An omega-stable countable theory is totally transcendental. The resulting definability and finite-base theorem for types says that every type over a parameter set $A$ is based on a finite tuple from $A$ and is determined by a countable choice of formulas over that tuple. For infinite $|A|=\kappa$, there are only $\kappa$ finite tuples from $A$ and only countably many formulas, so
$$
|S_n(A)|\leq\kappa\cdot\aleph_0=\kappa.
$$
Thus $T$ is $\kappa$-stable for every infinite cardinal $\kappa$. This is <omega-stability implies stability in every infinite cardinal>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 144 / 7 / Solution

For a countable theory, omega-stability implies $|S_n(A)|\leq|A|+\aleph_0$ for every parameter set $A$. Total transcendence gives each type a finite base and a definition over it; counting finite bases and formulas proves the bound.


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 Incoming links: Omega-stability implies stability in every infinite cardinal