# Equidecomposable Nested Sets

## Theorem

Let $A, B, C$ be sets such that $A$ and $C$ are equidecomposable and $A \subseteq B \subseteq C$.

Then $B$ and $C$ are equidecomposable.

## Proof

Let $\left\{{X_k}\right\}_{k \mathop = 1}^n$ be a decomposition of $A$ and $C$, so that there are isometries $\left\{{\phi_k}\right\}_{k \mathop = 1}^n$ and $\left\{{\psi_k}\right\}_{k \mathop = 1}^n$ such that:

$\displaystyle A = \bigcup_{k \mathop = 1}^n \phi_k \left({X_k}\right)$

$\displaystyle C = \bigcup_{k \mathop = 1}^n \psi_k \left({X_k}\right)$

Let $Y_k = \psi_k^{-1} \left({ B \cap \psi_k \left({X_k}\right)}\right)$.

Then:

$\displaystyle \bigcup_{k \mathop = 1}^n \psi_k \left({Y_k}\right) = B \cap \bigcup_{k \mathop = 1}^n \psi_k \left({X_k}\right) = B \cap C = B$

Let $Z_k = \phi^{-1} \left({ A \cap \phi_k \left({X_k}\right)}\right)$.

Then:

$\displaystyle \bigcup_{k \mathop = 1}^n \phi_k \left({Z_k}\right) = A \cap \bigcup_{k \mathop = 1}^n \phi_k \left({X_k}\right) = A \cap C = A$