Theorem (Banach-Tarski Paradox): The unit ball $\mathbb{D}^3 \subset \mathbb{R}^3$ is equi-decomposable 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 $\mathbb R^3$ such that $\mathbb D^3 \cap D^3 = \varnothing$.
Let $\mathbb S^2 = \partial \mathbb D^3$ (Its boundary $\partial \mathbb D^3$ is the surface of that ball, which is a sphere. It is the bounding surface of the ball. It can be said it is the $2$-dimensional unit sphere of $\mathbb R^3$).
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 $r \in \mathbb R_{>0}=(0,+ \infty)$, define a function $r^{*}: \mathbb R^3 \to \mathbb R^3$ as $r^{*}(\mathbf x ) = r \mathbf x$, and define the sets:
$$ \displaystyle W = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(A)$$ $$ \displaystyle X = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(B)$$ $$\displaystyle Y = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(C)$$ $$\displaystyle Z = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(D)$$
Let $T = W \cup Z \cup \left\{{ \mathbf 0 }\right\}$.
Since $A$ and $B \cup C$ are congruent, then this means $W$ and $X \cup Y$ are clearly congruent, 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 $D$ is only countable, but $\mathbb{SO}(3)$ is not, we have:
$$ \exists \phi \in \mathbb{SO} (3): \phi (D) \subset A \cup B \cup C $$
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:
$Y$, $W \cup X \cup Y$
$H$, $Z$
$\left\{{0}\right\}$, $\left\{{p}\right\}$
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.
This is a complete proof of the BT paradox which i have a lot of questions about. I think i understand everything until the (*).
What does the $\mathbb{SO}(3)$ mean???
Also if two sets are congruent, does that mean they are equidecomposable? Because it seems like that is what it is above (*).