A question about indiscernible sets Let $\cal{M}$ be a $\cal{L}$-structure. Definition from chapter 5 in David Marker:
"Let $I$ be an infinite set and suppose that $X=\{x_i:i\in I\}$ is a set of distinct elements of $\cal{M}$. We say that $X$ is an indiscernible set if whenever $i_1,...,i_m$ and $j_1,...,j_m$ are two sequence of $m$ distinct elements of $I$, then $\cal{M}\models\phi(x_{i_1},...,x_{i_m})\leftrightarrow\phi(x_{j_1},...,x_{j_m})$."
I am trying to think about adding constant symbols to $\mathcal{L}$. Of course, if we add a random constant symbol, then $X$ might not be indiscernible any more (if the constant symbol is $x$ for some $x\in X$). But what about if we know that two constant symbols already work? My question is, if $X$ is indiscernible when we add a constant $a$ to language, and $X$ is indiscernible when we add a constant $b$ to the language, then is $X$ indiscernible when we add the constants $a$ and $b$ to the language?
 A: The answer is no, and a counterexample can actually be found in one of the motivating examples of indiscernible sets! First, a bit of terminology to describe the situation you're asking about: if $X$ is an indiscernible set, and it remains indiscernible after adding constant symbols to the language for all elements of some subset $A\subset\mathcal{M}$, then we say that $X$ is "$A$-indiscernible" or "indiscernible over $A$".
Okay, onto the example. Let $T$ be the theory of torsion-free, divisible Abelian groups in the language $\mathcal{L}=\{0,+\}$. This theory is complete, and its models are precisely the $\mathbb{Q}$-vector spaces. Let $V\models T$ be any model of $T$ such that $\dim_{\mathbb{Q}}V\geqslant\aleph_0$, and let $E=\{e_i:i\in\mathbb{N}\}$ be any infinite linearly independent subset of $V$ whose span is not all of $V$. Then $E$ is an indiscernible set; can you see why? (Hint: extend $E$ to a basis of $V$ and consider automorphisms.)
Now, let $f\in V$ be any element such that $E\cup\{f\}$ is linearly independent; then $E$ is still indiscernible over $\{f\}$. (Why?) Similarly, the set $E\cup\{e_0+f\}$ is also linearly independent, so $E$ is also indiscernible over $\{e_0+f\}$. But $E$ is not indiscernible over $A:=\{f,e_0+f\}$, since $e_0$ is the only element of $X$ satisfying the $A$-formula $v+f=e_0+f$. So this gives the desired counterexample.
