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David Marker's Model Theory:
Theorem 4.2.11 Suppose that $T$ is a complete theory in a countable language and $A \subseteq M \models T$ is countable. If $|S_n^M(A)| < 2^{\aleph_{0}}$, then $|S_n^M(A)| < {\aleph_{0}}$.
Proof. Suppose that $|S_n^M(A)| > {\aleph_{0}}$. We claim that $|S_n^M(A)| = 2^{\aleph_{0}}$ . Because $|S_n^M(A)| > {\aleph_{0}}$ and there are only countably many $L_A$-formulas, there is a formula $\phi$ such that $|[\phi]| > {\aleph_{0}}$.
Claim If $|[\phi]| > {\aleph_{0}}$, there is an $L_A$-formula $\psi$ such that $|[\phi \wedge \psi]| > {\aleph_{0}}$ and $|[\phi \wedge \neg\psi]| > {\aleph_{0}}$. Suppose not. Let $p = \big\{\psi(\bar{v}):|[\phi \wedge \psi]| > {\aleph_{0}}\big\} $. Clearly, for each $\psi$ either $\psi \in p$ or $\neg \psi \in p$ but not both. We claim that $p$ is satisfiable. Suppose that $\psi_1,...,\psi_m \in p$. Either $\psi_1\wedge...\wedge\psi_m \in p$, in which case $\big\{\psi_1,...,\psi_m \big\} ∪ Th_A(M)$ is satisfiable, or $\neg\psi_1 ∨ . . . ∨ \neg\psi_m \in p$. Because $[\neg\psi_1 ∨ . . . ∨ \neg\psi_m]=[\neg\psi_1]∪...∪[\neg\psi_m]$, we must have $|[\neg\psi_i]| > {\aleph_{0}}$ for some $1\leq i\leq m$, a contradiction. Thus $p \in S_n^M(A)$.

My question is about this part of the proof which says if $\psi_1\wedge...\wedge\psi_m \in p$ then $\big\{\psi_1,...,\psi_m \big\} ∪ Th_A(M)$ is satisfiable!
I can't see why that's true. Would be thankful for your help.

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  • $\begingroup$ Does "$[\chi]$" mean "the number of models of $T\cup\{\chi\}$"? If not, what does it mean? $\endgroup$ Dec 14, 2016 at 16:37
  • $\begingroup$ @NoahSchweber It means the set of all types containing $\chi$. It's Marker's notation for the basic clopens in the Stone space. $\endgroup$ Dec 14, 2016 at 17:25

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If $\psi_1\land \ldots\land \psi_m\in p$, then by definition $|[\psi_1\land \ldots\land \psi_m]|>\aleph_0$. In particular, this set is nonempty. So there's some type $q\in S^M_n(A)$ containing $\psi_1\land\ldots\land \psi_m$. Since types are consistent, $\{\psi_1,\dots,\psi_m\}\cup \mathrm{Th}_A(M)$ is satisfiable.

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  • $\begingroup$ Thank you Alex, it was simply beautiful! $\endgroup$
    – Aref
    Dec 14, 2016 at 18:00

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