Banach-Tarski Paradox/Proof 1

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

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

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

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

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

$W$ 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 Equidecomposability is Equivalence Relation.

So $T$ and $\mathbb D^3$ are equidecomposable, from Equidecomposability Unaffected by Union.

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

Since $Q$ is only countable, but $\map {\operatorname {SO} } 3$ is not, we have:
 * $\exists \phi \in \map {\operatorname {SO} } 3: \map \phi Q \subset A \cup B \cup C$

so that $I = \map \phi Q \subset W \cup X \cup Y$.

Since $X$ and $W \cup X \cup Y$ are equidecomposable, by Subsets of Equidecomposable Subsets are Equidecomposable, $\exists H \subseteq X$ such that $H$ and $I$ are equidecomposable.

Finally, let $p \in X - H$ be a point and define $S = Y \cup H \cup \set p$.

Since:
 * $Y$ and $W \cup X \cup Y$
 * $H$ and $Z$
 * $\set 0$ and $\set p$

are all equidecomposable in pairs, $S$ and $\mathbb B^3$ are equidecomposable by Equidecomposability Unaffected by Union.

Since $D^3$ and $\mathbb D^3$ are congruent, $D^3$ and $S$ are equidecomposable, from Equidecomposability is Equivalence Relation.

By Equidecomposability Unaffected by Union, $T \cup S$ 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.