Indiscernible to create descending chain of elementary models

Question

Let $M$ an infinite structure such that $\mid M \mid \ge \mid L(M) \mid $. Show that exists a proper elementary extension $N$ and a chain $\langle N_{i} \mid i < \omega \rangle $ such that $$N=N_{0}, \ \ N_{i} \succ N_{i+1} $$ for all $i< \omega $ and $M=\bigcap_{i< \omega} N_{i}$

That this exercise comes as an application of sequence of indiscernible, the procedure would be to find a sequence of indiscernible and with this sequence and with $M$ build the first model $N_{0}$ of the chain of elementary extensions of $M$. I'm not sure if that is the correct. Thanks for any hint.

Answer 1

Expand the language with Skolem functions. Let $\{a_i: i<\omega\}$ be some elements such that $a_n\notin\operatorname{dcl}(\{a_i: n<i<\omega\}\cup M)$. (The definability closure is relative to the Skolemized language.) Any indiscernible sequence certainly meets this requirement. Let $N_n$ be (the reduct to $L$ of) the structure generated by $M\cup\{a_i: n<i<\omega\}$.

Question. Is there an answer that does not use Skolem functions?

Answer 2

Primo Petri's answer gives the right idea, but it doesn't work exactly as stated. The problem is that there might be some element $b\notin M$ such that $b\in N_n = \text{dcl}(\{a_i\mid n < i < \omega\}\cup M)$ (in the Skolemized language) for all $n$.

For example, let $L = \{E,f\}$, and let $T$ be the theory asserting that $E$ is an equivalence relation with infinitely many infinite classes and $f$ picks out a representative from each class. That is, $aEf(a)$ and $aEb\rightarrow f(a) = f(b)$. Now let $M$ be a model, and let $\{a_i\}_{i\in\omega}$ be an $M$-indiscernible sequence in a single new equivalence class (so $a_iEa_j$ for all $i$ and $j$, but $\lnot a_i E m$ for all $i$ and $m\in M$). Now the element $b = f(a_n)$ is in $N_n$ for all $n$, but not in $M$.

To fix this, we can pick our sequence more carefully. Here are some definitions:

• A type $p(x)$ over $A$ is a coheir of its restriction to $M\subseteq A$ if for every formula $\varphi(x,\overline{a})$, there exists $m\in M$ such that $\models \varphi(m,\overline{a})$.
• Let $p(x)$ be a global type (i.e. a type the monster model $\mathcal{U}$, though if you're not comfortable with monster models, $|M|^+$-saturated will do), which is a coheir of its restriction to $M$. A coheir sequence for $p(x)$ is a sequence $\{a_i\}_{i\in\omega}$ from $\mathcal{U}$ such that $a_n\models p\restriction Ma_1\dots a_{n-1}$ for all $n$. Note that $\text{tp}(a_n/Ma_1\dots a_{n-1})$ is a coheir of its restriction to $M$.

And here are some things to check:

• For any type $p(x)$ over a model $M$, and any larger set $A$, there is an extension of $p$ to a type $q(x)\supseteq p(x)$ over $A$, such that $q$ is a coheir of $p$.
• Any coheir sequence is indiscernible.
• If we do Primo Petri's construction with a coheir sequence over $M$ (in the Skolemized language), we get a solution to the problem.

I've seen this problem given as an exercise on more than one occasion by people who think it will be an easy application of EM-models, without realizing that not any indiscernible sequence works and a more advanced topic like coheir sequences is necessary for the solution.