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.

• 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
• @wythagoras: I highly doubt that the BT paradox needs a separate tag. – Asaf Karagila May 23 '15 at 12:35
• @wythagoras: It's not just a question of volume. It's a question of classification. All questions about the BT paradox contain the keywords "Banach-Tarski" making them easily searchable with the crummy internal search engine of the site. I suggest that you bring this up on meta, where an actual discussion on the merits of this tag can take place. (Also, if you search for "Banach Tarski" is:question you will see that the number is closer to 64 questions, not 200.) – Asaf Karagila May 23 '15 at 12:37

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.)

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.