Subsets of Equidecomposable Subsets are Equidecomposable

Theorem
Let $A, B \subseteq \R^n$ be equidecomposable.

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 $\mu_1, \ldots, \mu_m, \nu_1, \ldots, \nu_m: \R^n \to \R^n$ such that:


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

and


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

Define:


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

Then:

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

But for each $i$:


 * $\left({S \cap \mu_i \left({X_i}\right)}\right) \subseteq \mu_i \left({X_i}\right)$

and so:

Hence:
 * $\nu_i \left({Y_i}\right) \subseteq \nu_i \left({X_i}\right)$

and so:

Define:
 * $\displaystyle \bigcup_{i \mathop = 1}^m \nu_i \left({Y_i}\right) = T$

Hence the result.