Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a set of ordinals $A$ we say that $A^\#$ exists if there exists a closed and unbounded class of indiscernibles, $I\subseteq\operatorname{Ord}$, for $L[A]$. Formally, if such class exists we define $A^\#=\{\varphi\mid L_{i_\omega}[A]\models\varphi(i_1,\ldots,i_n)\}$, where $\varphi$ is a formula in the language $\{\in,A\}$. If we consider the Godel encoding of $\varphi$ we can think of $A^\#$ as a subset of $\omega$ or as a real number.

Suppose $V$ is a model of $ZFC$, $A$ a set of ordinals and $M$ is a transitive inner model such that $A\in M$ and $M\models A^\#\text{ exists}$.

Does that imply $V\models A^\#\text{ exists}$? What about the other direction (i.e. $A^\#$ exists in $V$, and $A\in M$)?

My intuition says the answer is yes in both cases, but my intuition was completely obliterated and needs to be rebuilt.

share|improve this question
1  
Asaf: $A^\sharp$ is more complicated than your description, which only captures $0^\sharp$. You want $L_{i_\omega}[A]$ rather than $L_{i_\omega}$, for example, and you want to specify your language so that you have access to $A$ and its elements. –  Andres Caicedo Jul 27 '11 at 18:55
    
@Andres: Yes. You are correct. I did not specify that I meant $L_{i_\omega}[A]$ and that $\varphi$ is in the language of $\{\in, A\}$. I will fix that right away. –  Asaf Karagila Jul 27 '11 at 19:06
    
I think by the other direction you mean that if $A^\sharp\in M$ then $M$ thinks that $A^\sharp$ exists, but perhaps you mean that if $A\in M$ and $A^\sharp$ exists in $V$ then it does in $M$. These questions have different answers. I tried to address both below. –  Andres Caicedo Jul 27 '11 at 19:23
add comment

1 Answer

up vote 6 down vote accepted

One can use the class of indiscernibles to generate elementary embeddings of $L[A]$ to itself with critical point above the supremum of $A$, in much the same way as the indiscernibles of $L$ give us embeddings of $L$ into itself.

By results of Kunen (for $L$, but they generalize straightforwardly), the existence of an embedding $j:L[A]\to L[A]$ that has critical point above the supremum of $A$ implies the existence of $A^\sharp$, so both statements are equivalent.

So, suppose $M$ is an inner model and $M$ thinks that $A^\sharp$ exists. Then $M$ sees an elementary embedding of $L[A]$ into itself. This is clearly seen in $V$ as well. So $A^\sharp$ exists in $V$. The converse does not hold. After all, $L[A]$ sees $A$ but never $A^\sharp$. (However, if $M$ is an inner model that thinks that $A^\sharp$ exists, then $A^\sharp$ exists and coincides with $M$'s idea of it. This can be seen descriptive set theoretically (see below), or using the Kunen characterization. Also, if $A^\sharp$ belongs to $M$, then $M$ thinks that $A^\sharp$ exists, because from the sharp one can easily produce an $L[A]$-normal measure on the indiscernible $i_0$, and one can use this to recover the elementary embeddings.)

There is a better way of thinking about sharps, though, namely in terms of mice. This is probably folklore by now, but a nice reference is Schimmerling's "The ABC's of mice", Bulletin of Symbolic Logic 7 (2001) 485-503.

A sharp is then a model of the form $M=(L_\alpha[A],\in,A,U)$ where $U$ is an (external) measure on some $L_\alpha[A]$ cardinal $\kappa$. We add certain requirements that depend on your fine structural taste (say, $\alpha=\kappa^{++}$ in the sense of $L_\alpha[A]$), $M$ has the same size as $A$ and is sound and iterable. Soundness is a technical condition which essentially ensures $M$ is as small as possible.

Iterability is more complicated but it means that (as you imagine) iterating the process of forming ultrapowers by $U$ and its images only produces well-founded models. All the conditions describing the mouse $A^\sharp$ except for iterability are clearly absolute between inner models and $V$.

Iterability is as well, but this requires an argument. But then absoluteness gives us a positive answer to your question. (This means that if an inner model thinks that the mouse $A^\sharp$ exists, then it does in $V$, and if $A^\sharp$ exists and belongs to an inner model, then the inner model knows that it is $A^\sharp$.)

Consider first the case where $A$ is a subset of $\omega$. Then if one iterates $M$ some countable ordinal number of times, the whole iteration is codable by reals, and we can see that the statement that a real $x$ coding a model is iterable is $\Pi^1_2(x)$ (you basically have to say: every real coding an iteration of $x$ gives a well-founded model). The point is that if some iterate of $x$ is ill-founded, then some countable iterate is ill-founded, so you only need to describe iterations "accessible" by reals.

Finally, if $A$ is not a real, we can pretend that it is by working on a collapse of a sufficiently large initial segment of the universe to verify absoluteness. Or we can form the tree of attempts to build $A^\sharp$ and check that if $M$ thinks that $A^\sharp$ exists, then the tree is ill-founded, so $A^\sharp$ exists in $V$, and if $A^\sharp$ is in $M$, hen $M$ has a witness to the ill-foundedness of the tree. You can see a more detailed sketch of that idea in my paper with Schindler, "Projective well-orderings of the reals", Archive for Mathematical Logic 45 (7) (2006), 783-793, available at my page.

(Of course, one can also go as Jech does and verify the absoluteness of sharps in terms of blueprints, but I prefer the two other approaches I sketched.)

share|improve this answer
    
Many thanks, Andres. I do have two questions regarding the answer. First is what does "is probably folklore" mean? I think I have seen that twice today, in the same context. The other is the first sentence of the seventh paragraph is unclear, it is as though it refers to a previous sentence, but there is none. Lastly, I have been skimming through inner model introductory texts for this assignment (this question is a byproduct of a byproduct of a question, and you did not reveal any actual part of the solution, so no worries there), I have made plans with a friend to study inner models soon :-) –  Asaf Karagila Jul 27 '11 at 20:17
    
Hi Asaf. "Folklore" typically refers to the body of knowledge in the field that is mostly oral tradition and whose origins seem lost in the fog of time; it is hard to find precise references in the literature, but people know of it. –  Andres Caicedo Jul 27 '11 at 20:23
    
Oh, seventh paragraph: Iterability of a mouse $m$ is (absolute) as well (between $V$ and inner models $M$ with $m\in M$). One word of caution is that for larger mice the appropriate versions of iterability are no longer absolute. –  Andres Caicedo Jul 27 '11 at 20:25
    
Nice to hear about inner model theory. It takes some work to go through the preliminaries, but I think it is worth the effort. Let me know by email if you need suggestions regarding references. –  Andres Caicedo Jul 27 '11 at 20:27
    
The folklore thing sounds like the proof of transitivity for the Mitchell order. It seems that no one ever published one :-) as for the suggestions about reading references I would be delighted to have some. I'll let you know when I'm planning to start, and ask for your opinion. Thanks! –  Asaf Karagila Jul 27 '11 at 20:32
add comment

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.