# Types realized in an atomic model

Suppose that $M$ and $N$ are two atomic models of a complete theory $T$ in a countable language.

David Marker's Model Theory book:
Because the types in $S_n(T)$ realized in an atomic model are exactly the isolated types, M and N realize the same types.

I have this question that why the types in $S_n(T)$ realized in an atomic model are exactly the isolated types and if so why those models realize the same types?

If a complete type $p$ is realized by a $n$-tuple $\vec{a}$ in an atomic model, then $p$ is isolated by definition of atomic model.
If a type $p$ is isolated, say by formula $\varphi$, then since $T$ is complete $T\models\exists\vec{x}\varphi(\vec{x})$, so $p$ is realized in any model of $T$.
Finally, since $T$ is complete, $M\models\varphi$ is and only if $N\models\varphi$, which proves that they realize the same isolated types, and hence the same types by what we just proved.
• Thank you for your answer, but could you say why "If a type $p$ is isolated, say by formula $φ$, then since $T$ is complete $T\models\exists\vec{x}\varphi(\vec{x})$"? and why if we accept they realize the same isolated types, then they realize the same types? – Aref Dec 10 '16 at 15:34
• Since $p\in S_n(T)$, $p\cup T$ is satisfiable, and since $T$ is complete, $T\models\exists\vec{x}\varphi(\vec{x})$, where $\varphi$ is the formula isolating $p$, as it cannot be that $T\models\neg\exists\vec{x}\varphi(\vec{x})$. – Leo163 Dec 10 '16 at 17:16
• More or less. Suppose that $p$ is a type realized in $M$ but not in $N$. By what we already proved, $p$ is isolated, say by $\varphi$, so that $M\models\exists\vec{x}\varphi(\vec{x})$, but then, since $T$ is complete, $T\models\exists\vec{x}\varphi(\vec{x})$, but then $N\models\exists\vec{x}\varphi(\vec{x})$, a contradiction. – Leo163 Dec 11 '16 at 18:15