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.

[Added Feb 4 at 5:50] And finally it is established, there are two distinct but OD-indiscernible populations of Sacks-generic reals over $L$, arXiv . In fact this is an unpublished result of Solovay dated back to 2002. A similar result holds for $E_0$-large forcing, via a known ``canonization'' theorem, but the problem is open for other popular forcing notions.

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    $\begingroup$ That's a really nice question! $\endgroup$
    – Asaf Karagila
    Dec 28, 2019 at 13:45
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    $\begingroup$ In my paper jdh.hamkins.org/ehrenfeuchts-lemma-in-set-theory, we prove a negative answer in the extension with two Sacks reals. $\endgroup$ Dec 28, 2019 at 14:03
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    $\begingroup$ Yes thm 4.6 in 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. $\endgroup$ Dec 28, 2019 at 15:48

1 Answer 1


The question is now answered in this recent paper, where it is shown that in the extension $L(s)$ of the constructible universe $L$ by a single Sacks real $s$, there is a definable pair $\{a,b\}$ of objects (where $a$ and $b$ are subsets of real numbers) such that neither $a$ nor $b$ is ordinal definable.

This shows that $a$ and $b$ are indiscernible in $L(s)$ in the following strong sense: for any formula of set theory $\phi(x,y,\alpha)$, where $\alpha$ is an ordinal parameter, $\phi(a,b,\alpha) \leftrightarrow\phi(b,a,\alpha)$ holds in $L(s)$.


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