---
tag: FH250K068Z
kind: statement
role: theorem
name: Banach–Tarski paradox
msc: ["03E25", 51M25, 43A07]
uses: [0HTGB13FE2, 0878DN3WCS, 0XPR1E85W0, 020Z5SEN27, 0CGYNDCWRR, PG8Z6F6DV9]
about: [0HTGB13FE2]
sources:
  - url: https://en.wikipedia.org/wiki/Banach–Tarski_paradox
same_as:
  wikidata: Q737851
  wikipedia: Banach–Tarski paradox
---

## Statement

Assume ZFC, that is, ZF with the [[axiom of choice|0HTGB13FE2]].
Let $B = \set{x \in \R^3 : \norm{x} \le 1}$ be the closed unit ball in [[Euclidean space|0XPR1E85W0]] $\R^3$, and let $B'$ be a translate of $B$ that is disjoint from $B$.
Then there are an integer $n \ge 1$ and a partition of $B$ into disjoint sets $A_1, \ldots, A_n$ with the following property.
There are [[isometries|0878DN3WCS]] $g_1, \ldots, g_n$ of $\R^3$ such that $g_1(A_1), \ldots, g_n(A_n)$ are pairwise disjoint with union $B \cup B'$.

## Remarks

The proof starts with two rotations that generate a [[free group|020Z5SEN27]] $F_2$ of rank $2$ inside the [[special orthogonal group|0CGYNDCWRR]] $SO(3)$.
Apart from the identity, the group $F_2$ splits into four pieces.
Left multiplication by the generators turns two of them into a copy of $F_2$, and the other two into another copy.
The axiom of choice picks one point from each orbit of $F_2$ on the sphere.
This transfers the splitting from the group to the sphere, outside a countable set of fixed points.
Radial segments carry it to the ball minus its centre, and the centre and the countable set are absorbed by a further rotation argument.
The pieces cannot all be Lebesgue measurable, since isometries preserve volume.
Five pieces suffice, and no fewer than five work.
The analogue fails in $\R^1$ and $\R^2$, because their isometry groups are [[amenable|PG8Z6F6DV9]].
So there are finitely additive isometry-invariant measures on all bounded subsets that give the unit ball a positive value.
A stronger form holds in $\R^n$ for each $n \ge 3$: any two bounded subsets with nonempty interior can be cut into finitely many pieces that match by isometries.

## Explanations

### L1 human

You can cut a solid ball into a few pieces.
Moving the pieces only by rotations and translations, you can put them back together as two balls of the same size as the first.
This seems to break the conservation of volume.
There is no contradiction, because the pieces are so scattered that they have no volume at all.
Such pieces can only be shown to exist with the axiom of choice; nobody can draw them.

### L2 human

The theorem shows that there is no finitely additive measure on all subsets of $\R^3$ that is invariant under isometries and gives the unit ball a positive finite value.
This is one reason why measure theory works with a restricted family of measurable sets.
The key algebraic input is a free subgroup of $SO(3)$.
The general idea behind the paradox is the failure of amenability.
A group that contains a free subgroup of rank $2$ is not amenable, and its actions can be paradoxical.