, where $\lambda$ is some sufficiently nice ordinal to ensure that $L[U]_\lambda$ is a model of a sufficiently decent fragment $T$ of ZFC. (That $\lambda$ exists is a consequence of the reflection theorem.) Then (by the nonfine structural version of condensation) the transitive collapse of $X$ is $\bar X=s(L[D],\gamma)$ for some countable $\gamma$ and $D$ a set that, in $\bar X$, seems to be a normal measure on some cardinal $\tau$. In particular, there is a real $x\in\bar X$ such that $\bar X\models x=0^\sharp$ (because we can assume $T$ strong enough that the existence of $0^\sharp$ is provable in $T$ from the existence of measurable cardinals). Since the collapse map is the identity on reals, it follows that actually $0^\sharp=x$. This means that $0^\sharp$ is ("quickly") definable from $D$. But then $D$ cannot be in $s(L[U],\kappa)$ or else $0^\sharp$ would also be there, which contradicts that $L[U]_\kappa$ is just an initial segment of $L$. This means that, not only is $\bar X$ not an initial segment of the constructibility hierarchy of $L[U]$, but it is not even a subset of a very large initial segment of $L[U]$. Hence, if we want a strong version of condensation to hold, where the structures $\bar X$ not only ``have the right shape'' but are also initial segments, then we necessarily must use a different hierarchy, meaning we cannot simply form $L[U]$ by constructing from $U$. (Note that, as classes, $L[U]=L[A]$ for many sets $A\in L[U]$.) This suggests (almost forces on us) the approach that fine structure takes, of considering a more elaborate predicate than just $U$ but rather one of the form $(U_\alpha\mid\alpha<\tau)$ where each $U_\alpha$ is a measure (such as $U$) or a "small" measure (such as a sharp): In this sequence we would add $0^\sharp$, something like the set $D$ in the example above, and many others. The result is that in a sense it takes us longer to build the stage of the construction where we finally add $U$, since we will be adding more and more sets along the way. But the payoff is that we get back a strong version of condensation. This also has additional advantages, of course, although one needs to understand a bit of fine structure to appreciate them. For example, it is a popular question in the Qual exams in Set Theory at UC Berkeley, to ask for a proof of diamond in $L[U]$. Once one understands that $L[U]$ can be reorganized in the way hinted at above, one can then prove diamond rather easily, essentially by the same argument as in $L$ (using the now available strong version of condensation). 

