Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi all,

In most set-theory accounts dealing with $ 0 ^ \sharp $ and its related effects on the constructible sets, a proof of the following theorem is crucial:

(*) Assuming there is a measurable cardinals, for uncountable limit cardinals $ \alpha , \beta $ $ L_\alpha $ is elementary equivalent to $ L_\beta $.

Few accounts are Jech, Kanamori, Devlin and Silver's papers.

The proof goes through a heavy abstract machinery of the Silver Indiscernibles in model theory, relying on a Ramsey cardinal to find a "remarkable well founded E.M set". My question - does anyone know of a shorter proof for (*)? something along the line of Godel condensation theorem for L?

share|improve this question

2 Answers 2

up vote 5 down vote accepted

A short proof of $(*)$ uses the characterization of $0^\sharp$ in terms of mice. A nice account of this version is in Ernest Schimmerling's paper "The ABC of mice".

The idea is that a sharp is a kind of "local measurable cardinal". More formally, we can think of a set equivalent to $0^\sharp$ (in a strong sense) to be of the form $M=(L_{(\kappa^+)^L},\in,{\mathcal U})$ for some $L$-cardinal $\kappa$ and some $L$-measure ${\mathcal U}$ on $\kappa$. We ask some definability requirements of $M$ that imply that $M$ is countable (in $V$), and we require that we can iterate ultrapowers by ${\mathcal U}$ and obtain well-founded models.

(I say "local measurable" because as soon as we continue the constructible hierarchy over $M$, we are able to define a map from $\omega$ onto $L_{(\kappa^+)^L}$, so not only is $M$ countable, but ${\mathcal U}$ is very far from being a measure in $L[M]$. On the other hand, ${\mathcal U}$ measures all the subsets of $\kappa$ in $L$, so ultrapowers using ${\mathcal U}$ of $L$ or initial segments of $L$ make sense.)

It follows at once from continuity properties of embeddings that if $\alpha<\beta$ are uncountable cardinals in $V$, then iterating the ultrapower embedding by ${\mathcal U}$ will send $L_\kappa$ eventually to $L_\alpha$ and later to $L_\beta$, and so in particular $L_\alpha\equiv L_\beta$.

What we are doing here is casting Kunen's characterization of $0^\sharp$ as the key defining property. All of this, and the way to recover the usual version of $0^\sharp$ from this one, is explained in very nice detail in Ernest paper, that I recommend you study.

The presentation itself is part of the folklore of the subject. I learned it from John Steel. It is essential to understand this presentation if one is to make sense of current fine structure or inner model theory.

That being said, the EM-blueprints approach is not outdated. It is essentially the only way we can make sense of sharps of sets that are not 'internally well-ordered'. This idea was first developed by Robert Solovay when he introduced ${\mathbb R}^\sharp$ in the course of his work on determinacy. It has been much refined by Hugh Woodin during his study of the Chang model.

share|improve this answer
    
thank you so much for the answer and the reference. –  Eran Jan 22 '11 at 16:56
    
Re the reference - can you kindly suggest the best one for core model construction (up to a measurable cardinal)? most references I know require deep fine structure pre-requisites. The best one I think is either the original Jensen's and Dodd's paper, or Dodd's book, but I was wondering if you can recommend something else. –  Eran May 26 '11 at 20:44

I think you are just asking for a direct proof from a measurable cardinal. In this case, here is one way to do it. If there is a measurable cardinal $\kappa$ with normal measure $\mu$, then by considering the collapse of a countable elementary substructure $X\prec V_\theta$ one finds a countable transitive set $M_0$ with a measurable cardinal $\kappa_0$ and normal measure $\mu_0$. Since the iterated ultrapower of $M_0$ by $\mu_0$ maps directly into the iteration of $V$ by $\mu$, we see that all such iterations are well-founded. In other words, $\langle M_0,\in,\mu_0\rangle$ is iterable. Let $j_\alpha:M_0\to M_\alpha$ be the $\alpha$-th iteration, where at successor steps we take the ultrapower by the image of $\mu_0$, and at limit stages we take the direct limit. Since the internal ultrapowers do not increase cardinality and the maps are continuous at limits, it follows that the corresponding images $\{j_\alpha(\kappa_0) \mid \alpha\in ORD\}$ is a closed unbounded class of ordinals, and contains all uncountable cardinals of $V$, since the process is continuous at limits.

Now, just restrict to $L$ and observe that $M_{\kappa_\alpha}\prec M_{\kappa_\beta}$ whenever $\alpha\lt\beta$, because the iteration between $\alpha$ and $\beta$ has critical point $\kappa_\alpha$. In particular, $L_{\kappa_\alpha}\prec L_{\kappa_\beta}$, and this includes all the uncountable cardinals, so your conclusion is obtained. One can weaken measurability just to having a countable iterable structure and use the same argument.

(But I see now that Andres has just posted an answer with the same idea.)

share|improve this answer
1  
Hi Joel. Let me add that the argument in your answer is usually presented as part of the Gaifman-Rowbottom theorem of 1965: If there is a measurable, then the set of reals in L is countable. Although this can be proved somewhat more directly, The presentation in terms of embeddings ("linear iterations") is key. Actually, I now remember I once wrote a short note on this: caicedoteaching.files.wordpress.com/2011/01/118-handout6.pdf –  Andres Caicedo Jan 21 '11 at 23:40
    
Thanks ! –  Joel David Hamkins Jan 21 '11 at 23:44

Your Answer

 
discard

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.