# Using the compactness theorem to disprove axiomatizability

Another model-theoretic exercise from Smirnov's book.

Problem: Construct infinite family of varieties such that their union is not axiomatizable.

My solution: Denote by $\mathcal{A}_n$ the variety of all groups satisfying the sentence $\varphi_n : (\forall x) x^n = e.$ Let $\mathcal{A} = \bigcup_{i = 1}^{\infty}\mathcal{A_i}.$ In order to prove that $\mathcal{A}$ is not axiomatizable class I am trying to prove that it is not closed under ultraproducts.

There is a statement in the book:

Assume a class $\mathcal{K}$ is closed under ultraproducts and satisfies infinite disjunction $\bigvee_{i = 1}^{\infty} \varphi_n$, then it satisfies some finite disjunction $\varphi_{i_1} \vee \dots \vee \varphi_{i_n}.$

$\mathcal{A}$ satisfies $\bigvee_{i = 1}^{\infty} \varphi_n$, but for every finite disjunction $\varphi_{i_1} \vee \dots \vee \varphi_{i_n}$ we can choose $\mathbb{Z}_p$, where $p$ is relatively prime to each $i_k$, $k = \overline{1, n}$, which does not satisfy this disjunction. So $\mathcal{A}$ satisfies no finite disjunction and by the proposition above it is not closed under ultraproducts and hence is not axiomatizable.

Question: Is this proof correct? I am new to model theory, but I am convinced that there is more elegant proof which uses the compactness theorem directly. Can you explain me on this example how to use the compactness theorem for disproving axiomatizability? Thank you!

• I'm not sure why you are getting downvoted. This seems to be a fine question to me. Aug 2, 2014 at 20:33

Your proof is correct, given the statement from Smirnov. Actually, you don't even need to take $p$ relatively prime to the $i_k$, taking $p$ greater than all the $i_k$ will do!
Also, you're right that you can use the compactness theorem directly. For contradiction, assume the theory $T$ (in the language of groups) axiomatizes the class $\mathcal{A}$. Now consider the theory $T' = T \cup \{\lnot \phi_n\mid n\geq 1\}$. Let $\Delta$ be a finite subset of this theory. Since $\Delta$ contains only finitely many $\phi_n$, we can pick $m$ such that $m > n$ for all $n$ such that $\phi_n \in \Delta$. Then $\mathbb{Z}_m \models T$ (since $\mathbb{Z}_m\in\mathcal{A}$), and $\mathbb{Z}_m\models \lnot\phi_n$ whenever $n<m$ (since $1$ has order greater than $n$), so $\mathbb{Z}_m\models \Delta$, and $T'$ is consistent by compactness.
But if $M\models T'$, then $M\models T$, so $M\in \mathcal{A}$, but $M\notin \mathcal{A}_i$ for any $i$, since $M\models \lnot \phi_i$, contradiction.