The Banach–Tarski Paradox

Is the axiom of choice to blame? This is the standard reading, and it is roughly right, but the details matter.

Robert Solovay showed in 1970 that — assuming the consistency of an inaccessible cardinal — there is a model of ZF plus dependent choice in which every set of reals is Lebesgue measurable. In that world there are no non-measurable sets and the Banach–Tarski decomposition does not exist. So the paradox genuinely requires more choice than ZF + DC provides; you cannot get it from the countable choice that ordinary analysis leans on.

But it does not need the full axiom either. Janusz Pawlikowski proved in 1991 that the Banach–Tarski paradox follows from the Hahn–Banach theorem, which is strictly weaker than AC. Hahn–Banach is one of the load-bearing results of functional analysis and almost nobody proposes discarding it. So "just drop choice and the paradox goes away" costs more than it first appears.

Is non-measurability the real culprit? Not entirely, and this is the result that most often surprises people who think they have the paradox filed away. In 1994 Randall Dougherty and Matthew Foreman proved that the pieces can be chosen to have the Baire property — solving a problem of Marczewski from 1930 — and, more strikingly, that the ball can be split using open sets whose union is dense. So the decomposition does not require pathology in every sense at once. It requires non-measurability, but topological respectability is achievable alongside it.

Can the pieces be moved continuously? The usual reassurance is that the motions are abstract rearrangements rather than physical journeys. That reassurance is wrong. Trevor Wilson proved in 2005 that the pieces can be moved along continuous paths, never overlapping at any moment, from the initial ball to the two final balls. This settles a problem of de Groot. The pieces really can be "carried" — you simply cannot describe, build or see them.

How few pieces? Raphael Robinson showed in 1947 that five pieces suffice and that four do not. One of the five is a single point, needed to handle the centre of the ball; if that irritating detail is set aside, four pieces do the work.

What working mathematicians conclude. The overwhelming majority view is that the theorem is a fact about the interaction of the axiom of choice with the group of Euclidean motions, and that the correct response was the one measure theory actually took: define measure on a σ-algebra rather than on all subsets. A minority — constructivists in the tradition of Brouwer and Bishop, and some predicativists — regard the theorem as a symptom of a set theory that permits objects with no content, and reject the framework rather than accommodate the result. Nobody disputes the derivation.

A note on the word "paradox". Banach and Tarski did not call their result a paradox in the 1924 paper; they called it a decomposition theorem, and presented it as part of a systematic study of the measure problem, which was a central topic of the Lwów and Warsaw schools. The name accumulated afterwards. There is no contradiction anywhere in the mathematics, which is why this entry is filed as a counterintuitive result rather than a logical one.

1904. Ernst Zermelo published the axiom of choice, and the well-ordering theorem it delivers, provoking an immediate international argument about non-constructive existence proofs.

1905. Giuseppe Vitali constructed the first non-measurable set, choosing one representative from each coset of the rationals inside the reals. This is the prototype for everything that follows: the whole line splits into countably many congruent copies of the Vitali set, whose measures would have to sum to a finite positive number while all being equal — impossible.

1914. Felix Hausdorff proved the Hausdorff paradox in Mathematische Annalen: the sphere, with a countable set of points removed, can be divided into three pieces $A$, $B$, $C$ such that $A$, $B$, $C$ and $B \cup C$ are all congruent. The consequence he drew was the one he cared about — there is no finitely additive, rotation-invariant measure defined on all subsets of the sphere. Everything essential in Banach–Tarski is already present here, including the free group of rotations.

1923. Banach proved the complementary positive result, in Fundamenta Mathematicae: on the line and in the plane, finitely additive, isometry-invariant measures defined on all bounded sets do exist, extending Lebesgue measure. This is why the paradox is a three-dimensional phenomenon, and it was known a year before the paradox itself.

1924. Banach and Tarski published "Sur la décomposition des ensembles de points en parties respectivement congruentes" in Fundamenta Mathematicae 6, pages 244–277. They removed Hausdorff's countable exceptional set, extended the result from the sphere to the solid ball, and proved the general statement: any two bounded subsets of $\mathbb{R}^n$ with non-empty interior, for $n \ge 3$, are equidecomposable. The pea and the Sun is a corollary, not a metaphor.

1929. John von Neumann isolated the property that actually matters. A group admitting a finitely additive, left-invariant probability measure — later called amenable, a term coined by Mahlon Day in 1949 — cannot act paradoxically. The isometry groups of $\mathbb{R}$ and $\mathbb{R}^2$ are solvable and therefore amenable; that of $\mathbb{R}^3$ is not, because $SO(3)$ contains a free group on two generators. The dimensional divide is a group-theoretic fact, not a geometrical one.

1938. Tarski proved the converse in full generality: a group action is paradoxical if and only if no finitely additive invariant measure exists on all subsets. The paradox and the failure of measure are the same statement.

1947. Robinson established that five pieces are necessary and sufficient.

1970 and after. Solovay's model, Pawlikowski's Hahn–Banach result (1991), Laczkovich's solution of Tarski's circle-squaring problem (1990), Dougherty–Foreman on the Baire property (1994) and Wilson's continuous motion (2005) mapped out exactly how much is needed and what the pieces can be made to look like.

It settled the measure problem. Before 1924, an open question was whether Lebesgue measure could be extended to a countably additive, isometry-invariant measure on all subsets of $\mathbb{R}^n$ that gives the unit cube measure 1. Banach–Tarski answers no for $n \ge 3$, and does so without any appeal to countable additivity — even finite additivity fails. The response is visible in every measure theory course: measure is defined on a σ-algebra of measurable sets, and the restriction is not fastidiousness but necessity. Probability theory inherits it directly. An event, in Kolmogorov's axioms, is a measurable set, and the reason is here.

It created amenability. Von Neumann's analysis of the paradox produced one of the most productive concepts in twentieth-century mathematics. Amenability is now central to geometric group theory, operator algebras, ergodic theory and random walks. The Følner condition, Tarski numbers, the Grigorchuk groups of intermediate growth and the von Neumann conjecture — refuted by Ol'shanskii in 1980, who built a non-amenable group with no free subgroup — all descend from a question first asked about a decomposed sphere.

It is the reference point in the argument over the axiom of choice. AC has consequences that are pleasant (every vector space has a basis, Tychonoff's theorem, Hahn–Banach) and consequences that are unpleasant, and this is the canonical unpleasant one. Any philosophical position on non-constructive existence has to say something about it. Gödel's 1938 consistency proof and Cohen's 1963 independence proof mean the choice is genuinely a choice.

It disciplines intuition about infinite sets. The instinct that a set "has a size", that cutting and reassembling preserves it, and that pieces of a solid are themselves solids are all reliable in ordinary experience and all false in general. This theorem is the sharpest available demonstration that geometric intuition is a guide to measurable sets specifically, not to sets.

Measure theory and probability, daily. Every time a proof says "let $A$ be measurable", this theorem is the reason the qualifier is there. Stochastic processes, integration and functional analysis are all built on a σ-algebra rather than a power set because of what happens when you do not.

Geometric group theory. Amenability governs whether a group admits invariant means, whether random walks on it return, whether its Cayley graph has exponential growth. Growth, Følner sequences, the Tarski number of a group and paradoxical decompositions of graphs are an active research area whose founding example is the ball.

Measurable combinatorics. A recent and thriving line of work asks which combinatorial constructions on graphs and group actions can be done measurably — with Borel or measurable pieces rather than arbitrary ones. Marks and Unger's measurable solution to Tarski's circle-squaring problem (2017) is a direct descendant, and the field's guiding question is precisely where the Banach–Tarski phenomenon can and cannot be tamed.

A caution about popularisation. The theorem is regularly misreported as showing that matter can be duplicated, that physics is broken, or that mathematics is unreliable. None of these follow. The most useful thing a reader can take away is the discipline of asking what the words in a theorem actually denote — here, that "piece" means an arbitrary set of points and "volume" means a function defined only where it can be.

Statement

Theorem (Banach–Tarski, 1924). Let $B \subset \mathbb{R}^3$ be a closed ball. There is a partition $B = A_1 \sqcup \cdots \sqcup A_5$ and isometries $g_1, \dots, g_5 \in E(3)$ such that

$$g_1 A_1 \sqcup g_2 A_2 ;=; B, \qquad g_3 A_3 \sqcup g_4 A_4 \sqcup g_5 A_5 ;=; B$$

Strong form. If $X, Y \subseteq \mathbb{R}^n$ are bounded with non-empty interior and $n \ge 3$, then $X$ and $Y$ are equidecomposable: $X \sim Y$.

Write $X \sim Y$ when $X$ and $Y$ admit finite partitions into pairwise congruent pieces. Equidecomposability is an equivalence relation, and the Banach–Schröder–Bernstein theorem holds for it: if $X$ is equidecomposable with a subset of $Y$ and $Y$ with a subset of $X$, then $X \sim Y$. This is what reduces piece counts to the minimum.

A set $X$ is $G$-paradoxical for a group $G$ acting on it if there are disjoint $A_1, \dots, A_m, B_1, \dots, B_n \subseteq X$ and elements $g_i, h_j \in G$ with

$$X = \bigcup_i g_i A_i = \bigcup_j h_j B_j$$

Step 1 — the free group $F_2$ is paradoxical

Let $F_2 = \langle a, b \rangle$ be the free group of rank 2: elements are reduced words in $a, a^{-1}, b, b^{-1}$, with no letter adjacent to its inverse. For a letter $x$ let $W(x)$ be the set of reduced words beginning with $x$. Then

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

Every $w \in W(a^{-1})$ has the form $a^{-1}u$ with $u$ reduced and not beginning with $a$, so

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

and therefore

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

Four disjoint sets, two group elements, two copies of $F_2$. No choice is used here and nothing infinite is invoked beyond the group itself. This is the paradox. Everything else is transport.

Note why nothing like this can happen in an abelian, or even a solvable, group: a free subgroup of rank 2 is the source of the duplication, and amenable groups contain none.

Step 2 — embedding $F_2$ in $SO(3)$

Let $\theta = \arccos(1/3)$ and take rotations by $\theta$ about the $x$- and $z$-axes:

$$A = \begin{pmatrix} 1 & 0 & 0 \ 0 & \tfrac13 & -\tfrac{2\sqrt2}{3} \ 0 & \tfrac{2\sqrt2}{3} & \tfrac13 \end{pmatrix}, \qquad B = \begin{pmatrix} \tfrac13 & -\tfrac{2\sqrt2}{3} & 0 \ \tfrac{2\sqrt2}{3} & \tfrac13 & 0 \ 0 & 0 & 1 \end{pmatrix}$$

A short induction on word length shows that any non-empty reduced word of length $k$ in $A^{\pm 1}, B^{\pm 1}$, applied to the vector $(1,0,0)$, yields

$$\frac{1}{3^{k}}\left(p,\ q\sqrt{2},\ r\right) \quad \text{with } p,q,r \in \mathbb{Z} \text{ and } 3 \nmid q$$

Since $q \neq 0$, the result is never $(1,0,0)$, so no non-empty reduced word can be the identity matrix. Hence

$$\langle A, B \rangle \cong F_2 \le SO(3)$$

This is the geometric content of the theorem, and it is why three dimensions are required: $SO(2)$ is abelian and the isometry groups $E(1)$ and $E(2)$ are solvable, so neither contains a copy of $F_2$.

Step 3 — from the group to the sphere, using choice

$F_2$ acts on $S^2$. Each non-identity rotation fixes exactly two points, and $F_2$ is countable, so the set

$$D = {, x \in S^2 : gx = x \text{ for some } g \neq e ,}$$

is countable. On $S^2 \setminus D$ the action is free, so the orbits partition it and every orbit is a faithful copy of $F_2$.

Here the axiom of choice enters, and only here. Choose one point from each orbit, giving a set $M \subseteq S^2 \setminus D$. Then

$$S^2 \setminus D = \bigsqcup_{g \in F_2} gM$$

There are $2^{\aleph_0}$ orbits and no definable rule selecting representatives; $M$ exists by fiat. Transporting the decomposition of Step 1:

$$\mathcal{A} = \bigsqcup_{g \in W(a)} gM, \quad \mathcal{A}' = \bigsqcup_{g \in W(a^{-1})} gM, \quad \mathcal{B} = \bigsqcup_{g \in W(b)} gM, \quad \mathcal{B}' = \bigsqcup_{g \in W(b^{-1})} gM$$

gives $S^2 \setminus D = \mathcal{A} \sqcup A\mathcal{A}'$ and $S^2 \setminus D = \mathcal{B} \sqcup B\mathcal{B}'$, up to the orbit of $M$ itself. This is Hausdorff's 1914 result.

Step 4 — absorbing the exceptional set (the hotel again)

$D$ is countable, so there is an axis missing $D$ and an angle such that the rotation $\rho$ about it satisfies $\rho^m D \cap \rho^n D = \emptyset$ for all $m \neq n \ge 0$ — only countably many angles are excluded, and there are uncountably many available. Put

$$D^* = \bigcup_{n \ge 0} \rho^n D$$

Then $\rho D^* = D^* \setminus D$, so

$$S^2 = (S^2 \setminus D^) \sqcup D^ \ \sim\ (S^2 \setminus D^) \sqcup \rho D^ = S^2 \setminus D$$

using two pieces and one rotation. This is Hilbert's hotel executed on a sphere: an infinite set absorbs a countable annoyance by shifting along one notch. Hence $S^2 \sim S^2 \setminus D$ and $S^2$ is $SO(3)$-paradoxical.

Step 5 — sphere to ball

Extend radially: each $x \in S^2$ carries the open segment ${tx : 0 < t \le 1}$, so a paradoxical decomposition of $S^2$ lifts to one of $B \setminus {0}$.

The centre is handled by one more shift. Choose a rotation $\sigma$ about an axis through a point at distance $1/2$ from the origin, by an angle that is an irrational multiple of $\pi$. Then ${\sigma^n 0 : n \ge 0}$ is an infinite set of distinct points inside $B$, and

$$B \setminus {0} \sim B$$

by the same absorption. Applying Banach–Schröder–Bernstein and optimising yields Robinson's five pieces, of which one is ${0}$ alone.

Non-measurability is forced

Suppose the pieces $A_1, \dots, A_5$ were Lebesgue measurable. Lebesgue measure $\lambda$ on $\mathbb{R}^3$ is countably additive and invariant under $E(3)$, so

$$\lambda(B) = \sum_{i=1}^5 \lambda(A_i) = \lambda(g_1A_1) + \lambda(g_2A_2) + \lambda(g_3A_3) + \lambda(g_4A_4) + \lambda(g_5A_5) = \lambda(B) + \lambda(B)$$

giving $\lambda(B) = 2\lambda(B)$ with $0 < \lambda(B) = \tfrac43\pi r^3 < \infty$. Contradiction. At least one piece is non-measurable, and in fact they cannot all be measurable at any stage of the construction.

The sets $gM$ are exactly Vitali sets transplanted to the sphere: $M$ is a choice of coset representatives for the $F_2$-action, just as Vitali's set is a choice of coset representatives for $\mathbb{Q}$ acting on $\mathbb{R}$ by translation.

There is a converse, due to Tarski (1938): a group $G$ acting on $X$ makes $X$ paradoxical if and only if there is no finitely additive $G$-invariant measure $\mu$ on all subsets of $X$ with $\mu(X) = 1$. Paradoxicality and the failure of measure are not merely related; they are the same condition.

Why dimensions 1 and 2 are safe

Von Neumann's 1929 analysis gives the clean criterion. A group $G$ is amenable if there is a finitely additive, left-invariant probability measure on all subsets of $G$; equivalently, an invariant mean on $\ell^\infty(G)$.

Facts:

1. Abelian groups are amenable, and amenability is preserved by subgroups, quotients, extensions and directed unions. Hence every solvable group is amenable.
2. $F_2$ is not amenable — Step 1 exhibits precisely the paradoxical decomposition that an invariant mean forbids.
3. An amenable group acting on a set can never make it paradoxical, since the invariant mean pushes forward to an invariant finitely additive measure.

Now compare the isometry groups. $E(1) \cong \mathbb{R} \rtimes \mathbb{Z}/2$ and $E(2) \cong \mathbb{R}^2 \rtimes O(2)$ are both solvable — in $E(2)$ the translations form an abelian normal subgroup with abelian quotient $O(2)$ — hence amenable. Banach's 1923 theorem is the constructive expression of this: on $\mathbb{R}$ and $\mathbb{R}^2$ there exist finitely additive, isometry-invariant measures defined on every bounded set and extending Lebesgue measure. With such a measure in hand, no paradoxical decomposition can exist.

$E(3) \supseteq SO(3) \supseteq F_2$ is not amenable, and Steps 1–5 realise the consequence. Dimension is not the operative variable. Amenability is. For a sharp illustration: the free group $F_2$ acts on the hyperbolic plane $\mathbb{H}^2$ by isometries, and the hyperbolic plane does admit paradoxical decompositions despite being two-dimensional.

One caveat on the general theory: the von Neumann conjecture — that every non-amenable group contains a free subgroup of rank 2 — is false, refuted by Ol'shanskii in 1980. Non-amenability is the right criterion; a free subgroup is merely the most convenient sufficient condition, and the one Euclidean space happens to supply.

Exactly how much choice

• ZF alone does not suffice. Solovay (1970) built a model of ZF + DC in which every set of reals is Lebesgue measurable, assuming the consistency of an inaccessible cardinal; Shelah (1984) showed the large cardinal assumption cannot be dropped for that statement. In Solovay's model there is no Banach–Tarski decomposition.
• Full AC is not needed. Pawlikowski (1991) derived the paradox from the Hahn–Banach theorem, which is strictly weaker than AC and follows from the Boolean prime ideal theorem.
• So the paradox sits strictly between ZF + DC and ZFC. Rejecting it means rejecting Hahn–Banach.

What is not being claimed

The theorem makes no assertion about physical objects, and the following are all misreadings.

Points are not atoms. $B$ has $2^{\aleph_0}$ points, each of zero extent. A physical ball has on the order of $10^{23}$ atoms of positive size. There is no map from one situation to the other, and the theorem's hypotheses are false of matter at the first line.

No volume is created. Volume is a function $\lambda$ defined on the Lebesgue measurable sets. The pieces $A_i$ are not in its domain. The equation "volume in = volume out" is not violated; it is not even expressible for the intermediate objects. Asking where the extra volume came from is like asking for the temperature of a number.

Additivity is not violated. $\lambda$ is countably additive on the σ-algebra of measurable sets. The decomposition is a finite union of non-measurable sets, which the additivity axiom says nothing about. The theorem is what tells you the σ-algebra restriction was mandatory rather than conventional.

No procedure exists. $M$ is produced by the axiom of choice from uncountably many orbits. It is not definable, not Borel, not computable, and no algorithm approximates it. Nothing physical or computational instantiates the construction, in this universe or any other.

Conservation laws are untouched. Mass, energy and momentum are properties of physical systems obeying physical dynamics. The theorem concerns rigid motions of abstract point sets, a setting in which none of these quantities is defined. It neither supports nor threatens any claim in physics.

What the theorem is about is the interaction between the axiom of choice and the group $E(3)$ — and, correctly understood, it is one of the most informative facts in mathematics about the limits of the word "size".