Proof that there is no Banach-Tarski paradox in $\Bbb R^2$ using finitely additive invariant set functions?

Question

I am wondering if anyone is familiar with the above topic? I have found a proof that it is possible to define a finitely additive invariant set function in $\mathbb{R}^2$ on the circle in Lax's book "Functional Analysis". He follows the proof up by saying that this proves that there is no banach-tarski paradox in the plane but I don't see why. Is it obvious? If it isn't can anyone tell me where I can find a proof of this using the existence of such functions? Cheers

Answer 1

What Lax shows is that there is a finitely additive, rotationally invariant set function $m \colon P(S^1) \to [0,1]$ on the circle such that $m(S^1) = 1$:

Theorem 4. One can define a nonnegative finitely additive set function $m(P)$, for all subsets $P$ of the circle, that is invariant under rotation.

This implies in particular that it is impossible to decompose $S^1$ paradoxically into a disjoint union of finitely many pieces $A_1,\dots,A_n$ in such a way that $S^1$ can be written as disjoint union of rotated versions of $A_1,\dots,A_k$ as well as $A_{k+1},\dots,A_n$, i.e., $r_1 A_1 \cup \cdots \cup r_k A_k = S^1$ and $r_{k+1} A_{k+1} \cup \cdots \cup r_{n} A_{k+1} = S^1$ where $r_1,\dots,r_n$ are some rotations.

Indeed, we would have $$ 1 = m(S^1) = m(A_1 \cup \cdots \cup A_n) = m(A_1) + \dots +m(A_n) $$ as well as $$ \begin{align*} 1 & = m(S^1) = m(r_1 A_1 \cup \dots \cup r_k A_k) = m(A_1) + \dots + m(A_k) \cr 1 & = m(S^1) = m(r_{k+1} A_{k+1} \cup \dots \cup r_{n}A_n) = m(A_{k+1}) + \dots + m(A_{n}) \end{align*} $$ by finite additivity and invariance of $m$ under rotations. In particular, $$1 = m(A_1) + \dots +m(A_n) = [m(A_1) + \dots + m(A_k)] + [m(A_{k+1}) + \dots + m(A_{n})] = 2$$ which is absurd.

If you want to show that there is no Banach-Tarski paradox in the plane, you would need a finitely additive set-function $P(\mathbb{R}^2) \to [0,1]$ invariant under isometries and argue as above. Banach showed that such a set function does exist (and the proof is slightly harder because the group of affine isometries of $\mathbb{R}^2$ is not commutative; the key-word here is amenability). Lax does not establish that fact, but neither does he claim to disprove Banach-Tarski in the plane, he only talks about the circle in $\mathbb{R}^2$ and mentions that Hausdorff disproved the existence of a finitely additive rotationally invariant set function on the $2$-sphere:

NOTE. Rotations of the circle commute with each other, and so the operators $\mathbf{A}_\rho$ commute; this was needed in invoking theorem 7 of chapter 3. Rotations of the three-dimensional sphere do not commute, and neither do the corresponding operators $\mathbf{A}_\rho$. Therefore the above proof cannot be used to extended theorem 4 to three dimensions. In fact Hausdorff has shown that the three-dimensional analogue of theorem 4 is false; there is no rotational invariant, finitely additive set function on the $2$-sphere. The proof is based on a finite decomposition of the $2$-sphere, sometimes called the Banach-Tarski paradox.

Note that there is a bit of confusion about the dimensions: Lax talks about the unit sphere in the three-dimensional space and calls that thing both the three-dimensional sphere and the $2$-sphere.

All this and much more is very readably explained in Stan Wagon's book The Banach-Tarski Paradox.