Subsets of Equidecomposable Subsets are Equidecomposable
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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:
\(\ds \bigcup_{i \mathop = 1}^m \map {\mu_i} {Y_i}\) | \(=\) | \(\ds \bigcup_{i \mathop = 1}^m \map {\mu_i} { \mu_i^{-1} \paren {S \cap \map {\mu_i} {X_i} } }\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{i \mathop = 1}^m \paren {S \cap \map {\mu_i} {X_i} }\) | Image of Preimage under Mapping: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds S \cap \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}\) | Intersection Distributes over Union of Family | |||||||||||
\(\ds \) | \(=\) | \(\ds S \cap A\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds S\) | Intersection with Subset is Subset |
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:
\(\ds Y_i\) | \(=\) | \(\ds \map {\mu_i^{-1} } {S \cap \map {\mu_i} {X_i} }\) | by hypothesis | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \map {\mu_i^{-1} } {\map {\mu_i} {X_i} }\) | Preimage of Subset is Subset of Preimage | |||||||||||
\(\ds \) | \(=\) | \(\ds X_i\) | Definition of Isometry (Metric Spaces) and Definition of Bijection |
Hence:
- $\map {\nu_i} {Y_i} \subseteq \map {\nu_i} {X_i}$
and so:
\(\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i}\) | \(\subseteq\) | \(\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}\) | Set Union Preserves Subsets | |||||||||||
\(\ds \) | \(=\) | \(\ds B\) | by hypothesis |
Define:
- $\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i} = T$
Hence the result.
$\blacksquare$