I am going through this paper by Shelah and there is a statement in theorem 4.4 that I don’t seem to understand:
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[…] (page 199)
Clearly $B \subset |M|$ and $|B| < |M|$. Let $q$ be the type which $d$ realizes over $B$. Now $q$ cannot be algebraic, for if it were, everything satisfying it would be in $M$, since $B \subset |M|$ but $d \notin |M|$.
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Why can’t $q$ be algebraic? Of course this is not true in general, but I don’t see what in the definition of $B$ allows us to come to this conclusion.
Answering my own question before putting any edits to this. The reason that $q$ cannot be algebraic is as follows:
Assume not, $q$ is algebraic, then $q$ is isolated by some $\Phi \in q$, thus we have $\vDash \exists^{n} y \ \Phi$ for some $n < \omega$. By Tarski-Vaught then, $M \vDash \exists^{n} y \ \Phi$.
In other words, every realization of $q$ must be in $|M|$, but $d \vDash q$ and $d \notin |N|$, a contradiction.