Banach-Tarski Paradox: The unit ball $\mathbb{D}^3 \subset \mathbb{R}^3$ is equi-decomposable to the union of two unit balls.
First part of 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 $\mathbb S^2 = \partial \mathbb D^3$.
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 $r \in \mathbb R_{>0}$, define a function $r^{*}: \mathbb R^3 \to \mathbb R^3$ as $r^{*}(\mathbf x ) = r \mathbf x$, and define the sets:
$$ \displaystyle W = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(A)$$ $$ \displaystyle X = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(B)$$ $$\displaystyle Y = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(C)$$ $$\displaystyle Z = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(D)$$
Can someone please explain what $\mathbb S^2 = \partial \mathbb D^3$ means. Is it just partially differentiating a $3$ dimensional unit ball?
Also, what does $r \in \mathbb R_{>0}$ mean? I have absolutely no idea!!
Let $(X,\tau)$ be a topological space and $A\subseteq X$, one defines: $$\partial A:=\overline{A}\setminus\mathring{A}.$$ One can show that: $$\partial A=\overline{A}\cap\overline{X\setminus A}.$$ In your case, $\partial\mathbb{D}^3$ is indeed the $2$-dimensional unit sphere of $\mathbb{R}^3$.