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Let $T$ be a stable $L$-theory with elimination of imaginaries. We work in the monster model $\mathfrak C$ of $T$. Let $A$ be a small (infinite) set of the monster, $\phi(x,y)$ be a $L(A)$-formula and $a_i:i\in \omega$ be a Morley sequence over $A$ such that $\mathfrak C\models \phi(a_0,a_1)$.

Can we find a Morley sequence $b_i:i\in \omega$ over a finite set $A_0$ such that $\mathfrak C\models \phi(b_0,b_1)$?

Note:This is of course true for a superstable theory and if one replaces $\phi$ by a type then this fails in non-superstable theories.

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Your question comes down to asking whether given a formula $\phi(x,b)$ which does not fork over $A$, there is a finite subset of $A$, $A_0$, such that $\phi(x,b)$ does not fork over $A_0$.

The answer is yes. The reason depends on your preferred definition of forking, of course. Using the definability of types definition, we just have to take $a$ independent of $b$ over $A$ and take $A_0$ to be the parameters appearing in the $\phi$-definition of $\text{tp}^\phi(a/Ab)$.

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  • $\begingroup$ Thanks, in my notation take the $\phi$-type $tp^\phi(a_0/Aa_1)$. It is definable over some finite set $A_0$ hence non-forking over $A_0$. Now any non-forking extension of this $\phi$-type does the job, we choose a Morley sequence $/A_0$ in one of this nf-extension. $\endgroup$
    – TimZ
    Mar 6, 2014 at 11:33
  • $\begingroup$ Exactly. I just rephrased it because I wanted to emphasize that the nice property of superstable theories (nonforking always descends to a finite base) is true for stable theories if you only care about a single (or finitely many) formulas. $\endgroup$
    – Alex Kruckman
    Mar 6, 2014 at 17:11

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