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A. Makelov

What does the Banach-Tarski theorem have to do with the axiom of choice?

`What’s an anagram of Banach-Tarski?’

`Banach-Tarski Banach-Tarski.’

Banach and Tarski

The Banach-Tarski theorem says the following:

Theorem 1 [Banach-Tarski theorem] Given a solid ball in 3-dimensional Euclidean space {\mathbb{R}^3}, we can partition it into a finite number of pieces, so that we can rearrange them to get two solid balls congruent to the first ball.

This clearly goes against people’s intuitions about volume. Often, it is noted that the Banach-Tarski theorem is a consequence of the axiom of choice; this is inherent, since it is in fact equivalent to the axiom of choice. In one of its most strikingly obvious formulations the latter says:

[Axiom of choice] The product of a collection of nonempty sets is nonempty.

That is, for any collection of sets {\{S_i\}_{i\in I}}, where the index set {I} can be an arbitrary set, one can find an indexed family {S = (s_i)_{i\in I}} such that {s_i\in S_i} for all {i\in I}.

So, we have these two things which are equivalent, and one is completely, obviously true, whereas the other is completely, obviously false. OK.

What do these two things have to do with each other? Here’s the beginning of an answer. Our intuitions about concepts like volume and area are formalized in mathematics through what is called a measure. Here’s a definition that summarizes the intuitively desirable properties of such a measure:

Definition 2 A measure on {\mathbb{R}^n} is a non-negative, translation-invariant, countably additive function {\mu:\mathcal{P}(\mathbb{R}^n)\to\mathbb{R}} that assigns to each parallelepiped its volume. That is to say,

  1. {\forall S\subset\mathbb{R}^n, \mu(S)\geq0}.
  2. {\forall S\subset\mathbb{R}^n, v\in\mathbb{R}^n, \mu(S) = \mu(S+v)}
  3. {\forall S_1,S_2,\ldots \subset \mathbb{R}^n} disjoint,

    \displaystyle \begin{array}{rcl} \mu\left(\displaystyle\bigcup_{i=1}^\infty S_i\right)=\displaystyle\sum_{i=1}^\infty \mu(S_i) \end{array}

  4. {\forall a_i,b_i\in\mathbb{R}, \mu( \prod_{i=1}^{n} [a_i,b_i])= \prod_{i=1}^{n} (b_i-a_i)}.

We would then hope to be able to build up complicated sets from many parallelepipeds. Or something like that in any case.

Why do we allow only countably many sets in (2)? Well, the (only!) alternative is to allow at least uncountably many sets, which would then imply that the measure of the entire {\mathbb{R}^n} is equal to the sum of the measures of the points; but points should have zero volume! So our condition (3) is actually pretty liberal.

However, it turns out that in the above definition we wanted too much, i.e. it is inconsistent:

Theorem 3 There exists a subset {A\subset\mathbb{R}} for which {\mu(A)} doesn’t exist.

Proof: Consider the equivalence relation on {[0,1]} given by

\displaystyle \alpha\sim\beta \iff \alpha-\beta\in\mathbb{Q}

By the axiom of choice, we can pick an indexed collection of representatives for each equivalence class; call the set underlying this collection of representatives {A}. To really convince yourself that this is indeed a set (because that’s tricky business), you should play around with the axioms of Zermelo-Fraenkel set theory and the definition of an indexed collection. We shall show that {\mu(A)} doesn’t exist.

Assume the opposite, and consider, for each {r\in\mathbb{Q}\cap[-1,1]}, the sets

\displaystyle A_r = \{ a + r \ \big| \ a\in A\}

Observe that the {A_r} are disjoint, for otherwise we would have {a_1\pm r_1=a_2\pm r_2} for {a_1\neq a_2\in A} and {r_1,r_2\in\mathbb{Q}\cap[0,1]}, and hence {a_1-a_2\in\mathbb{Q}}, contradicting the fact that {a_1,a_2} are in different equivalence classes.

On the other hand, {[0,1]\subset\displaystyle\bigcup_{r} A_r\subset[-1,2]}. Thus, by the properties of measure, we have

\displaystyle 1 \leq \displaystyle\sum_{r} \mu(A)\leq 3

The first inequality gives us {\mu(A)>0}, whereas the second gives us {\mu(A)=0}, thus the contradiction with the assumption that {\mu(A)} exists. \Box

This is where our intuition about volume breaks: it’s impossible to formalize it so that it works for all sets. Now, it’s kind of clear that at least one of the pieces in the decomposition of the ball in the Banach-Tarski paradox has to be similar to the set {A} above, and that is where `volume conservation’ fails.

What people do to define the measure in a consistent way is to be very careful about the sets for which the measure applies. This leads to the ideas of {\sigma}-algebras and Lebesgue measure, which are the established formalisms of measure theory. There still exist sets that are not Lebesgue-measurable (the one constructed above is an example), but this is no longer an inconsistency of the theory; it’s a `weirdness’ of math.

5 responses to “What does the Banach-Tarski theorem have to do with the axiom of choice?

  1. robertdodier June 27, 2014 at 6:27 pm

    Hi, thanks for the interesting article. How do you compose your articles? I have tried a couple of different ways to compose mathematically-oriented blog articles and the results are always kind of ugly. I’d be grateful for any info about the set-up you use.


  2. Victor July 31, 2014 at 9:49 pm



  3. tarball2 December 13, 2014 at 10:42 pm

    I don’t understand this step: “The first inequality gives us mu(A)>0, whereas the second gives mu(A) = 0. I’m probably just missing something simple. Could you explain this?


  4. Billy February 25, 2016 at 5:09 pm

    Why are you saying that Banach-Tarsky is equivalent to AC? I can’t find a proof of BT => AC, could you point me to one?


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