Definition 1: Let $M$ be a $\mathcal{L}$ - structure and $A\subseteq Dom(M)$. Define:

$Def_{A}(M):=\{X\subseteq Dom(M)~|~\exists n\in \omega~~\exists \varphi (x,y_1,...,y_n)\in \mathcal{L}-Form~~\exists b_1,...,b_n \in A;X=\{a\in Dom(M)~|~M\vDash \varphi (a,b_1,...,b_n)\}\}$

Definition 2: Let $\kappa >\lambda\geq \aleph_{0}$ be two cardinals and $M$ a $\mathcal{L}$ - structure and $A\subseteq Dom(M)$. We say the model $M$ is $(\kappa , \lambda)$ - minimal over $A$ if $~~\forall X\in Def_{A}(M)~~~~~|X|\geq \kappa~\vee~|X|\leq \lambda$.

The following result is proved by Rowbottom:

Theorem: If $\kappa$ is a measurable cardinal and $\mu$ a normal measure on it then:

Every $\mathcal{L}$ - structure $M$ with the following properties:

(a) $\kappa\subseteq Dom(M)$

(b) $|\mathcal{L}|+\aleph_{0}<\kappa$

has an elementary substructure like $N$ with following properties:

(c) $\mu(Dom(N)\cap \kappa)=1$

(d) $N$ is a $(\kappa ,|\mathcal{L}|+\aleph_{0})$ - minimal model over $\emptyset$.

Question: Is there any large cardinal $\kappa$ (larger than measurables) such that Rowbottom's theorem be true by extending the parameter set $\emptyset$ in the statement (d) to $Dom(N)$?

  • 8
    $\begingroup$ As a courtesy, instead of outright deleting your comment, I removed the excessive question marks and exclamation points. Next time though, such a comment will be deleted. The noise level is becoming excessive. $\endgroup$ – Todd Trimble Dec 1 '13 at 18:45

The answer is no, because there are structures $M$ having subsets that are definable from parameters of arbitrary size up to $\kappa$, and this will include sizes in the forbidden region of the gap between $|\mathcal{L}+\aleph_0|$ and $\kappa$.

For example, let $M=\langle\kappa,\lt\rangle$ be the usual order on the cardinal $\kappa$. If $N\prec M$ and has $\text{Dom}(N)\in\mu$, then $N$ must be unbounded in $\kappa$. But in this case, for any cardinal $\delta\leq\kappa$, there will be an initial segment of $N$ of size $\delta$ that is definable from parameters. Since $|\mathcal{L}+\aleph_0|\lt\kappa$, there will be such cardinals $\delta$ in the forbidden region, and so there can be no such $N$ that is $(\kappa,|\mathcal{L}+\aleph_0|)$-minimal over $\text{Dom}(N)$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.