Banach–Tarski paradox

Remarks

The proof starts with two rotations that generate a free group $F_2$ of rank $2$ inside the special orthogonal group $SO(3)$. 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. 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.

Going deeper

For a student

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.