Banach-Tarski and What the Theorem Actually Says, math that surprises you lesson

Can a solid ball be split into a handful of pieces and put back together, without stretching or shrinking anything, into two balls each exactly like the first? For a ball in the mathematical sense, an infinite set of points, the Banach-Tarski theorem says yes. That seems to contradict volume, since moving a piece does not change its volume and two balls have twice the volume of one. This page states what the theorem actually says, with every condition, finds the step of the volume argument that fails, and explains why the theorem gives no way to duplicate a physical object.

Key terms

Banach-Tarski paradox

Definition Let $B$ be a solid ball in three-dimensional space, the set of all points whose distance from a center point is at most $r$ , for some $r > 0$ . The Banach-Tarski theorem, proved by Stefan Banach and Alfred Tarski in 1924, says that $B$ can be split into finitely many pieces $A_1, \dots, A_k$ , sets of points such that every point of $B$ lies in exactly one of them, and that there are rigid motions $g_1, \dots, g_k$ such that no two of the moved pieces $g_1(A_1), \dots, g_k(A_k)$ share a point and together they make up exactly two disjoint balls of radius $r$ . Each piece is used once, and none is stretched, shrunk or copied. Five pieces are enough, and fewer than five are not. The result seems to double the volume of the ball, which is why it is called a paradox. The resolution is that some of the pieces are nonmeasurable sets, which have no volume at all, so the argument that adds up the volumes of the pieces cannot be applied to them. Those pieces cannot be regions that a knife could cut out, since every such region has a volume, and their existence is proved with the axiom of choice, without any description of what they look like. The theorem holds in space of three or more dimensions, but on a line or in the plane no segment or disk can be doubled in this way. It is a theorem about ideal sets of points. A physical object is not an infinite set of ideal points that could be sorted into such sets, so the theorem gives no way to duplicate one.

In plain words A mathematical ball is an infinite set of points, and the theorem sorts those points into a few sets so unlike ordinary pieces that moving each set as a whole produces two complete balls. What fails is not the bookkeeping, since every point is used exactly once, but the assumption that every set of points has a volume. Compare The Missing Square. There the picture hid a bent edge, and the apparent paradox was an error. Here the theorem is correct, and the hidden assumption is that volume applies to every piece.

Example Take the ball of radius $1$ centered at the origin. Its volume is

\frac{4}{3}\pi \cdot 1^3 = \frac{4\pi}{3} \approx 4.19.

The theorem gives pieces $A_1, \dots, A_k$ and rigid motions after which the pieces fill exactly the original ball and the ball of radius $1$ centered at $(3, 0, 0)$ . These two balls are disjoint, because their centers are $3$ apart, more than the sum $1 + 1$ of the radii, and together they have volume $8\pi/3 \approx 8.38$ . Suppose every piece had a volume, say $v_1, \dots, v_k$ . The pieces fill the original ball without overlapping, so their volumes add up to $4\pi/3$ . Rigid motions do not change volumes, and the moved pieces fill the two balls without overlapping, so the same numbers add up to $8\pi/3$ . That would mean

\begin{aligned} v_1 + \cdots + v_k &= \frac{4\pi}{3}, \\[0.4em] v_1 + \cdots + v_k &= \frac{8\pi}{3}, \end{aligned}

which is impossible. So at least one of the pieces has no volume.

Rigid motion

Definition A rigid motion of space is a map $g$ that sends each point $x$ to a point $g(x)$ so that all distances stay the same,

\lvert g(x) - g(y) \rvert = \lvert x - y \rvert

for all points $x$ and $y$ , where $\lvert x - y \rvert$ is the distance between $x$ and $y$ , and that does not turn space into its mirror image, so a left hand stays a left hand. Every rigid motion is a rotation about an axis through the origin followed by a translation, which shifts every point by the same amount in the same direction. A rigid motion moves a set as one solid whole. It carries a ball to a ball of the same radius, and a set that has a volume to a set with the same volume. Reflections also keep distances, but they turn left hands into right hands, and the Banach-Tarski theorem does not need them.

In plain words A rigid motion picks a set up and puts it down somewhere else, perhaps turned around, without bending, stretching or tearing it. All points of the set move together.

Example The quarter turn about the vertical axis sends each point $(x, y, z)$ to $(-y, x, z)$ . It sends $(1, 0, 0)$ to $(0, 1, 0)$ and $(0, 1, 0)$ to $(-1, 0, 0)$ , and the distance $\sqrt{2}$ between the first two points equals the distance between their images. The translation that sends $(x, y, z)$ to $(x + 3, y, z)$ carries the ball of radius $1$ centered at the origin onto the ball of radius $1$ centered at $(3, 0, 0)$ . Doing the quarter turn and then this translation sends $(1, 0, 0)$ to $(3, 1, 0)$ . By contrast, the map that sends every point $x$ to $2x$ doubles every distance, so it is not a rigid motion, and the mirror map that sends $(x, y, z)$ to $(-x, y, z)$ keeps distances but reverses handedness.

Nonmeasurable set

Definition Volume in space, in the precise form called Lebesgue measure, is defined for a very large family of sets called the measurable sets. They include every point, box and ball, every open set and every closed set, and every set built from these by countably many unions, intersections and complements, so in particular every solid that could be cut out with a knife. On the measurable sets, volume gives a box its usual volume, does not change under rigid motions, and is countably additive, which means that for finitely many measurable sets, or an infinite sequence of them, no two of which share a point, the volume of their union is the sum of their volumes. Length on a line and area in the plane follow the same rules. A nonmeasurable set is a set of points that is not measurable, so it has no volume in this sense. This is not the same as having volume zero. A single point, a flat square and any countable set are measurable with volume zero, and looking complicated does not make a set nonmeasurable, since every closed set is measurable, including familiar fractals such as the Cantor set and the Sierpinski triangle. The Banach-Tarski theorem shows that in space this gap cannot be closed. No rule that gives every set of points a volume can be finitely additive and unchanged by rigid motions while giving a ball a positive finite volume. On a line and in the plane such rules do exist, which is why no segment or disk can be doubled there.

In plain words A nonmeasurable set is not tiny or thin. It is a set for which the rules of volume have no answer at all, so any argument that adds up its volume cannot even start.

Example The first nonmeasurable set, found by Giuseppe Vitali in 1905, lives on the interval from $0$ to $1$ , including $0$ but not $1$ , with shifts that wrap around, so that a shift by $q$ sends $x$ to $x + q$ if that is less than $1$ and to $x + q - 1$ otherwise. Such a shift does not change lengths. Put two numbers of the interval in the same class when their difference is a rational number, choose one number from each class, a step that needs the axiom of choice described below, and let $V$ be the set of chosen numbers. For each rational $q$ from $0$ up to but not including $1$ , shift $V$ by $q$ . Every number $x$ of the interval lies in exactly one of these shifted copies, the one for which $q$ is the difference between $x$ and the chosen number of its class, plus $1$ if that difference is negative. There are countably many rational $q$ , so if $V$ had a length $L$ , countable additivity would give

1 = L + L + L + \cdots.

The right side is $0$ when $L = 0$ and infinite when $L > 0$ , so it is never $1$ . Hence $V$ has no length.

Axiom of choice

Definition The axiom of choice is one of the basic assumptions of set theory, the standard foundation of mathematics. It says that for every collection of nonempty sets there is a way to pick one element from each set in the collection, even when the collection is infinite and no rule says which element to pick. In other words, there is a choice function, which assigns to each set of the collection one of its own elements. The axiom says that the selection exists, not how to make it. The usual proof of the Banach-Tarski theorem uses it, and some form of choice is needed. In 1970 Robert Solovay showed that, if set theory with the existence of an inaccessible cardinal, a kind of very large infinite size, is free of contradictions, then so is set theory with only a weaker form of choice in which every set of points is measurable, and there the volume argument rules out the duplication. The theorem does not need the full axiom, since it also follows from weaker principles such as the Hahn-Banach theorem of analysis, so it is not equivalent to the axiom of choice.

In plain words You may pick one item from each of infinitely many bins at once, even when nothing tells the items in a bin apart. The axiom promises that such a selection exists without saying how to find it, so sets built this way usually cannot be described in any more explicit way.

Example A classic comparison uses shoes and socks. From infinitely many pairs of shoes you can pick one shoe from each pair by a rule, always the left shoe, and the axiom of choice is not needed. From infinitely many pairs of socks in which the two socks of a pair are exactly alike, no rule singles one out, and the axiom is what guarantees a selection. The Banach-Tarski proof needs a selection of the sock kind. It uses two rotations about the center of the ball, chosen with care, combined with each other and their reverses in every possible order. Two points of the sphere that bounds the ball go into the same class when one of these combined rotations carries one point to the other. Each class is countable, there are uncountably many classes, and the axiom of choice provides a set containing exactly one point from each class. The Vitali set is built the same way, from classes of numbers that differ by rational numbers.

Cardinality

Definition Two sets have the same cardinality when there is a one-to-one correspondence between them, a pairing in which every element of each set has exactly one partner in the other. For finite sets this means having the same number of elements. An infinite set can have the same cardinality as a part of itself, as in Hilbert’s Hotel. A solid ball, two disjoint balls, a line segment of positive length and all of space have the same cardinality, that of the set of real numbers. Having the same cardinality is a much weaker property than what Banach-Tarski requires. A pairing may send each point anywhere, independently of its neighbors, while the theorem demands finitely many pieces, each moved as a whole by one rigid motion.

In plain words Cardinality counts by pairing and ignores size, shape and distance. In that sense one ball and two balls have equally many points, so the surprise of Banach-Tarski is not the number of points but that finitely many rigid pieces are enough.

Example The map that sends each point $x$ to $\sqrt[3]{2}\,x$ pairs the points of a ball of volume $V$ one to one with the points of a ball of volume $2V$ , but it stretches every distance by the factor $\sqrt[3]{2} \approx 1.26$ , so it is not a rigid motion. A rigid version of Hilbert’s Hotel works on a circle. Measure positions on the circle by their angle in radians from a fixed point $P$ , and mark the points at angles $1, 2, 3, \dots$ radians. These points are all different, and none of them is $P$ , at angle $0$ . Two different whole numbers of radians could give the same point only if their difference were a nonzero whole multiple of $2\pi$ , which would make $\pi$ a fraction, and $\pi$ is irrational. Now remove $P$ and split what is left into two pieces, the marked points and everything else. Rotate the marked points backward by $1$ radian, so the point at angle $n$ moves to angle $n - 1$ , and leave the rest in place. The point at angle $1$ lands on $P$ and each other marked point lands on the previous one, so the circle without $P$ becomes the whole circle, using two pieces and two rigid motions, one of which leaves its piece where it is. The gain is a single point, which has length zero, so no rule of length is broken. Banach-Tarski gains a whole ball.

Check yourself

In the circle example, remove two points, $P$ at angle $0$ and the point at angle $1$ radian. Find two pieces and two rotations that reassemble what is left into the whole circle.

Show the answer

Mark the points at angles $2, 3, 4, \dots$ radians. As before they are all different, and none of them is at angle $0$ or angle $1$ , since an angle of $n$ radians with $n \ge 2$ can match $0$ or $1$ only if $n$ or $n - 1$ is a positive multiple of $2\pi$ , which would make $\pi$ rational. Rotate the marked points backward by $2$ radians, so the point at angle $n$ moves to angle $n - 2$ , and leave everything else in place. The points at angles $2$ and $3$ land on the two removed points, at angles $0$ and $1$ , and each later marked point lands on the marked point two steps earlier. The result is the whole circle, from two pieces, one rotated by $2$ radians backward and the other rotated by $0$ , which leaves it in place.

Someone argues that the Banach-Tarski pieces must be like fine dust with volume zero. What is wrong with this, and could exactly one of the pieces be nonmeasurable?

Show the answer

A set with volume zero is measurable. If every piece had volume zero, the volumes of the pieces would add up to $0$ , but they would have to add up to the volume of the ball, which is $4\pi r^3/3 > 0$ . So the pieces are not all of volume zero, and some of them have no volume at all.

Exactly one nonmeasurable piece is also impossible. Suppose $A_1, \dots, A_{k-1}$ were all measurable. The last piece is what remains of the ball after removing their union $U$ ,

\begin{aligned} U &= A_1 \cup \cdots \cup A_{k-1}, \\[0.4em] A_k &= B \setminus U, \end{aligned}

and a set built from finitely many measurable sets by unions, intersections and complements is measurable. Then every piece would be measurable, and the volume argument would give a contradiction. So at least two of the pieces are nonmeasurable.

Change the sizes. Can a ball of radius $1$ be split into finitely many pieces that rigid motions reassemble into a single ball of radius $2$ , or into two disjoint balls of radius $2$ ?

Show the answer

Yes to both. Banach and Tarski proved a stronger form of their theorem. If two bounded sets in space each contain a ball of positive radius, then either one can be split into finitely many pieces that rigid motions reassemble into the other. A ball of radius $1$ and a ball of radius $2$ are such sets, and so is the union of two disjoint balls of radius $2$ . The volumes are

\begin{gathered} \frac{4\pi}{3} \approx 4.19, \\[0.4em] \frac{32\pi}{3} \approx 33.51, \end{gathered}

for the small ball and one large ball, and $64\pi/3 \approx 67.02$ for the two large balls, so the rearrangement multiplies the volume by $8$ or by $16$ . The volume argument applies just as before and shows that some pieces must be nonmeasurable. The map that sends $x$ to $2x$ also pairs the small ball with the large one, but it stretches distances, while the theorem uses only rigid motions.

Banach-Tarski and What the Theorem Actually Says

Video 1, chapter 23

Key terms

Banach-Tarski paradox

Rigid motion

Nonmeasurable set

Axiom of choice

Cardinality

Check yourself

How this lesson connects

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