In most people’s minds, or at least in the minds of those with a mathematical bent, every set of points
in
(a fancy way of writing three-dimensional Euclidean space) has a volume. Some sets, such as any set consisting of a finite number of points, has zero volume, whereas other sets, such as the set of all points less than one unit away from the origin, have finite but non-zero volume, and still other sets, such as
itself, have infinite volume. But does every set necessarily have a volume? It turns out the answer is no, even in one-dimensional Euclidean space; I showed this in a previous post. Instead, I’ll show you the Banach-Tarski paradox, which states that it is possible, via translations and rotations, to dissect a three-dimensional sphere (technically a 3-ball) and rearrange it into two spheres of the same radius as the original.
For simplicity, we’ll start by only looking at the surface of the sphere,
. Let
refer to rotation by
(that is, pi degrees) about the
-axis and let
refer to rotation by
about the
-axis. The exact degree of the rotation isn’t important, only that it should be impossible via any combination of
and
to go back to the origin. Let
be the set of all rotations that you can get by combining
,
, and their inverses. Then
and
form what’s known as the ”free group on two generators“, which is composed of all strings using
,
,
, and
as symbols, with the provision that
and
cannot appear together for
. The multiplication for this group is just writing the strings together, subject to the rule that
and
cancel, as do
and
. So, for example,
. The name arises from the fact that there are two ‘fundamental’ symbols,
and
, and it is ‘free’, since they don’t commute (that is,
).
Now, one very weird fact about the free group on two generators (written
) is that you can break it into four pieces, rearrange those four pieces, and then reassemble them into two copies of it. To elaborate, let
denote the set of strings in
that start with
, and similarly for the other three symbols; let
denote the empty string. Then obviously

in
) 
doesn’t start with
, then it’s the same string as
and so is in
; otherwise, it’s in
. We also have, for the same reason, 
How is that relevant? Well,
partitions
into orbits, where an orbit is a collection of points such that each element of
moves points from the orbit into the orbit, and points not in the orbit into another point not in the orbit. For each orbit, we can pick a point
inside it; let the set of all these points be
. Then we can turn the decomposition of
into a decomposition of
; using
to indicate the set of all rotations in
that start with
, we have

These four sets together make up the sphere (except for
itself; we’ll get back to that), but if we pick two of them and rotate one of those two, we also get back the sphere. Therefore, we can break down the sphere into four pieces and recompose them into two spheres. If we then draw lines joining the sphere to the origin, we can turn this into a decomposition of two solid spheres.
Now, there are two omissions in this proof: first off, I ignored the origin. It turns out that you can cut the sphere without the origin up, rearrange it, and then put it back together so that you have the origin. Second, I ignored the axes of the rotations in
; they are the fixed points of these rotations, and so might cause problems; it turns out that if we call those points
; we can again cut up
and rearrange it into
.
So what does this mean? If we want to have any reasonable definition of volume, then we want translations and rotations to leave it the same. Then, we must have that some of these sets don’t have a well-defined volume; much like how you can construct a subset of the real line that doesn’t have one. We break the sphere down into a number of sets, some of which don’t have a defined volume; then we move them around and reassemble them. There’s no contradiction here, since two sets that don’t have a well-defined volume (or non-measurable sets, as they’re known) might have a union that has a volume. Of course, this is physically meaningless, since you can’t actually perform the required divisions; atoms are not infinitely small, after all.
is true when S is empty, or else we would lose the theorem that
. While it could certainly be special-cased to take into account the cases where A or B is the empty set, it would detract from its simplicity, and would require proofs involving it to first show that neither of the two is empty (which can be a challenge, and might even require additional axioms if dealing with infinite sets, choice functions, etc.)
, even when
. But returning to the linguistics theme, it seems rather odd; this is because the quantifier ‘all’ has an implication that the entities that are being talked about exist; in other words, the statement ‘all x are y’ implies that at least one x exists. If this is known to be false by the speaker, it sounds odd, much like the statement ‘John has two children’ sounds odd if the listener knows that John actually has exactly four children; the statement ‘John has two children’ implicates that he has exactly two.
. 
into
:
. Even though the imaginary part isn’t physically meaningful, it turns out that adding it makes the mathematics a lot simpler to manipulate, because exponentials are easy to multiply, whereas individual cosines and sines are not.
. This is the sequence that goes through
complete cosine waves from 0 to
, with a matching real component. What’s
for this sequence? It’s just
, then this is just
. Less obvious, though, is the fact that if
is an integer, then
. So essentially by transforming the sequence in this fashion, you can ‘decompose’ it into cosine waves. (Sine waves will manifest themselves as imaginary components). So if you have a very large
, you can get a large number of different frequencies, ranging from 1 hertz all the way up to N hertz. There are various intricacies, such as the fact that signals are not, in general, complex numbers, but the core explanation here is enough.
, we obviously should have that the length of
signifies the length of
,
for every
and then every set would have length zero! Not exactly useful.
such that for all
, there is a
. Again, written out, this becomes
be the set of irrational numbers in
, restricted so that if
, then
. Oh, and we also have to include 0 in
,
. We can do this since the rational numbers are all countable. Then, we define
is disjoint from each other one, since if
was in
and
, then we would have
, for some
or
, an impossibility by the definition of
. Obviously,
, but we also have
To show this, pick some
. If
, then we’re done, else it was excluded since there is some
such that
. So we have
. We know that
. But since each
moved around on the real line, we have
, which will either be infinity or zero depending on whether V has zero measure or not. But by the above relation, we have
. So therefore,
, is defined by
whenever 
. Then obviously
. Your homework for today is to prove that
. It’s not enough to show that those two sets are uncountable; there exist uncountable sets of measure 0, such as the Cantor set.