Suppose we have an action $G \curvearrowright X$. We say that two subsets $A, B \in X$ are equidecomposable (written $A \sim B$) if there exist a disjoint partition $(A_i)_{i=1}^n$ of $A$ and elements $(g_i)_{i=1}^n$ of $G$ such that $B = \bigcup_{i=1}^n{g_iA_i}$.

Suppose that $A, B, D_1, D_2 \subseteq X$ are subsets such that

$A \cap D_1 = \emptyset, \qquad B \cap D_2 = \emptyset, \qquad A \sim B, \qquad (A \cup D_1) \sim (B \cup D_2)$

then we can conclude that $D_1 \sim D_2$.

This is a kind of cancellation law for equidecomposability used in Tarski's theorem, which shows equivalence of amenability and non-paradoxicality. I didn't see the result stated this way, but I believe I wrote down all the necessary assumptions. Is there a straightforward way, mostly playing with partitions, to prove it.

EDIT: This is wrong, see the counterexample below.

  • $\begingroup$ I believe that we usually mean a disjoint union when we speak of a partition, but I might be wrong, still I will precise that in the definition. I am surprised that the statement itself is wrong, do you have a counterexample that I totally missed? $\endgroup$ – Olivier Melançon Nov 14 '15 at 4:59

Here is an example: Let $X$ denote the set of integers, $G$ the group of integer translations, $D_1=\{1\}$, $D_2=\{0, 1\}$, $A= [2, \infty)\cap X, B=[2, \infty)\cap X$. Then $D_1\cup A$ and $D_2\cup B$ are equidecomposable (with $n=1$, $g_1$ is the unit translation), but $D_1$ cannot be equidecomposable to $D_2$ since they have different cardinalities.

And just for the record: It should be $B=\bigsqcup_{i=1}^n g_i A_i$ in you definition.

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  • $\begingroup$ I feel a little dumb right now. Thank you for enlightening me! $\endgroup$ – Olivier Melançon Nov 14 '15 at 5:16

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