Recently, in a set theory class, I learned the proof of the Banach-Tarski paradox. Contrary to what I expected, the construction of the decomposition is quite clean. I am going to try and be as concise as possible in this post, so excuse me for being a bit hand-wavy in a way that can be easily formalized.
Definition 1. For subset \( A, B \subseteq \mathbb{R}^3 \), we write \( A \cong B \) if \( A \) can decomposed into finitely many pieces and rearranged to make \( B \). Formally, \( A \cong B \) if there exist partitions \( \{A _ 1, \ldots, A _ n\} \) of \( A \) and \( \{B _ 1, \ldots, B _ n\} \) and rigid motions \( g _ 1, \ldots, g _ n \) of \( \mathbb{R}^3 \) such that \( g _ i(A _ i) = B _ i \).
Theorem 2 (Banach-Tarski paradox). Denote by \( B^3 \) the closed unit ball in \( \mathbb{R}^3 \). Then \( B^3 \cong B^3 \amalg B^3 \). (Technically, \( B^3 \amalg B^3 \) is not a subset of \( \mathbb{R}^3 \), so we may write \( B^3 \cong B^3 \cup (B^3 + (3,0,0)) \). But we don’t care too much about these issues.)
The following proposition will be incredibly useful in simplifying the proof of the Banach-Tarski paradox. With the proposition, we don’t have to explicitly name and write down all the elements of the partition.
Proposition 3. The relation \( \cong \) is an equivalence relation.
Proof. Reflexivity and symmetry are obvious. To show \( A \cong C \) from \( A \cong B \) and \( B \cong C \), you can partition \( B \) in one way and then partition in the other way so that they can be rearranged into both \( A \) and \( C \). ▨
1. Throwing away countable sets#
We are first going to prove \( S^2 \cong S^2 \amalg S^2 \). But in doing this, we may run into technical problems like two sets having countable intersections. To take care of these problems, we first prove that countable sets really don’t matter.
Proposition 4. If \( X \subseteq S^2 \) is a countable set, then \( S^2 \cong S^2 \setminus X \).
Proof. The idea is that if we have \( M = \{ e^{i n \alpha} : n \in \mathbb{Z} _ {\ge 0} \} \subseteq S^1 \) with \( \alpha / \pi \notin \mathbb{Q} \), then rotating \( M \) by \( \alpha \) gives \( M \setminus \{1\} \). So we can easily add or remove a single point. Take a unit vector \( v \in S^2 \) such that \( v, -v \notin X \). This is possible because \( X \) is countable. The subgroup of \( \mathrm{SO}(3) \) that fixes \( v \) is isomorphic to \( \mathrm{SO}(2) \cong \mathbb{R}/\mathbb{Z} \). There exists a \( g \in \mathrm{SO}(2) \subseteq \mathrm{SO}(3) \) such that \( g^n X \cap X = \emptyset \) for all \( n \ge 1 \), again because of the countability of \( X \). Let \[ \displaystyle Y = \bigcup _ {n = 0}^\infty g^n X. \] Then \( g Y = Y \setminus X \). Now we can split \( S^2 \) into \( S^2 \setminus Y \) and \( Y \), rotate \( Y \) by \( g \), and get \( (S^2 \setminus Y) \cup (Y \setminus X) = S^2 \setminus X \). Therefore \( S^2 \cong S^2 \setminus X \). ▨
2. Banach-Tarski for \( S^2 \)#
Let us denote by \( F _ 2 \) the free group generated by \( a \) and \( b \). Denote by \( S _ a, S _ b, S _ {a^{-1}}, S _ {b^{-1}} \subseteq F _ 2 \) the set of words that start with \( a, b, a^{-1}, b^{-1} \) respectively in the reduced representation. Then there are partitions \[ \displaystyle F _ 2 = \{e\} \cup S _ a \cup S _ b \cup S _ {a^{-1}} \cup S _ {b^{-1}} = S _ a \cup a S^{a^{-1}} = S _ b \cup b S _ {b^{-1}}. \] This suggests that \( F _ 2 \setminus \{e\} \) can be decomposed and "rearranged" to form two copies of \( F _ 2 \). So in some sense, we have \( F _ 2 \setminus \{e\} \cong F _ 2 \amalg F _ 2 \). Next, we can cover up the \( e \) by taking \( \{a^n : n \ge 0\} \) and multiplying it by \( a \). This shows that \( F _ 2 \cong F _ 2 \setminus \{e\} \), and using Proposition 3, we see that \( F _ 2 \cong F _ 2 \amalg F _ 2 \). We are going to use this decomposition to obtain \( S^2 \cong S^2 \amalg S^2 \).
Lemma 5. There is an embedding \( F _ 2 \hookrightarrow \mathrm{SO}(3) \).
Proof. Well, this is the most non-trivial part in the whole proof. There are a lot of ways you can see this, and this post by David Speyer and this Mathoverflow topic has a very interesting discussions. For a more elementary proof, you can take a look at here. Because I don’t really see any better way, I am going to skip this part. ▨
Now we have a copy of \( F _ 2 \) in \( \mathrm{SO}(3) \). Define \[ \displaystyle D = \{ x \in S^2 : g x = x \text{ for some } 1 \neq g \in F _ 2 \subseteq \mathrm{SO}(3) \}. \] Then \( D \) is countable because \( F _ 2 \) is countable and each rotation has exactly two fixed points. Thus by Proposition 4, \( S^2 \cong S^2 \setminus D \). On the other hand, the action of \( F _ 2 \subseteq \mathrm{SO}(3) \) on \( S^2 \) is now free. If we pick a set of representatives \( P \subseteq S^2 \setminus D \) of the orbits of \( S^2 \) under the \( F _ 2 \) action, (here we are finally using the Axiom of Choice) then \[ \displaystyle S^2 \setminus D = \coprod _ {g \in F _ 2} g P. \] Applying the rearrangement \( F _ 2 \cong F _ 2 \amalg F _ 2 \) on \( P \) immediately gives \( S^2 \setminus D \cong (S^2 \setminus D) \amalg (S^2 \setminus D) \). Therefore \[ \displaystyle S^2 \cong S^2 \setminus D \cong (S^2 \setminus D) \amalg (S^2 \setminus D) \cong S^2 \amalg S^2. \]
3. Filling in the interior#
The final step is to obtain \( B^3 \cong B^3 \amalg B^3 \) from this. Because all actions we have used are rotations fixing the origin, we can replace each point \( p \in S^2 \) with the segment \( \{ t p : 0 < t \le 1 \} \). This gives \[ \displaystyle B^3 \setminus \{0\} \cong (B^3 \setminus \{0\}) \amalg (B^3 \setminus \{0\}). \] Now it suffices to show that \( B^3 \setminus \{0\} \cong B^3 \). Take a small copy of \( S^1 \) that passes through \( 0 \) and is contained in \( B^3 \). We can then use \( S^1 \cong S^1 \setminus \{1\} \) to get \( B^3 \cong B^3 \setminus \{0\} \).