Banach-Tarski Paradox/Proof 2

Lemmata
Let $U$ denote the closed ball in real Euclidean space of $3$ dimensions defined as:


 * $U = \set {\mathbf x \in \R^3 : \size {\mathbf x - \mathbf c} \le r}$

where:
 * $r$ is the radius of $U$
 * $\mathbf x$ is the position vector of a point $\tuple {x_1, x_2, x_3} \in \R^3$
 * $\mathbf c$ is the position vector of the center $\tuple {c_1, c_2, c_3}$ of $U$
 * $\size {\mathbf x - \mathbf c}$ denotes the Euclidean distance:
 * $\size {\mathbf x - \mathbf c} = \sqrt { {x_1 - c_1}^2 + {x_2 - c_2}^2 + {x_3 - c_3}^2}$

Let $\Bbb S^2$ denote the sphere which is the boundary of $U$:
 * $\Bbb S^2 = \set {\mathbf x \in \R^3 : \size {\mathbf x - \mathbf c} = r}$

Let:
 * $(1): \quad \Bbb S^2 = A \cup B \cup C \cup Q$

be the partition of $\Bbb S^2$ from Hausdorff Paradox.

Recall:
 * $A$, $B$ and $C$ are disjoint
 * $A$, $B$ and $C$ are congruent to each other
 * $B \cup C$ is congruent to each of $A$, $B$ and $C$

Hence :
 * $A \approx B \approx C \approx B \cup C$

For a given subset $T \subseteq \Bbb S^2$, let $\overline T \subseteq U$ be defined as the set of all $\mathbb x \in U$ apart from $\mathbf c$ whose projection onto $\Bbb S^2$ from $\mathbf c$ is $T$.

Hence using Vector Equation of Straight Line in Space:
 * $\overline T = \set {\mathbf y \in U: \mathbf y = \paren {1 - s} \mathbf c + s \mathbf x: \mathbf x \in T, s \in \hointl 0 1}$

That is, for all points $\mathbf x$ in $T$, all the points in $U$ along the radius of $U$ to $\mathbf x$ but not including the center of $U$.

We have that:
 * $U = \overline A \cup \overline B \cup \overline C \cup \overline Q \cup \set {\mathbb c}$

Because we have:
 * $A \approx B \approx C \approx B \cup C$

it follows that:
 * $(2): \quad \overline A \approx \overline B \approx \overline C \approx \overline B \cup \overline C$

Let:

From $(2)$ and :


 * $\overline A \approx \overline A \cup \overline B \cup \overline C$

and so:
 * $(3): \quad X \approx U$

Let $\alpha$ be a rotation of $U$ such that $\alpha$ is not in the original $G$ that was used to construct $(1)$ with the property that:
 * $Q \cap \alpha Q = \O$

Hence using:
 * $\overline C \approx \overline A \cup \overline B \cup \overline C$

there exists $S \subseteq C$ such that:
 * $\overline S \approx \overline Q$

Let $p$ be an arbitrary point in $\overline C \setminus \overline S$.

We have that:
 * $(4): \quad \overline A \cup \overline Q \cup \set {\mathbb c} \approx \overline B \cup \overline S \cup \set p$

Because:
 * $\overline B \cup \overline S \cup \set p \subseteq Y \subseteq U$

we can use $(3)$ and $(4)$ along with to get:
 * $Y \approx U$

Hence the result.