First, let's consider the definitions:
Let $\lambda$ be an ordinal. A model $M$ is $\lambda-$saturated if for all A⊂M, |A|<$\lambda$, every 1-type p in A (over $T(M)=\{φ \in L(A)-sent:M⊨φ\}$) is realized in M, i.e. there exists a∈M such that M⊨φ(a),$\forall \phi \in p$.
A model $M$ is saturated if it is $|M|-$saturated.
If $\lambda$ is an ordinal, let $\lambda^+$ be the successor of $\lambda$.
Im trying to solve the following question:
Suppose that $M$ is a model of cardinality $\lambda$, a $\textbf{regular}$ cardinal, such that $M$ is the union of an elementary chain, $(M_{\beta})_{\beta<\lambda}$, i.e $M = \bigcup_{\beta<\lambda} M_{\beta}$, and $M_{\beta}$ is $\beta^+-$saturated for all $\beta$. Prove that $M$ is saturated.
My attempt:
Let $A⊂M$ with $|A|<\lambda$ and $p$ a $1-$type in $A$ over $T(M)$. For every $a \in A$, choose $i_a < \lambda$ in a way that $a \in M_{i_a}$. Now, if $I=\{i_a:a \in A\}$, then $|I|<|A|<\lambda$ and since $\lambda$ is regular, there exists $j<\lambda$ such that $i_a\leq j$ for all $a \in A$. Therefore, $A \subset M_j$. The only thing that remains to show is that $M_j$ is $\beta-$saturated for some $\beta>|A|$, knowing that it is $j^+-$saturated. I think that it can be done if $|I|=|A|$, since in this case $i_a < j^+$ for all $a \in A$, hence $|A|<j^+$. Is there a way to choose $i_a$ in order to $|I|=|A|$? Or there are another approach?
Best Answer
You have essentially the right idea, but there are some imprecise formulations in your proof. For example, you say $|I| < |A|$, but this should be $|I| \leq |A|$ (luckily, that one does not really influence the rest of your proof). Also, when you should replace "ordinal" in your definition by "cardinal", and we really want $\lambda^+$ to be the successor cardinal and not the successor ordinal.
Anyway, given that your proof already captures the important ideas, I will just post a full proof as an answer.
Let $A \subset M$ with $|A| < \lambda$, and let $p(x)$ be a type over $A$. For each $a \in A$ there must be $\beta_a < \lambda$ such that $a \in M_{\beta_a}$. Then since $\lambda$ is regular, taking $\beta = \sup_{a \in A} \beta_a$ will still be such that $\beta < \lambda$. Now we have $A \subset M_\beta$. Furthermore, we may assume $|A| \leq \beta$ (because otherwise, we replace $\beta$ by $|A|$ and we still have $A \subset M_\beta$ and $\beta < \lambda$). Then since $M_\beta$ is $\beta^+$-saturated, there we find $b \in M_\beta$ that is a realisation of $p(x)$, and since $M_\beta \preceq M$ we have that $b \in M$ is a realisation of $p(x)$.