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Banach-Tarski Paradox

Date: 07/20/2002 at 02:38:00
From: Tony Zhu
Subject: Axiom of choice and Banach-Tarski paradox

The following is an excerpt from _The Mystery of the Aleph_ by Amir 
Aczel.

  The Banach-Tarski paradox starts with an application of the 
  axiom of choice.  By mathematical derivations in Euclidean 
  space (the usual space of three or more dimensions in which 
  geometry is studied), the two mathematicians have shown that 
  a sphere of a fixed radius can be decomposed into a finite 
  number of parts and then put together again to form two 
  spheres each with the same radius as the original sphere.

What's described in the excerpt is very paradoxical to me.  It seems 
that something is created out of nothing. Can someone shed light on 
what is happening?

Thanks.


Date: 07/20/2002 at 17:39:11
From: Doctor Tom
Subject: Re: Axiom of choice and Banach-Tarski paradox

Hi Tony,

Maybe I can shed a little light.  Banach-Tarski in fact splits the 
sphere into 5 pieces, one of which is the point at the center of the 
sphere. The other sets are weird sets of points that can only be 
constructed with the help of the axiom of choice.

The reason it seems paradoxical is because the sphere itself is 
measurable - it has a volume given by the formula V = (4/3)pi*r^3, 
where r is the radius.

But the other four sets of B-T are not measurable. In other words, 
there is no sensible way to assign a meaningful measure to those sets.

One can show that any number you assign as a measure will lead to some 
sort of contradiction, just in the same way that you could define 1/0 
to be any number, but if you did so, many other rules of arithmetic 
would fail, so usually, 1/0 is simply called "undefined."  In the same
way, the measures of some sets derived using the axiom of choice must 
be "undefined."

Measure is a VERY slippery subject, and a great deal of time is taken 
in courses on measure theory to decide exactly which sets are 
measurable, and which are not. The non-measurable sets are usually 
ignored, since you can, in a sense, make them behave any way you want.

Here is a relatively simple example of a non-measurable set, which 
should show you basically why you get into trouble.

Start with the set of real numbers that are greater than or equal to 
zero and less than 1. The measure (length, in this case) of the set is 
clearly 1.

Now take the set Q of the rational numbers in this set, and for every 
real number r in the set, define the set:

  Q+r = {q+r mod 1, where q is a member of Q}

by "mod 1" I mean that if q+r is bigger than 1, then subtract 1 to get 
it back in the range.

If r1 and r2 are any rational numbers, then

  Q+r1 = Q+r2.

If r is a real number and q is a rational number, then

  Q+r = Q+(r+q).

In this way, we can divide all the real numbers into equivalence 
classes, and we say that r1 == r2 (r1 is equivalent to r2) if and only 
if:

  Q+r1 = Q+r2

Now, use the axiom of choice to pick one member from each equivalence 
class, and call this set S.

For every two different rational numbers q1 and q2, S+q1 is not equal 
to S+q2; but for any number in [0,1) it is a member of S+q, for some 
rational q.

If you take the union of all the sets S+q for all rational numbers q, 
they make the interval [0,1). But they are all the same size, since 
they are basically translations of each other. And there are a 
countably infinite number of them. If the measure of S is zero, the 
measure of [0,1) is 0 times a countably infinite number, or zero.

If S has some non-zero measure, then the measure of [0,1) is infinite.  
There is no reasonable way to assign a measure to S.

Maybe this isn't so simple, but it is the simplest example of a 
non-measurable set, so if you do want to understand the B-T "paradox," 
you must begin by understanding this.

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
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