Suppose $\mathcal L$ is the empty language (i.e., no constant, function, or relation symbols), and $C$ is the class of $\mathcal L$-structures whose underlying set is of the form $\{1, ..., n\}$ for some integer $n \geq 1$. Show that $\text{Th}_a(C)$ is complete.
Here are some defintions: a sentence is a formula without any free variables (i.e. all variables are bound). A theory of $C$ is the set $\text{Th}(C)$ of $\mathcal L$-sentences $\sigma$ with $\textbf{A} \models \sigma$ for all $\mathcal A \in C$, and the asymptotic theory of $C$ is the set $\text{Th}_a(C)$ of $\mathcal L$-sentences $\sigma$ with $\mathcal A \models \sigma$ for all but finitely many $\mathcal A$ in $C$. A set of $\mathcal L$ sentences $T$ is called consistent if there doesn't exist an $\mathcal L$ sentence $\sigma$ such that $T \vdash \sigma$ and $T \vdash \lnot \sigma$.
I have proven that the models of $\text{Th}_a(C)$ are exactly the infinite models of $\text{Th}(C)$. Based on this, it suffices to show that every infinite model of $\text{Th}(C)$ is complete. Here is what I have so far: any sentence in the empty language $\mathcal L$ consists of $\forall, \exists, \lnot, \lor, \land$, and $v_i = v_j$ for some variables $v_i$ and $v_j$.
I am to prove that for every sentence $\tau$ of $\mathcal L$, exactly one of $\text{Th}_a(C) \vdash \tau$ and $\text{Th}_a(C) \vdash \lnot \tau$ holds true. I know that if neither holds true, then $\text{Th}_a(C) \not\vdash \tau$ implies $\text{Th}_a(C) \cup \{\tau\}$ is consistent; similarly, $\text{Th}_a(C) \not\vdash \lnot \tau$ implies $\text{Th}_a(C) \cup \{\lnot \tau\}$ is consistent.
We are also allowed to use the fact that consistent theories in empty languages have at most countable models.
I am not sure what to do. Can you help?
The first observation is that the structures of this language are extremely simple to categorize: there is exactly one (up to isomorphism) for every finite or infinite cardinality. The next thing is that the complete theories are also very simple to categorize: there is one for each finite sized model, and then all of the infinite structures have the same complete theory, regardless of cardinality (why?).
As you've seen, any model of $\operatorname{Th}_a(C)$ must be infinite, so it follows that if there are any they are all elementarily equivalent. The fact that it has has a model at all follows from compactness (easy to see any finite subtheory has a finite model). So it is just the complete theory of the infinite structures.