Why if $T$ is NIP $T^{eq}$ is NIP too?

Question

I'm working on excercice 2.14 of Pierre's book "A guide to NIP theories" but I have not idea what to do. This excercie asks to prove that if $T$ is a NIP (dependent) theory then $T^{eq}$ is NIP too.

Answer 1

Hint. Let $(a_i)_{i < \omega}$ be an $\emptyset$-indiscernible sequence of tuples in $T^{\operatorname{eq}}$, and $\varphi(x)$ a formula with parameters in $T^{\operatorname{eq}}$. Assume that the truth of $\varphi(a_i)$ alternates infinitely. Can you find an indiscernible sequence of real (not imaginary) tuples and formula with parameters whose truth alternates infinitely?

Further hint:

The sequence $(a_i)_{i < \omega}$ is actually a sequence of $E$-equivalence classes of real tuples; that is, we have a definable equivalence relation $E$ over $\emptyset$, and each $a_i$ is really some $[b_i]_E$. The sequence $(b_i)_{i < \omega}$ need not already be indiscernible, but you can use compactness + Ramsey's Theorem to correct that flaw. What does that get you?

---

As per the discussion in the comments, Alex's answer is more direct because it does not require building an indiscernible sequence. Either way will work, however, and it may (or may not) be pedagogically enlightening to try it both ways. Even though it's not necessary for this exercise in particular, building indiscernible sequences with various properties is an important technique for working with NIP theories.

Answer 2

It looks like A guide to NIP theories doesn't include a definition of $T^{eq}$, so it's worth recalling the basics.

If $M\models T^{eq}$, and $M'$ is the reduct of $M$ to the language of $T$ (taking the reduct here includes throwing away all the imaginary sorts), then $M'\models T$ and $M = (M')^{eq}$. That is, any $a\in M\setminus M'$ is an element of an imaginary sort $S_E$, and the preimage of $a$ under $f_E$ is an $E$-equivalence class. We'll say $\overline{b}$ is a real lift of $a$ if $f_E(\overline{b}) = a$. If $a\in M'$, then $a$ is a real lift of itself.

Now for every formula $\varphi$ in the language of $T^{eq}$, there is an associated formula $\varphi'$ in the language of $T$ such that the following are equivalent:

1. $M\models\varphi(a_1,\dots,a_n)$
2. $M'\models\varphi'((\overline{b})_1,\dots,(\overline{b})_n)$ for some real lifts $(\overline{b})_i$ of the $a_i$
3. $M'\models\varphi'((\overline{b})_1,\dots,(\overline{b})_n)$ for all real lifts $(\overline{b})_i$ of the $a_i$.

If these facts aren't clear to you, you should try to prove them. Once you know them, the exercise is straightforward: Given an instance of the independence property in $T^{eq}$, witnessed by a formula $\varphi$ and tuples $(a_i)_{i\in I}$ and $(b_X)_{X\subseteq I}$, show that the associated formula $\varphi'$ and real lifts of all these tuples give an instance of the independence property in $T$.