Marker's "Model Theory" exercise 5.5.4: indiscernibles in a $\kappa$-saturated model I am solving exercises in Model Theory: An Introduction from David Marker. I think I have an idea, but I'd like to know if my intuition is correct. In this exercise, we work on a countable complete theory $T$.
Exercise 5.5.4: Show that if $\mathcal{M}$ is $\kappa$-saturated, then there is $I \subseteq M$, a sequence of order indiscernibles with $|I| = \kappa$.
Attempt: Let $p_0 \in S_n^\mathcal{M}(T)$, i.e. a type over $\emptyset$. As $\mathcal{M}$ is $\kappa$-saturated, there is $a_0 \in M$ such that $a_0 \vDash p_0$. Now consider $p_1 \in S_n^\mathcal{M}(\{ a_0 \})$. Again, as $\mathcal{M}$ is $\kappa$-saturated, then there is $a_1 \in M$ such that $a_1 \vDash p_1$. Following this argument, I can find a set $J := (a_\alpha)_{\alpha < \kappa}$.
My question is: can I conclude that this set $J$ is the sequence of order indiscernibles that I was looking for? If not, how can I find this sequences of indiscernibles?
 A: The easiest solution to this exercise is just putting together some facts that you probably already know. 


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*Pick a linear order $(O,<)$ of cardinality $\kappa$. Then there is some model $\mathcal{N}\models T$ and $J\subseteq \mathcal{N}$ a sequence of order indiscernibles of order type $O$ (Marker Theorem 5.2.3).

*By Löwenheim-Skolem, we can assume $|N| = \kappa$. 

*Now $\kappa$-saturated models are $\kappa^+$-universal (Marker Lemma 4.3.17), so there is an elementary embedding $f\colon \mathcal{N}\to \mathcal{M}$. And $I = f(J)$ is a sequence of order indiscernibles of order type $O$. 
You could also rephrase this proof as (1) checking that for any linear order $(O,<)$, the partial type in $\kappa$-many free variables saying that those elements form an indiscernible sequence of order type $O$ is consistent, and then (2) using the fact that $\kappa$-saturated models realizes all consistent types in $(\leq \kappa)$-many variables over parameters sets of size $(<\kappa)$. The proof of (1) is essentially the same as the proof of Marker's Theorem 5.2.3, and the proof of (2) is essentially the same as the proof of Marker's Lemma 4.3.17. 

Your proof definitely doesn't work as stated, since it seems like at every step of the induction you're just picking an arbitrary type $p_\alpha\in S(\{a_\beta\mid \beta<\alpha\})$. So for example there's no reason why $a_0$ and $a_1$ should even have the same type over $\emptyset$, much less start an indiscernible sequence. 
But a version of the same idea can be made to work. The first step is to start with a type $p$ in $S(M)$ and build a sequence $(a_\alpha)_{\alpha<\kappa}$ by induction, letting $a_\alpha$ realize $p|_{\{a_\beta\mid \beta<\alpha\}}$. This way at least you know that for any $\alpha\leq \lambda$, $a_\alpha$ and $a_\lambda$ will realize the same type over $\{a_\beta\mid \beta < \alpha\}$. 
But it turns out that this still isn't quite enough to ensure that $(a_\alpha)_{\alpha<\kappa}$ is indiscernible. For this, you'll need to start with not just any type $p\in S(M)$: you want $p$ to be invariant (over some set $A\subseteq \mathcal{M}$ with $|A|<\kappa$). 
A type $p\in S(M)$ is invariant over A if for every formula $\varphi(x,y)$ and all tuples $b,b'\in M^y$, if $\text{tp}(b/A) = \text{tp}(b'/A)$, then $\varphi(x,b)\in p$ if and only if $\varphi(x,b')\in p$. [Aside: If $\mathcal{M}$ is saturated, not just $\kappa$-saturated, then this is equivalent to saying that $p$ is invariant under the natural action of $\text{Aut}(\mathcal{M})$ on $S(M)$.]
Now it's a good exercise to prove that if $p\in S(M)$ is invariant over $A$, with $|A|<\kappa$, and you build a sequence $(a_\alpha)_{\alpha<\kappa}$ by induction, letting $a_\alpha$ realize $p|_{A\cup \{a_\beta\mid \beta<\alpha\}}$, then the sequence is indiscernible over $A$. 
This kind of sequence is called a Morley sequence in the type $p$ over $A$. You might be familiar with another definition of Morley sequence, in stability theory: an indiscernible sequence where each element is forking independent from the previous elements of the sequence. The reason for the overloading of this name is that if a type $p\in S(M)$ doesn't fork over $A$, and $p|_A$ is stationary, then $p$ is invariant over $A$ (and a Morley sequence in $p$ in the invariant types sense is the same as a Morley sequence in $p$ in the forking sense). Indeed, these were the first kinds of invariant types to be studied in model theory, so the stability theory notion of Morley sequence predates the invariant types notion. 
