[proofplan] The proof proceeds in three stages. First, we exhibit two rotations in $SO(3)$ that generate a free group $F_2$ of rank $2$ (this requires a careful algebraic argument showing that certain rotation matrices produce no nontrivial relations). Second, we apply the Hausdorff paradox: the free group $F_2$ acts on the unit sphere $S^2$, and by choosing representatives of orbits via the Axiom of Choice, we decompose $S^2$ (minus a countable set) into four pieces that can be rearranged into two copies of $S^2$ (minus a countable set). We then absorb the countable exceptional set using a rotation with no fixed points in it. Third, we extend the sphere decomposition to the full ball by radial projection and handle the centre separately. [/proofplan] [step:Find two rotations in $SO(3)$ that generate a free group] Let $\phi := \arccos(1/3)$. Define the rotation matrices \begin{align*} \rho_a &:= \begin{pmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad \rho_b := \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\phi & -\sin\phi \\ 0 & \sin\phi & \cos\phi \end{pmatrix}. \end{align*} These are rotations by angle $\phi$ about the $z$-axis and $x$-axis respectively. We claim that $\rho_a$ and $\rho_b$ generate a free subgroup of $SO(3)$ — that is, no nontrivial reduced word in $\rho_a, \rho_a^{-1}, \rho_b, \rho_b^{-1}$ equals the identity matrix. To verify this, one shows by induction on word length that for any nontrivial reduced word $w$ in $\{\rho_a^{\pm 1}, \rho_b^{\pm 1}\}$, the matrix entries of $w$ are rational polynomials in $\cos\phi = 1/3$ and $\sin\phi = 2\sqrt{2}/3$ of a specific form that prevents $w$ from equalling the identity. Concretely, when $w$ begins with a power of $\rho_a$, the entry $(w)_{31}$ has the form $p \cdot (2\sqrt{2}/3)^n / 3^m$ where $p$ is an integer not divisible by $3$ and $n \ge 1$, so $(w)_{31} \neq 0$ and $w \neq I$. The case where $w$ begins with a power of $\rho_b$ is handled by symmetry (examining the entry $(w)_{12}$). Let $F_2 = \langle \rho_a, \rho_b \rangle \subset SO(3)$ denote this free group of rank $2$. [guided] The existence of a free subgroup in $SO(3)$ is the algebraic foundation of the entire paradox. The angle $\phi = \arccos(1/3)$ is chosen so that $\cos\phi = 1/3$ is rational while $\sin\phi = 2\sqrt{2}/3$ is irrational, and the powers of $3$ in the denominators interact with the $\sqrt{2}$ factors in a way that prevents cancellation. Why is this specific to dimension $\geq 3$? The rotation group $SO(2)$ is abelian, so it contains no free subgroup of rank $2$ (every subgroup of an abelian group is abelian, but $F_2$ is not abelian). This is why the Banach-Tarski paradox fails in $\mathbb{R}^2$: the isometry group is too small to support the paradoxical decomposition. [/guided] [/step] [step:Partition the free group $F_2$ into four pieces and verify the paradoxical decomposition] Denote the generators by $a = \rho_a$ and $b = \rho_b$. Every element of $F_2$ is a unique reduced word in $\{a, a^{-1}, b, b^{-1}\}$. Partition $F_2$ into five sets: \begin{align*} W(a) &:= \{\text{reduced words beginning with } a\}, \\ W(a^{-1}) &:= \{\text{reduced words beginning with } a^{-1}\}, \\ W(b) &:= \{\text{reduced words beginning with } b\}, \\ W(b^{-1}) &:= \{\text{reduced words beginning with } b^{-1}\}, \\ \{e\} &:= \{\text{the identity}\}. \end{align*} The key algebraic identities are: \begin{align*} a \cdot W(a^{-1}) &= F_2 \setminus W(a), \\ b \cdot W(b^{-1}) &= F_2 \setminus W(b). \end{align*} The first holds because left-multiplying a reduced word beginning with $a^{-1}$ by $a$ cancels the leading $a^{-1}$, producing either the identity or a word beginning with $a^{-1}$, $b$, or $b^{-1}$. Conversely, every word not beginning with $a$ becomes a word beginning with $a^{-1}$ after left-multiplication by $a^{-1}$. Therefore: \begin{align*} F_2 &= W(a) \cup \underbrace{a \cdot W(a^{-1})}_{= F_2 \setminus W(a)} = W(a) \cup a \cdot W(a^{-1}), \\ F_2 &= W(b) \cup b \cdot W(b^{-1}). \end{align*} The group $F_2$ is "paradoxical": the four sets $W(a), W(a^{-1}), W(b), W(b^{-1})$ (together with $\{e\}$, which we absorb into $W(a)$ by replacing $W(a)$ with $W(a) \cup \{e\}$) can be rearranged to form two copies of $F_2$. [/step] [step:Transfer the paradox from $F_2$ to the sphere $S^2$] The group $F_2 \subset SO(3)$ acts on $S^2$ by rotation. Define the exceptional set \begin{align*} D := \{p \in S^2 : \exists\, w \in F_2 \setminus \{e\}, \, w(p) = p\}, \end{align*} the set of points fixed by some nontrivial element of $F_2$. Each nontrivial rotation $w \in SO(3)$ fixes exactly two antipodal points (the intersections of its rotation axis with $S^2$), so $D$ is a union of countably many two-point sets (one pair per nontrivial word), hence $D$ is countable. On $S^2 \setminus D$, the action of $F_2$ is free: if $w(p) = p$ for $w \in F_2$ and $p \in S^2 \setminus D$, then $w = e$. By the Axiom of Choice, choose a set $M \subset S^2 \setminus D$ containing exactly one point from each $F_2$-orbit. Then $S^2 \setminus D = \bigsqcup_{w \in F_2} w(M)$, a disjoint union. Define \begin{align*} S_a &:= \bigcup_{w \in W(a) \cup \{e\}} w(M), \quad S_{a^{-1}} := \bigcup_{w \in W(a^{-1})} w(M), \\ S_b &:= \bigcup_{w \in W(b)} w(M), \quad S_{b^{-1}} := \bigcup_{w \in W(b^{-1})} w(M). \end{align*} These four sets partition $S^2 \setminus D$. From the group-level decomposition: \begin{align*} S_a \cup \rho_a(S_{a^{-1}}) &= \bigcup_{w \in W(a) \cup \{e\}} w(M) \cup \bigcup_{w \in a \cdot W(a^{-1})} w(M) = \bigcup_{w \in F_2} w(M) = S^2 \setminus D, \\ S_b \cup \rho_b(S_{b^{-1}}) &= S^2 \setminus D. \end{align*} So $S^2 \setminus D$ is decomposed into four pieces that, after rotation, form two copies of $S^2 \setminus D$. [/step] [step:Absorb the countable exceptional set $D$] The set $D$ is countable. Choose a rotation $\sigma \in SO(3)$ such that $\sigma^n(D) \cap D = \varnothing$ for all $n \ge 1$ (such a rotation exists: since the set of rotation axes that map any point of $D$ back into $D$ after finitely many applications is countable, and the set of all rotation axes is uncountable, a generic axis avoids all conflicts). Define \begin{align*} D' := \bigcup_{n=0}^{\infty} \sigma^n(D). \end{align*} Then $\sigma(D') = \bigcup_{n=1}^{\infty} \sigma^n(D) = D' \setminus D$, so $S^2 = (S^2 \setminus D') \cup D'$ and $\sigma$ maps $D'$ bijectively onto $D' \setminus D$. Therefore \begin{align*} S^2 \setminus D = (S^2 \setminus D') \cup (D' \setminus D) = (S^2 \setminus D') \cup \sigma(D'). \end{align*} This shows $S^2 \setminus D$ is equidecomposable with $S^2$ (using two pieces: $S^2 \setminus D'$ stays fixed and $D'$ is rotated by $\sigma$). Composing with the four-piece decomposition from the previous step, $S^2$ itself admits a paradoxical decomposition into five pieces that form two copies of $S^2$. [/step] [step:Extend from the sphere to the solid ball] For $p \in S^2$ and $0 < t \le 1$, the point $tp$ lies in $B \setminus \{0\}$. Every point of $B \setminus \{0\}$ is uniquely written as $tp$ with $p \in S^2$ and $t \in (0, 1]$. The radial projection \begin{align*} \pi: B \setminus \{0\} &\to S^2 \\ x &\mapsto x/\|x\| \end{align*} is a surjection, and the decomposition of $S^2$ lifts to $B \setminus \{0\}$: if $S^2 = C_1 \sqcup \cdots \sqcup C_5$ and rotations $\sigma_i$ reassemble $C_1, C_2, C_3$ into $S^2$ and $C_4, C_5$ into $S^2$, then defining $\hat{C}_i := \{tp : p \in C_i, \, t \in (0,1]\} = \pi^{-1}(C_i)$, the sets $\hat{C}_1, \ldots, \hat{C}_5$ partition $B \setminus \{0\}$, and the same rotations reassemble them into two copies of $B \setminus \{0\}$. To handle the centre $\{0\}$: absorb it into one of the five pieces by the same orbit-absorption trick used for $D$. Choose a line $\ell$ through $0$ that does not pass through any point of $D$. The intersection $\ell \cap B$ is a diameter. A rotation about an axis perpendicular to $\ell$ by an irrational multiple of $\pi$ maps $0$ into a sequence of distinct points on a circle, allowing absorption of $\{0\}$ into the piece containing that circle. The five sets $A_1, \ldots, A_5$ and five rigid motions $\sigma_1, \ldots, \sigma_5$ so obtained satisfy $B = A_1 \cup \cdots \cup A_5$ (disjoint) and $B = \sigma_1(A_1) \cup \sigma_2(A_2) \cup \sigma_3(A_3) = \sigma_4(A_4) \cup \sigma_5(A_5)$. [/step]