Equidecomposability Unaffected by Union

Theorem

Suppose $\left\{{S_1, \dots, S_m}\right\}, \left\{{T_1, \dots, T_m }\right\}$ are collections of point sets in $\R^n$ such that for each $k \in \left\{{1, \dots, m}\right\}, S_k$ and $T_k$ are equidecomposable.

Then the set $\displaystyle S = \bigcup_{i=1}^m S_i$ is equidecomposable with $\displaystyle T = \bigcup_{i=1}^m T_i$.

Proof

We have for each $k \in \left\{{1, \dots, m}\right\}$ a decomposition $\left\{{S_{k,1}, \dots, S_{k,l_k}}\right\}$ and set of isometries $\phi_{i,j}:\R^n \to \R^n$ such that:

$\displaystyle S_k = \bigcup_{a=1}^{l_k} \phi_{k,a}(S_{k,a})$

and similarly for $T_k$ and some isometries $\theta_{i,j}:\R^n \to \R^n$.

So then

$\displaystyle S = \bigcup_{k=1}^m \bigcup_{i=1}^{l_k} \phi_{k,i}(S_{k,i})$

and

$\displaystyle T = \bigcup_{k=1}^m \bigcup_{i=1}^{l_k} \theta_{k,i}(T_{k,i})$

but since be definition, each of the $S_{a,b}, T_{a,b}$ are congruent, they yield equivalent decompositions of $S$ and $T$.

$\blacksquare$