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Work in a saturated model $\cal U$ of sufficiently large cardinality.

Are there assumptions on $\kappa<|\cal U|$ that guarantee that any sequence $\langle a_i:i<\kappa\rangle$ has an subsequence of length $\kappa$ that is indiscernible?

(If the answer is yes could you please help me with a reference.)

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  • $\begingroup$ If the theory has order, then the pairs $\{a_i, a_j\}$ are divided into to parts depending on whether $a_i < a_j$ or not. An indiscernible sequence must be homogeneous, showing that $\kappa \to (\kappa)^2_2$. Thus $\kappa$ must be weakly compact. I think using a similar argument we can show that $\kappa$ is Ramsey. I don't know if being Ramsey is sufficient. $\endgroup$ Commented Sep 15, 2014 at 10:27
  • $\begingroup$ Indeed $\kappa \to (\kappa)^2_\mu$ for $\mu>2^{|L|}$ may be sufficient. Then we obtain the sequence of indiscernibles by the usual method only, in place of proceeding by induction and applying compactness, $\kappa \to (\kappa)^2_\mu$ yields the required sequence in one step. $\endgroup$ Commented Sep 15, 2014 at 10:49

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Actually your question follows easily from the existence of a Ramsey cardinal $\kappa$, if you only ask your sequence to be indescernible over a subset of size $<\kappa$ and $|L|<\kappa$.

Recall that a Ramsey cardinal is an infinite cardinal satisfying that for each function $f:[\kappa]^{<\omega}\rightarrow\tau$, $2\leq\tau<\kappa$ there is $H\subseteq\kappa$ with $|H|=\kappa$ such that for all $n<\omega, f\upharpoonleft [H]^n$ is constant.

Given any subsets $A,B\subseteq U$ with $|B|=\kappa$ and $|A|<\kappa$, consider the function $f:[B]^{<\omega}\rightarrow \bigcup_{n<\omega} S^n(A)$ given by $(a_1,\ldots, a_n)\mapsto tp(a_1,\ldots, a_n)$. Then as $|\bigcup_{n<\omega} S^n(A)|\leq 2^{|L|+|A|}<\kappa$; $\kappa$ is inaccesible, there is $B'\subseteq B$ with $|B'|=\kappa$ such that $f\upharpoonleft [B']^n$ is constant for all $n$. This clearly implies in particular that any sequence in $U$ of length $\kappa$, has a subsequence of the same length indiscernible over $A$.

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