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 $\Gamma$.
If we suppose $\mathcal{A} = \{\mathcal{B}_i:i\in I\}$ is a class of $EC_{\Delta}$ classes for some index set $I$, and $\mathcal{B}_i$ is defined by a set $\Gamma_i$, meaning that $\mathcal{B}_i$ is the set or class of models such that each model $\mathcal{B}_i\in \mathcal{B}_i$ is such that $\vDash_{\mathcal{B}} \varphi$ for each $\varphi \in \Gamma_i$. Suppose $\mathcal{D}$ is an EC class generated by some formula $\beta$. I am wondering how to show that $\bigcap \mathcal{A} \subseteq \mathcal{D}$ iff $\bigcup_i \Gamma_i \vDash \beta$
Here are my thoughts: I know that each class $\mathcal{B}_i$ must entail ($\vDash$) every formula in $\Gamma_i$ for some $i\in I$, which means that if the intersection of all of these $\mathcal{B}_i$'s is a subset of $\mathcal{D}$ contains models which are all in the EC class $\mathcal{D}$, then since the union all $\Gamma_i$'s is entailed by the intersection of each $\mathcal{B}_i$ (*still trying to reason why formally even though it seems intuitive), and we have that $\mathcal{D} \vDash \beta$, we must have $\bigcap \mathcal{A} \subseteq \mathcal{D}$ iff $\bigcup_i \Gamma_i \vDash \beta$ by composing the two parts of our last sentence.
Am I approaching this correctly? How do I make asterisk* step more formal. Thanks!
This comes down to the fact that $\bigcap \mathcal{A}$ is an EC$_\Delta$ class, axiomatized by $\bigcup_i \Gamma_i$. Indeed, ($M\in \bigcap\mathcal{A}$) if and only if ($M\in \mathcal{B}_i$ for all $i$) if and only if ($M\models \Gamma_i$ for all $i$) if and only if ($M\models \bigcup_i \Gamma_i$).
Now the definition of $\bigcup_i\Gamma_i \models \beta$ is that every model of $\bigcup_i\Gamma_i$ satisfies $\beta$. Since $\bigcap\mathcal{A}$ is the class of models of $\bigcup_i\Gamma_i$ and $\mathcal{D}$ is the class of models of $\beta$, this is exactly the statement $\bigcap\mathcal{A}\subseteq \mathcal{D}$.