Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 144 / 2

The Fundamental theorem of stability says that, for a complete theory with infinite models, the following are equivalent:

Let $\lambda =|T|$ and choose an infinite cardinal $\kappa$ satisfying ${\kappa }^{\lambda }=\kappa$, for example $\kappa =2^{\lambda }$. Take $M\models T$ of cardinality $\kappa$. By assumption, every $p\in S_1(M)$ is definable.

For each parameter-free formula scheme $\phi (x;y)$, the definition $d_p\phi (y)$ is a formula with finitely many parameters from $M$. There are at most $\kappa$ such formulas. A definable type is completely determined by choosing one definition for each of the at most $\lambda$ formula schemes, soThe same count applies to every model of cardinality $\kappa$, and therefore $T$ is $\kappa$-stable. Hence $T$ is a stable theory.

Choose $b\in N^m$ and an $L$-formula $\phi (\bar{x};y)$ such thatLet $p(y)=\mathrm{tp}(b/M)$. Stability and the Fundamental theorem of stability make this a definable type. Apply its definition to the formula $\psi (y;\bar{x})=\phi (\bar{x};y)$. There is an $M$-formula $d_p\psi (\bar{x})$ such that, for every $\bar{a}\in M^n$,Thus $d_p\psi$ defines $X\cap M^n$ in $M$, which is the stable trace of a definable set property.