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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. $A\models\exists\ _{<n}\bar x\phi(\bar x)$ 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!


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What's your definition of algebraic type? This is the definition I'm familiar with... – Zhen Lin Dec 10 '12 at 8:11
I second Zhen's request for your definition of an algebraic type. I have seen precisely the property you want to prove used for that. – Levon Haykazyan Dec 10 '12 at 10:36
Clarification: A type Φ(x) over a structure A is algebraic if any tuple realising it is algebraic, but this includes any tuple in any elementary extension B of A. – Saeed Dec 10 '12 at 23:33
And what is the definition of an algebraic tuple? I would say a tuple is algebraic if it realizes an algebraic type! – Alex Kruckman Dec 11 '12 at 9:37
@Saeed: You probably wanted to say that the type is algebraic if every tuple realising it is algebraic. – tomasz Dec 11 '12 at 21:17

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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@Alex Kruckman: By a tuple being algebraic it is meant that every element of the tuple is algebraic – Saeed Dec 12 '12 at 20:30
Yes, by "any" I meant "every" – Saeed Dec 12 '12 at 20:31
@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... – tomasz Dec 12 '12 at 21:05

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