Regular probability comparisons imply the Banach–Tarski Paradox

Abstract

Consider the regularity thesis that each possible event has non-zero probability. Hájek challenges this in two ways: (a) there can be nonmeasurable events that have no probability at all and (b) on a large enough sample space, some probabilities will have to be zero. But arguments for the existence of nonmeasurable events depend on the axiom of choice (AC). We shall show that the existence of anything like regular probabilities is by itself enough to imply a weak version of AC sufficient to prove the Banach–Tarski Paradox on the decomposition of a ball into two equally sized balls, and hence to show the existence of nonmeasurable events. This provides a powerful argument against unrestricted orthodox Bayesianism that works even without AC. A corollary of our formal result is that if every partial order extends to a total preorder while maintaining strict comparisons, then the Banach–Tarski Paradox holds. This yields an argument that incommensurability cannot be avoided in ambitious versions of decision theory.

Appendix: Sketch of proof of the Theorem

Pawlikowski (1991, Lemma and Note 1) essentially showed how to prove in ZF:

Lemma 1

Suppose that for every pairwise disjoint collection $\fancyscript{B}$ of countable subsets of $\mathbb R^3$ there is a function $\phi$ that assigns to each member $B\in \fancyscript{B}$ a collection $\phi (B)$ of subsets of $B$ with the properties:

• (i) If $B\in \fancyscript{B}$, $A\in \phi (B)$ and $A\subseteq A'\subseteq B$, then $A'\in \phi (B)$

• (ii) If $B_1,B_2,B_3,B_4$ are disjoint non-empty subsets of a $B\in \fancyscript{B}$, then (a) at most one of the sets $B_i$ is a member of $\phi (B)$ and (b) at least one of the sets $B-B_i$ is a member of $\phi (B)$.

Then the Banach–Tarski Paradox holds.

A few things need to be noted. First, Pawlikowski in his paper works with a general set $X$ rather than a subset of $\mathbb R^3$, but it is clear from the methods of Wagon (1994, Chapters 1–3) that only the case where $X$ is a sphere in $\mathbb R^3$ matters for BT. Second, the collection corresponding to our $\fancyscript{B}$ that Pawlikowski works with is the set of orbits under a free group $F$ on two generators. Since the group $F$ is obviously countable, these orbits are all countable sets.¹³ Third, Pawlikowski formulates his proof and remarks in terms of a direct sum of boolean algebras. His work can all be easily reformulated in the setting of our Lemma 1, or else one can apply Pawlikowski Note 1 after letting $\mathbf B$ be the boolean direct sum of the boolean algebras $\mathcal {P}B$ as $B$ ranges over the members of $\fancyscript{B}$, and letting his set $\mathbf D$ be the set of elements of $\mathbf B$ of the form $\bigvee _{i=1}^n \langle D_i \rangle$ for $D_i \in \bigcup _{B \in \fancyscript{B}} \phi (B)$.

Our Theorem 1 follows immediately from Lemma 1 and:

Lemma 2

Suppose that there is a total preorder $\lesssim$ on the countable members of $\mathcal {P}\mathbb R^3$ such that: (a) if $A\subseteq B$, then $A\lesssim B$, and (b) if $A$ and $B$ are disjoint sets with $B$ non-empty and $A\lesssim B$, then $A < A\cup B$. Then the function $\phi$ defined by $\phi (B) = \{ A \subseteq B : B-A < A \}$ satisfies conditions (i) and (ii) of Lemma 1.

For there is a bijection of $\mathbb R^3$ with $[0,1]$ (this is provable in ZF) and so the existence of the preorder in Lemma 2 follows from the existence of the preorder assumed in the Theorem, and hence BT follows from Lemma 1.

And since the methods of Wagon (1994) that Pawlikowski invokes prove BT by first constructing a paradoxical decomposition of the sphere, we also get Theorem 2.

It remains to prove Lemma 2. Write $X\sim Y$ whenever $X\lesssim Y$ and $Y\lesssim X$.

Proof of Lemma 2

Suppose $A\subseteq A' \subseteq B$ and $A\in \phi (B)$. Then $A\lesssim A'$. Moreover, we have $B-A' \subseteq B-A$ and so $B-A' \lesssim B-A$. Thus, $B-A' \lesssim B-A < A \lesssim A'$ and so $A'\in \phi (B)$. We thus have (i) from Lemma 1.

Observe that for each $A\subseteq B$, at most one of $A$ and $B-A$ is in $\phi (B)$.

Now suppose that $B_1,B_2,B_3,B_4$ are disjoint non-empty subsets of $B$. Suppose first that $B_i \in \phi (B)$ and $j\ne i$. Then $B_i \subseteq B-B_j$ by disjointness and so $B-B_j \in \phi (B)$, and hence $B_j \notin \phi (B)$. This yields (ii)(a) in Lemma 1.

It remains to show (ii)(b) in Lemma 1. Suppose $i$ is such that . By totality, we have $B-B_i < B_i$ and so $B_i\in \phi (B)$. We know this can happen for at most one of the $i$. Thus, there are at least three distinct indices $i$ such that $B_{i} \lesssim B-B_{i}$. Let $I$ be a set of three such indices. If any one of the $B_i \lesssim B-B_i$ inequalities is strict, we have (ii)(b).

So suppose that none of the inequalities are strict. Thus, $B_i \sim B-B_i$ for $i\in I$. Fix distinct $i$ and $j$ in $I$. Then $B_i \sim B-B_i$ and $B_j \sim B-B_j$. By disjointness $B_i \subseteq B-B_j$, so $B_i \lesssim B-B_j \lesssim B_j$. By the same token $B_j \lesssim B_i$, and so $B_j\sim B_i$. But $i$ and $j$ were arbitrary distinct indices in $I$. Thus, if now we write $I=\{ i,j,k \}$, we have $B_i\sim B_j \sim B_k$. But by property (b) of our preorder, we have $B_i < B_i \cup B_j$, since $B_j$ is non-empty. Since $B_i \cup B_j \subseteq B - B_k$ by disjointness, we have $B_k \sim B_i < B_i \cup B_j \lesssim B-B_k$ and it follows that $B_k < B-B_k$, contrary to the assumption that none of the three inequalities were strict.¹⁴