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Suppose that $\mathsf E$ is a countable Borel equivalence relation on the reals, and $\mathsf B$ is a finer equivalence of order 2, so that each $\mathsf E$-class consists of precisely two $\mathsf B$-classes.

Is there a large Borel set $X$ with the property that it intersects not more than one $\mathsf B$-class within each $\mathsf E$-class?

This depends on what large means, of course. If large = perfect, super-perfect, or $\mathsf E_0$-large then the answer is yes (but only modulo some nontrivial canonization theorems in descriptive set theory). What about large = positive measure or large = non-meager? Just interesting.

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For non-meager or positive measure the answer is no. Let $B$ be $E_0$, and let $E$ be given by $x E y$ iff $x E_0 y$ or $f(x) E_0 y$, where $f(x)$ switches all digits of $x$ from 0 to 1 and vice versa. If there were such a Borel set $X$ which was non-meager, then its $E_0$-saturation has the same property and is comeager; however, since $f$ is a homeomorphism we would have $f[X]$ comeager and disjoint from $X$. A similar argument shows no such $X$ of positive measure exists.

(I have heard this relation called the "mismatched socks").

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  • $\begingroup$ Yes very clear thanks. My own example is $E=E_0$ and $B$ being to have a finite even-number symmetric difference $\endgroup$
    – Vladimir Kanovei
    Commented May 1, 2021 at 21:04

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