I'm studying Model Theory with the book of Chang-Keisler, and I'm trying to solve the following problem stated in Chapter 5:
The union of an elementary chain $(M_i)_{i < \alpha}$ of $\alpha$-saturated models is also $\alpha$-saturated, if $\alpha$ is regular.
I'm considering the following definition:
$M$ is a $\alpha$-saturated model if for all $A \subset M$, $|A| < \alpha$, every $1$-type $p$ in $A$ over $T(M) = \left \{ \varphi: M \models \varphi \right \}$ is realized in $M$, i.e. there exists $a \in M$ such that $M \models \varphi(a), \, \forall \varphi \in p.$
My attempt:
Let $(M_i)_{i < \alpha}$ be an elementary chain of $\alpha$-saturated models and let $M = \bigcup_{i < \alpha} M_i$. Given $A \subset M$ with $|A| < \alpha$ and $p$ a $1$-type in $A$ over $T(M)$, if I prove that there exists $i < \alpha$ such that $p$ is a type over $T(M_i)$, we could conclude that there exists $a$ in $M_i$, with $M_i \models \varphi(a)$ for all $\varphi \in p$. Hence $a \in M$ and $M \models \varphi(a)$, since $M_i$ is an elementary substructure of $M$.
So, how can I proof the existence of such $i < \alpha$?
Best Answer
This follows directly from the definition of regular cardinal.
For each $a\in A$, let $i_a$ be the least ordinal such that $a\in M_{i_a}$. Then $I = \{i_a\mid a\in A\}$ is a subset of $\alpha$ of cardinality at most $|A|<\alpha$. By regularity, $I$ is not cofinal in $\alpha$, so $j = \sup I < \alpha$. Thus $A\subseteq M_j$, and $p$ is a type over $M_j$, as desired.