The Banach-Tarski paradox

Let $B \subseteq \mathbb{R}^3$ be a closed ball. There are a partition of $B$ into finitely many pieces $B = A_1 \cup \cdots \cup A_n$ and rigid motions $g_1, \dots, g_n$ of $\mathbb{R}^3$ such that $g_1(A_1), \dots, g_n(A_n)$ partition two disjoint balls each congruent to $B$. Five pieces suffice, and four is impossible. More generally, any two bounded subsets of $\mathbb{R}^3$ with nonempty interior are equidecomposable.

The pieces are necessarily non-measurable, so the statement contradicts no theorem about volume; what it refutes is the existence of a finitely additive, isometry-invariant extension of Lebesgue measure to all subsets of $\mathbb{R}^3$. In $\mathbb{R}^1$ and $\mathbb{R}^2$ such extensions do exist, by a Banach construction using the amenability of the isometry groups there, so the paradox is a fact about $n \ge 3$, where the rotation group contains a free subgroup of rank two.

It needs choice beyond ZF + DC. If ZFC together with "there exists an inaccessible cardinal" is consistent, then ZF + DC does not prove the paradox: in Solovay's model of ZF + DC every set of reals is Lebesgue measurable, and the same holds in $\mathbb{R}^3$, while the pieces above are necessarily non-measurable, so no such decomposition exists there. That model is built by collapsing an inaccessible, so the conclusion drawn from it is conditional on that large-cardinal hypothesis and not merely on the consistency of ZF. The paradox does not need the full Axiom of Choice either: it follows from the Hahn-Banach theorem, hence from the Boolean prime ideal theorem, which the Axiom of Choice implies outright and which, if ZF is consistent, does not imply the Axiom of Choice in return.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. The free group of rank two inside $SO(3)$, its paradoxical decomposition, transfer of that decomposition to the sphere by choosing one point from each orbit (The Axiom of Choice ↗), the handling of the countably many fixed points, and the Banach-Schroder-Bernstein theorem for equidecomposability. The group theory is elementary and reachable; the statement is deferred because it is about measure, and because its interest is precisely that the pieces are not measurable (Lebesgue measure and the Lebesgue integral ‡, The Vitali set: a non-measurable subset of $\mathbb{R}$ ‡).

Which page it serves. The order, Zorn and the Axiom of Choice examples page, as the most dramatic consequence of choice, and any future measure page as the reason finite additivity plus isometry invariance cannot be had on all sets in dimension three. It is also the standard corrective to the belief that choice is harmless: it is a theorem of ZFC and cannot be blamed on a defect of the measure.

A frequent misstatement. The pieces are not "infinitely thin" or "of measure zero": they have no measure at all, and there are finitely many of them. The word paradox records a conflict with intuition, not an inconsistency.