Finite Set of Models This is only directed towards logicians, model theorists etc.. I am reading "Model Theory" by Keisler and Chang and have encountered the following question.
Let $\Psi = \{M_1,...,M_n \}$ be a finite set of models (also known as interpretations) of a given language $L$. Let me also mention that $L$ is generated from a countably infinite alphabet $S$. Prove that $\exists$ a set $\Gamma$ of sentences (or well-formed formulas) such that $\Psi$ is the set of all models for $\Gamma$. 
 A: (Using the notation from Chang-Keisler.) Let the models be $M_1 , \ldots , M_n \subseteq \mathscr{S}$.
For each sentence symbol $S$ of $\mathscr{S}$ and each $i \leq n$, by $S_i$ denote
$$
\begin{cases}
S, &\text{if }S \in M_i \\
\neg S, &\text{if }S \notin M_i.
\end{cases}
$$
Let $\Gamma$ be the set of all sentences of the form
$$
(S^1_1 \wedge \cdots \wedge S^m_1 ) \vee \cdots \vee (S^1_n \wedge \cdots \wedge S^m_n )
$$
where $S^1 , \ldots , S^m \in \mathscr{S}$.
Clearly each $M_i$ is a model of $\Gamma$.
Suppose now that $N \subseteq \mathscr{S}$ is some model distinct from each $M_i$.
For each $i \leq n$ there is a sentence symbol, $S^i$ such that either $S^i \in M_i$ and $S^i \notin N$, or the opposite.  It is easily seen that $N$ is not a model of the sentence
$$
(S^1_1 \wedge \cdots \wedge S^n_1 ) \vee \cdots \vee (S^1_n \wedge \cdots \wedge S^n_n )
$$
($N$ cannot model $(S^1_i \wedge \cdots \wedge S^n_i )$ since by construction it does not model $S^i_i$.)
A: Let $\Gamma$ be the set of sentences that are true for all of the $M_i$.
If $M_i\ne M_j$, there is a sentence $\phi_{i,j}$ that is true in $M_i$ and false in $M_j$.
Let $\phi_i=\phi_{i,1}\land\cdots\land\phi_{i,n}$ (with $\phi_{i,i}$ omitted). Then $\phi_i$ is true in $M_i$ and false in all $M_j$ with $j\ne i$.
Let $\phi=\phi_1\lor\cdots\lor\phi_n$. Then $\phi$ holds in all $M_i$, hene $\phi\in\Gamma$.
Let $M$ be a model of $\Gamma$. Then $\phi$ is true in $M$, hence (at least) one $\phi_i$ is true in $M$. If $\psi$ is a sentence that is true in $M_i$ then  $\phi_i\to \psi$ is in $\Gamma$ because $\phi_i$ is false in $M_j$ with $j\ne i$ and $\psi$ is true in $M_i$. It follows that $\phi_i\to \psi$ is true in $M$ and finally $\psi$ is true in $M$. If on the other hand, $\psi$ is false in $M_i$, then $\neg\psi$ is true in $M_i$, hence true in $M$, hence $\psi$ false in $M$. We conclude that $M=M_i$.
