# A Bird in the Hand is Worth…

..two in the hand?

There are many counter-intuitive, bizzare theorems in mathematics. One of my favourites has to be the Banach-Tarski Paradox. A result due to Stefen Banach and Alfred Tarski in 1924, the paradox states that given a ball, there is a way you can cut the ball into a finite number of pieces and rearrange these pieces to form two balls that are exactly the same size as the original. This seemingly defies our intuition. I will start with an ‘intuitive’ approach and then dive straight into the deep end with some technical stuff.

I won’t be proving the theorem here as it requires a high level understanding of group theory, measure theory and geometric set theory. What I will be doing is illustrating problems with infinity that hopefully can give you an inkling as to how the Banach-Tarski might be true.

# Meet The HyperWebster

This thought experiment was put forward by Dr. Ian Stewart. We consider a dictionary with every possible letter combination. Start of with the ‘empty word’ (think of it as a word with no letters — this is important and we will come back to it later) we will denote the empty word as ‘ ’. Then we move on with *a, aa, aaa, … *until we have an infinite string of *a*’s. Then move on to *ab, aba, abaa, … *and proceed in this manner all the way to an infinite string of *z*’s.

In this dictionary, you will find the entirety of War and Peace (without spaces and punctuation of course), the full script of Hamlet and every possible variation of what happened to Amelia Earhart, as well as. Every. Possible. Combination. Ever.

Every single road you will live on. As well as every single road you won’t live one. It also contains lots of garbage words as well.

This dictionary is obviously large.

Like stupidly large.

Well it’s **infinite**.

In any case, let’s look at this HyperWebster in a bit more detail.

# Volumes

As with most encyclopediae or other large books, we will break the HyperWebster up into volumes.

These volumes will be categorised alphabetically. So all the words beginning with *a* will be in the volume entitled *A*. All the words starting with *b* in the volume entitled *B* and so on and so forth.

We have now classified every single word imaginable. Except one; the empty word. We will leave that to one side in it’s own special volume but remember that it is still there.

# Printing cost

The publishers of the HyperWebster decide that they want to cut costs of printing. They make the observation that they can remove the first letter of each word and tell the reader to look at the volume title and add that letter to the front.

For example the word *aadfff…* in the *K *volume will in fact be the word *kaadfff…*

If we consider the *A* volume, then it contains every single letter combination that starts with the letter *a*. So when we remove the letter *a*, we get every single word combination.

We recover the entire HyperWebster.

What’s even neater is that when we remove *a* from the word *a*, we recover the empty word (told you we would need it — and it turns out to be important when considering the sphere). So we truly do get the whole dictionary back again.

Doing this to every volume, we recover 26 times the original HyperWebster.

**To summarise**, we split up the dictionary into volumes. Then by changing these volumes slightly (definitely not increasing them, whatever increasing means in this scenario) we get 26 times the original copy.

# Back to Banach-Tarski

Hopefully you can see that if we can somehow label each point on a sphere with a sequence of letters, then split them into volumes and manipulate those carefully, then we could create twice the original sphere.

There are a few twists and turns down the path of that proof but that is the skeleton of it.

Once we have made two spheres, we can make three, four, five,… As many as we want.

# The Deep End

Banach-Tarski formally states that given two bounded subsets *A* and *B *of ℝ*ⁿ*, for *n≥3 *with non-empty interior, then *A *and *B *are equidecomposable with respect to the isometries of ℝ*ⁿ.*

Two sets *A *and *B *are equidecomposable with respect to a group *G *that acts on *A *and* B *if there exists a partition *A*₁,…,*A*ₖ and *B*₁,…,*B*ₖ and group action *g*₁*,…g*ₖ such that *g*ᵢ*(A*ᵢ*) = B*ᵢ.

This reduces to the Banach-Tarski Paradox by letting *A *be a closed ball of unit radius and *B *be two closed balls of unit radius. These are both bounded subsets of ℝ*³ *and have non-empty interiors

A similar result holds for *n = 2 *with the slight difference that the equidecomposition is via area preserving affine transformations and it is called the Von Neumann Paradox.

The proof of Banach-Tarski is dependent on the assumption of the axiom of choice and in fact cannot be proved if we do not assume it. Recall the axiom of choice states that for a (possibly infinite) collection of sets, there exists a choice function that chooses some (possibly no) elements of each set and puts it all into a new set. As a result, the proof is non-constructive and most pieces are impossible to make in the real world.

Note that we do not require the pieces of the partition to be Lebesgue measurable since isometries acting on sets preserves the measure and so if each piece of the partition were measurable then *A *and *B *would be forced to have the same measure. This is quite a strong restriction in comparison to the original statement.

This is actually a necesary and sufficient condition for the pieces to be measurable: a recent result due to Grabowski, Mathé, and Pikhurko (2019) states that provided *A *and *B *satisfy the conditions of the Banach-Tarski Theorem and they (are measurable) and of equal measure then they are equidecomposable with respect to isometries of ℝ*ⁿ *with the added requirement that each piece of the partition is (Lebesgue) measurable.

Another interesting result is that the Banach-Tarski (and also the 2019 result) is independent of the geometry; it holds true in hyperbolic and spherical geometries of dimension 3 or higher.

I will finish on what a mathematician would consider a ‘joke’.

Q. What is an anagram of Banach-Tarski?

A. Banach-Tarski Banach-Tarski.