Banach-Tarski ParadoxDate: 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/ |
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