Equidecomposable Subsets

Theorem

Let $A, B \subseteq \R^n$ be equidecomposable and let $S \subseteq A$.

Then there exists $T \subseteq B$ such that $S$ and $T$ are equidecomposable.

Proof

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

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

Define

$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i) )$

Then

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

and so $\left\{{Y_i}\right\}_{i=1}^m$ forms a decomposition of $S$.

But for each $i$:

$( S \cap \mu_i(X_i) ) \subseteq \mu_i(X_i)$

and so:

$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i)) \subseteq \mu_i^{-1}(\mu_i(X_i)) = X_i$

Hence

$\nu_i(Y_i) \subseteq \nu_i(X_i)$

and so

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

Define $\displaystyle \bigcup_{i=1}^m \nu_i(Y_i) = T$.

$\blacksquare$