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