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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?

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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.

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    $\begingroup$ 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? $\endgroup$ – Aref Dec 10 '16 at 15:34
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    $\begingroup$ 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 $T\models\neg\exists\vec{x}\varphi(\vec{x})$. $\endgroup$ – Leo163 Dec 10 '16 at 17:16
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    $\begingroup$ Thank you that's right now I see! but now that they realize the same isolated types, how we can say they realize the same types? Is it because they realize the same complete types and hence they realize all types? $\endgroup$ – Aref Dec 10 '16 at 18:20
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    $\begingroup$ 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. $\endgroup$ – Leo163 Dec 11 '16 at 18:15

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