## 502 – The Banach-Tarski paradox

1. Non-measurable sets

In these notes I want to present a proof of the Banach-Tarski paradox, a consequence of the axiom of choice that shows us that a naive understanding of the concept of volume can lead to contradictions. A good reference for this topic is the very nice book The Banach-Tarski paradox by Stan Wagon.

The result is one of a family of theorems indicating limitations of any reasonable notion of measure on the real numbers or in Euclidean space, and in this section I include a few examples. We will work with Lebesgue measure. The version of this measure for ${{\mathbb R}^n}$ we denote ${\lambda_n.}$ Actually, we do not need to know much about Lebesgue measure. It suffices that ${\lambda_n}$ has the following properties (and, really, we will only look at ${n=1,2,}$ or ${3}$):

• ${\lambda_n}$ is a measure, so ${\lambda_n}$ is a function with domain a ${\sigma}$-algebra of subsets of ${{\mathbb R}^n}$ and range ${[0,\infty],}$ ${\lambda_n(\emptyset)=0,}$ and ${\lambda_n}$ is ${\sigma}$-additive, i.e., if ${(A_k\mid k<\omega)}$ is a sequence of pairwise disjoint elements of the domain of ${\lambda_n}$ (i.e., measurable sets), then

$\displaystyle \lambda_n\left(\bigcup_k A_k\right)=\sum_k\lambda_n(A_k).$

• If ${{\mathbb I}_n}$ is the unit ${n}$-cube, i.e.,

$\displaystyle {\mathbb I}_n=\{(a_1,\dots,a_n)\in{\mathbb R}^n\mid 0\le a_i\le 1\mbox{\ for all\ }i\},$

then ${\lambda_n({\mathbb I}_n)=1.}$ (Other than ${\lambda_n({\mathbb I}_n)\ne0,}$ we do not really need this fact.)

• ${\lambda_n}$ is non-atomic, in the sense that each singleton is measurable, and ${\lambda_n(\{a\})=0}$ for all ${a\in{\mathbb R}^n.}$
• ${\lambda_n}$ is invariant under the group of isometries of ${{\mathbb R}^n.}$ For example, if ${n=1,}$ for ${A\subseteq{\mathbb R}}$ measurable and ${r\in{\mathbb R},}$ let

$\displaystyle A+r=\{a+r\mid a\in A\}.$

Then ${A+r}$ is measurable, and ${\lambda_1(A)=\lambda_1(A+r).}$ We say that ${\lambda_1}$ is translation invariant. Also, if

$\displaystyle -A=\{-a\mid a\in A\},$

then ${-A}$ is measurable, and ${\lambda_1(A)=\lambda_1(-A).}$ In ${{\mathbb R}^2,}$ ${\lambda_2}$ is not only translation invariant but also invariant under rotations and reflections. In fact, if

$\displaystyle rA=\{ra\mid a\in A\}$

for ${r\in{\mathbb R}}$ and ${A\subseteq {\mathbb R}^n}$ measurable, then ${rA}$ is measurable, and ${\lambda_n(rA)=|r|^n\lambda_n(A),}$ but we won’t be needing this fact.

${\lambda_1}$ is our formalization of the notion of length, similarly, ${\lambda_2}$ and ${\lambda_3}$ formalize the notions of area and volume.

Theorem 1 There is a subset of ${{\mathbb R}}$ that is not measurable.

Proof: We present an argument due to Vitali. On ${{}[0,1],}$ say that ${r\sim s}$ iff ${r-s\in{\mathbb Q}.}$ This is an equivalence relation. Let ${M}$ be a choice set, so ${M\cap A}$ is a singleton for each ${\sim}$-equivalence class ${A.}$ Then

$\displaystyle {}[0,1]=\bigcup_{q\in{\mathbb Q}}(M+q)\cap[0,1].$

This is a countable union of disjoint measurable sets. By translation invariance (and monotonicity, a consequence of ${\sigma}$-additivity), if ${\lambda_1(M)=0,}$ then each of the sets in the union has measure zero as well, which leads to the contradiction ${\lambda_1([0,1])=0.}$

Similarly,

$\displaystyle \bigcup_{q\in{\mathbb Q}\cap[0,1]}(M+q)\subseteq[0,2],$

and therefore, if ${\lambda_1(M)>0,}$ then ${\lambda_1([0,2])=\infty,}$ again a contradiction.

It follows that ${M}$ is not Lebesgue measurable. $\Box$

Remark 1 Translation invariance is essential here. It is consistent with ${{\sf ZFC}}$ that Lebesgue measure can be extended to a measure defined on all subsets of ${{\mathbb R}.}$

The example above used the axiom of choice explicitly, guaranteeing the existence of the set ${M.}$ Another typical use of choice is the existence of a well-ordering of ${{\mathbb R}.}$ A set of measure zero is called null. Otherwise, it is non-null. Note that a non-null set needs not be measurable. The example below uses a bit more of the properties of ${\lambda_1}$ and ${\lambda_2,}$ namely, Fubini’s theorem.

Theorem 2 No well-ordering of a non-null sets of reals is measurable (as a subset of ${{\mathbb R}^2}$).

Proof: We claim that whenever ${S\subseteq{\mathbb R}}$ is non-null and ${W\subseteq{\mathbb R}^2}$ is a well-ordering of ${S,}$ then ${W}$ is non-measurable. We proceed by contradiction. Say that ${\tau}$ is the least ordinal for which there is an enumeration ${(r_\alpha\mid\alpha<\tau)}$ of a non-null set ${S\subseteq{\mathbb R}}$ such that ${W=\{(r_\alpha,r_\beta)\mid\alpha<\beta<\tau\}}$ is measurable.

For ${\alpha<\tau}$ let ${S_\alpha=\{r_\beta\mid\beta<\alpha\}}$ and ${S^\alpha=\{r_\beta\mid\alpha<\beta\}.}$ For ${x\in S}$ let ${r^{-1}x}$ denote the unique ${\alpha<\tau}$ such that ${x=r_\alpha.}$

Note that ${S}$ is measurable: By Fubini’s theorem, for almost all ${x\in S}$ and almost all ${y\in S,}$ both ${S^{r^{-1}x}}$ and ${S_{r^{-1}y}}$ are measurable. Since ${S=S_{\alpha+1}\cup S^\gamma}$ for any ${\alpha\le\gamma<\tau,}$ then ${S}$ must be measurable as well.

It suffices to show that for some ${\gamma<\tau,}$ ${S_\gamma}$ is non-null and measurable. If so, ${W\cap(S_\gamma\times S_\gamma)}$ is a measurable well-ordering of ${S_\gamma}$ is order-type ${\gamma<\tau,}$ contradicting the minimality of ${\tau.}$

Now, for almost all ${y\in S,}$ ${S_{r^{-1}y}}$ is measurable. Thus if no ${S_\gamma}$ is as claimed, then for almost all ${y\in S}$ we have that ${S_{r^{-1}y}}$ is null. Hence,

$\displaystyle 0=\lambda_2(W)=\int_S\lambda_1(S_{r^{-1}y})\,dy=\int_S\lambda_1(S^{r^{-1}x})\,dx.$

But for almost all ${x\in S,}$ we have ${S^{r^{-1}x}=S\setminus(S_{r^{-1}x}\cup\{x\})}$ has positive measure. Contradiction. $\Box$

Remark 2 The argument shows that, even if there is an extension ${\mu_1}$ of Lebesgue measure to all subsets of ${{\mathbb R},}$ the completion of ${\mu_1\times\mu_1}$ is not defined on all of ${{\mathcal P}({\mathbb R}^2).}$

2. Tarski’s circle-squaring problem

In order to understand the statement of the Banach-Tarski result, it is convenient to first explain why it is stated for ${{\mathbb R}^3}$ rather than the plane.

Theorem 3 (Banach, von Neumann) For ${i=1}$ or ${2,}$ there is a finitely additive “measure” ${\sigma}$ extending Lebesgue measure, defined on all of ${{\mathcal P}({\mathbb R}^i),}$ and invariant under isometries. ${\Box}$

I won’t prove this result here. It is a consequence of a more general extension theorem, related to the fact that the group of isometries of the plane is amenable. Its proof requires choice. An obvious consequence of this result is that if a (measurable) figure ${A}$ in the plane has some area ${a,}$ and it is cut out into finitely many pieces that are then rearranged into another measurable figure ${B,}$ then ${\lambda_2(A)=\lambda_2(B).}$

In contrast, the Banach-Tarski paradox allows us to divide a sphere the size of a pea into pieces that, when rearranged, make up a sphere the size of the sun. Clearly, this precludes the existence of an extension of Lebesgue measure in ${{\mathbb R}^3}$ as in the Banach-von Neumann result. On the other hand, a version of the paradox can be obtained for ${{\mathbb R}^2}$. In order to discuss this, I need a few notions.

Definition 4 Two sets in ${{\mathbb R}^n}$ are equidecomposable if one can be partitioned into finitely many disjoint sets (called pieces) that can be rearranged (via isometries) to form a partition of the other.

There is a well-known particular instance of equidecomposability that has been studied in some detail in the context of classical Euclidean geometry.

Definition 5 Two polygons in ${{\mathbb R}^2}$ are equidissectable if, allowing overlap on boundaries, one can be split into finitely many triangles that can be rearranged (via isometries) into the other.

Sometimes one says that the polygons are congruent by dissection.

Theorem 6 (Bolyai-Gerwien) Two polygons are equidissectable iff they have the same area.

Proof: I only present a sketch. One direction is obvious. For the other, first one argues that a polygon is dissectable into triangles. Then, that any triangle is equidissectable with a square. This is done in two stages. First, one easily shows that a triangle is equidissectable with a rectangle. Some care is then needed to see that a rectangle is equidissectable with a square; but this is classical, it only goes a bit beyond constructing the geometric mean of two given numbers. Finally, one shows that two squares are equidissectable with a single square, and then induction completes the proof. The argument for two squares amounts to one of the well-known proofs of the Pythagorean theorem. $\Box$

The version of the Banach-Tarski paradox in ${{\mathbb R}^2}$ is a generalization of the following strengthening of the Bolyai-Gerwien result:

Theorem 7 (Banach-Tarski) Two polygons in ${{\mathbb R}^2}$ are equidecomposable iff they have the same area. ${\Box}$

The proof (that I omit) follows the same outline as the paradox in ${{\mathbb R}^3}$.

Tarski then asked for the possibility of extending Theorem 7 to other regions. In particular, he asked whether the circle and a square of the same area are equidecomposable. This question remained open until 1990, when M. Laczkovich solved it affirmatively, in his paper Equidecomposability and discrepancy; a solution of Tarski’s circle-squaring problem, Journal für die reine und angewandte Mathematik, 404 (1990), 77-117. Among his results, I highlight:

Theorem 8 (Laczkovich)

1. Any two polygons of the same area are equidecomposable using only translations.
2. Let ${J}$ be a piecewise smooth Jordan curve for which there are two positive constants ${a such that the curvature at each point of ${J}$ is between ${a}$ and ${b.}$ Then the domain enclosed by ${J}$ is equidecomposable to a square via translations alone. ${\Box}$

The number of pieces required in the decomposition is rather large. Laczkovich computes that about ${10^{50}}$ pieces are required in the equidecomposition of an arbitrary isosceles right triangle and a square. In contrast, 5 pieces are required for the Banach-Tarski paradox in ${{\mathbb R}^3}$.

The pieces of the decomposition are also not explicitly definable; in particular, Dubins,Hirsch, and Karush showed that if only Jordan domains are used, then the circle and the square are not equidecomposable.

The full statement of the Banach-Tarski paradox in ${{\mathbb R}^3}$ is as follows:

Theorem 9 Any two bounded subsets of ${{\mathbb R}^3}$ with nonempty interior are equidecomposable. ${\Box}$

We will prove a weaker result, namely, that a sphere is equidecomposable with two copies of itself. Recall that ${S^2}$ denotes the unit sphere in ${{\mathbb R}^3.}$ The argument takes three steps. First, we talk about paradoxical group actions and show that ${{\mathbb F}_2,}$ the free group on two generators, acts paradoxically on itself. This is then used to show Hausdorff’s paradox, that there is a countable set ${D}$ such that ${{\mathbb F}_2}$ acts paradoxically on ${S^2\setminus D.}$ We conclude by showing the Banach-Tarski paradox itself, that ${{\mathbb F}_2}$ acts paradoxically on ${S^2.}$

Recall that a group action is a map ${\varphi}$ from a group ${G}$ into the group ${{\rm Bij}(X)}$ of bijections of a set ${X}$ into itself, ${\varphi:G\rightarrow{\rm Bij}(X),}$ such that:

• ${\varphi(e)=id,}$ where ${e}$ is the identity of ${G.}$
• For all ${g,h\in G}$ and ${x\in X,}$ ${\varphi(g\cdot h)(x)=(\varphi(g)\circ\varphi(h))(x).}$

To ease readability, we will follow the usual convention of writing ${g}$ rather than ${\varphi(g),}$ and ${g\cdot x}$ or even ${gx}$ rather than ${\varphi(g)(x)=g(x).}$ Also, given ${g\in G}$ and ${A\subseteq X,}$ write ${g\cdot A}$ or ${gA}$ for ${\{ga\mid a\in A\}.}$

Definition 10 Let the group ${G}$ act on the set ${X.}$ We say that ${G}$ acts paradoxically on ${E\subseteq X}$ iff there are pairwise disjoint subsets of ${E,}$

$\displaystyle A_1,\dots,A_n,B_1,\dots,B_m,$

and corresponding elements of ${G,}$

$\displaystyle g_1,\dots,g_n,h_1,\dots,h_m,$

such that

$\displaystyle E=\bigcup_i g_i\cdot A_i=\bigcup_j h_j\cdot B_j.$

It is easy to check that if ${G}$ acts paradoxically on ${E,}$ we can assume moreover that the pieces ${A_i,B_j}$ satisfy that ${E=\bigcup_iA_i\cup\bigcup_jB_j,}$ see Fact 1. This explains the name: Note that then we have that ${E}$ is “equidecomposable” with two copies of itself, since ${\bigcup_iA_i}$ and ${\bigcup_j B_j}$ are both equidecomposable with ${E,}$ where we are abusing notation, using subsets of ${X}$ rather than of ${{\mathbb R}^n}$ and elements of ${G}$ rather than isometries.

Given a group ${G}$ acting on a set ${X,}$ write ${A\sim_G B}$ to denote that ${A,B\subseteq X,}$ that ${A}$ can be partitioned into finitely many pieces,

$\displaystyle A=\bigcup_{i

and that there are (not necessarily distinct) elements ${g_0,\dots,g_{n-1}}$ of ${G}$ such that, letting ${B_i=gA_i}$ for ${i then the ${B_i}$ are pairwise disjoint and partition ${B.}$

It is easy to verify that ${\sim_G}$ is an equivalence relation.

Fact 1 Given a group ${G}$ acting on a set ${X,}$ we have that ${G}$ acts paradoxically on ${E\subseteq X}$ iff there are disjoint sets ${A,B\subseteq E}$ such that ${E=A\cup B}$ and ${A\sim_G E\sim_G B.}$

Proof: Verify that the argument for the Schröder-Bernstein theorem gives that if ${A\sim_G B\subseteq C\subseteq A,}$ then ${A\sim_G C.}$ $\Box$

A trivial example of a paradoxical action is given by the group ${G={\rm Bij}(X)}$ acting on ${X}$ via the action ${fx=f(x)}$ for ${X}$ an infinite set. This action is paradoxical, since any infinite ${X}$ is in bijection with two copies of itself. A more interesting example, essential to the argument, is as follows:

Theorem 11 ${{\mathbb F}_2}$ acts paradoxically on itself by left multiplication.

Proof: Let ${\sigma,\tau}$ be generators of ${{\mathbb F}_2,}$ and consider the following 4 subsets:

• ${A_1=\tau{\mathbb F}_2,}$
• ${A_2=\tau^{-1}{\mathbb F}_2,}$
• ${A_3=\sigma{\mathbb F}_2\cup\{\sigma^{-n}\mid n\in\omega\},}$ and
• ${A_4=\sigma^{-1}{\mathbb F}_2\setminus\{\sigma^{-n}\mid 0

Note that ${A_1,\dots,A_4}$ are disjoint and partition ${{\mathbb F}_2.}$ Moreover,

$\displaystyle \sigma A_2=A_2\cup A_3\cup A_4,$

and

$\displaystyle \tau A_4=A_1\cup A_2\cup A_4.$

This gives the result, as ${{\mathbb F}_2=A_1\cup\sigma A_2=A_3\cup\tau A_4.}$ $\Box$

The key tool we use to show that ${{\mathbb F}_2}$ acts paradoxically on the sets in ${{\mathbb R}^3}$ that interest us is the following result. Say that an action of a group ${G}$ on a set ${X}$ is without nontrivial fixed points iff the only element of ${G}$ that fixes a point of ${X}$ is the identity ${e.}$

Theorem 12 If ${{\mathbb F}_2}$ acts on ${X}$ without nontrivial fixed points, then ${{\mathbb F}_2}$ acts paradoxically on ${X.}$

Proof: Fix an action of ${{\mathbb F}_2}$ on ${X}$ without nontrivial fixed points. Using the axiom of choice, let ${M}$ be a choice set picking an element of each orbit ${{\mathbb F}_2a.}$ Note that for any ${y\in X}$ there is a (unique) ${m\in M}$ such that ${y\in{\mathbb F}_2m,}$ and therefore there is a ${g\in{\mathbb F}_2}$ such that ${y\in gM.}$ In fact, there is a unique such ${g.}$ Otherwise, there are ${g_1,g_2\in G}$ and ${m_1,m_2\in M}$ such that

$\displaystyle g_1m_1=y=g_2m_2,$

and therefore ${m_1=g_1^{-1}g_2m_2\in{\mathbb F}_2m_2.}$ Since ${M}$ is a choice set, then ${m_1=m_2.}$ But then ${g_1=g_2,}$ since the action of ${{\mathbb F}_2}$ is without nontrivial fixed points.

Let ${A\cup B}$ be a paradoxical partition of ${{\mathbb F}_2,}$ so ${A\sim_{{\mathbb F}_2}{\mathbb F}_2\sim_{{\mathbb F}_2} B.}$ Let ${A^*=\bigcup_{g\in A}gM}$ and ${B^*=\bigcup_{g\in B}gM.}$ By our observation on the paragraph above, ${A^*\cap B^*=\emptyset}$ and ${A\cup B=X.}$ Since ${A\sim_{{\mathbb F}_2}{\mathbb F}_2\sim_{{\mathbb F}_2} B,}$ we immediately get ${A^*\sim_{{\mathbb F}_2}X\sim_{{\mathbb F}_2} B^*.}$ $\Box$

In this section, I prove the following result:

Theorem 13 There is a countable set ${D}$ such that the natural action of the group of isometries of ${{\mathbb R}^3}$ on ${S^2\setminus D}$ is paradoxical.

In light of the results from last section, it suffices to find two isometries ${\rho,\phi}$ such that the group they generate is free, and acts on ${S^2\setminus D}$ without nontrivial fixed points, for some appropriate countable set ${D.}$

There are many ways of doing this. Following Wagon’s suggestion, let ${\phi}$ be the counterclockwise rotation around the ${z}$-axis by an angle of ${\cos^{-1}(1/3).}$ The matrix associated to ${\phi}$ is easily found to be:

$\displaystyle \phi=\begin{pmatrix}\vspace{1mm}\displaystyle\frac13&\displaystyle -\frac{2\sqrt2}3&0\\ \vspace{1mm}\displaystyle\frac{2\sqrt2}3&\displaystyle\frac13&0\\ 0&0&1\end{pmatrix}.$

Note that ${\phi^{-1}=\phi^T}$ is the transpose of ${\phi.}$

Now let ${\rho}$ be the counterclockwise rotation around the ${x}$-axis by an angle of ${\cos^{-1}(1/3).}$ As with ${\phi,}$ it is easy to see that the matrix associated to ${\rho}$ is:

$\displaystyle \rho=\begin{pmatrix}\vspace{1mm}1&0&0\\\vspace{1mm}0&\displaystyle\frac13&\displaystyle -\frac{2\sqrt2}3\\ 0&\displaystyle\frac{2\sqrt2}3&\displaystyle\frac13\end{pmatrix}.$

As with ${\phi,}$ ${\rho^{-1}=\rho^T.}$

Lemma 14 The group generated by ${\rho,\phi}$ is free.

Proof: I sketch the argument. Let ${w}$ be a reduced word in the alphabet ${\{\rho,\rho^{-1},\phi,\phi^{-1}.}$ We need to show that if ${w}$ is nontrivial (as a word), then it is not the identity (as an isometry). Since ${w=id}$ iff ${\phi w\phi^{-1}=id,}$ we may assume that ${w}$ ends in either ${\phi}$ or ${\phi^{-1}.}$ We say that ${w}$ is valid.

I claim that

$\displaystyle w\begin{pmatrix}1\\\end{pmatrix}=\begin{pmatrix}a\\b\sqrt2\\c\end{pmatrix}/3^k$

for some integers ${a,b,c,k}$ with ${k\ge0}$ and ${b\not\equiv0\pmod3.}$ In particular, this shows that ${w}$ does not map ${\begin{pmatrix}1&0&0\end{pmatrix}^T}$ into itself, so ${w\ne id.}$

The claim is proved by induction on the length ${lh(w)}$ of ${w.}$ The result is clear if ${lh(w)=1,}$ since ${w=\phi^{\pm1}}$ and ${w\begin{pmatrix}1&0&0\end{pmatrix}^T}$ is just the first column of ${\phi^{\pm1},}$ that has the required form with ${a=1,}$ ${b=\pm2,}$ ${c=0,}$ and ${k=1.}$

Suppose now ${lh(w)>1}$ and we know the result for all shorter lengths. We have that ${w=\phi^{\pm1}w'}$ or ${w=\rho^{\pm1}w'}$ for some valid word ${w'.}$ We have that for some appropriate integers ${a',b',c',k',}$

$\displaystyle w'\begin{pmatrix}1\\\end{pmatrix}=\begin{pmatrix}a'\\b'\sqrt2\\c'\end{pmatrix}/3^{k'}.$

Then

$\displaystyle w\begin{pmatrix}1\\\end{pmatrix}=\begin{pmatrix}a\\b\sqrt2\\c\end{pmatrix}/3^k,$

where ${a=a'\mp4b',}$ ${b=b'\pm2a',}$ ${c=3c',}$ and ${k=k'+1}$ in the first case, or ${a=3a',}$ ${b=b'\mp2c',}$ ${c=c'\pm4b',}$ and ${k=k'+1}$ in the second case.

In both cases, we have that ${a,b,c,k}$ are integers, and that ${k>0.}$ All that remains is to argue that ${b}$ is not divisible by 3. For this, we consider the word ${v}$ such that ${w=\phi^{\pm1}\rho^{\pm1}v,}$ or ${\phi^{\pm1}\phi^{\pm1}v,}$ or ${\phi^{\pm2}v,}$ or ${\rho^{\pm2}v.}$ Note that ${v}$ is valid in the first and fourth cases. In the second and third cases, ${v}$ is valid unless it is the empty word. In any case, note that for some integers ${a'',b'',c'',k''}$ with ${k''\ge0,}$ we have

$\displaystyle v\begin{pmatrix}1\\\end{pmatrix}=\begin{pmatrix}a''\\b''\sqrt2\\c''\end{pmatrix}/3^{k''},$

and ${b''\not\equiv0\pmod3}$ unless ${v}$ is the empty word, in which case ${a''=1,}$ ${b''=c''=k''=0.}$

From the formulas above, we see respectively that ${b=b'\mp2c'}$ with ${c'\equiv0\pmod3,}$ or ${b=b'\pm2a'}$ with ${a'\equiv0\pmod3,}$ or ${b=b'\pm2a'=b'\pm2(a''\mp4b'')=b'+b''\pm2a''-9b''=2b'-9b'',}$ or ${b=b'\mp2c'=b'\mp2(c''\pm4b'')=b'+b''\mp2c''-9b''=2b'-9b''.}$ In all cases, that ${b\not\equiv0\pmod3}$ now follows from the induction hypothesis, and we are done. $\Box$

The Hausdorff paradox follows immediately: It suffices to show that there is a countable set ${D}$ such that the natural action of the group ${G=\langle\phi,\rho\rangle}$ on ${S^2\setminus D}$ is without nontrivial fixed points. But for any matrix ${w\in G}$ other than the identity, either ${w}$ fixes no vector in ${S^2,}$ or else it fixes exactly two, of the form ${v}$ and ${-v.}$ This is because either ${1}$ is not an eigenvalue of ${w,}$ or else it is an eigenvalue of multiplicity 1: Note that if ${1}$ has multiplicity 3 as an eigenvalue, ${w=id,}$ and it cannot have multiplicity 2, because ${\det(w)=1}$ is the product of the eigenvalues of ${w.}$

Now take as ${D}$ the set of all vectors in ${S^2}$ that are fixed by some ${w\in G\setminus\{id\}.}$ Since ${G}$ is countable, and each ${w\in G\setminus\{id\}}$ contributes at most two vectors to ${D,}$ ${D}$ is countable.

We now conclude the proof of the following result:

Theorem 15 (Banach-Tarski) The group of isometries of ${{\mathbb R}^3}$ acts paradoxically on ${S^2.}$

As mentioned earlier, there is a stronger form of the paradox where any two bounded sets with nonempty interior and equidecomposable. Several additional strengthenings are possible. For example, only 5 pieces are needed to witness the equidecomposition (and 4 do not suffice). Let ${G}$ be the group of isometries of ${{\mathbb R}^3.}$ Very recently, Trevor Wilson, then an undergraduate at Caltech, showed that the equidecomposition can be carried out continuously, i.e., say that ${A}$ and ${B}$ are bounded and with nonempty interior. Then there is a partition

$\displaystyle A=\bigcup_{i<5}A_i$

and continuous functions

$\displaystyle \gamma^i:[0,1]\rightarrow G$

for ${i<5,}$ such that ${\gamma^i(0)=id,}$

$\displaystyle \gamma^i(t)A_i\cap\gamma^j(t)A_j=\emptyset$

whenever ${i and ${t\in[0,1],}$ and

$\displaystyle B=\bigcup_{i<5}\gamma^i(1)A_i.$

For the strong version of the Banach-Tarski paradox and the fact that precisely 5 pieces are needed, see Wagon’s book. For the continuous version, see Trevor Wilson, A continuous movement version of the Banach-Tarski paradox: A solution to de Groot’s problem, The Journal of Symbolic Logic, 70 (3), 2005, 946-952.

Proof: In view of the Hausdorff paradox, it suffices to show that, with ${G}$ the group of isometries of ${{\mathbb R}^3,}$ we have that ${S^2\sim_G S^2\setminus D}$ whenever ${D}$ is countable. To see this, it suffices to check that there is a rotation ${\rho}$ such that ${D,\rho D,\rho^2D,\dots}$ are pairwise disjoint. because if that’s the case, letting

$\displaystyle \bar D=\bigcup_{n\in\omega}\rho^nD,$

we have

$\displaystyle S^2=\bar D\cup(S^2\setminus \bar D)\sim_G\rho\bar D\cup(S^2\setminus\bar D)=S^2\setminus D.$

To find ${\rho,}$ note that there is a line ${\ell}$ that goes through the origin and misses ${D.}$ Let ${A}$ be the set of angles ${\theta}$ such that for some ${n>0}$ there is a point ${P\in D}$ such that ${\rho_{n\theta} P\in D,}$ where ${\rho_\alpha}$ is the rotation around ${\ell}$ of ${\alpha}$ radians.

Clearly, ${A}$ is countable, so there is an angle ${\theta}$ not in ${A,}$ and we can take ${\rho=\rho_\theta.}$ In effect, we have that ${D\cap\rho_{n\theta}D=\emptyset}$ for all ${n>0.}$ But then ${\rho_{m\theta}D\cap\rho_{n\theta}D=\emptyset}$ for all ${n as well, and we are done. $\Box$

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