MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Thanks a lot.

share|cite|improve this question

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 non-fine 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 Jensen-Dodd, 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, non-fine 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 non-fine 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).

share|cite|improve this answer
Dear Andres, By L[U] I mean a k-model (k measurable) as is defined in Kanamori's section 20. I do refer to the "the non-fine structural version". The fine-structure version is indeed available in few places (best is Devlin) but I would like to know if there is a simpler one, without going into the morass of the J_alphas. Something along the lines of the L-version? – Eran Oct 8 '10 at 18:38
Hi Eran. Then definitely the argument in the middle of the proof of Lemma 20.2 (with the reference to 3.3.(b)) is the fastest route. – Andrés E. Caicedo Oct 8 '10 at 18:43
OK, so the proof relies on existence of this sentence sigma_1 from 3.3(b) (then by elementarity every submodel of L[U] belongs to the hierarchy). Now - where can I find this sentence (Without using as a parameter L[U]) ? does this follow from the well ordering of L[U]? – Eran Oct 8 '10 at 19:27
3.3.(b) is not difficult, actually. (And anyway, Kanamori refers to specific sections in Devlin's "Constructibility" and Moschovakis's "Descriptive Set Theory", and I'm pretty sure it can also be found somewhere in Jech's.) In any case, a more subtle issue is that we cannot get for this version of the construction of L that the small L[U] like model you get by condensing is actually an initial segment of L[U]. I'm running late to catch a plane, but I hope I'll remember (once I'm over the jet lag) to write a remark on this issue. – Andrés E. Caicedo Oct 9 '10 at 2:16
In Jech Set Theory, Theorem 19.3 (page 340 6'th line from the top) he claims that each premice is an initial segment for $<_{L[D]}$ – Eran Oct 9 '10 at 7:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.