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A set $$S \subset \mathbb{R}^n \ $$ is decomposable in $$m \ $$ sets $$A_1,\dots, A_m \subset \mathbb{R}^n \ $$ if there exist isometries $$\phi_1,\ldots,\phi_m: \mathbb{R}^n \to \mathbb{R}^n \ $$ such that


 * 1) $$S = \bigcup_{k=1}^m \phi_k(A_k) \ $$
 * 2) For $$i \neq j, \phi_i(A_i) \cap \phi_j(A_j) = \varnothing \ $$

Two sets $$S, T \subset \mathbb{R}^n \ $$ are said to be equidecomposable if there exists a set

$$X = \left\{{A_1, \dots, A_m }\right\} \subset \mathcal{P} \left({\mathbb{R}^n}\right) $$,

the latter being the power set of $$\mathbb{R}^n \ $$, such that both $$S \ $$ and $$T \ $$ are decomposable into the elements of $$X \ $$; that is, there are two sets of isometries, one of which will take the pieces of X and "assemble" them into S, and another set which takes the same sets and "assembles" them into T.

We now begin with the theorems:

Equidecomposable Subsets
Let $$A, B \subseteq \mathbb{R}^n \ $$ be equidecomposable and 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 $$\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\mathbb{R}^n \to \mathbb{R}^n \ $$ such that

$$A = \bigcup_{i=1}^m \mu_i(X_i) \ $$

and

$$B = \bigcup_{i=1}^m \nu_i(X_i) \ $$.

Define

$$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i) ) \ $$

Then

$$\bigcup_{i=1}^m \mu_i(Y_i) = \bigcup_{i=1}^m ( S \cap \mu_i(X_i) ) = S \cap \bigcup_{i=1}^m \mu_i(X_i) = S \cap A = S \ $$

and so $$\left\{{Y_i}\right\}_{i=1}^m \ $$ forms a decomposition of $$S \ $$. But for each $$i \ $$,

$$( S \cap \mu_i(X_i) ) \subseteq \mu_i(X_i) \ $$,

and so,

$$Y_i = \mu_i^{-1} ( S \cap \mu_i(X_i)) \subseteq \mu_i^{-1}(\mu_i(X_i)) = X_i \ $$

Hence

$$\nu_i(Y_i) \subseteq \nu_i(X_i) \ $$,

and so

$$\bigcup_{i=1}^m \nu_i(Y_i) \subseteq \bigcup_{i=1}^m \nu_i(X_i) = B \ $$.

Define $$\bigcup_{i=1}^m \nu_i(Y_i) = T \ $$.

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

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

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 $$\phi_{i,j}:\mathbb{R}^n \to \mathbb{R}^n \ $$ such that

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

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

So then

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

and

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

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

Equidecomposability is an Equivalence Relation
The property of being equidecomposable is an equivalence relation on the power set $$\mathcal{P} \left({\mathbb{R}^n}\right) \ $$.

Relexivity
A set is necessarily equidecomposable with itself; the same decomposition and set of isometries suffice for $$A \ $$ as for $$A \ $$.

Symmetry
There is no order to the relation of being equidecomposable; symmetry follows.

Transitivity
Suppose $$A, B, C \subset \mathbb{R}^n \ $$ are sets such that $$A, B \ $$ are equidecomposable and $$B, C \ $$ are equidecomposable. Let $$X_1, \dots, X_m \ $$ be a decomposition of $$A, B \ $$ together with isometries $$\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\mathbb{R}^n \to \mathbb{R}^n \ $$ such that

$$A = \bigcup_{i=1}^m \mu_i(X_i) \ $$

and

$$B = \bigcup_{i=1}^m \nu_i(X_i) \ $$.

Further let $$Y_1, \dots, Y_p \ $$ together with $$\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p \ $$ be sets and isometries such that

$$B = \bigcup_{i=1}^p \xi_i(Y_i) \ $$

and

$$C = \bigcup_{i=1}^p \tau_i(Y_i) \ $$.

Consider the sets

$$Z_{i,j} = \nu_i(X_i) \cap \xi_j(Y_j) \ $$

where $$ 1 \leq i \leq m \ $$ and $$ 1 \leq j \leq p \ $$.

We have

$$\bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1})(Z_{i,j}) = \bigcup_{i=1}^m \bigcup_{j=1}^p (\mu_i \circ \nu_i^{-1} )( \nu_i(X_i) \cap \xi_j(Y_j) ) = \bigcup_{i=1}^m (\mu_i \circ \nu_i^{-1} \circ \nu_i) (X_i) = \bigcup_{i=1}^m \mu_i(X_i) = A \ $$,

$$\bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1})(Z_{i,j}) = \bigcup_{j=1}^p \bigcup_{i=1}^m (\tau_j \circ \xi_j^{-1}) (\nu_i(X_i) \cap \xi_j(Y_j)) = \bigcup_{j=1}^p (\tau_j \circ \xi_j^{-1} \circ \xi_j) (Y_j) = \bigcup_{j=1}^p \tau_j(Y_j) = C \ $$,

so $$Z_{i,j} \ $$ together with the isometries $$\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1} \ $$ as a decomposition of $$A \ $$ and $$C \ $$, hence these two are equidecomposable.

Nested Equidecomposability
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
I'm working on this still, but it's true!

Hausdorff Paradox
There is a disjoint decomposition of the sphere $$\mathbb{S}^2 \ $$ into four sets $$A, B, C, D \ $$ such that $$A, B, C, B \cup C \ $$ are all congruent and $$D \ $$ is countable.

Proof
Let $$R \subset \mathbb{SO}(3) \ $$ be the group generated by the $$\pi \ $$ and $$\tfrac{2\pi}{3} \ $$ rotations around different axes.

The elements

$$\psi = \begin{pmatrix} -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} & 0 \\ -\tfrac{\sqrt{3}}{2} & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, \phi = \begin{pmatrix} -\text{cos}(\vartheta) & 0 & \text{sin}(\vartheta) \\ 0 & -1 & 0 \\ \text{sin}(\vartheta) & 0 & \text{cos}(\vartheta) \\ \end{pmatrix} \ $$

form a basis for $$R \ $$ for some $$\vartheta \ $$. We have $$\psi^3 = \phi^2 = \mathbf{I}_3 \ $$, so $$\forall r \in R | r \neq \mathbf{I}_3, \psi, \phi, \psi^2 \ $$ some natural number $$n \ $$ and some set of numbers $$m_k \in \left\{{1,2}\right\}, 1 \leq k \leq n \ $$ such that $$r \ $$ can written as one of the following:

$$ \text{a}:

r = \prod_{k=1}^n \phi \psi^{m_k} $$

$$

\text{b}:

r = \psi^{m_1} \left( \prod_{k=2}^n \phi \psi^{m_k} \right)  \phi $$

$$

\text{c}:

r = \left( \prod_{k=1}^n \phi \psi^{m_k} \right) \phi

$$

$$ \text{d}:

r = \psi^{m_1} \left( \prod_{k=2}^n \phi \psi^{m_k} \right) $$

Now we fix $$\vartheta \ $$ such that $$\mathbf{I}_3 \ $$ cannot be written in any of the ways A, B, C, D.

The action of $$R \ $$ on $$\mathbb{S}^2 \ $$ will leave two points unchanged for each element of $$R \ $$ (the intersection of the axis of rotation and the sphere, to be exact).

Since $$R \ $$ is finitely generated, it is a countable group, and so the set of points of $$\mathbb{S}^2 \ $$ which are unchanged by at least one element of $$R \ $$ is also countable.

We call this set $$D \subset \mathbb{S}^2 \ $$, so that $$R \ $$ acts freely on $$\mathbb{S}^2 - D \ $$. This partitions $$\mathbb{S}^2 - D \ $$ into orbits. By the Axiom of Choice, there is a set $$X \ $$ containing one element of each orbit.

For any $$r \in R \ $$, let $$X_r \ $$ be the action of $$r \ $$ on $$X \ $$. We have

$$\mathbb{S}^2 - D = \bigcup_{r \in R} X_r \ $$

Define the sets $$A, B, C \ $$ to be the smallest sets satisfying

$$X \subseteq A \ $$

$$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi} \subset B, A, C, \text{respectively.} \ $$

$$\text{If } X_r \subset A, B, C, \text{ then } X_{r\psi} \subset B, C, A, \text{respectively.} \ $$

$$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi^2} \subset C, A, B, \text{respectively.} \ $$

These sets are defined due to the uniqueness of the properties a-d, and $$A, B, C, B \cup C \ $$ are congruent since they are rotations of each other; namely,

$$A_{\psi} = B, B_{\psi^2} = C, A_{\phi} = B \cup C \ $$.

Hence we have constructed the sets $$A, B, C, D \ $$ of the theorem.

= Banach-Tarski Paradox =

The unit ball $$\mathbb{D}^3 \subset \R^3 \ $$ is equidecomposable to the union of two unit balls.

Proof
Let $$\mathbb{D}^3 \ $$ be centered at the origin, and $$D^3 \ $$ be some other unit ball in $$\R^3 \ $$ such that $$\mathbb{D}^3 \cap D^3 = \varnothing \ $$.

Let $$\mathbb{S}^2 = \partial \mathbb{D}^3 \ $$.

By the Hausdorff Paradox, there exists a decomposition of $$ \mathbb{S}^2 \ $$ into four sets $$A, B, C, D \ $$ such that $$A, B, C, \ $$ and $$B \cup C \ $$ are congruent, and $$D \ $$ is countable.

For $$r \in \R_{>0} \ $$, define a function $$r^{*} : \R^3 \to \R^3 \ $$ as $$r^{*}(\mathbf x ) = r \mathbf x \ $$, and define the sets

$$ W = \bigcup_{0<r\leq 1} r^{*}(A) \ $$

$$ X = \bigcup_{0<r \leq 1} r^{*}(B) \ $$

$$ Y = \bigcup_{0<r \leq 1} r^{*}(C) \ $$

$$ Z = \bigcup_{0<r \leq 1} r^{*}(D) \ $$

Let $$T = W \cup Z \cup \left\{{ \mathbf 0 }\right\} \ $$.

$$W \ $$ and $$X \cup Y \ $$ are clearly congruent by the congruency of $$A \ $$ with $$B \cup C \ $$, hence $$W \ $$ and $$X \cup Y \ $$ are equidecomposable.

Since $$X \ $$ and $$Y \ $$ are congruent, and $$W \ $$ and $$X \ $$ are congruent, $$X \cup Y \ $$ and $$W \cup X \ $$ are equidecomposable.

$$W \ $$ and $$X \cup Y \ $$ as well as $$X \ $$ and $$W \ $$ are congruent, so $$W \cup X \ $$ and $$W \cup X \cup Y \ $$ are equidecomposable.

Hence $$W \ $$ and $$W \cup X \cup Y \ $$ are equidecomposable, by the fact that equidecomposability is an equivalence relation and hence $$T \ $$ and $$\mathbb{D}^3 \ $$ are equidecomposable, since unions of equidecomposable sets are equidecomposable.

Similarly we find $$X \ $$, $$Y \ $$, and $$W \cup X \cup Y \ $$ are equidecomposable.

Since $$D \ $$ is only countable, but $$\mathbb{SO}(3) \ $$ is not, $$\exists \phi \in \mathbb{SO}(3) \ $$ such that $$\phi(D) \subset A \cup B \cup C \ $$ so that $$I = \phi (D) \subset W \cup X \cup Y \ $$. Since $$X \ $$ and $$W \cup X \cup Y \ $$ are equidecomposable, by a theorem on equidecomposability and subsets, $$\exists H \subseteq X \ $$ such that $$H \ $$ and $$I \ $$ are equidecomposable.

Finally, let $$p \in X - H \ $$ be a point and define $$S = Y \cup H \cup \left\{{p}\right\} \ $$. Since $$Y \ $$ and $$W \cup X \cup Y \ $$, $$H \ $$ and $$Z \ $$, $$\left\{{0}\right\} \ $$ and $$\left\{{p}\right\} \ $$ are all equidecomposable in pairs, $$S \ $$ and $$\mathbb{B}^3 \ $$ are equidecomposable by the equidecomposability of unions.Since $$D^3 \ $$ and $$\mathbb{D}^3 \ $$ are congruent, $$D^3 \ $$ and $$S \ $$ are equidecomposable, since equidecomposability is an equivalence relation.

By the equidecomposability of unions, $$T \cup S \ $$ and $$\mathbb{D}^3 \cup D^3 \ $$ are equidecomposable. Hence $$T \cup S \subseteq \mathbb{D}^3 \subset \mathbb{D}^3 \cup D^3 \ $$ are equidecomposable and so, by the "nested" property of equidecomposability, $$\mathbb{D}^3 \ $$ and $$\mathbb{D}^3 \cup D^3 \ $$ are equidecomposable.