---
tag: 9G9G8ED97D
kind: counterexample
name: Hausdorff paradox
msc: [28A12, "03E25", 43A07]
uses: [0HTGB13FE2, 0HFREYRSZP, 01TQB88H4C, 0CGYNDCWRR]
sources:
  - url: https://en.wikipedia.org/wiki/Hausdorff_paradox
same_as:
  wikidata: Q1959890
  wikipedia: Hausdorff paradox
---

## Counterexample

Let $S^2 = \set{x \in \R^3 : \norm{x} = 1}$ be the unit sphere.
Assume the [[axiom of choice|0HTGB13FE2]].
There are a [[countable set|01TQB88H4C]] $D \subseteq S^2$ and pairwise disjoint sets $A$, $B$, $C$ with $S^2 \setminus D = A \cup B \cup C$.
They can be chosen so that the four sets $A$, $B$, $C$ and $B \cup C$ are congruent to each other by [[rotations|0HFREYRSZP]] of $\R^3$.
Hence there is no finitely additive, rotation-invariant function $\mu$ on all subsets of $S^2$ with values in $[0, \infty)$ and $\mu(S^2) = 1$.

## Remarks

The proof uses two rotations $\varphi$ and $\psi$ in the [[special orthogonal group|0CGYNDCWRR]] $SO(3)$ with $\varphi^2 = \psi^3 = \id$.
They can be chosen to generate a free product of a group of order $2$ and a group of order $3$.
The set $D$ collects the fixed points of the non-identity elements of this group, and the group acts freely on the rest of the sphere.
Choosing one point in each orbit transfers a paradoxical splitting of the group to the sphere.
For the consequence, finite additivity and invariance give $\mu(A) = \mu(B) = \mu(C) = \mu(B) + \mu(C)$.
So all three are $0$, and $\mu(D) = 1$.
But for a countable $D$ there is a rotation $\rho$ such that the sets $D, \rho(D), \rho^2(D), \ldots$ are pairwise disjoint, which forces $\mu(D) \le 1/n$ for every $n$.
Hausdorff used this to show that the measure problem has a negative answer in dimension $3$ and higher; the Banach–Tarski paradox later strengthened the result.

## Explanations

### L1 human

Apart from a countable set of points, the sphere can be cut into three pieces $A$, $B$, $C$ with a strange property.
Rotating $A$ gives $B$, rotating it again gives $C$, and some rotation also turns $A$ into $B$ and $C$ together.
So $A$ looks like one third of the sphere and also like one half of it.
This shows that no notion of area can be defined for all subsets of the sphere, if it adds up correctly and does not change under rotation.
The pieces cannot be drawn; their existence uses the axiom of choice.