First some definitions: For a set Σ of L-sentences, Mod(Σ) denotes the class of all models that satisfy Σ. For a class M of models, we say it is EC if M=Mod(σ) for some sentence σ and ECΔ if M=Mod(Σ).
Problem. Let T be a theory having arbitrary large finite models. (For example, T can be axioms for groups, or fields, or linear orderings.) Let Kinf={M : M⊨T, Card(|M|) is infinite.} and Kfin={M : M⊨T, Card(|M|) is finite}.
Question.
(a) Kinf is ECΔ
(b) Kfin is not ECΔ
(c) Kinf is not EC
Best Answer
I'll just give hints, the rest is up to you :
a) Consider the theory $$T_{inf} := T \cup \{\exists x_1 \exists x_2 \dots \exists x_n \bigwedge_{1\leqslant i < j \leqslant n} x_i \neq x_j \ \big| \ n \in \mathbb{N} \}$$ The models of $T_{inf}$ are precisely the infinite models of $T$.
b) assume for contradiction that Kfin is ECΔ. Now consider an ultraproduct, say $U$, of arbitrarily large finite models of $T$ : $U$ is an infinite model of $T$. Since an ECΔ class is closed under ultraproducts, $U$ is also in Kfin, a contradiction.
c) assume for contradiction that Kinf is EC, ie there is a sentence $\sigma$ such that Kinf $ = \operatorname{Mod}(\sigma)$. Then the models in Kfin are precisely the models of $T \cup \{\neg \sigma\}$ : hence Kfin is ECΔ, a contradiction to point b).