# 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:

$\ds A = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$

and

$\ds B = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$

Define:

$Y_i = \mu_i^{-1} \paren {S \cap \map {\mu_i} {X_i} }$

Then:

 $\ds \bigcup_{i \mathop = 1}^m \map {\mu_i} {Y_i}$ $=$ $\ds \bigcup_{i \mathop = 1}^m \paren {S \cap \map {\mu_i} {X_i} }$ $\ds$ $=$ $\ds S \cap \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$ $\ds$ $=$ $\ds S \cap A$ $\ds$ $=$ $\ds S$

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

But for each $i$:

$\paren {S \cap \map {\mu_i} {X_i} } \subseteq \map {\mu_i} {X_i}$

and so:

 $\ds Y_i$ $=$ $\ds \map {\mu_i^{-1} } {S \cap \map {\mu_i} {X_i} }$ $\ds$ $\subseteq$ $\ds \map {\mu_i^{-1} } {\map {\mu_i} {X_i} }$ $\ds$ $=$ $\ds X_i$

Hence:

$\map {\nu_i} {Y_i} \subseteq \map {\nu_i} {X_i}$

and so:

 $\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i}$ $\subseteq$ $\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$ $\ds$ $=$ $\ds B$

Define:

$\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i} = T$

Hence the result.

$\blacksquare$