Equidecomposability is an Equivalence Relation

Theorem

The property of being equidecomposable is an equivalence relation on the power set $\mathcal P \left({\R^n}\right)$.

Proof

Relexivity

A set is necessarily equidecomposable with itself; the same decomposition and set of isometries suffice for $A$ as for $A$.

Symmetry

There is no order to the relation of being equidecomposable; symmetry follows.

Transitivity

Suppose $A, B, C \subset \R^n$ are sets such that $A, B$ are equidecomposable and $B, C$ are equidecomposable.

Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\R^n \to \R^n$ such that:

$\displaystyle A = \bigcup_{i=1}^m \mu_i(X_i)$

and

$\displaystyle B = \bigcup_{i=1}^m \nu_i(X_i)$

Further let $Y_1, \dots, Y_p$ together with $\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p$ be sets and isometries such that:

$\displaystyle B = \bigcup_{i=1}^p \xi_i(Y_i)$

and:

$\displaystyle C = \bigcup_{i=1}^p \tau_i(Y_i)$

Consider the sets

$Z_{i,j} = \nu_i(X_i) \cap \xi_j(Y_j)$

where $ 1 \leq i \leq m$ and $ 1 \leq j \leq p$.

We have:

$\displaystyle \bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1})(Z_{i,j}) = \bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1} )( \nu_i(X_i) \cap \xi_j(Y_j) ) = \bigcup_{i=1}^m (\mu_i \circ \nu_i^{-1} \circ \nu_i) (X_i) = \bigcup_{i=1}^m \mu_i(X_i) = A$,

$\displaystyle \bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1})(Z_{i,j}) = \bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1}) (\nu_i(X_i) \cap \xi_j(Y_j)) = \bigcup_{j=1}^p (\tau_j \circ \xi_j^{-1} \circ \xi_j) (Y_j) = \bigcup_{j=1}^p \tau_j(Y_j) = C$,

so $Z_{i,j}$ together with the isometries $\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1}$ as a decomposition of $A$ and $C$, hence these two are equidecomposable.

$\blacksquare$