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Let $\mathbb{I}=\langle a_{i} \mid i< \omega + \omega\rangle$ be a sequence of indiscernibles over a set $A$. Then $\langle a_{i} \mid \omega \le i< \omega + \omega\rangle$ is an indiscernible sequence over $A \cup \langle a_{i} \mid i< \omega \rangle$.

This is a specific case of a stronger result: Let $(I, <)$ be an infinite linearly ordered set and suppose that exists $p \in I$ such that $\{q \in I \mid p<q\}$ is infinite, if $\langle a_{i} \mid i \in I \rangle$ is a sequence of indiscernible over $A$ then $\langle a_{i} \mid i \in I, p<i\rangle$ is a sequence of indiscernibles over $A \cup \{a_{i} \mid i \in I, i<p \}$.

I do not understand why the sets $\{q \in I \mid p<q\}$ and $\{q \in I \mid p>q\}$ must be both infinite. At first $I$ thought of using a permutation of the set I to order ascending elements of $\langle a_{i} \mid i \in I, p<i\rangle$ and use the hypothesis that $\langle a_{i} \mid i \in I \rangle$ is a sequence of indiscernible over $A$. But apparently is not the right way.

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You don't need either set to be infinite. The full result should be

Let $I$ be linearly ordered, $I = I_0 \cup I_1$ a disjoint union, $I_0 < I_1$. Furthermore, let $M$ be a model, $A \subseteq M$, and $\{b_i\}_{i\in I}$ be tuples from $M$. If $\{b_i\}_{i\in I}$ is indiscernible over $A$, then both

  • $\{b_i\}_{i\in I_0}$ is indiscernible over $A \cup \{b_i\}_{i\in I_1}$ and
  • $\{b_i\}_{i\in I_1}$ is indiscernible over $A \cup \{b_i\}_{i\in I_0}$.

Proof sketch. We need to show that the truth of a formula about $\{b_i\}_{i\in I_0}$ with parameters from $\{b_i\}_{i\in I_1}$ doesn't change when we move the $\{b_i\}_{i\in I_0}$ pieces (preserving order). (The other part is the same argument with the roles of $I_0$ and $I_1$ switched.) This result follows immediately by the indiscernibility of the full sequence, together with the fact that $I_0 < I_1$ means we can't have moved any of the terms "past" the parameters (in $I$'s order).

(Exercise: find an example where the converse fails.)

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