proof of Banach-Tarski paradox

We deal with some technicalities first, mainly concerning theproperties of equi-decomposability. We can then prove the paradox in aclear and unencumbered line of argument: we show that, given two unitballs $U$ and $U^{\prime }$ with arbitrary origin, $U$ and $U\cup U^{\prime }$ areequi-decomposable, regardless whether $U$ and $U^{\prime }$ are disjoint ornot. The original proof can be found in [BT].

Technicalities

Theorem 1.

Equi-decomposability gives rise to an equivalence relation on thesubsets of Euclidean space.

Proof.

The only non-obvious part is transitivity. So let $A$, $B$, $C$ besets such that $A$ and $B$ as well as $B$ and $C$ areequi-decomposable. Then there exist disjoint decompositions$A_1,\mathrm{\dots }{A}_n$ of $A$ and $B_1,\mathrm{\dots }{B}_n$ of $B$ such that $A_k$and $B_k$ are congruent for $1\le k\le n$. Furthermore, there existdisjoint decompositions $B_1^{\prime },\mathrm{\dots },B_m^{\prime }$ of $B$ and $C_1,\mathrm{\dots }{C}_m$of $C$ such that $B_l^{\prime }$ and $C_l$ are congruent for $1\le l\le m$. Define

$$B_{k,l}:=B_k\cap B_l^{\prime }\text{ for }1\le k\le n,1\le l\le m.$$

Now if $A_k$ and $B_k$ are congruent via some isometry $\theta :A_k\to B_k$, we obtain a disjoint decomposition of $A_k$ by setting$A_{k,l}:={\theta }^{-1}(B_{k,l})$. Likewise, if $B_l^{\prime }$ and $C_l$ arecongruent via some isometry ${\theta }^{\prime }:B_l^{\prime }\to C_l$, we obtain adisjoint decomposition of $C_l$ by setting$C_{k,l}:={\theta }^{\prime }(B_{k,l})$. Clearly, the $A_{k,l}$ and $C_{k,l}$define a disjoint decomposition of $A$ and $C$, respectively, into$nm$ parts. By transitivity of congruence, $A_{k,l}$ and $C_{k,l}$ arecongruent for $1\le k\le n$ and $1\le l\le m$. Therefore, $A$ and$C$ are equi-decomposable.∎

Theorem 2.

Given disjoint sets $A_{\mathrm{1}}\mathrm{,}\mathrm{,}\mathrm{\dots }{A}_n$, $B_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathit{}{B}_n$ such that$A_k$ and $B_k$ are equi-decomposable for $\mathrm{1}\mathrm{\le }k\mathrm{\le }n$, theirunions $A\mathrm{=}\mathrm{\bigcup }_{k\mathrm{=}\mathrm{1}}^n{A}_k$ and $B\mathrm{=}\mathrm{\bigcup }_{k\mathrm{=}\mathrm{1}}^n{B}_k$ areequi-decomposable as well.

Proof.

By definition, there exists for every $k$, $1\le k\le n$ an integer$l_k$ such that there are disjoint decompositions

$$A_k=\bigcup _{i=1}^{l_k}{A}_{k,i},B_k=\bigcup _{i=1}^{l_k}{B}_{k,i}$$

such that $A_{k,i}$ and $B_{k,i}$ are congruent for $1\le i\le l_k$. Rewriting $A$ and $B$ in the form

$$A=\bigcup _{k=1}^n\bigcup _{i=1}^{l_k}{A}_{k,i},B=\bigcup _{k=1}^n\bigcup _{i=1}^{l_k}{B}_{k,i}$$

gives the result.∎

Theorem 3.

Let $A$, $B$, $C$ be sets such that $A$ and $B$ are equi-decomposableand $C\mathrm{⊊}A$, then there exists $D\mathrm{⊊}B$ such that $C$and $D$ are equi-decomposable.

Proof.

Let $A_1,\mathrm{\dots },A_n$ and $B_1,\mathrm{\dots }{B}_n$ be disjoint decompositions of$A$ and $B$, respectively, such that $A_k$ and $B_k$ are congruent viaan isometry ${\theta }_k:A_k\to B_k$ for all $1\le k\le n$. Let$\theta :A\to B$ a map such that $\theta (x)={\theta }_k(x)$ if $x\in A_k$. Since the $A_k$ are disjoint, $\theta$ is well-definedeverywhere. Furthermore, $\theta$ is obviously bijective. Now set$C_k:=C\cap A_k$ and define $D_k:={\theta }_k(C_k)$, so that $C_k$ and$D_k$ are congruent for $1\le k\le n$, so the disjoint union$D:=D_1\cup \mathrm{\cdots }\cup D_n$ and $C$ are equi-decomposable. Byconstruction, $\theta (C)=D$. Since $C$ is a proper subset of $A$ and$\theta$ is bijective, $D$ is a proper subset of $B$.∎

Theorem 4.

Let $A$, $B$ and $C$ be sets such that $A$ and $C$ areequi-decomposable and $A\mathrm{\subseteq }B\mathrm{\subseteq }C$. Then $B$ and $C$ areequi-decomposable.

Proof.

Let $A_1,\mathrm{\dots },A_n$ and $C_1,\mathrm{\dots }{C}_n$ disjoint decompositions of$A$ and $C$, respectively, such that $A_k$ and $C_k$ are congruent viaan isometry ${\theta }_k$ for $1\le k\le n$. Like in the proof oftheorem 3, let $\theta :A\to C$ be the well-defined,bijective map such that $\theta (x)={\theta }_k(x)$ if $x\in A_k$. Now,for every $b\in B$, let $𝒞(b)$ be the intersection of all sets$X\subseteq B$ satisfying

• $b\in X$,

• for all $x\in X$, the preimage ${\theta }^{-1}(x)$ lies in $X$,

• for all $x\in X\cap A$, the image $\theta (x)$ lies in $X$.

Let $b_1,b_2\in B$ such that $𝒞(b_1)$ and $𝒞(b_2)$ are notdisjoint. Then there is a $b\in 𝒞(b_1)\cap 𝒞(b_2)$ such that$b={\theta }^r(b_1)={\theta }^s(b_2)$ for suitable integers $s$ and$r$. Given $b^{\prime }\in 𝒞(b_1)$, we have$b^{\prime }={\theta }^t(b_1)={\theta }^{t+s-r}(b_2)$ for a suitable integer $t$,that is $b^{\prime }\in 𝒞(b_2)$, so that$𝒞(b_1)\subseteq 𝒞(b_2)$. The reverse inclusion followslikewise, and we see that for arbitrary $b_1,b_2\in B$ either$𝒞(b_1)=𝒞(b_2)$ or $𝒞(b_1)$ and $𝒞(b_2)$ aredisjoint. Now set

$$D:=\{b\in B\mid 𝒞(b)\subseteq A\},$$

then obviously $D\subseteq A$. If for $b\in B$, $𝒞(b)$ consistsof the sequence of elements$\mathrm{\dots },{\theta }^{-1}(b),b,\theta (b),\mathrm{\dots }$ which is infinite in bothdirections, then $b\in D$. If the sequence is infinite in only onedirection, but the final element lies in $A$, then $b\in D$ as well.Let $E:=\theta (D)$ and $F:=B\setminus D$, then clearly $E\cup F\subseteq C$.

Now let $c\in C$. If $c\notin E$, then ${\theta }^{-1}(c)\notin D$, so$𝒞({\theta }^{-1}(c))$ consists of a sequence$\mathrm{\dots },{\theta }^{-2}(c),{\theta }^{-1}(c),\mathrm{\dots }$ which is infinite inonly one direction and the final element does not lie in $A$. Now${\theta }^{-1}(c)\in 𝒞({\theta }^{-1}(c))$, but since ${\theta }^{-1}(c)$does lie in $A$, it is not the final element. Therefore thesubsequent element $c$ lies in $𝒞({\theta }^{-1}(c))$, in particular$c\in B$ and $𝒞(c)=𝒞({\theta }^{-1}(c))⊈A$, so$c\in B\setminus D=F$. It follows that $C=E\cup F$, and furthermore $E$and $F$ are disjoint.

It now follows similarly as in the preceding proofs that $D$ and$E=\theta (D)$ are equi-decomposable. By theorem 2, $B=D\cup F$ and $C=E\cup F$ are equi-decomposable.∎

The proof

We may assume that the unit ball $U$ is centered at the origin, thatis $U={𝔹}_3$, while the other unit ball $U^{\prime }$ has an arbitraryorigin.Let $S$ be the surface of $U$, that is, the unit sphere. By theHausdorff paradox, there exists a disjoint decomposition

$$S=B^{\prime }\cup C^{\prime }\cup D^{\prime }\cup E^{\prime }$$

such that $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ and $C^{\prime }\cup D^{\prime }$ are congruent, and $E^{\prime }$ iscountable. For $r>0$ and $A\subseteq {ℝ}^3$, let $rA$ be the set ofall vectors of $A$ multiplied by $r$. Set

These sets give a disjoint decomposition of the unit ball with theorigin deleted, and obviously $B$, $C$, $D$ and $C\cup D$are congruent (but $E$ is of course not countable). Set

$$A_1:=B\cup E\cup \{0\}$$

where $0$ is the origin. $B$ and $C\cup D$ are triviallyequi-decomposable. Since $C$ and $B$ as well as $D$ and $C$ arecongruent, $C\cup D$ and $B\cup C$ are equi-decomposable. Finally, $B$and $C\cup D$ as well as $C$ and $B$ are congruent, so $B\cup C$ and$B\cup C\cup D$ are equi-decomposable. In total, $B$ and $B\cup C\cup D$ are equi-decomposable by theorem 1, so $A_1$ and $U$ areequi-decomposable by theorem 2. Similarly, we conclude that$C$, $D$ and $B\cup C\cup D$ are equi-decomposable.

Since $E^{\prime }$ is only countable but there are uncountably many rotationsof $S$, there exists a rotation $\theta$ such that$\theta (E^{\prime })⊊B^{\prime }\cup C^{\prime }\cup D^{\prime }$, so $F:=\theta (E)$ is a propersubset of $B\cup C\cup D$. Since $C$ and $B\cup C\cup D$ areequi-decomposable, there exists by theorem 3 a proper subset$G⊊C$ such that $G$ and $F$ (and thus $E$) areequi-decomposable. Finally, let $p\in C\setminus G$ an arbitrary pointand set

$$A_2:=D\cup G\cup \{p\},$$

a disjoint union by construction. Since $D$ and $B\cup C\cup D$, $G$and $E$ as well as $\{0\}$ and $\{p\}$ are equi-decomposable, $A_2$and $U$ are equi-decomposable by theorem 2. $A_1$ and $A_2$are disjoint (but $A_1\cup A_2\ne U$!).

Now $U$ and $U^{\prime }$ are congruent, so $U^{\prime }$ and $A_2$ areequi-decomposable by theorem 1. If $U$ and $U^{\prime }$ aredisjoint, set $H:=A_2$. Otherwise we may use theorem 3 andchoose $H⊊A_2$ such that $H$ and $U^{\prime }\setminus U$ areequi-decomposable. By theorem 2, $A_1\cup H$ and $U\cup (U^{\prime }\setminus U)=U\cup U^{\prime }$ are equi-decomposable. Now we have

$$A_1\cup H\subseteq U\subseteq U\cup U^{\prime },$$

so by theorem 4, $U$ and $U\cup U^{\prime }$ are equi-decomposable.