It is a common occurrence in mathematics that when something does go wrong, it goes terribly wrong. This exact phenomenon occurs with the Banach-Tarski Paradox. Informally, it says that one can take a sphere (in 3 or more dimensional space)  can be split into finitely many pieces and, using only rigid motions, can be rearranged to form two spheres.

Alternatively, it says that we can take a pea, break it up, and rearrange it to be the size of the sun. By now, one might ask themselves how something such as mathematics could support or even prove this outrageous claim! The answer lies with the Axiom of Choice, which we now state.

AXIOM: For every collection of sets $\mathbb{X}$ there is a function $f: \mathbb{X} \to \cup_{A\in \mathbb{X}} \mathbb{X}$ so that $f(A) \in A$ for every $A\in f(A)$.

The axiom simply says that we can assign to every set, an element of itself. This doesn’t seem that far-fetched, which is why it is generally accepted by mathematicians. In fact it has equivalent forms, such as Zorn’s Lemma, Well-Ordering Principle, and the Hausdorff Maximal Principle, which have been used to prove some of the most important theorems in mathematics. Such examples include:

• Every vector space has a (Hamel) basis.
• Every commutative ring with unit contains a maximal ideal
• The Cartesian product of of compact topological spaces forms a topological space with the product topology.
• Banach’s extension theorem which is used to prove the Hahn-Banach Theorem.

These are but a few of the consequences of the Axiom of Choice. One such consequence is the subject of this text. We will now formally state the Banach-Tarski Theorem.

THEOREM: Let $A$ and $B$ be bounded subsets of $\mathbb{R}^n$, for $n \geq 3$. Suppose further that the interiors of $A$ and $B$ are not empty. Then there exists some $k$ and partitions of $A$ and $B$ that can be written as $A = A_1\sqcup A_2\sqcup ... \sqcup A_k$ and $B = B_1\sqcup B_2 \sqcup ... \sqcup B_k$  where $A_i$ is congruent to $B_i$. In this case we write $A \cong B$.

Put simply, if we have two bounded sets with nonempty interior in Euclidean space of three dimensions or greater,  then we can break them up into sets so that their pieces are congruent. Now, there are some words that we should define just so we are all on the same page. We will go through those now. For the purposes of this post, $A \subset \mathbb{R}^n$.

A set is bounded if it is in some sense of finite size. More formally,

DEFINITION: A set $A$ in Euclidean space is bounded if $\sup\{\|a\|\mid a\in A\} < \infty$, where $\|a\|$ is the Euclidean norm.

A partition of a set is a set of pieces which can be put together to make the set. Or,

DEFINITION: Given a set $A$, a partition of $A$ is a finite collection of subsets $A_1, ..., A_k$ so that

1. $A = \bigcup_{i=1}^k A_i$; and
2. $A_i \cap A_j = \emptyset$ if $i \neq j$.

The interior of a set is the set of all points that do not lie on the boundary (or edge). Formally,

DEFINITION: A point $a$ is interior to a set $A$ if there is some $\epsilon > 0$ for which $\{x \in A\mid d(x,a) < \epsilon\} \subset A$. The interior of $A$ is the collection of all such points.

Finally, two sets are congruent in Euclidean space if there is some way to fit one over the other without stretching, i.e. if they “look the same”.  We define this formally as,

DEFINITION: Two sets are said to be congruent if there is an isometry (distance preserving, injective map) between them.

We will now prove the Banach-Tarski Theorem for a solid ball in $\mathbb{R}^3$ in 4 steps.

STEP 1: Let $F$ be the (free) group of all strings containing 4 different symbols $a, a_{-1}, b, b^{-1}$ so that $a$ does not appear next to $a^{-1}$ and no
$b$ appears next to a $b^{-1}$. In vSauce’s video, he calls these 4 elements $U, D, L, R$. Now we will define a set $S(a)$ to be the set of all allowable strings that begin with $a$. We will define the sets $S(a^{-1}), S(b),$ and $S(b^{-1})$ similarly. If $e$ denotes the identity then we can write $F$ as $F = \{e\} \cup S(a) \cup S(a^{-1}) \cup S(b) \cup S(b^{-1})$. This is because every string that is not the identity, must start with one of the four elements. Note that also the set $a^{-1}S(a)$ is the set of all strings that do not start with $a$. Hence we can write $F$ as $F = a^{-1}S(a) \cup S(a)$. We can do this with the b’s as well. $F = b^{-1}S(b)\cup S(b)$. So, what we’ve done here is taken the group $F$, split it up into 4 pieces, then shift these pieces to create two identical copies of the original group. Now we have to find a structure in $\mathbb{R}^3$ that is isomorphic to the $F$.

STEP 2: We will consider our motions as rotations about an axis. To be more precise let $\alpha$ be a rotation of $\theta = arccos(1/3)$ about the $x$-axis, and let $\beta$ be a rotation of $\theta$ about the $z$-axis. There are other choices for $\theta$ but we shall content ourselves with this one. Now we let $G$ be the group generated by these rotations $\alpha$ and $\beta$. Note that this is almost the same as $F$ except we need to know that there are no nontrivial combinations of $\alpha$ and $\beta$ that returns us to where we started. Let $\rho \in S(\beta)$. Then it can be shown (see [2]), by using elementary linear algebra, that $\rho$ maps $(1,0,0)$ to $(\frac{a}{3^n}, \frac{b\sqrt{2}}{3^n}, \frac{c}{3^n})$ where $a, b, c \in \mathbb{Z}$ and $b \neq 0$. This means that $\rho \neq e$. Hence what we have is that $G$ is isomorphic to $F$.

It is at step 2 that we see why we must be in dimension 3 or higher as we cannot form this group in 2 dimensions. We will now start to see how the sphere fits in to this picture. This may get a little tough after this point.

STEP 3Let $S^2$ be the unit sphere with a partition generated by $G$. Each set in the partition is called an orbit.  We now apply the axiom of choice to choose exactly one point from each orbit. We collect all of these points in a set $M$. Now we define four sets arising from $G$.

• $A_1 = \{\rho(x) \mid x\in M, \rho\in S(\alpha)\}$;
• $A_2 = \{\rho(x) \mid x\in M, \rho\in S(\alpha^{-1}\}$;
• $B_1 = \{\rho(x) \mid x\in M, \rho\in S(\beta)\}$ and;
• $B_2 = \{\rho(x) \mid x\in M, \rho\in S(\beta^{-1}\}$.

Now we have that $A = A_1 \cup \alpha(A_2) = S^2 - D$ where $D$ is the set of points that are fixed by rotations of $G$. Similarly $B = B_1 \cup \beta(B_2) = S^2 - D$. Notice that any rotation has exactly two fixed points, and since $G$ is countable, there are only countably many points fixed by an element of $G$, i.e. $D$ is countable. Now let $\ell$ be a line through the center of the ball that misses $D$. Let $\tau$ be an angle so that $\tau^n(D) \cap D = \emptyset$, for all $n\in \mathbb{N}$. Let $\overline{D} = \cup_n \tau^n(D)$. Then we see $S = \overline{D} \cup (S^2-\overline{D}) \cong \tau(\overline{D})\cup (S^2-\overline{D}) = S^2-D$.

This tells us then that $A \cong S^2-D \cong S^2$ and $B \cong S^2-D \cong S^2$.

We have made it out of the thick part of the woods now. We will now extend this result to the entire sphere.

STEP 4: We now extend this result to the solid sphere, $D^3$ minus its center simply by extending each point to a segment emanating from the center but not including the center (this is called radial extension). To finish, we consider the line $\lambda$ obtained by shifting the $x$-axis up by half the radius of the sphere. We next consider a rotation $\sigma$ about $\lambda$ with angle $arccos(1/3)$. It can be seen that $\sigma^n((0,0,0)) \neq \sigma^m((0,0,0))$ whenever $n\neq m$. Thus we collect all of these into a set $H$. We conclude by the same reasoning as before that $D^3 = H\cup (D^3 - H) \cong \sigma(H) \cup (D^3 - H) = D^3 - \{(0,0,0)\}$.

This is no trivial theorem by any means. Its consequence is indeed known as a paradox. However, note that we also relied on the sphere containing uncountably many points. It is not at all clear that there are uncountably many points in real life. But if we do, then it should be possible to duplicate objects. And if we don’t it would seem that Zeno’s Paradox should hold. It is interesting to think about the consequences of seemingly simple things. I encourage the interested reader to explore the references listed below.

[1] Wikipedia

[3] S. Wagon, The Banach-Tarski Paradox, Cambridge University Press, Cambridge (1985).