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I'm working on the following problem:

Assume that every model of a sentence $\varphi$ satisfies a sentence from $\Sigma$. Show that there is a finite $\Delta \subseteq \Sigma$ such that every model of $\varphi$ satisfies a sentence in $\Delta$.

The quantifiers in this problem are throwing me off; besides some kind of compactness application I'm not sure where to go with it (hence the very poor title). Any hint?

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You can produce from that a very neat proof of why $ZFC$ is not finitely axiomatizable by setting $\Sigma$ to be: $\varphi_0=2^{\aleph_0}>\aleph_\omega; \varphi_n=2^{\aleph_0}=\aleph_n$ for $n>0$. Since we can produce a model in which each of these happen, there is no single sentence in which we can write $ZFC$. –  Asaf Karagila Aug 4 '11 at 23:35

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up vote 13 down vote accepted

Cute, in a twisted sort of way. You are right, the quantifier structure is the main hurdle to solving the problem.

We can assume that $\varphi$ has a model, else the result is trivially true.

Suppose that there is no finite $\Delta\subseteq \Sigma$ with the desired property.

Then for every finite $\Delta \subseteq \Sigma$, the set $\{\varphi, \Delta'\}$ has a model. (For any set $\Gamma$ of sentences, $\Gamma'$ will denote the set of negations of sentences in $\Gamma$.)

By the Compactness Theorem, we conclude that $\{\varphi, \Sigma'\}$ has a model $M$.

This model $M$ is a model of $\varphi$ in which no sentence in $\Sigma$ is true, contradicting the fact that every model of $\varphi$ satisfies a sentence from $\Sigma$.

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