The Sierpinski-Mazurkiewicz paradox yields a nonempty rigid-motion paradoxical subset $S$ of the Euclidean plane: $S$ is the disjoint union of $A$ and $B$, each of which is $G$-equidecomposable with $S$, for a group $G$ of rigid motions.

(Here sets $U$ and $V$ are $G$-equidecomposable for a group of $G$ acting on a space containing $U$ and $V$ iff $U$ can be written as the disjoint union of $U_1,...,U_n$, $V$ as the disjoint union of $V_1,...,V_n$ and $V_i = g_iU_i$ for some sequence $g_1,...,g_n\in G$.)

Question: Can one do this with $S$ itself being $G$-invariant? I.e., is there is a group $G$ of rigid motions (variant: isometries) and a subset $S$ of the plane such that $S$ is $G$-paradoxical and $gS=S$ for all $g\in G$?


1 Answer 1


No, it's not possible because the isometry group of the Euclidean plane is amenable (as discrete group, since it's solvable), so every $G$-set admits an invariant mean defined on all subsets. If $S$ is $G$-invariant, this applies to $S$ as $G$-set and an immediate contradiction follows.

  • $\begingroup$ Ooops. I had forgotten this group was amenable. $\endgroup$ Oct 21, 2020 at 18:57

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