# $(\kappa , \lambda)$ - Minimal Models of $\text{ZF}$

The notion of minimality in model theory is related to the existence of a gap in the size of definable subsets of a model. Now consider the following generalization:

Definition 1: Let $M$ be a $\mathcal{L}$ - structure. Define:

$Def(M):=\{X\subseteq Dom(M)~|~\exists n \in \omega~~~\exists \varphi (x ,y_1 , ...,y_n)\in \mathcal{L}-Form~~~$

$\exists b_1 , ...,b_n\in Dom(M);~~~~~X=\{a\in Dom(M)~|~M\vDash \varphi(a, b_1,...,b_n)\}\}$

Definition 2: Let $\kappa >\lambda \geq \aleph_{0}$ be two cardinals. A model $M$ called $(\kappa , \lambda)$ - minimal if $\forall X\in Def(M);~~~~ |X|\geq \kappa~\vee~~|X|\leq \lambda$.

Assuming consistency of $\text{ZF}$:

Question 1: Does $\text{ZF}$ have a $(\kappa , \lambda)$ - minimal model for each $\kappa >\lambda \geq \aleph_{0}$?

If the answer of the above question is negative:

Question 2: Does $\text{ZF}$ have an extension like $T$ such that $T$ has a $(\kappa , \lambda)$ - minimal model for each $\kappa >\lambda \geq \aleph_{0}$?

Question 3: For which (probably large) cardinals like $\kappa >\lambda \geq \aleph_{0}$ does $\text{ZF}$ have a $(\kappa , \lambda)$ - minimal model?

Theorem. For every $\kappa\gt\lambda\geq\aleph_0$, and for any consistent theory with an infinite model in a countable language, there is a $(\kappa,\lambda)$-minimal model.
Proof. Let $M$ be any $\kappa$-saturated model of the theory. It follows that every infinite definable subset of $M$ has size at least $\kappa$, since by saturation we may always find an additional satisfying instance of the definition different than any fewer than $\kappa$ many known instances. Thus, $M$ is $(\kappa,\aleph_0)$-minimal. QED
Indeed, this argument shows that one may attain the sharper separation property that $|X|\geq\kappa$ or $|X|\lt\lambda$, since in a $\kappa$-saturated model every definable class has size at least $\kappa$ or is finite.