Mathematics

The Banach-Tarski Paradox: Doubling a Sphere with Pure Mathematics

We explain the Banach-Tarski paradox — how a solid ball in three-dimensional space can be decomposed into finitely many pieces and reassembled into two balls identical to the original — and explore the role of the Axiom of Choice.

Set Theory Geometry

The Theorem

The Banach-Tarski Paradox (1924)

A solid ball $B^3 \subset \mathbb{R}^3$ can be partitioned into finitely many pieces which can be rearranged — using only rotations and translations — to form two solid balls, each identical in size to the original.

More precisely: there exist disjoint sets $A_1, \ldots, A_k, B_1, \ldots, B_m$ with $B^3 = A_1 \sqcup \cdots \sqcup A_k \sqcup B_1 \sqcup \cdots \sqcup B_m$ , and rigid motions $g_1, \ldots, g_k, h_1, \ldots, h_m$ such that:

$g_1(A_1) \sqcup \cdots \sqcup g_k(A_k) = B^3 \quad \text{and} \quad h_1(B_1) \sqcup \cdots \sqcup h_m(B_m) = B^3$

The minimum number of pieces required is five (proved by Raphael Robinson in 1947).

Why It Sounds Impossible

Our physical intuition rebels: how can you break a ball into pieces and get two balls of the same size? The answer lies in the nature of the pieces.

The pieces in the Banach-Tarski decomposition are not measurable in the sense of Lebesgue measure. They are so wildly scattered and fragmented that they have no well-defined volume. The "paradox" is not that volume is doubled — it is that the pieces have no volume at all, so the usual conservation of volume does not apply.

Paradoxical Decompositions and Free Groups

The proof begins with a paradoxical decomposition of the free group on two generators.

The Free Group $F_2$

Let $F_2 = \langle a, b \rangle$ be the free group on two generators. Every element is a reduced word in $\{a, a^{-1}, b, b^{-1}\}$ , for example $aba^{-1}b^2$ .

Lemma (Paradoxical decomposition of $F_2$ ).

Define:

$W(a)$ = set of reduced words starting with $a$
$W(a^{-1})$ = set of reduced words starting with $a^{-1}$
$W(b)$ = set of reduced words starting with $b$
$W(b^{-1})$ = set of reduced words starting with $b^{-1}$

Then $F_2 = \{e\} \sqcup W(a) \sqcup W(a^{-1}) \sqcup W(b) \sqcup W(b^{-1})$ , and:

$F_2 = W(a) \sqcup a \cdot W(a^{-1}) \qquad \text{and} \qquad F_2 = W(b) \sqcup b \cdot W(b^{-1})$

Proof. For the first equation: every reduced word either starts with $a$ (so it is in $W(a)$ ) or does not. If it does not start with $a$ , then prepending $a^{-1}$ gives a word starting with $a^{-1}$ , so the original word is $a \cdot w$ for some $w \in W(a^{-1})$ . The identity and words starting with $b^{\pm 1}$ are all captured by $a \cdot W(a^{-1})$ . $\square$

This means $F_2$ can be "cut" into four pieces and reassembled (using left multiplication) into two disjoint copies of $F_2$ .

From $F_2$ to $\mathrm{SO}(3)$

The next step is to embed $F_2$ into the rotation group of $\mathbb{R}^3$ .

Lemma. There exist rotations $A, B \in \mathrm{SO}(3)$ that generate a free subgroup. For example:

$A = \begin{pmatrix} 1/3 & -2\sqrt{2}/3 & 0 \\ 2\sqrt{2}/3 & 1/3 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/3 & -2\sqrt{2}/3 \\ 0 & 2\sqrt{2}/3 & 1/3 \end{pmatrix}$

These are rotations by $\arccos(1/3)$ about the $z$ -axis and $x$ -axis respectively.

The fact that $\mathrm{SO}(3)$ contains a free subgroup is what makes the paradox possible in three dimensions. In one and two dimensions, the rotation group is abelian, and no such embedding exists — which is why Banach-Tarski fails in dimensions 1 and 2.

Proof Outline

Step 1 — Paradoxical decomposition of $S^2 \setminus D$ .

Let $\langle A, B \rangle \cong F_2 \subset \mathrm{SO}(3)$ and let $D \subset S^2$ be the countable set of points with a non-trivial stabilizer under the action of $\langle A, B \rangle$ . The group acts freely on $S^2 \setminus D$ , so the paradoxical decomposition of $F_2$ transfers to $S^2 \setminus D$ .

Step 2 — Absorb the countable set $D$ .

Using the Axiom of Choice, choose a rotation $\rho$ of infinite order whose axis avoids $D$ . Then $D \subset \rho(S^2 \setminus D) \cup \rho^2(S^2 \setminus D) \cup \cdots$ , allowing us to absorb $D$ into the decomposition.

Step 3 — Extend from $S^2$ to $B^3$ .

Each point of the solid ball (except the center) lies on a unique ray from the origin through a point of $S^2$ . The paradoxical decomposition of $S^2$ extends radially to all of $B^3 \setminus \{0\}$ . The center is absorbed similarly to Step 2.

Step 4 — Conclusion.

We have partitioned $B^3$ into finitely many pieces and reassembled them into two copies of $B^3$ using rigid motions. $\square$

The Role of the Axiom of Choice

The Axiom of Choice is essential for the Banach-Tarski paradox. In models of set theory where the Axiom of Choice fails (e.g., Solovay's model, 1970), every subset of $\mathbb{R}^n$ is Lebesgue measurable, and paradoxical decompositions are impossible.

Specifically, the Axiom of Choice is used to:

Select orbit representatives in Step 2,
Construct the non-measurable pieces in Step 1.

This is one of the most dramatic consequences of the Axiom of Choice and a major reason why some mathematicians are wary of accepting it.

Why Dimension Matters

Dimension	Rotation Group	Contains $F_2$ ?	Banach-Tarski?
1	$\{1, -1\}$	No	No
2	$\mathrm{SO}(2) \cong S^1$	No (abelian)	No
$\geq 3$	$\mathrm{SO}(n)$	Yes	Yes

In dimensions 1 and 2, the Banach-Tarski paradox does not hold. The key difference: $\mathrm{SO}(2)$ is abelian, while $\mathrm{SO}(3)$ is not. The Hausdorff paradox (1914) — a predecessor that decomposes $S^2$ minus a countable set — was the first indication that dimension 3 behaves differently.

Consequences for Measure Theory

The Banach-Tarski paradox proves a foundational negative result:

There is no finitely additive, rotation-invariant measure defined on all subsets of $\mathbb{R}^3$ that assigns the unit ball a positive finite value.

This is why Lebesgue measure cannot be extended to all subsets of $\mathbb{R}^n$ for $n \geq 3$ — non-measurable sets are unavoidable.

In dimensions 1 and 2, the situation is different: Banach (1923) showed that a finitely additive, isometry-invariant measure on all subsets does exist (though it is not countably additive).

The Von Neumann Perspective

John von Neumann (1929) clarified the paradox by identifying the algebraic root cause: the group of isometries. A group $G$ is amenable if it admits a finitely additive, left-invariant probability measure on all its subsets.

Abelian groups are amenable — no paradox in dimensions 1 and 2.
$F_2$ is not amenable — paradox in dimension $\geq 3$ .

The study of amenable groups, initiated by von Neumann, has become a major area in geometric group theory and operator algebras.

Summary

\begin{aligned} &F_2 = W(a) \sqcup aW(a^{-1}) = W(b) \sqcup bW(b^{-1}) \\[6pt] &\Downarrow \quad \text{embed } F_2 \hookrightarrow \mathrm{SO}(3) \\[6pt] &S^2 \setminus D \text{ inherits paradoxical decomposition} \\[6pt] &\Downarrow \quad \text{Axiom of Choice absorbs } D \\[6pt] &S^2 \text{ is paradoxical} \implies B^3 \text{ is paradoxical} \\[6pt] &\Downarrow \\[6pt] &\text{One ball} = \text{Two balls (using 5 pieces)} \end{aligned}

References

Wagon, S., The Banach-Tarski Paradox, Cambridge University Press, 1993.
Tomkowicz, G. and Wagon, S., The Banach-Tarski Paradox, 2nd edition, Cambridge University Press, 2016.
Wikipedia — Banach-Tarski paradox
Vsauce — "The Banach-Tarski Paradox" (YouTube)
Wikipedia — Amenable group
Paterson, A., Amenability, AMS Mathematical Surveys and Monographs, 1988.

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