Theorem 7: the Banach-Tarski paradox

Let’s try a thought experiment.  Take a (theoretical) snooker ball.  Is it possible to break it up into one hundred parts in such a way that the parts can be put together to form two snooker balls?  The balls have to be solid, with no holes, the pieces may not be stretched or squashed, and the new balls must be the same size as the original one (or it would make for a bad game of snooker!).

Snooker balls

Obviously, it’s not possible, right?

Well…

Why is it obviously not possible?  Let’s examine our intuition.  Here’s an argument that seems compelling to me.

When we split the ball into several pieces, we don’t change the total mass of the pieces: it’s still the mass of the original ball.  (This is one advantage of a theoretical snooker ball, where we can break it without losing any splinters on the way!)  When we rearrange those pieces, in any way at all, we don’t change the total mass.  But the total mass of two snooker balls is (obviously) twice the mass of a single snooker ball.  Since cutting the balls and rearranging pieces can’t increase the mass, we can’t make two balls from one.

Are you convinced?

Casting doubt

Why did I cast doubts on this earlier?  I’m allowing us to break up the snooker ball in any way we like (as long as it’s into one hundred pieces).  In particular, we’re allowed to do things that aren’t possible in practice.  For example, one piece could consist of a single point, if we wanted.  What implications does that have for the argument I gave above?  Well, what’s the mass of a single point?

If a single point has a positive mass, then a snooker ball (which consists of infinitely many points) has infinite mass.  Since a snooker ball plainly does not have infinite mass, we’d better decide that a single point has zero mass.  Did you expect that?  It suggests to me that perhaps some interesting things can happen when we break up a snooker ball mathematically, rather than with a chisel and hammer.  But it’s going to get stranger yet…

As mathematicians, it’s starting to look a bit concerning.  We have an intuitive idea of what it means to talk about the mass of an object, but do we have a precise, mathematical definition?  Can we assign a mass to any subset of a snooker ball?

A proper definition of mass

It turns out that it is possible to come up with a precise definition of mass.  (I hasten to add that this is mass as viewed by a pure mathematician.  I make no claims that this is relevant to real physical objects!)  This leads to an area of maths called measure theory.  We need an object called a measure.  As with many mathematical definitions, the definition of a measure is that it is a particular type of object (in this case, a function) that satisfies certain conditions (the properties that we expect any reasonable notion of mass to satisfy).

We need a function: the function takes a subset of three-dimensional space (a chunk of our snooker ball), and gives as output the mass of that set, which should be a non-negative real number.  (In general, it could also be infinity, but our snooker ball definitely has finite mass!)  What properties do we want this function to have?  I’m not going to get into the precise details here, but let me mention two ideas.

One is that the mass of the empty set (the set with no members) is 0.  Well, any reasonable definition of mass ought to insist that the mass of nothing is nothing!

The other is that if we take two disjoint objects (they have no points in common), then the mass of both objects together is the sum of the masses of the individual objects.  For example, earlier we said that the mass of two snooker balls is twice the mass of an individual ball.  (In fact, the relevant property of a measure is more general, but I don’t want to be side-tracked.)

Right, so we can define a proper notion of mass.  So what’s the problem?  I rather glossed over the domain of the function.  That is, I didn’t tell you which sets have masses.  The measure that mathematicians use as a model of ordinary mass is the Lebesgue measure.  It turns out that no matter how hard one tries*, there are subsets that don’t have a Lebesgue measure: it’s not possible to give them a mass!  There’s a direct proof of this (rather extraordinary) fact, but one can also deduce it from this week’s theorem — so I’d better tell you what that result is.

Theorem (the BanachTarski paradox)  It is possible to break a (mathematical) snooker ball into one hundred pieces in such a way that those pieces can be reassembled to form two snooker balls.

(I’ve been rather generous in writing “one hundred” here: a somewhat smaller number would do.  The point is really that it’s a small finite number.)

The existence of non-measurable sets

 Our earlier argument (the one at the start) shows that if the pieces of the snooker ball are always measurable (have masses), then this wouldn’t work, so there must be a non-measurable set!

(It’s easier to prove the existence of a non-measurable set than it is to prove the Banach-Tarski paradox, but if one knows about the “paradox” then the existence of a non-measurable set follows directly.)

* I’m assuming the Axiom of Choice in this post.  It’s a particular assumption that mathematicians may or may not choose to make in their work.  It’s not always relevant, but there are some results that really depend on it, so it can be a good idea to mention when one is using it.  The moral of this post is that it can lead to some apparently bizarre consequences (despite sounding very plausible)!

Further reading

Here’s the Wikipedia page about the Banach-Tarski paradox.  The classic book on the subject is The Banach-Tarski Paradox by Stan Wagon, published by Cambridge University Press in the series Encyclopedia of mathematics and its applications.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: