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Is it possible to reverse the operation of the Banach-Tarski paradox? That is, I have two three-dimensional balls, and is it possible to combine them into one ball that is identical to one of the balls (using the axiom of choice)?

Thank you very much.

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Could you explain the asymmetry you seem to be seeing? Under the only meaning of "combine" that I can think of, two congruent balls being combinable into one ball that's congruent to them is synonymous with one ball being decomposable into two balls congruent to the one ball. –  joriki Mar 20 '12 at 10:00
    
@joriki I was asking whether the axiom of choice can be used to do the process in reverse. So, is it the same? I know that I can use the axiom of choice make one ball decompose into two. I am asking whether I can use the axiom of choice to combine two into one. that's all. nothing special into the meaning of "combine". –  user25148 Mar 20 '12 at 10:19
    
joriki's point is that you should convince yourself that if you can decompose $A$ into pieces and rearrange the pieces to get $B$ then you can do this process in the reverse direction and transform $B$ into $A$ in this way. As Dejan points out in the answer below, a much better results holds. –  t.b. Mar 20 '12 at 10:35
    
@user25148: If you want to decompose two balls into one from scratch, you basically need to do the same construction as if you were decomposing one ball into two. So the axiom of choice is needed in this case. But if you already have a decomposition of one ball into two, you need no further axiom of choice, since as I describe below, you can just invert the isometries the axiom of choice has given you before. –  Dejan Govc Mar 20 '12 at 10:36
    
I'm not sure that the tag [axiom-of-choice] fits here... –  Asaf Karagila Mar 20 '12 at 11:31

2 Answers 2

Yes. In fact, the strong form of Banach-Tarski paradox is as follows:

Theorem. Let $n\geq3$ and let $A,B\subseteq\mathbb{R}^n$ be arbitrary bounded sets with non-empty interior. Then $A$ and $B$ are equidecomposable.

This means that we can cut any such $A$ into finitely many pieces and using isometries (distance-preserving maps; i.e. linear maps & translations) put them back together to form $B$.

Note, that equidecomposability is easily seen to be an equivalence relation, from which it already follows that: if we can cut one ball into finitely many pieces and create two, we can also reverse the process.

To see that equidecomposability is a symmetric relation, just note that isometries of any metric space (including $\mathbb R^n$) form a group. So if you have decomposed $A$ into sets $A_1, A_2, ..., A_n$ and used isometries $g_1, g_2, ..., g_n$ to get sets $B_1, B_2,...,B_n$ which together form $B$ (we usually write $g_iA_i=B_i$ to denote the isometry $g_i$ is acting on a set $A_i$, i.e. $g_iA_i$ is the set we get from $A_i$ when we use the isometry $g_i$ on it), you can just use the inverses of these isometries to arrive from $B$ back to $A$. (So in the notation from above: if $g_iA_i = B_i$ for $i=1,2,...,n$, then we also have $A_i=g_i^{-1}B_i$. Put simply: if I have moved a set and rotated it a bit, I can always rotate and move it back, to get the original set back.)

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The short answer is, of course.

The Banach-Tarski paradox tells you how to take a ball and cut it in such way that you can move the pieces around and create two balls. The process is completely symmetric.

What does the Banach-Tarski paradox say? It says that given a ball $B$ we can write it as $B=A_1\cup\ldots\cup A_n$ which are all disjoint, and using very nice maps $f_1,\ldots,f_n$ (rotation and moving then pieces without stretching them) we get that $f_1(A_1)\cup\ldots\cup f_n(A_n)$ is actually two balls! These $f_i$ are so nice that we can reverse them, that is rotate back and shift around the other direction - again without stretching.

Given two disjoint balls we can consider a very canonical map $g$ the two balls of the Banach-Tarski recomposition two our new ones. Now we have that for $i=1,\ldots,n$ we have that $B_i=g(f_i(A_i))$ is the part which essentially decomposed the unit ball. Reverse the process that $f_i$ did to each $B_i$ and the result will recompose a single ball.

The trivial and obvious example is when your two balls are the ones you get from the Banach-Tarski paradox itself, in which case $g$ is the identity.

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