# $(\kappa,\lambda)$ - Minimal Models & Stronger Version of Rowbottom's Theorem

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 $Dom(N)$?

• 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. – 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 $\text{Dom}(N)$.