Double for Nothing: the Banach-Tarski Paradox

Though those stodgy engineers and accountants may disagree with me, one of the most exciting parts of mathematics is learning about a result so bizarre, so against your intuition built up by years of experience, that it must obviously be false. These are often labeled “paradoxes.”


But this is mathematics we’re talking about. These results may assault the intuition, but it’s not because they’re wrong or the result of faulty logic, like so many other paradoxes. We come with proofs and irrefutable logic. Sure, you might be able to argue against the assumptions (axioms) we start with, but given those perfectly reasonable and seemingly innocent rules, we can create truly bizarre things.


No matter how much those accountants dislike it.


And today, we get to talk about a doozy of a paradox. We’re going to take a ball, cut it up into 5 or so pieces, move them apart, rotate them around a bit, and end up with two balls, exactly identical to the original.


This is the infamous Banach-Tarski paradox.


Of course, creating something from nothing is an old trick. For instance, if you take all the numbers between 0 and 1, which has length 1, and you multiply each of those numbers by 2, you end up with all the numbers between 0 and 2, which has length 2! Pretty much everyone’s okay with this one.

But the Banach-Tarski paradox is much weirder. We turn one ball into two, but there is no stretching involved. The only things we do is cut the ball into pieces, move them apart, rotate the pieces a bit, then move them back together. And just moving parts of a ball shouldn’t create a new ball out of thin air.


But if we do it just right, we can.

So, how do we start this black magic?


While this paradox seems to be about geometry and measuring things, the key step, the key observation, has nothing to do with either. It’s all about a dictionary with every possible word in it.


For simplicity (and to line up nicely with what we’ll do later), let’s pretend our alphabet only has 4 letters — B, F, L, and R. So, the first word in our dictionary would be B, followed by BB, then BBB, then BBBB, then BBBBB, then…

And, finally, after all those exciting words, we’d finally get to BF, then BFB, then BFBB, then…

Of course, after you finish all the B words, you get to start the F words! F, FB, FBB, … FFB, FFBB, … FL, FLB, FLBB, …

I think you get where this is going.


Now, we have all of these infinitely many words, every word that could every be thought of or written. How are we going to print this dictionary?

Well, the first volume of our dictionary could contain all the “B” words. Of course, if everyone knows it’s the “B” volume, we could save space by not printing the first letter of each word. That means the first word would be ___, then B, then BB, then BBB, then …, then F, then FB, then FBB, then …

But wait a second. Those words F, FB, FBB, etc. which represent BF, BFB, BFBB, etc. are actually all the words that begin with F!

In fact, since every word could be extended to a “B” word by just adding a B as the first letter, by taking off that first B, our “B” volume is really a dictionary of all the possible words!1 One volume of our dictionary can take the place of four!


See what we did there?

No? Well, let’s look at it in a more graphical way, and even closer to what we’ll actually do for the ball.

Grab a ball.

Instead of letters in a word, we can think of B, F, L, and R as being directions to rotate the ball a few degrees. “B” means to rotate the ball a bit backwards, towards yourself, “F” means to rotate the ball a bit forwards, away from yourself, “L” a bit left, and “R” a bit right. Then, all the one letter “words” represent rotating the ball exactly once, and we could graphically represent them like this:



The center intersection, which we’ve labeled “N” is not rotating at all. In our dictionary it was the ____ “word.”

Two letter “words” would represent rotating twice. For words, though, we could have something like “BF” or “FB,” but when thinking about rotations, we don’t want to undo the rotation we just did, so we’re not going to allow rotations that undo the one we just had. In fact, we can think of them cancelling out, so that “BF” is really the same as “N.”

But the remaining ones, like “FL” and “RF” are fine. We can add those to our graph like this:


We make each additional step half the size so that we can keep all these words straight, and when we say “FL,” we should think of that word representing doing a left rotation, then a forward rotation. Thus, all the series of rotations that end with a left rotation are on the left hand side of our graph.

After the two letter words, we get the 3 letter words (3 rotations), then the 4 letter words (4 rotations), etc.. This graph contains all possible order of rotations we could have made! Each intersection on the graph represents a different order of rotations.



So, how can we create something out of nothing?

Look closely at, say, the left part of the graph.


All of the points in this part of the graph are represented by words starting with L, meaning the last rotation was to the left. But, like with the dictionary, by removing the first letter (the last rotation), we can have other words appear!

For words, it made sense to just remove the first letter. But since now we’re doing series of rotations, we can remove the first L instead by taking the ball and rotating right to undo that last left rotation!

If we do that, for instance, L becomes RL, which cancel out and give us the center point, N. The rotation LF becomes RLF, but the R and L cancel out and this becomes just F. The rotation LBR becomes RLBR, which is really just BR.


If you keep track of each set of rotations, after undoing that last L, the set of rotations on the left (which was about 1/4 of the graph) becomes the entire top, left, and bottom parts of the graph (about 3/4 of the graph!)


The only points we don’t get are the ones on the right, but that’s because those words start with R, and we couldn’t have had “LR…” words in the left part of the graph, since we didn’t allow rotations that would cancel each other out like that.

So, by undoing a rotation, we seemingly create points out of thin air! This is the key trick that will let us duplicate a ball.

But for right now, let’s use it just on this graph. I’m going to cut the graph up into 5 pieces, then, using this “undoing a rotation” trick, make two complete copies of the graph, with one piece left over!

To do this, we’ll call S(L) to be the set of all rotations, where the last rotation was to the left. This is the set we were playing with earlier. We’ll define S(R), S(B), and S(F) similarly. The fifth and final set is the odd one out, just the center point N.

To undo the final left rotation of S(L), we can rotate to the right, which we could write as R\,S(L). As we’ve already said, R\,S(L) is already the entire graph, except for the right part, S(R). But that was one of the pieces we have lying around. So, the first copy of the graph is R\,S(L) plus S(R).


To make the second copy, we can take all the rotations that end with a back rotation, S(B), and then undo it with a forward rotation. Then, like before, F\,S(B) plus S(F) make a second copy of the graph.


Two graphs for the price of one!


This trick is the true heart of the Banach-Tarski paradox. By using rotations to split up the ball in a very careful way, we can create points out of nothing by undoing the last rotation. And then we can very carefully put them back together, creating two balls out of one!

But let’s leave those details till next time!

<– Previous Post: Non-measurable Sets That Go Bump in the Night
First post in this series: How Long is Infinity?
–> Next Post: Double for Nothing, part 2

  1. Plus that extra ____ word. 

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