1
$\begingroup$

Pillay's Introduction to Stability Theory has in Chapter 1 the following exercise: if $T$ is stable, and $M \prec N$ are models, then for all $N$-definable $X \subset N$, the set $X \cap M$ is an $M$-definable subset of $M$. He also says that this is just another way of saying that all types are definable.

(The above is not a verbatim statement of the exercise; I might replace it with a verbatim copy later.)

I have been trying to do the exercise by following his hint, but I do not understand the relationship between definable types and definable subsets. The following are my thoughts: If $\phi(x, \bar n)$ is an $L(N)$-formula defining $X$ in $N$, for each $x$ in $X$ we have $\bar m_x$ in $M$ such that $\models \phi(x, \bar m_x)$ by elementarity. Since $\bar m_x$ are, at least prima facie, infinitely many parameters, so we have an infinite set of $L(M)$ formulas, i.e., an incomplete $M$-type. However, this seems useless. The right interpretation of the set of formulas is an infinite disjunction, where as a type usually is an infinite conjunction, so to speak. In addition, even though a type is definable, the defining schema $d$ does not have to be "definable" (in an informal sense), I do not know how $d$ helps us remove those infinite conjunctions or disjunction.

What is the relationship between definable types and definable sets that is useful here?

$\endgroup$
3
$\begingroup$

Can you use that in a stable theory types over models are definable? If the answer is yes, here is how you can do it:

Suppose $X\subseteq N^n$ is definable by the formula $\phi(\overline{x},\overline{b})$ for some tuple $\overline{b}$ from $N$, and consider the type $p=tp(\overline{b}/M)$ and consider the formula $\phi^*(\overline{y};\overline{x})=\phi(\overline{x};\overline{y})$.

The fact that $p$ is definable over $M$ means that there is a formula $\psi(\overline{x})=d_p\phi^*(\overline{x})$ with parameters in $M$ such that, for every $\overline{a}\in M^{n}$, $\phi^*(\overline{y},\overline{a})\in p$ if and only if $M\models \psi(\overline{a})$.

Then, for $\overline{a}\in M^{n}$ we have

\begin{align*} M\models \psi(\overline{a}) &\Leftrightarrow \phi^*(\overline{y};\overline{a})\in p\\ &\Leftrightarrow \phi(\overline{a},\overline{y})\in tp(\overline{b}/M)\\ &\Leftrightarrow N\models \phi(\overline{a},\overline{b})\\ &\Leftrightarrow \overline{a}\in X \end{align*}

This shows that $X\cap M$ is defined precisely by the formula $\psi(\overline{x})$, with parameters from $M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.