# Exchange property of infinite minimal structures

Here's an exercise from the really nice Hinman's book "Foundamentals of Mathematical Logic". The minimal structures are those whose definable sets are finite or cofinite. The algebraic closure (acl) of $$X$$ is a set with elements from some finite set that is definable with parameters from $$X$$. I've stuck with the last sentence from the hint. I've proved the union contains at most $$nm$$ elements but can't see how to get a contradiction from it. Will be greatful for all your tips.

P.S. $$\exists^{=k}xF$$ means there are exactly $$k$$ values of $$x$$ s.t. $$F$$ is true. Since $$\mathfrak{A}$$ is infinite, because $$\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$$ is finite its complement in $$\mathfrak{A}$$ must be infinite. Let $$b'\in\mathfrak{A}\setminus\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$$ be arbitrary; then $$b'$$ satisfies $$\psi(a'_i,y,c)$$ for each $$i\leqslant m$$. Recall that $$\psi(a'_i,y,c)\equiv\phi(a'_i,y,c)\wedge\exists^{=m}x\phi(x,y,c).$$ In particular, $$\phi(a'_i,b',c)$$ holds for each $$i\in\{0,\dots,m\}$$, and so $$\phi_{b',c}^\mathfrak{A}$$ has size at least $$m+1$$. On the other hand, $$\exists^{=m}x\phi(x,b',c)$$ holds as well, so this gives the desired contradiction.
• Thanks a lot! The answer is really nice. Now I see why we need the infinity of $\mathfrak{A}$ here Dec 19, 2020 at 17:58
• @ElmarGuseinov (as a small comment, this result does still hold in finite structures, for the simple reason that every element is algebraic in a finite structure! and in particular every element is algebraic over $\emptyset$, just witnessed by the formula $v=v$.) Dec 19, 2020 at 18:13
• @AtticusStonestorm yes, then the implication holds trivially. Another fact is that $b\in acl_{\mathfrak{A}}(X\cup \{a\})$ but $b\notin acl_{\mathfrak{A}} (X)$. This shows the symmetry of the expression :-) Dec 19, 2020 at 18:31