Banach-Tarski Paradox

Theorem
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 chain property of equidecomposability, $$\mathbb{D}^3 \ $$ and $$\mathbb{D}^3 \cup D^3 \ $$ are equidecomposable.