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.

$\blacksquare$