Given a chain $M_n \prec M_{n+1}$, construct $M$ such that $M_n \prec M$

Question

Let $(M_n \mid n \in \mathbb N)$ be a sequence of $L$-structures for some given language $L$ such that $$ M_n \prec M_{n+1} $$ for all $n \in \mathbb N$. (I.e. $M_n$ is an elementary substructure of $M_{n+1}$.)

I want to show that there is an $L$-structure $M$ such that $M_n \prec M$ for all $n \in \mathbb N$. Right now, I don't know how to approach this task and am looking for a hint to get me going.

Answer 1

Hint: It suffices to let $M = \bigcup_{n \in \mathbb N} M_n$. More precisely:

• The universe of $M$ is the union of all the universes of $M_n$, for $n \in \mathbb N$.
• If $c \in L$ is a constant symbol, let $c^M = c^{M_0}$.
• If $f \in L$ is a $k$-ary function symbol and $x_1, \ldots, x_k \in M$, then fix $m \in \mathbb N$ such that $x_1, \ldots, x_k \in M_m$ and let $$ f^M(x_1, \ldots, x_k) = f^{M_m}(x_1, \ldots, x_k).$$
• If $r \in L$ is a $k$-ary relation symbol and $x_1, \ldots, x_k \in M$, fix $m \in \mathbb N$ such that $x_1, \ldots, x_k \in M_m$ and let $$ r^M(x_1, \ldots, x_k) \iff r^{M_m}(x_1, \ldots, x_k). $$

To show that $M_n \prec M$, use the Tarski-Vaught Test and proceed by induction on the complexity of $L$-formulae.