I want to prove this exercise from David Marker's model theory book. Let $L = {E}$, where $E$ is a binary relation symbol, and let $T$ be the theory of an equivalence relation with infinitely many classes each of which is infinite. Show that in any $M |= T$ we can find infinite sets of indiscernibles $I_0$ and $I_1$ such that $tp(I_0)\neq tp(I_1)$, but if $J$ is any other infinite set of indiscernibles, then $tp(J) = tp(I_i)$ for $i = 0$ or $1$.
I don't really understand what the difference between $I_0, I_1$ and $J$ is. I mean they're all infinite set of indiscernibles. Also, any hint for the solution is appreciated.
Let $M\models T$. Take $I_0$ to consist of $\aleph_0$-many elements from a single equivalence class, and take $I_1$ to consist of $\aleph_0$-many elements from pairwise distinct equivalence classes.
Certainly $\operatorname{tp}(I_0)\neq\operatorname{tp}(I_1)$, as the former contains the formula $E(v_i,v_j)$ for any $i\neq j\in\omega$ and the latter contains the formula $\neg E(v_i,v_j)$ for any $i\neq j\in\omega$. Furthermore, the theory of an equivalence relation with infinitely many infinite equivalence classes has quantifier elimination, so showing that $I_0$ and $I_1$ are indiscernible sets amounts to considering atomic formulas, all of which are the form $E(v,w)$ (or $v=w$), whence by the first sentence of this paragraph $I_0$ and $I_1$ are indeed indiscernible.
Finally, let $J$ be any other infinite set of indiscernibles. To be pedantic, for $J$ to have the same type as $I_0$ or $I_1$ it needs to have the same number of variables as $I_0$ and $I_1$, so we assume this is the case. If any two elements of $J$ come from pairwise distinct equivalence classes, then all elements of $J$ do, in which case $\operatorname{tp}(J)=\operatorname{tp}(I_1)$. Similarly, if any two elements of $J$ come from the same equivalence class, then all elements of $J$ do, in which case $\operatorname{tp}(J)=\operatorname{tp}(I_0)$.