PlanetMath: proof of Banach-Tarski paradox

We deal with some technicalities first, mainly concering the properties of equi-decomposability. We can then prove the paradox in a clear and unencumbered line of argument: we show that, given two unit balls $U$ and $U'$ with arbitrary origin, $U$ and $U\cup U'$ are equi-decomposable, regardless whether $U$ and $U'$ are disjoint or not. The original proof can be found in [BT].

Technicalities

Proof. The only non-obvious part is transitivity. So let $A$ , $B$ , $C$ be sets such that $A$ and $B$ as well as $B$ and $C$ are equi-decomposable. Then there exist disjoint decompositions $A_1,\ldots A_n$ of $A$ and $B_1,\ldots B_n$ of $B$ such that $A_k$ and $B_k$ are congruent for $1\leq k\leq n$ . Furthermore, there exist disjoint decompositions $B_1',\ldots,B_m'$ of $B$ and $C_1,\ldots C_m$ of $C$ such that $B_l'$ and $C_l$ are congruent for $1\leq l\leq m$ . Define \begin{equation*} B_{k,l}:=B_k\cap B_l'\text{ for }1\leq k\leq n, 1\leq l\leq m. \end{equation*}Now if $A_k$ and $B_k$ are congruent via some isometry $\theta\colon A_k\to B_k$ , we obtain a disjoint decomoposition of $A_k$ by setting $A_{k,l}:=\theta^{-1}(B_{k,l})$ . Likewise, if $B_l'$ and $C_l$ are congruent via some ismometry $\theta'\colon B_l'\to C_l$ , we obtain a disjoint decomposition of $C_l$ by setting $C_{k,l}:=\theta'(B_{k,l})$ . 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 $C_{k,l}$ are congruent for $1\leq k\leq n$ and $1\leq l\leq m$ . Therefore, $A$ and $C$ are equi-decomposable. $\qedsymbol$

Proof. Let $A_1,\ldots,A_n$ and $C_1,\ldots C_n$ disjoint decompositions of $A$ and $C$ , respectively, such that $A_k$ and $C_k$ are congruent via an isometry $\theta_k$ for $1\leq k\leq n$ . Like in the proof of theorem 3, let $\theta\colon A\to C$ 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 $\mc{C}(b)$ be the intersection of all sets $X\subseteq B$ satisfying -125-JG ,

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 $\mc{C}(b_1)$ and $\mc{C}(b_2)$ are not disjoint. Then there is a $b\in\mc{C}(b_1)\cap\mc{C}(b_2)$ such that $b=\theta^r(b_1)=\theta^s(b_2)$ for suitable integers $s$ and $r$ . Given $b'\in\mc{C}(b_1)$ , we have $b'=\theta^t(b_1)=\theta^{t+s-r}(b_2)$ for a suitable integer $t$ , that is $b'\in\mc{C}(b_2)$ , so that $\mc{C}(b_1)\subseteq\mc{C}(b_2)$ . The reverse inclusion follows likewise, and we see that for arbitrary $b_1,b_2\in B$ either $\mc{C}(b_1)=\mc{C}(b_2)$ or $\mc{C}(b_1)$ and $\mc{C}(b_2)$ are disjoint. Now set \begin{equation*} D:=\{b\in B\mid\mc{C}(b)\subseteq A\}, \end{equation*}then obviously $D\subseteq A$ . If for $b\in B$ , $\mc{C}(b)$ consists of the sequence of elements $\ldots,\theta^{-1}(b),b,\theta(b),\ldots$ which is infinite in both directions, then $b\in D$ . If the sequence is infinite in only one direction, 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 $\mc{C}(\theta^{-1}(c))$ consists of a sequence $\ldots,\theta^{-2}(c),\theta^{-1}(c),\ldots$ which is infinite in only one direction and the final element does not lie in $A$ . Now $\theta^{-1}(c)\in\mc{C}(\theta^{-1}(c))$ , but since $\theta^{-1}(c)$ does lie in $A$ , it is not the final element. Therefore the subsequent element $c$ lies in $\mc{C}(\theta^{-1}(c))$ , in particular $c\in B$ and $\mc{C}(c)=\mc{C}(\theta^{-1}(c))\not\subseteq A$ , 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. $\qedsymbol$

The proof

Since $E'$ is only countable but there are uncountably many rotations of $S$ , there exists a rotation $\theta$ such that $\theta(E')\subsetneq B'\cup C'\cup D'$ , so $F:=\theta(E)$ is a proper subset of $B\cup C\cup D$ . Since $C$ and $B\cup C\cup D$ are equi-decomposable, there exists by theorem 3 a proper subset $G\subsetneq C$ such that $G$ and $F$ (and thus $E$ ) are equi-decomposable. Finally, let $p\in C\setminus G$ an arbitrary point and set \begin{equation*} A_2:=D\cup G\cup\{p\}, \end{equation*}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 $A_1\cup A_2\neq U$ !).

Now $U$ and $U'$ are congruent, so $U'$ and $A_2$ are equi-decomposable by theorem 1. If $U$ and $U'$ are disjoint, set $H:=A_2$ . Otherwise we may use theorem 3 and choose $H\subsetneq A_2$ such that $H$ and $U'\setminus U$ are equi-decomposable. By theorem 2, $A_1\cup H$ and $U\cup (U'\setminus U)=U\cup U'$ are equi-decomposable. Now we have \begin{equation*} A_1\cup H\subseteq U\subseteq U\cup U', \end{equation*}so by theorem

, $U$ and $U\cup U'$ are equi-decomposable.

Technicalities

The proof

Bibliography