# Equidecomposable Nested Sets

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

$\ds A = \bigcup_{k \mathop = 1}^n \map {\phi_k} {X_k}$

$\ds C = \bigcup_{k \mathop = 1}^n \map {\psi_k} {X_k}$

Let $Y_k = \map {\psi_k^{-1} } {B \cap \map {\psi_k} {X_k} }$.

Then:

- $\ds \bigcup_{k \mathop = 1}^n \map {\psi_k} {Y_k} = B \cap \bigcup_{k \mathop = 1}^n \map {\psi_k} {X_k} = B \cap C = B$

Let $Z_k = \map {\phi^{-1} } {A \cap \map {\phi_k} {X_k} }$.

Then:

- $\ds \bigcup_{k \mathop = 1}^n \map {\phi_k} {Z_k} = A \cap \bigcup_{k \mathop = 1}^n \map {\phi_k} {X_k} = A \cap C = A$

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