In Los-Vaught test as stated in Enderton's, it's said that the theory should not have any finite models as a condition. As I was reading the proof, I didn't figure out where exactly this condition is used. Here is his proof with different notation:
It suffices to show that for any two models $M_1 , M_2$ of T we have $M_1\equiv M_2$. Since $M_1,M_2$ are infinite , there exists structures $N_1,N_2$ having cardinalitiy $k$ s.t. $M_2\cong N_2$, $M_2\cong N_2$(by Löwnhein-Skolem theorem). So we have $M_1\equiv N_1 \cong N_2 \equiv M_2$. Thus, $M_1 \equiv M_2$.
Where is the condition that there is not finite models used implicitly in the proof? Could anyone point this out?
Let $L$ be the language with a single binary predicate symbol $\lt$. Let $\varphi$ be the conjunction of the axioms of the usual theory of densely ordered sets with no first or last element. Let $T$ be the theory with single axiom $\forall x\forall y(x=y)\lor\varphi$. Then $T$ is $\omega$-categorical but not complete.