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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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.
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.)