Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
Comment have been deleted; the full thread can be found here: tea.mathoverflow.net/discussion/1629/… –  François G. Dorais Dec 3 '13 at 3:22

1 Answer 1

up vote 9 down vote accepted

The answer is yes, one can always find models with as large a gap in their definable classes as desired.

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.

share|improve this answer
Those precious moments where model theory is used in set theory like that. –  Asaf Karagila Dec 1 '13 at 14:55

Your Answer


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.