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Given a theory $T$ and a formula $\phi(x)$ we say that they admit a $(\kappa, \lambda)$ model if there is a model $M$ such that $|M| = \kappa$ and $|\phi(M)| = \lambda$.

In all examples that I know of that do not admit $(\kappa, \lambda)$ models for some pairs (such as $(\mathbb R, \mathbb Q, <)$) there is a formula expressing that different elements realise different types over $\phi$ (such as $\forall x, y (x < y \to \exists z \in Q (x < z < y))$ for $(\mathbb R, \mathbb Q, <)$). Or one can iterate it finitely many times (such as in $(\cal P(P(\omega)), P(\omega), \omega, \in)$).

I am wondering if this is the only obstruction to admitting all pairs of cardinals. More concretely a question can be posed as follows. Assume that there is a model $M$ such that for every chain $\phi(x) = \psi_0(x), ..., \psi_n(x) = ``x=x\text{"}$ there is some $i < n$ with $\psi_{i+1}(M)$ containing infinitely many elements that realise the same type over $\psi_i(M)$. Does it follow that $T$ and $\phi$ admit a $(\kappa, \lambda)$ model for every $\kappa \ge\lambda$?

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    $\begingroup$ You might want to adjust the question to allow for adding finitely many constants at each step in the chain. Reason: Internality sometimes requires parameters. For example, consider the theory consisting of three disjoint copies of $\mathbb{Q}$, together with a ternary relation, where $R(a_1,a_2,a_3)$ holds iff each $a_i$ is in the $i$th copy of $\mathbb{Q}$ and $a_1+a_2 = a_3$. Fixing any element puts the other two copies of $\mathbb{Q}$ in bijection, so there's no $(\kappa,\lambda)$ model for any $\kappa\neq \lambda$, but ... $\endgroup$ Jan 4, 2017 at 21:08
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    $\begingroup$ without fixing any parameters outside of the first copy of $\mathbb{Q}$, all elements have the same type over the first copy of $\mathbb{Q}$. $\endgroup$ Jan 4, 2017 at 21:09
  • $\begingroup$ It seems that parameters can be essential not only for internality. E.g. in the example $(\mathbb R, \mathbb Q, <)$ one can replace $<$ with a sort $O$ of all linear orderings of $\mathbb R$ where $\mathbb Q$ is dense-codense (and an evaluation map). Then any element of $O$ can be used to express the fact that $\mathbb Q$ is dense in $\mathbb R$. However without $O$ every element of $\mathbb R$ realises the same type over $\mathbb Q$. $\endgroup$ Jan 22, 2017 at 2:46
  • $\begingroup$ I am not sure what is the best way to fix the question though. If say parameters realise a nonisolated type, then it can be omitted. However omitting a type in uncountable models is a more delicate business. So it may be possible to construct an example where parameters realise a nonisolated type which can't be omitted in larger models. $\endgroup$ Jan 22, 2017 at 2:55
  • $\begingroup$ I like your structure $(\mathbb{R}, \mathbb{Q}, O, x<_o y)$, it's a nice example! My suggestion for how to fix the question was to say "Assume there is a model $M$ such that for every chain... and every finite set $B$ of parameters, there is some $i<n$ with $\psi_{i+1}(M)$ containing infinitely many elements that realize the same type over $\psi_i(M)\cup B$." But I would guess that the answer is probably no in that case, too. 2-cardinal problems for general theories are notoriously subtle... $\endgroup$ Jan 22, 2017 at 4:01

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