Banach-Tarski Paradox

Motivation

Recall that by Vitali's Theorem, there is no function $\mu :2^{\mathbb {R} }\to [0,+\infty ]$ satisfying all three of the following properties:

• $\mu$ is countably additive,
• $\mu$ is translation invariant, and
• for each interval $(a,b)$, we have $\mu ((a,b))=b-a$.

This result can be easily be generalized to higher-dimensional Euclidean spaces.

To obtain a function that can reasonably "measure" Euclidean space, one might try to weaken some of the above properties. However, if we weaken only the requirement that $\mu$ be countably additive, namely by requiring that $\mu$ be merely finitely additive, we still run into problems in higher dimensions. The Banach-Tarski Paradox illustrates some of these problems.

The Banach-Tarski Paradox

In 1924, Stefan Banach and Alfred Tarski proved the following result:

Let $U$ and $V$ be arbitrary bounded open sets in $\mathbb {R} ^{n}$ for $n\geq 3$. There exist $k\in \mathbb {N}$ and subsets $E_{1},\dots ,E_{k},F_{1},\dots ,F_{k}$ of $\mathbb {R} ^{n}$ such that

• the $E_{j}$'s are disjoint and their union is $U$,
• the $F_{j}$'s are disjoint and their union is $V$, and
• for each $j\in \{1,\dots ,k\}$, $E_{j}$ is congruent to $F_{j}$, i.e. $E_{j}$ can be transformed into $F_{j}$ by translations, rotations, and reflections.

Hence, any function $\mu :2^{\mathbb {R} ^{n}}\to [0,+\infty ]$ fails to satisfy the desired properties, because we can take two bounded open sets $U$ and $V$ that should have different measures (based on the fact that the measure of cubes is forced), split each of these two sets into $k$ pieces, then translate/rotate/reflect these pieces to match each other. Since the pieces of $U$ are congruent to the corresponding pieces of $V$, these pieces should have the same measure, i.e. $\mu (E_{j})=\mu (F_{j})$. By finite additivity of $\mu$, this forces $U$ and $V$ to have the same measure, which contradicts our earlier assumption that $U$ and $V$ have different measures.

Rough Sketch of Proof for Special Case

The following is a rough sketch of how to decompose $D^{3}$ into two translated copies of $D^{3}$ plus a bit more using only translations and rotations. Without too much more trouble one could adapt this to eliminate the extra material, but we shall omit this for brevity and clarity.

Choose an angle $\alpha$ which is incommensurable to $2\pi$. Let $A$ be rotation centered at the origin by $\alpha$ counter-clockwise along the xy-plane. Let $B$ be rotation centered at the origin by $\alpha$ counter-clockwise along the yz-plane. Let $G$ be the free group generated by $A$ and $B$. We then can define a set M, using the axiom of choice, which selects one representative of each orbit defined by the action of $G$ on $S^{2}$. Thus each point on $S^{2}$ is uniquely represented by a group element of $G$ and a point in $M$. We then define $S_{r}=\{gx\in S^{2}|g\in G,x\in M$, and $g$, as a word, begins with $r\}$ for $r\in \{A,B,A^{-1},B^{-1}\}$. Observe then that $S^{2}=S_{A}\sqcup S_{B}\sqcup S_{A^{-1}}\sqcup S_{B^{-1}}\sqcup M$. We may then observe that $A^{-1}S_{A}=S_{A}\sqcup S_{B}\sqcup S_{B^{-1}}\sqcup M$ and $AS_{A^{-1}}=S_{B}\sqcup S_{A^{-1}}\sqcup S_{B^{-1}}\sqcup M$ and $B^{-1}S_{B}=S_{A}\sqcup S_{B}\sqcup S_{A^{-1}}\sqcup M$. So by rotating three pieces of $S^{2}$, we now have the pieces $\{S_{B^{-1}},M,S_{A},S_{B},S_{B^{-1}},M,S_{B},S_{A^{-1}},S_{B^{-1}},M,S_{A},S_{B},S_{A^{-1}},M\}$, which is more than sufficient to produce two copies of $S^{2}$ with translations. The argument extends to $D^{3}$ by instead of considering a point on $S^{2}$, one considers the half-open line segment from the origin to $S^{2}$. Then after some more work that needs to be done with the origin, one has more than doubled the unit sphere through rigid transformations.

Geometric Implications

The Banach-Tarski Paradox violates our geometric intuition of how volume is preserved when a physical object is broken into pieces and reassembled. Indeed, if the Banach-Tarski Paradox is to be believed, then it is possible to take a small ball (such as a pea); decompose the ball into finitely many pieces; move these pieces around in space via translations, rotations, and reflections; and reassemble these pieces into a new ball as large as the Sun. Of course, the ball may have to be decomposed into very strange pieces to accomplish this task.

One may note that the proof of the Banach-Tarski Paradox relies on the Axiom of Choice. Because of the many unintuitive implications of the Axiom of Choice, including the Banach-Tarski Paradox and the existence of unmeasurable subsets of Euclidean space, some mathematicians choose to reject the Axiom of Choice.