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Let $\Phi(\bar x)$ be a type over a set $X$ with respect to a structure $A$. Show that if $\Phi$ is algebraic, then $\Phi$ contains a formula $\phi$ s.t. $A\models\exists\ _{ for some $\ n<\omega$.

I've really hit a wall with this one; I can only deal with the case $\Phi(\bar x)$ is a complete type. Any help is appreciated!

-Thanks

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    @Saeed: You probably wanted to say that the type is algebraic if *every* tuple realising it is algebraic.2012-12-11

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A type $\Phi$ in a complete theory corresponds to a closed subset $[\Phi]$ of the space of types $S_n(X)$.

According to your definition, a type is algebraic if the closed subset defined by it is covered by a family of open subsets $[\varphi]$ corresponding to algebraic formulas $\varphi$.

But the Stone space is compact, and the type is a closed subset, so $[\Phi]$ is covered by finitely many of these, so there are algebraic $\varphi_1,\ldots,\varphi_n$ such that $[\Phi]\subseteq \bigcup_{j=1}^n [\varphi_j]$, but the latter is just equal to $[\bigvee_{j=1}^n \varphi_j]$, so $\Phi\vdash \bigvee_{j=1}^n \varphi_j$, and $\bigvee_{j=1}^n \varphi_j$ is of course algebraic.

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    @Saeed: I think the more natural description is that a tuple is algebraic if its type is algebraic; it is equivalent to what you said, but...2012-12-12