**Preliminaries**

First of all, let's nail down what exactly
we're talking about so that we're all on the same
page.

First and foremost, we're talking about a
*mathematical* sphere, not a *physical
sphere*, although I'd like to use an analogy
with physical spheres to describe one possible
way to intuit the Banach-Tarski Paradox. By
*mathematical sphere,* I mean the set of
points that lie within a 3-dimensional spherical
area in ℜ^{3}, where ℜ is
the set of all real numbers. For simplicity,
let's assume a radius of 1, so our sphere would
be the set:

S = {(x,y,z) |
x^{2}+y^{2}+z^{2} <= 1 }

One important difference between S and a real,
physical sphere is that S is *infinitely*
divisible. Mathematically speaking, S contains
an infinite number^{2} of points. This
is not true of a physical sphere, as there are
a finite number of atoms in any given physical
sphere; so a physical sphere is *not*
infinitely divisible.

^{2}*Or, to be precise, *`c`
points, where `c` is the cardinality of
the continuum.

In fact, if we assume that spheres are *not* infinitely divisible, then the
Banach-Tarski paradox doesn't apply, because each
of the "pieces" in the paradox is so infinitely
complex that they are not "measurable" (in
human language, they do not have a well-defined
volume; it is impossible to measure their volume).
Immeasurable pieces can only exist if the sphere
can be cut into infinitely-detailed pieces; this
obviously isn't true for real spheres, since you
cannot cut atoms into arbitrary shapes, especially
not into infinitely complex shapes.

Now let's move on to the paradox itself.

**The Banach-Tarski Paradox**

The Banach-Tarski Paradox states, basically,
that it is possible to take S (as we've defined
above), and cut it up into n disjoint pieces,
which we shall call A_{1}, A_{2},
... A_{n}, whose union is S itself, where n is a finite number, such that if we perform
some (finite) sequence of rotations and translations on each of the
A_{i}'s, we will end up with two copies
of S. Or, to be mathematically precise:

- A
_{i} ∩ A_{j} = ∅ for each i and j between 1 and n such
that i≠j
(no two pieces overlap each other)
- A
_{1} ∪ A_{2} ∪
... ∪ A_{n} = S
(assembling all the pieces yields the original sphere S)
- There exist T
_{1}, T_{2}, ...
T_{n}, where each T_{i}
represents some finite sequence of rotations and
translations, such that if we apply each
T_{i} to each A_{i} (let's
call the result A_{i}'), then:
- A
_{1}' ∪ A_{2}' ∪
... ∪ A_{m}' = S
(a subset of the original pieces forms S)
- A
_{m+1}' ∪ A_{m+2}'
∪ ... A_{n} = S'
(the remaining pieces forms a copy of S)

where m is some number between 1 and n, and
S' is S translated by some finite amount so
that S and S' are disjoint.

It is interesting to note that one corollary
of this paradox is that you can take a sphere,
cut it into n pieces, *remove* some of
the pieces, and reassemble the remaining pieces
back into the original sphere without missing anything. Obviously, this is impossible with a
physical sphere; but it is quite possible with
mathematical spheres (which are infinitely divisible), if the Axiom of Choice is assumed.

*(Before you dismiss this notion outright, let
me state that mathematically infinite objects do
not always behave intuitively. As a comparison,
we use a more intuitive example of duplicating
the set of integers: given N, the (infinitely large) set containing all the
integers, we can split them up into two sets,
E containing all the even integers, and F containing all the odd integers. Are E and F
each
smaller than N, the set of all integers? Intuitively, it appears to be so; however, I will
convince you that they are, in fact, the same size.
First, we take E, and rename each member of E
so that a number x is renamed to x divided by two.
What do we get? We now find that E=N. Similarly,
we take each member y from F, and rename y to
(y-1)/2. Whoopie, we also find that F=N. We have
just duplicated the set of integers using nothing
more than just the original integers. We didn't
even need to use infinitely-divisible freak
objects to achieve this.)*

Coming back to spheres, it is helpful to
keep in mind that each of these pieces A_{i} are potentially
infinitely complex so that they do not have any
well-defined volume.
Now, this whole paradox may seem remotely possible if we had, say,
required 1,000,000 pieces to achieve our feat;
there's intuitively more room for a sleight of
hand if we had a million pieces to play with.
However, the kicker about this whole
paradox is that we don't need more than **five** pieces to achieve this feat. And
unlike our odd/even number example, we do *not* need to play tricks with renaming individual points of each piece; we can perform
the miracle by merely using well-behaved operations like rotations and translations.
*Furthermore,* one of these pieces only needs to
contain the single point at the centre of the
sphere.

In other words, it is mathematically possible
to cut S into a mere *four* pieces (if
we disregard the one center point), and to
reassemble two of these pieces into the original
S, and reassemble the other two into a copy of
S.

The catch, of course, is that each of these
four pieces are so complex that they do not have
any "measure" (i.e., their respective volumes are
not well-defined), and that we do not know how to
mathematically describe them other than the fact
that they exist and exhibit the strange re-assembly property. In fact, it is quite possible that each of those pieces consists of
isolated points spread out throughout the entire
volume of the original sphere S.

**How can this ***ever* be intuitive??!

I promised to convince you that this bizarre
mathematical phenomenon isn't all that strange,
after all. So here's my proposed "intuitive"
rationalization of it. I'll do it by way of an
analogy with a physical sphere.

Let's forget for the moment the mathematical
sphere S, which has infinite density. Let's consider a real, physical sphere B (for "ball"),
also of radius 1. B is identical to S except that
it consists of a finite (albeit large) number of
atoms. The way these atoms are laid out in B is
called the *crystalline structure* of B.
(I.e., if you take B, or any physical object for
that matter, and look at it under an electron
microscope, you will see the atoms laid out in
a fixed, regular pattern. That's called its
"crystal lattice".) Usually, the crystalline
structure is a simple geometric relationship
between neighbouring atoms.

Notice that although the geometric relationship
between atoms define its crystalline structure,
the *precise distance* between atoms may
vary. This leads to materials of different densities.

Now, we perform the equivalent of a Banach-Tarski decomposition on our physical sphere
B: we "atomize" B into four spherical clouds of
atoms, let's call them C_{1}, C_{2}, C_{3}, and C_{4}.
(We'll ignore the central atom in B, just as in
the mathematical version of this decomposition.)
Let's assume that each of these clouds are sparse
enough that they are gaseous, no longer solid by
themselves (imitating the immeasurability of the
mathematical pieces of S). Furthermore, let's say
that the atoms in each of these clouds are laid
out in a regular pattern, so that if we rotate
C_{1} by some angle G, and put it together
with C_{2} in the same spherical region,
the atoms in both clouds line up into the same
crystalline structure as B, except that now the
distance between atoms is greater (to account for
the missing atoms now in clouds C_{3} and
C_{4}). Similarly, assume we can do the
same with C_{3} and C_{4}: we
just translate them away from the original spherical region of B so that they don't interfere with C_{1} and C_{2}, and
reassemble them into another sphere.

Now, we have successfully built two (physical!)
spheres with the same radius as B, using only
material from B itself. Each of the two spheres
have the same crystalline structure as B. The only
difference between these spheres and B is that
they each have only half the density of B.

To bring this analogy back to the mathematical
sphere S: we can think of the infinitely complex
pieces A_{1}, ... A_{4} as the
equivalent of "atom clouds", which are non-solid
(immeasurable). Think of the crystalline structure
of S as the topological structure of points in
ℜ^{3}. These "clouds" lack this
"crystalline structure" (i.e., they are unmeasurable); but
by suitable rearrangement of them, we can form
them into two identical spheres, with half the
density of the original, so that they *do*
have the same "crystalline structure" (i.e., the
resultant two spheres are well-behaved, measurable
sets). These two spheres are identical to S,
except for having only half the density of S.
However, S is infinitely dense, and so are its
pieces A_{1}, ... A_{4}. This
means the two resultant spheres are *still* infinitely dense. That is to say, they are
*identical* to S.

**Bingo!** There is no
paradox here after all. We are merely seeing the
logical consequence of mathematical sets like S
being infinitely dense. In fact, if you think
about it, this is not any stranger than how we
managed to duplicate the set of all integers,
by splitting it up into two halves, and renaming
the members in each half so they each become
identical to the original set again. It is only
logical that we can continually extract more
volume out of an infinitely dense, mathematical
sphere S.

**Epilogue**

Now, having convinced you that the Banach-Tarski Paradox isn't really *that*
strange after all, I'd like to mention that the
derivation of this paradox depends on the Axiom
of Choice, and although most mathematicians
accept the Axiom of Choice, not all agree with it. There has
been much debate over the merit of adopting this
axiom, as well as research into the consequences
of choosing either way: it does simplify a lot of mathematical
proofs, but it also introduces strange results
like the Banach-Tarski paradox which
we just discussed.

If you're unfamiliar with the Axiom of Choice,
it basically goes like this: if you have a collection
of sets C (which may potentially contain an uncountably large number of sets), then there
exists a set H, called the *choice set*, which
contains precisely one element from each (non-empty) set in C. H is called the "choice set" because you
are essentially going through each set in C and
*choosing* one element from it.
One feature of the Axiom of Choice is that H is
simply assumed to *exist*; there is no algorithm given which might tell you how to construct
an example of H.

In the case of the Banach-Tarski
paradox, each of the infinitely complex "pieces" of
the sphere S is built from these choice sets. Since we do not know of any algorithm to actually
construct these sets, we can only indirectly infer some of the properties of the "pieces", such as their not having a
(Lebesgue-) measure (i.e., they have intractible geometric complexity).
Some of the debate surrounding the Axiom of Choice
revolves around whether these non-constructible sets
are mathematically admissible. The reader is encouraged to make good use to Google for more information about this debate; it is too vast a topic
to explore in this article. It suffices to say that
most mathematicians adopt the Axiom of Choice, simply
because of the usefulness of results that can be derived.

One might wonder, then, about what would happen if we
*didn't* assume the Axiom of Choice. We do
know that we would likely be unable to derive the
Banach-Tarski paradox; however, we *also* know
that paradoxical sets *do* exist even without
the Axiom of Choice. These paradoxical sets exhibit
the same "weird" behaviour of the Banach-Tarski spheres, in that you can decompose these sets into a
finite number of parts, and reassemble them into
multiple copies of the original.