Banach–Tarski paradox

The Banach–Tarski paradox is a theorem in set-theoretic geometry that shows something very strange. It says that you can take a solid ball in three-dimensional space and cut it into a few pieces. Then, by just moving and turning those pieces, you can put them back together to make two balls that are exactly the same size as the original one. This seems impossible because moving and turning things usually doesn’t change how much space they take up.

One famous version of this idea is called the "pea and the Sun paradox". It says you could take a tiny object, like a pea, cut it into pieces, and rearrange those pieces to make something huge, like the Sun. This sounds like magic, but it’s based on a real mathematical idea.

This paradox works because the pieces aren’t normal solids. They are made of infinite points that don’t have a usual “volume”. Because of this, the normal rules about size and space don’t apply. The proof of this idea depends on something called the axiom of choice, which lets mathematicians create very unusual collections of points.

Even though it seems to break our everyday ideas about size, the Banach–Tarski paradox doesn’t actually break any rules in advanced mathematics. It just shows that some things in math can be very surprising!

Banach and Tarski publication

In 1924, mathematicians Stefan Banach and Alfred Tarski wrote a paper about a surprising idea. They showed that a solid ball can be split into pieces and then rearranged to make two identical balls. This is called the Banach–Tarski paradox.

They used earlier work by other mathematicians and studied how shapes can be broken apart and moved in space. They found that this surprising result only works in three dimensions or higher, not in one or two dimensions. This is because the rules for moving shapes are different in higher dimensions.

Formal treatment

The Banach–Tarski paradox shows that a ball in space can be split into pieces and rearranged to make two balls that are exactly the same size as the original. This is done by cutting the ball into smaller parts, moving and turning these parts, and putting them back together in a new way.

Mathematicians study this idea by looking at how shapes can match each other after certain moves. They use special groups of moves and talk about sets that can be split and rearranged in these ways. This helps explain the surprising result in a clearer way.

Connection with earlier work and the role of the axiom of choice

Banach and Tarski recognized that earlier work by Giuseppe Vitali, Hausdorff, and Banach themselves helped shape their ideas. They depended on Zermelo's axiom of choice, called "AC", to prove their surprising results. This same idea is also needed to prove other true statements in geometry, so they argued that rejecting AC because of this paradox would also reject other sensible results.

Later, A. P. Morse showed that one of these geometry results could be proved without AC, and Paul Cohen proved that AC cannot be proven from the basic rules of set theory. Even a weaker version of AC called the axiom of dependent choice, or DC, is not enough to prove the Banach–Tarski paradox.

Despite this, the paradox has inspired important research in mathematics, particularly in the study of groups. In 1991, research by Matthew Foreman and Friedrich Wehrung helped Janusz Pawlikowski show that the paradox can also be proved using the Hahn–Banach theorem, which itself uses a weaker form of AC known as the ultrafilter lemma.

A sketch of the proof

Here a proof is sketched that is similar but not identical to that given by Banach and Tarski. Essentially, the paradoxical decomposition of the ball is achieved in four steps:

1. Find a paradoxical decomposition of the free group in two generators.
2. Find a group of rotations in 3-d space isomorphic to the free group in two generators.
3. Use the paradoxical decomposition of that group and the axiom of choice to produce a paradoxical decomposition of the hollow unit sphere.
4. Extend this decomposition of the sphere to a decomposition of the solid unit ball.

These steps are discussed in more detail below.

Step 1

The free group with two generators a and b consists of all finite strings that can be formed from the four symbols a, a⁻¹, b and b⁻¹ such that no a appears directly next to an a⁻¹ and no b appears directly next to a b⁻¹. Two such strings can be concatenated and converted into a string of this type by repeatedly replacing the "forbidden" substrings with the empty string. For instance: abab⁻¹a⁻¹ concatenated with abab⁻¹a yields abab⁻¹a⁻¹abab⁻¹a, which contains the substring a⁻¹a, and so gets reduced to abab⁻¹bab⁻¹a, which contains the substring b⁻¹b, which gets reduced to abaab⁻¹a. One can check that the set of those strings with this operation forms a group with identity element the empty string e. This group may be called F₂.

The group F₂ can be "paradoxically decomposed" as follows: Let S(a) be the subset of F₂ consisting of all strings that start with a, and define S(a⁻¹), S(b) and S(b⁻¹) similarly. Clearly,

but also

and

where the notation aS(a⁻¹) means take all the strings in S(a⁻¹) and concatenate them on the left with a.

This is at the core of the proof. For example, there may be a string a a − 1 b in the set a S ( a − 1 ) which, because of the rule that a must not appear next to a − 1, reduces to the string b . Similarly, a S ( a − 1 ) contains all the strings that start with a − 1 (for example, the string a a − 1 a − 1 which reduces to a − 1 ). In this way, a S ( a − 1 ) contains all the strings that start with b , b − 1 and a − 1, as well as the empty string e .

Group F₂ has been cut into four pieces (plus the singleton {e}), then two of them "shifted" by multiplying with a or b, then "reassembled" as two pieces to make one copy of F₂ and the other two to make another copy of F₂. That is exactly what is intended to do to the ball.

Step 2

In order to find a free group of rotations of 3D space, i.e., one that behaves just like (or "is isomorphic to") the free group F₂, two orthogonal axes are taken (e.g., the x and z axes). Then, A is taken to be a rotation of θ = arccos ⁡ ( 1 3 ) about the x axis, and B to be a rotation of θ about the z axis (many other suitable pairs of irrational multiples of π could also be used here).

The group of rotations generated by A and B will be called H. Let ω be an element of H that starts with a positive rotation about the z axis, that is, an element of the form ω = … b k 3 a k 2 b k 1 with k 1 > 0 ,   k 2 , k 3 , … , k n ≠ 0 ,   n ≥ 1 . It can be shown by induction that ω maps the point ( 1 , 0 , 0 ) to ( k 3 N , l 2 3 N , m 3 N ) , for some k , l , m ∈ Z , N ∈ N . Analyzing k , l, and m modulo 3, one can show that l ≠ 0 . The same argument repeated (by symmetry of the problem) is valid when ω starts with a negative rotation about the z axis, or a rotation about the x axis. This shows that if ω is given by a non-trivial word in A and B, then ω ≠ e . Therefore, the group H is a free group, isomorphic to F₂.

The two rotations behave just like the elements a and b in the group F₂: there is now a paradoxical decomposition of H.

This step cannot be performed in two dimensions since it involves rotations in three dimensions. If two nontrivial rotations are taken about the same axis, the resulting group is either Z (if the ratio between the two angles is rational) or the free abelian group over two elements; either way, it lacks the property required in step 1.

An alternative arithmetic proof of the existence of free groups in some special orthogonal groups using integral quaternions leads to paradoxical decompositions of the rotation group.

Step 3

The unit sphere S² is partitioned into orbits by the action of our group H: two points belong to the same orbit if and only if there is a rotation in H which moves the first point into the second. (Note that the orbit of a point is a dense set in S².) The axiom of choice can be used to pick exactly one point from every orbit; collect these points into a set M. The action of H on a given orbit is free and transitive and so each orbit can be identified with H (except for countably many points which lie on the axis of some rotation in H, see below for how to deal with this). In other words, every point in S² can be reached in exactly one way by applying the proper rotation from H to the proper element from M. Because of this, the paradoxical decomposition of H yields a paradoxical decomposition of S² into four pieces A₁, A₂, A₃, A₄ as follows:

A 1 = S ( a ) M ∪ M ∪ B

A 2 = S ( a − 1 ) M ∖ B

A 3 = S ( b ) M

A 4 = S ( b − 1 ) M

where we define

S ( a ) M = { s ( x ) | s ∈ S ( a ) , x ∈ M }

and likewise for the other sets, and where we define

B = a − 1 M ∪ a − 2 M ∪ …

(The five "paradoxical" parts of F₂ were not used directly, as they would leave M as an extra piece after doubling, owing to the presence of the singleton {e}.)

The (majority of the) sphere has now been divided into four sets (each one dense on the sphere), and when two of these are rotated, the result double what was had before:

a A 2 = A 2 ∪ A 3 ∪ A 4

b A 4 = A 1 ∪ A 2 ∪ A 4

Step 4

Finally, connect every point on S² with a half-open segment to the origin; the paradoxical decomposition of S² then yields a paradoxical decomposition of the solid unit ball minus the point at the ball's center. (This center point needs a bit more care; see below.)

Some details, fleshed out

In Step 3, the sphere was partitioned into orbits of our group H. To streamline the proof, the discussion of points that are fixed by some rotation was omitted; since the paradoxical decomposition of F₂ relies on shifting certain subsets, the fact that some points are fixed might cause some trouble. Since any rotation of S² (other than the null rotation) has exactly two fixed points, and since H, which is isomorphic to F₂, is countable, there are countably many points of S² that are fixed by some rotation in H. Denote this set of fixed points as D. Step 3 proves that S² − D admits a paradoxical decomposition.

What remains to be shown is the Claim: S² − D is equidecomposable with S².

Proof. Let λ be some line through the origin that does not intersect any point in D. This is possible since D is countable. Let J be the set of angles, α, such that for some natural number n, and some P in D, r(_n_α)P is also in D, where r(_n_α) is a rotation about λ of _n_α. Then J is countable. So there exists an angle θ not in J. Let ρ be the rotation about λ by θ. Then ρ acts on S² with no fixed points in D, i.e., ρⁿ(D) is disjoint from D, and for natural mⁿ(D) is disjoint from ρ^m(D). Let E be the disjoint union of ρⁿ(D) over n = 0, 1, 2, ... . Then ρ(E) is the disjoint union of ρⁿ(D) over n ≥ 1 and therefore ρ(E) = E − D. Now we can conclude

S² = E ∪ (S² − E) ~ ρ(E) ∪ (S² − E) = (E − D) ∪ (S² − E) = S² − D, where ~ denotes "is equidecomposable to".

For step 4, it has already been shown that the ball minus a point admits a paradoxical decomposition; it remains to be shown that the ball minus a point is equidecomposable with the ball. Consider a circle within the ball, containing the point at the center of the ball. Using an argument like that used to prove the Claim, one can see that the full circle is equidecomposable with the circle minus the point at the ball's center. (Basically, a countable set of points on the circle can be rotated to give itself plus one more point.) Note that this involves the rotation about a point other than the origin, so the Banach–Tarski paradox involves isometries of Euclidean 3-space rather than just SO(3).

Use is made of the fact that if A ~ B and B ~ C, then A ~ C. The decomposition of A into C can be done using number of pieces equal to the product of the numbers needed for taking A into B and for taking B into C.

The proof sketched above requires 2 × 4 × 2 + 8 = 24 pieces—a factor of 2 to remove fixed points, a factor 4 from step 1, a factor 2 to recreate fixed points, and 8 for the center point of the second ball. But in step 1, when moving {e} and all strings of the form aⁿ_ into S(a⁻¹), do this to all orbits except one. Move {e} of this last orbit to the center point of the second ball. This brings the total down to 16 + 1 pieces. With more algebra, one can also decompose fixed orbits into 4 sets as in step 1. This gives 5 pieces and is the best possible.

Obtaining infinitely many balls from one

Using the Banach–Tarski paradox, we can get as many copies of a ball as we want from just one ball. This works in spaces with three or more dimensions. The ball can be divided into pieces that can be rearranged to form the same ball again, or even many copies of it.

Because of special math properties, we can also split a sphere into countless pieces. Each piece can be turned and moved to match the whole sphere. Even more amazingly, a sphere can be split into as many pieces as there are numbers, and each small piece can still be turned into two pieces that match the whole sphere. These ideas also work for a ball missing just its center point.

Von Neumann paradox in the Euclidean plane

Main article: Von Neumann paradox

In flat space, two shapes that can be split and rearranged using only straight moves and turns must have the same size. Because of this, it's impossible to take a square or circle and split it into pieces that can be rearranged into two copies of the original shape.

Von Neumann studied why this works in flat space but not in three dimensions. He found that the rules for moving shapes in flat space allow for a special kind of measurement that always gives the same result, making such tricky rearrangements impossible. However, if we use more complex ways to move shapes — like stretching while keeping areas the same — it becomes possible to create similar surprising results in flat space. This was shown when he split a square into pieces that could be rearranged into two copies of the original square using these special moves.

Researchers have continued to explore these ideas, finding new ways to split shapes in different spaces and with different rules for moving them. These studies help us understand more about geometry and the ways shapes can be transformed.