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:
 * $\displaystyle W = \bigcup_{0 < r \le 1} r^{*}(A)$
 * $\displaystyle X = \bigcup_{0 < r \le 1} r^{*}(B)$
 * $\displaystyle Y = \bigcup_{0 < r \le 1} r^{*}(C)$
 * $\displaystyle Z = \bigcup_{0 < r \le 1} r^{*}(D)$

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.

Ao $T$ 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, we have:
 * $\exists \phi \in \mathbb{SO}(3): \phi \left({D}\right) \subset A \cup B \cup C$

so that $I = \phi \left({D}\right) \subset W \cup X \cup Y$.

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: are all equidecomposable in pairs, $S$ and $\mathbb B^3$ are equidecomposable by the equidecomposability of unions.
 * $Y$ and $W \cup X \cup Y$,
 * $H$ and $Z$,
 * $\left\{{0}\right\}$ and $\left\{{p}\right\}$

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.

Conclusion
Whether this is a veridical paradox or an antinomy is being hotly discussed to the present day. There are even Facebook groups anti and pro.

Pick your personal choice of philosophy and start ranting.

If you feel really adventurous, check out Rudy Rucker's novel White Light.

They raised this question in a collaborative paper in 1924.