Let $L$ be the ground model, and $a\in2^\omega$ be a Sacks-generic real over $L$. Note that any real $x\in S=(2^\omega\cap L[a])\setminus L$ is still Sacks-generic over $L$. Now assume that $\mathsf E$ is an OD (ordinal-definable) equivalence relation in $L[a]$ on the set $S$, with exactly two equivalence classes, say $M$ and $F$. (**Two genders** of the Sacks reals.) Are $M$ and $F$ necessarily OD themselves?

My idea of a **counterexample** is as follows. Let $F$ be the set of all continuous 1-1 maps $2^\omega$ onto $2^\omega$, coded in $L$. Then $F$ is a group under the superposition. Moreover if $x,y\in S$ then it is known that $y=f(x)$ for some $f\in F$. Now if $H\subseteq F$ is a subgroup coded in $L$ then the relation:

$x \mathrel{\mathsf E_H} y$ iff $y=f(x)$ for some $f\in H$

is an OD equivalence on $S$, and there is no immediate idea as how to OD-define the $\mathrel{\mathsf E_H}$-class of $a$ (w/o a reference to $a$). Now the goal is to define $H$ such that $\mathrel{\mathsf E_H}$ has exactly two equivalence classes on $S$ in $L[a]$. The principal non-commutativity of $F$ looks to be a huge obstacle though.

Ehrenfeucht’s Lemma in Set Theory, in fact the result can be improved so that the indiscernible sets of reals are countable and disjoint and the union is lightface $\varPi^1_2$ as in Golshani etc. MLQ 2017, 63, 1-2, 19-31. Those examples are designed to yield exactly that, the plain Sacks model is a different thing. $\endgroup$ – Vladimir Kanovei Dec 28 '19 at 15:48