Banach-Tarski Paradox on the middle third Cantor set

Question

In analysis and topology, the middle third Cantor set $C$ is often a very interesting topic of discussion. My question is that is it possible to have some sort of measure preserving bijection $f: C\rightarrow S^2$, where $S^2$ is the standard unit sphere, such that this produces a Banach-Tarski decomposition for $C$? Wagon's book $\textit{The Banach-Tarski Paradox}$ actually shows that it is quite possible to have a Banach-Tarski decomposition on intervals. If there really is some sort of measure preserving bijection as I asked in my question, what exactly would it be? This might just be an open problem that has yet to be answered...

Answer 1

As Henning says, whether or not you have a Banach-Tarski paradox depends on what symmetries of the Cantor set you allow. My understanding is that you want to allow all measure preserving continuous bijections. With this understanding, the Cantor has a Banach-Tarski decomposition.

Note that we can identify the Cantor set with the $2$-adic integers, $\mathbb{Z}_2$, and the $2$-adic measure is the standard measure on the Cantor set.

Let $F$ denote the free group on $2$ generators. Let $\Gamma(2)$ be the group of integer matrices $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ with $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) \equiv \left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right) \mod 2$ and let $P \Gamma(2)$ be the quotient of $\Gamma(2)$ by the cetnral order $2$ subgroup generated by $\left( \begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$. As is well known, $F$ is isomorphic to $P\Gamma(2)$.

Let $P\Gamma(2)$ act on $\mathbb{Z}_2$ by the Möbius transformation formula $z \mapsto \frac{az+b}{cz+d}$. Note that, since $c$ is even and $d$ is odd, the denominator is a unit of $\mathbb{Z}_2$, so the image of the map lands in $\mathbb{Z}_2$. I claim that this map is continuous and measure preserving. Specifically, I note that this map descends to a well defined map from $\mathbb{Z}/2^k \mathbb{Z}$ to itself. Thus, if we know $z$ with accuracy $2^{-k}$, we also know $\frac{az+b}{cz+d}$ with this accuracy (establishing continuity). And, for any open set of the form $U:=\{ u : u \equiv v \mod 2^k \}$, the image of $U$ is also of this form and thus has the same measure (establishing preservation of measure).

So we have a measure preserving action of $F$ on the Cantor set where each element of $F$ fixes at most two points. Now follow the standard proof of Banach-Tarski.

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ADDED: Sachin asks below for elaboration. I am going to assume that you are familiar with the $2$-adic numbers, which are roughly base $2$ numbers with expansions that run infinitely to the left. If you're not familiar, follow the Wikipedia link.

I believe you are describing an element of the Cantor set by a binary sequence $(a_0, a_1, a_2, \ldots, ) \in \{ 0,1 \}^{\mathbb{N}}$. One can identify this with the point $2 \sum a_i 3^{-1}$ on the real line, in the standard "middle thirds" description of the Cantor set. A basis of neighborhoods for the Cantor set is given by the sets $U(b_0 b_1 \cdots b_{k-1}) = \{ (a_0, a_1, \ldots) : a_i=b_i \ \mathrm{for}\ i<k \}$. As I understand it, you are using the measure where $U(b_0 b_1 \cdots b_{k-1})$ has measure $2^{-k}$.

I am identifying $(a_0, a_1, \ldots)$ with the $2$-adic number $\sum a_k 2^k$. This sum is of course very nonconvergent in the standard sense, but it does converge (basically by definition) $2$-adically. This identification shows that the Cantor set is homeomorphic with the $2$-adic integers. The measure I am using on the $2$-adics (which is the standard one) is the one where $U(b_0 b_1 \cdots b_{k-1})$ has measure $2^{-k}$.

What I get from the $2$-adics which is new is the ring structure which allows me to write down the maps $x \mapsto \frac{ax+b}{cx+d}$. These would be something very unmotivated looking without the apparatus of infinite base $2$ expansions.