I have to prove the following:
If a theory with equality $K$ has arbitrarily large finite normal models, then it has a denumerable normal model.
The setting is first-order logic. My attempt:
Assume $K$ has arbitrary large finite normal models $M_1, M_2,...$ with finite domains $D_1, D_2,..$, resepectively. Define $$D^* = \bigcup_{i=1}^\infty D_i.$$ Let the interpretations $M^*$ of $K$ have domain $D^*$. $M^*$ is again a normal model but it is infinite. The extension of the Skolem-Löwenheim Theorem says that any theory with equality that has an infinite normal model has a denumerable normal model. Hence, $K$ has a denumerable normal model.
I have the feeling that what I am doing is not really correct: I do not know that $K$ has infinitely many finite normal models. Also, how would I define $M^*$ such that it is again a normal model. Can someone help me with this proof?
EDIT
(Normal model in my book means the relation corresponding to the predicate = is an identitiy relation in the domain of the model).
Let $K$ be a theory with equality in the language $\mathcal{L}$. We add denumerably many new constant symbols $c_n$ for $n\in \mathbb{N}$ to $\mathcal{L}$. We also add denumerably many axioms $c_n \neq c_m$ for $n \neq m$. We call this new theory $K^*$. A normal model of $K^*$ is also a normal model of $K$, since each axiom of $K$ is also in $K^*$. Let $\Gamma$ be an arbitrary finite subset of axioms of $K^*$. We can write $\Gamma$ as the union of finitely many old axioms from $K$ and finitely many new axioms in $K^*$. Since $K$ has arbitrarily large finite normal models, we know there is an interpretation $M$ such that the finitely many old axioms are true. But how do I know the finitely many new axioms are true in $M$?
I do understand that if we have a normal model for $\Gamma$, by the compactness theorem there is a normal model $M^*$, which is denumerable, for $K^*$ and hence for $K$.