The Banach Tarski Paradox for dummies

The Banach Tarski Paradox is a historically significant paradox that shows that it is possible to take a ball, break it into pieces, and rearrange the pieces in order to form two new balls, therefore doubling the mass and volume.  (Note: the following is a sketch of the proof, it has many untied ends and is NOT airtight)(knowledge of sets, and countability are assummed.  Learn more at : http://en.wikipedia.org/wiki/Set_%28mathematics%29 and http://en.wikipedia.org/wiki/Countable_set)

We will prove this in 6 simple steps:

Step 1:  Background

The set F2 is the set of all simplified "words" of a and b.  In simpler terms, it is all combinations of a, a^-1 (which for simplicity of typing, I will call -a) b, b^-1.  A sample word looks like the following

                                                          a, b, b, -a, -b, -a, b 

Now, whenever one has an 'a' and a negative 'a' next to each other, they cancel, and the same goes for b and -b.  So, F2 doesn't include:

                                        -a, a, b, b            but it does include                 b, b

We also define S(x) as the subset of F2 that starts with x.

A good way to visualize this is the diagram below, where every word is defined as a point; to get to that point, one has to take the appropriate moves, where a is right, b is up, etc, and e is the empty word:

Therefore, S(a) is represented by the green circle, S(-a) by the red circle (the gaps in the circles were due to size limits in blogger uploads) , but how does aS(-a) represent the blue circle?  First of all, aS(x) = all strings that start with x with an 'a' concatenated in front.  We know that whenever we concatenate the 'a' with a word that starts with '-a', they cancel, and we get a word that starts with any letter except 'a', as if 'a' started our new word, that means our original word was -a, a, ..., which isn't simplified.  Therefore

                                                                 F2 - S(a) = aS(-a)

Also note:  b, b, b, b, b = b5 (5 equals number of repetitions)

We define the following:

W = S(a)                 X = S(-a)                     Y=S(b)         Z = S(-b)

The reason we defined the sets the following way was to get the following properties:

if x ϵ W, then ax ϵ W , and this holds for X with -a, Y with b, and Z with -b.

Also, the union of all 4 sets is F2-{empty}, and they have no elements in common.

WE ARE ALMOST DONE WITH THE BACKGROUND.  DON'T GIVE UP!

Two sets are "equidecomposable" (symbol: ~) if you can take one set, cut it into pieces, move them around and rotate them (no dilations, inversions, etc, otherwise the size of the pieces is changed) and get the other, as shown below:

For those of you with a greater mathematical background (I don't know why you are reading this then), this notion is an equivalence class.  Challenge yourself to prove that it is.

We are finally ready to actually begin the paradox.

Step 2: Prove that a "broken circle" ~ a circle

A broken circle is a circle with a missing point on the circumference, without loss of generality call it D.  We take the circle and "cut" it into two pieces

A = {All points that can be reached by taking point D and rotating it by rotation r x times}

A will look like this {D, Dr, Dr2, Dr3, etc.} (Note: this can be expressed as e^i* theta, but the r notation conveys the same result)

B = {All points on the circle that aren't in A}

If we rotate set A by r, all the elements of A are multiplied by r, as shown below

rA = {Dr, Dr2, Dr3, etc...}

We want to make sure that for any n and m > 0, Drn is not equal to Drm.  We can do so with the following argument:

There are a countable number of m's and n's that result in Drn = Drm.  However, there are an uncountable number of angles possible for rotation.  Therefore, there exists a rotation for which no rn = rm.

Therefore, by rotating A by r and keeping B the same, a point disappeared on the circle: D.

Step 3: A circle ~ A circle missing a countable number of points

We start with the circle.  We use Step two in order to remove a countable number of points by taking out one point at a time.  We must make sure that none of points in our procedure overlap.  We will do so with the following argument:

The number of points needed to remove 1 point using the procedure in Step 2 is countable.  The number of points removed is countable.  So, we know by the fact that the union of countably many sets is countable (For those of you who are bothered by my usage of the axiom of countable choice, the ride only gets rougher.  I suggest stopping now.) that the number of points that we have to guarantee don't overlap is countable. But the number of rotations is uncountable, therefore we can find a "good" angle to rotate by which satisfies all our demands.

Step 4: A sphere ~ A sphere missing countably many points

We use the same method as in Step 3 but we now rotate along an axis of the sphere

Step 5: A ball missing its center ~ 2 balls missing their centers (notice how the steps are progressively getting shorter.)

Theorem: rotations of a ball act like elements of F2

So, we can define W, X, Y, and Z as we did before, however a and b are now rotations, and -a and -b are their inverses.  If we rotate X by a and make it aX, we find the following

F2 - S(a) = aS(-a)   =>  F2 - W = aX  => F2 = W ∪ aX, and since a is a rotation, we know that those two pieces create a ball, as F2 is the entire ball.  We can use the same logic on Y and Z and say that they make a ball too.  We must do this ignoring the centers of each ball as the union of the W, X, Y, and Z doesn't include the center.

Step 6: Ball ~ 2 Balls

By Step two, we can insert a point into each ball.