Let $\mathcal T$ be a system of theories $T$ of first-order logic language $\mathcal L$, each $T$ in $\mathcal T$ containts finite number of formulas and $\mathcal T$ itself contains a finite amount of theories $T$. I need to understand if there exists theory $S$ of first-order logic language $\mathcal L$, such as interpretation $\mathcal M$ is model of theory $S$ if and only if it is a model of at least one of theories $T$ in $\mathcal T$.
By definition of model we can write: $$\mathcal M \vDash S \iff \exists T \in \mathcal T: \mathcal M \vDash T$$ and expand it: $$\forall \phi \in S:\mathcal M \vDash \phi \iff \exists T \in \mathcal T: \forall \psi \in T: \mathcal M \vDash \psi$$
How do I proceed? Intuitively I think that I need to take $S$ as union of all formulas in $\mathcal T$, but I can't understand how to show that $(\forall \phi \in S:\mathcal M \vDash \phi )\land (\forall T \in \mathcal T: \exists \phi \in T: \mathcal M \not\vDash \phi )$ is a contradiction. I'm supposed to use completeness theorem somehow, but I can't see how it applies here.
Since each theory is finite, we can show that they can be axiomatized by a single axiom consisting of the conjuction of all the axioms of the theory.
Now, we can axiomatize $S$ as the disjuction of all the single axiom $A$ axiomatizations of the theories in the system.
By definition of model, a model of $S$ has to satisfy $A$, and thus it has to satisfy at least one of its disjuncts. But if it satisfies a disjunct then it satisfies the whole theory that that disjunct axiomatizes.
The other direction is similarly evident.