I am wondering how to show that the intersection of a countable sequence of elementary classes is not an elementary class.
First I establish the following terminology: Using the definition of a delta elementary class ($EC_\Delta$) to be the set of structures which entail every formula in some set $\Gamma$. We define an elementary class ($EC$) to be such a class generated by a finite $\Gamma$.
Now consider the following setup: Suppose we have a countable sequence of $EC$ classes $\mathcal{A}_1, \mathcal{A}_2, \dots$ for each $i\in \mathbb{N}$ such that each $\mathcal{A}_{i+1} \subset \mathcal{A}_i$. That is to say that each $\mathcal{A}_{i+1}$ contains strictly fewer models than $\mathcal{A}_i$.
Now I'm trying to show that $\bigcap_{i\in \mathbb{N}} \mathcal{A}_i$ is not $EC$.
Here's the suggestion that I have been given: Use the result that for some class of $EC$ classes $\mathcal{K}$ and some $EC$ class $\mathcal{B}$ such that $\bigcap \mathcal{K} \subseteq \mathcal{B}$ there is some finite subcollection $\mathcal{K}'$ of $\mathcal{K}$ such that $\bigcap \mathcal{K}' \subseteq \mathcal{B}$
The second suggestion is to use the fact that $\bigcap_{i\in \mathbb{N}}\mathcal{A}_{i}\subseteq\bigcap_{i\in \mathbb{N}}\mathcal{A}_{i}.$
Here are my thoughts: We can assume for contradiction that $\bigcap_{i\in \mathbb{N}} \mathcal{A}_i$ is $EC$ and use it as our set $\mathcal{B}$ in the hint. Then we have $\bigcap_{i\in \mathbb{N}}\mathcal{K}' \subseteq\bigcap_{i\in \mathbb{N}}\mathcal{K}'$ for a finite class $\bigcap \mathcal{K}'$.
Am I on the right track? Any hints?
Thanks in advance!
The hint seems overly complicated to me.
First of all, note that by taking a finite conjunction if necessary every EC is in fact axiomatized by a single sentence.
Now pick sentences $\varphi_i$ such that $\mathcal{A}_i=\{M: M\models\varphi_i\}$. Suppose $\bigcap\mathcal{A}_i$ is elementary, and let $\psi$ be the corresponding axiom: $$\bigcap\mathcal{A}_i=\{M: M\models\psi\}.$$ This immediately gives us a contradiction with the Compactness Theorem: