# Banach Tarski Paradox

The **Banach-Tarski paradox** is a theorem in geometry and set theory which states that a $3$-dimensional ball may be decomposed into finitely many pieces, which can then be reassembled in a way that yields two copies of the original ball.

This is considered a paradox because it is contrary to geometric intuition that one can double the volume of an object by only cutting it up into pieces and rearranging these pieces rigidly. In fact, the pieces into which the ball is decomposed do not have well-defined volumes. The existence of such *nonmeasurable* subsets of Euclidean space depends on the axiom of choice.

## Proof Sketch

Instead of decomposing the solid $3$-dimensional ball, this article will present a decomposition of the $2$-sphere $S^2$ which can be rearranged into two copies of $S^2$. Recall that $S^2 = \{(x,y,z) \, : \, x^2 + y^2 + z^2 = 1\}$. From here, one can obtain the desired decomposition of the $3$-ball by taking unions of rays from the center of the $3$-ball to the pieces into which its boundary $S^2$ has been decomposed.

The idea of the proof is algebraic. One works with the group of rotations of $S^2$, which this article denotes $\mathbf{G}$. A subgroup of $\mathbf{G}$ is decomposed into self-similar subsets, and this subgroup is allowed to act on $S^2$. The orbits of this action will be the desired pieces of $S^2$.

Let $\mathbf{F}_2$ denote the free group on two generators $a$ and $b$. That is, $\mathbf{F}_2$ consists of words made up with the letters $a$ and $b$, with concatenation being the group operation. For example, some elements of $\mathbf{F}_2$ are $aba^{-1} b^{-1}$ and $b a^2 b^{-1}$; the concatenation of these elements is $aba^{-1} b^{-1} b a^2 b^{-1} = abab^{-1}$.

Let $S(a)$ denote the elements of $\mathbf{F}_2$ beginning with $a$, and similarly define $S(b)$, $S\big(a^{-1}\big)$, and $S\big(b^{-1}\big)$. Then $\mathbf{F}_2 = \{1\} \cup S(a) \cup S(b) \cup S\big(a^{-1}\big) \cup S\big(b^{-1}\big),$ where $1$ denotes the identity element of $\mathbf{F}^2$. However, one also has $\mathbf{F}_2 = a S\big(a^{-1}\big) \cup S(a)$ since $a S\big(a^{-1}\big)$ is precisely the elements of $\mathbf{F}_2$ beginning with $a^{-1}$, $b$, or $b^{-1}$. Similarly, $\mathbf{F}_2 = b S\big(b^{-1}\big) \cup S(b)$. Thus, one can decompose $\mathbf{F}_2$ into four pieces, "translate" two of them (where one considers group multiplication as sort of translation), and then reassemble the four pieces to obtain two copies of $\mathbf{F}_2$.

Now, one finds a subgroup $\mathbf{H} \le \mathbf{G}$ isomorphic to $\mathbf{F}_2$. Let $\theta = \arccos\big(\frac13\big)$, let $A$ be rotation by $\theta$ about the $x$-axis, and let $B$ be rotation by $\theta$ about the $z$-axis. One can show the group $\mathbf{H} = \langle A, B \rangle$ is isomorphic to $\mathbf{F}_2$. The orbits of $\mathbf{H}$ acting on $S^2$ are then demonstrated to give the desired paradoxical decomposition of $S^2$!

**Cite as:**Banach Tarski Paradox.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/banach-tarski-paradox/