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Background

I am interested in elementary embeddings from a model of set theory into itself. One way of producing such elementary embeddings is when the model is generated by indiscernibles; this idea is very closely related to the existence of sharps. Jech and Kanamori discuss $0^{\#}$ and $0^\dagger$ in detail but don't tell me much about other sharps. More advanced resources are difficult to understand without a lot of background knowledge.

Hypotheses

Let $\theta$ be an inaccessible cardinal, and suppose that some set $A$ of measurable cardinals below $\theta$ is a stationary subset of $\theta$. For each $\kappa \in A$, let $\mu_\kappa$ be a normal measure on $\kappa$, and let $\mathcal{U} = \{ \langle \kappa, \mu_\kappa \rangle : \kappa \in A \}$. Let $L[\mathcal{U}]_\theta$ denote those elements of $L[\mathcal{U}]$ of rank less than $\theta$.

Question statement

Do there exist large cardinal assumptions which imply the existence of a closed unbounded set of ordinal indiscernibles for $L[\mathcal{U}]_{\theta}$ such that every order-preserving map of these indiscernibles extends to an elementary embedding $j:L[\mathcal{U}]_{\theta} \to L[\mathcal{U}]_{\theta} \, \, ?$

Remarks

The large cardinal assumptions may be on $\theta$, the elements of $A$, or some other large cardinal. The values of $\theta$, $A$, and the $\mu_\kappa$ may be chosen in whatever way you like subject to the hypotheses above -- I just want this to work in some example, not in every example.

In The Core Model, Dodd mentions double mice, a generalization of $0^\dagger$. Maybe some version of these can be used to answer the question affirmatively, but I know nothing about them.

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It seems to me that a measurable cardinal above $\theta$ would give you indiscernibles of the sort you want, by the same argument that gives you $0^\#$ from one measurable cardinal. And my assumption of a measurable cardinal above $\theta$ is probably overkill; a Ramsey cardinal above $\theta$ should suffice, just as for $0^\#$. –  Andreas Blass Oct 9 '11 at 23:46
    
Thanks, Andreas. What is the intuition behind why this should work? Of course we can get indiscernibles from that assumption, but why should they be a club, and why should they generate the model? If I replaced $\mathcal{U}$ with an arbitrary subset of $V_\theta$, would it still work? –  Norman Lewis Perlmutter Oct 12 '11 at 15:20
    
My intuition here is only that I see no obstruction to transposing the usual proof for $0^\#$ into the new context. Note that you can't expect just any old class of indiscernibles to be a club, or any old Ehrenfeucht-Mostowski theory of indiscernibles to be satisfiable by a club. Even in the theory of $0^\#$, Silver needed to get the right E-M theory by taking a class of indiscernibles whose $\omega$-th element is as small as possible. –  Andreas Blass Oct 24 '11 at 21:29
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1 Answer 1

up vote 8 down vote accepted

My reading of this question was different from Andreas', because Norman asked for order preserving maps of the indiscernibles to extend to embeddings $j:L[U]_\theta \rightarrow$ $L[U] _\theta$

i.e. the indiscernibles should be below $\theta$ as the ordinal height of the structures mentioned is $\theta$?

In that case the measurable or Ramsey above $\theta$ only guarantees indiscernibles above $\theta$ and so "missing the target"?

In any case there are generalisations of "double mice" that you surmise that provide a positive answer. Let $M$ be the "least" in a certain canonical well-ordering of all such iterable structure that have a measurable cardinal $\kappa$ which is, in $M$, the limit of measurables cardinals below $\kappa$. (Such structures are called "mice".) By Loewenheim Skolem we may assume that $M$ is countable. `Iterability' here means we may form all iterated ultrapowers of the first structure $M=M_0$ using this top measure repeatedly; call the ultrapower structures $M_\tau$ for all ordinal $\tau$; then all the $M_\tau$ will be wellfounded. In the $\tau$'th model let $\kappa_\tau$ be the top-most measurable cardinal. Then $\tau ,\mu \rightarrow \kappa_\tau<\kappa_\mu$ and the class of all such $\kappa_\tau$ forms a closed unbounded class of indiscernibles for the model $W$ "left behind" by these iterates.

[Fornally $W = \bigcup _ \tau H(\kappa_\tau)^{M_\tau}$.]

Then $W$ is the `least' inner model with a proper class of measurables cardinals, (because as $\tau$ increases the order type of the measurables in the models $M_\tau$ increases; but these measures are left behind in the lower $H(\kappa_\tau)^{M_\tau}$ part of the model that goes into $W$). The $\kappa_\tau$ are indiscernibles for it, just as the Silver indiscernibles are for $L$.

The large cardinal assumption is then that "This iterable $M$ exists" just as a parallel to "$0$-sharp exists" is to $L$.

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You're right; I misread the question. –  Andreas Blass Feb 17 '12 at 0:27
    
Thanks for the answer. –  Norman Lewis Perlmutter Feb 17 '12 at 9:08
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