As I understand, the Banach-Tarski paradox says a ball in 3-space may be decomposed into finitely many pieces and reassembled into two balls each of the same size as the original. Despite being called a paradox it is of course a theorem.

Looking at the proof, it seems to rely heavily on the Axiom of Choice. However since the consequences of not accepting the Axiom of Choice seem even more weird, I am wondering whether the more experienced Mathematicians here find the implication of Banach-Tarski a perfectly acceptable Theorem, or whether it shows that ZF with Choice is actually ultimately pathological ? ( i.e. does it just seem a weird from a perspective that is not mathematically mature enough ?)

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    $\begingroup$ "Paradox" means "contrary to expectation, common sense, or received opinion" in its main meaning, though it is often mistaken to mean "logically contradictory". Nothing wrong with theorems being paradoxical. $\endgroup$ – Arturo Magidin Jan 30 '12 at 22:33
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    $\begingroup$ related question math.stackexchange.com/questions/103743/… $\endgroup$ – Emilio Ferrucci Jan 30 '12 at 22:33
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    $\begingroup$ Let me just note that, long before you get to Banach-Tarski, there are already results that the man-in-the-street might call paradoxical, e.g., there are as many even integers as integers, and as many integers as rationals, but fewer rationals than reals. That's math. $\endgroup$ – Gerry Myerson Jan 30 '12 at 22:51
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    $\begingroup$ The "sum" $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ can be rearranged to give an result we want, like $\pi^e$, or $-22/7$. Disturbing if our intuition is based on finite sums. Same with Banach-Tarski, if we think decomposition is like cutting up an apple. $\endgroup$ – André Nicolas Jan 30 '12 at 23:06
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    $\begingroup$ Note: you can't use banach-tarski to tease physicists because it would require subatomic splits of your apple/ball/other real world object $\endgroup$ – wim Jan 31 '12 at 6:03

The reason the Banach-Tarski paradox seems paradoxical is because of the following naive argument: surely the volume of a ball is the same as the sum of the volume of any possible decomposition of that ball into finitely many pieces, which is in turn not the same as the volume of two balls. More precisely, surely the total measure ought to remain invariant.

And the reason the Banach-Tarski paradox is a theorem is that the intermediate pieces it uses are very weird: in particular, they do not have volume. (More precisely, they are non-measurable.) So the naive argument breaks down completely, but naive arguments break down all the time in mathematics.

A more focused version of your question might be: how weird or pathological should I regard a non-measurable set as being? Well, of course they are weird, but they aren't weird to the point that they're a good reason (in my opinion) to reject the axiom of choice. One can construct non-measurable sets using the weaker ultrafilter lemma, which I happen to be extremely fond of, so I embrace them out of necessity.

Edit: You might also be interested in hearing Terence Tao's thoughts; he's written about Banach-Tarski several times and has enlightening things to say.

  • $\begingroup$ This particular "naive argument" does not break down, if you allow random picking, in other words, if you can choose real numbers at random in the interval [0,1]. The concept of picking real numbers at random is equivalent to saying all sets are measurable, and losing the ability to speak about random real numbers is not acceptable. This is a case where the mathematicians are stupid and brainwashed, not the critics. $\endgroup$ – Ron Maimon Apr 17 '14 at 15:31

It seems that the vast majority of mathematicians consider Banach-Tarski an acceptable price to pay for having the Axiom of Choice available. Its practical consequences are at least twofold, one general and one more specific:

  1. Be wary of your intuition about sets that are so weird that you need the axiom of choice to construct them. The general consensus seems to be that the Banach-Tarski components behave non-intuitively, but not paradoxically.

  2. In the presence of the Axiom of Choice one has to accept that there is no well-behaved measure on all subsets of $\mathbb R^n$ (where "well-behaved" means something like being invariant under isometries, at least finitely additive, and assigning a nonzero finite measure to an ball of finite radius). This entails a plethora of "assume such-and-such is measurable" premises in many theorems, but again this is generally treated as an acceptable price to pay for having choice.

Arguably, it even leads to a more general and useful theory that the main example of a measure is not defined on the full power set. Otherwise, the standard development of everything would just assume that all subsets are measurable, and then one would have trouble applying the theory to special measures that one wanted, for one reason or another, to be partial.

  • $\begingroup$ Borel sets are somewhat dependent on the axiom of choice, should we be wary of those? :-) $\endgroup$ – Asaf Karagila Jan 31 '12 at 0:10
  • $\begingroup$ Of course the Banach Tarski pieces behave non-paradoxically, because the axiom of choice is consistent! But this is not the same as saying that they are not paradoxical, because they contradict the concept of "randomly chosen real number", and require you to formulate the concept of "random real" differently, as a "random variable", for which you cannot do certain operations. For example, you cannot speak about the probability that a random chosen real lands in a non-measurable set, the concept is incoherent. So this means random reals are forbidden in your universe. $\endgroup$ – Ron Maimon Apr 17 '14 at 15:34
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    $\begingroup$ @AsafKaragila: Borel sets are not dependent on the kind of axiom of choice one is talking about here. They have nothing to do with uncountable continuum choice. $\endgroup$ – Ron Maimon Apr 17 '14 at 15:34
  • $\begingroup$ @Ron: It is consistent without the axiom of choice that every set of real numbers is Borel. So when talking about sets which are not Borel sets, we have to make appeal to the axiom of choice. Roughly countable choice suffices. $\endgroup$ – Asaf Karagila Apr 17 '14 at 18:40

Qiaochu's answer is great, and reflects exactly what a mathematician should feel about the Banach-Tarski theorem.

I would like to add another opinion, as someone who works mostly in a choiceless context I can assure you that mathematics has many surprises in store for you once you give up the choice needed for Banach-Tarski.

You might end up with the bizarre universe in which there are no free ultrafilters on $\mathbb N$; the real numbers might be a countable union of countable sets; or it might be possible to cut the real numbers into more non-empty parts than elements.

There is always a "paradox", which is really just a counter-intuitive theorem, hiding in the dark corners of the universe. It tells you, in the philosophical level, just one thing:

Our intuition is completely developed by history and the axioms we are used to work with. Once you are completely used to the axiom of choice there is no surprise in the Banach-Tarski theorem, much like there is no surprise in Gödel's incompleteness theorems or in Cantor's theorem about the uncountability of the real numbers.

These are all theorems that shook the foundations of mathematics and caused people to shake their heads in disbelief, but eventually these theorems were accepted and nowadays people don't fuss about Banach-Tarski because it's one of the first thing presented in a course about measure theory: You can't measure everything in a translation-invariant way and with countable completeness.

The main issue is that in any strong enough theory there will be unexpected results, which is why the Banach-Tarski theorem - while very surprising - should not deter you from the axiom of choice, which makes infinitary things easier to deal with.

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    $\begingroup$ I would consider a universe without ultrafilters less bizarre than one with ultrafilters. $\endgroup$ – Ben Crowell Jan 31 '12 at 1:22
  • $\begingroup$ @Ben: I'd hardly think this is the case for most mathematicians. $\endgroup$ – Asaf Karagila Jan 31 '12 at 7:59

The result points to the fact that it makes no sense to try to apply a geometric notion of measuring volume to all subsets of $\mathbb{R}^3$. But why should that be the case in the first place?

We measure volumes of sets by approximating them by sets that have a natural notion of volume, like disjoint unions of cubes. So we use mathematical methods to extend a notion of volume defined for simple objects to more complex objects. But there is no reason why the notion should apply to arbitrary sets of points.

And, intuitively, if there are sets of points somewhere that can not be approximated by objects with a natural notion of volume, the axiom of choice helps us to find them. Given that "set of points" and "geometric object" are in general quite different, it shouldn't be surprising that we can do that.


As something of an aside: it's important to remember that Banach--Tarski is not merely some artifact of choice. Assuming choice, the BT paradox holds for some group actions (e.g. $SO(3)$ acting on $\mathbb R^3$), but not for others (e.g. $SO(2)$ acting on $\mathbb R^2$). Indeed, there is an intrinsic group-theoretic property that governs whether or not the BT paradox can hold, namely amenability.

The study of group actions and their interaction with measure theory is an entire branch of mathematics; see e.g. Topics in orbit equivalence, by Kechris and Miller. There is much more to this topic than just the use of choice.

  • $\begingroup$ Yes, the results on amenability proper are independent of any choice-related issues (actually, I just checked, and the Wikipedia article does a more than decent role highlighting this). An interesting side comment: The continuum hypothesis is another useful tool when studying amenability, due to results of Mokobodzki (and another one of those abstract transfering tools, that allows us to eliminate the assumption of $\mathsf{CH}$ when the statements being proved are projective). $\endgroup$ – Andrés E. Caicedo Jan 15 '14 at 4:03

I am wondering whether the more experienced Mathematicians here find the implication of Banach-Tarski a perfectly acceptable Theorem

The Banach-Tarski paradox helps to illustrate the fact that a proof of existence within ZFC doesn't necessarily imply concrete existence in an intuitive sense. Stating the theorem in terms of real material objects like oranges or gold balls exaggerates this discrepancy between intuitive actual existence and existence within ZFC even further.

In a certain sense, already the existence of a Hammel basis of $\mathbb R$ as vector space over $\mathbb Q$ raises the question what existence within ZFC actually signifies. Banach-Tarski adds an intuition boost derived from the explicit constructive paradoxical decomposition of the free group on two generators to arrive at a theorem where even the layman can appreciate that this sort of existence is paradoxical.

However, the purpose of this answer is to state my opinion that it is a red herring to blame the axiom of choice for all undesirable features of ZFC. (Blaming the axiom of power set axiom instead wouldn't help either.) The paradox is derived from ZFC as a whole, not from the axiom of choice alone. I say this because the axiom of choice can be added (in an appropriate form) to most foundational systems without disturbing things. It is quite useful, and it basically expresses a property that we expect in a set universe allowing "arbitrary" sets. (It's a bit similar to adding a unit element to a semigroup. It can have some undesired side effects, but overall it just simplifies things.)

A feature of ZFC that I like much less than the axiom of choice is the impredicativity of the axiom schemes of comprehension and replacement. Hence I added to my initial answer that it might be a better idea to blame them instead of the axiom of choice. However, after reading a bit on Banach-Tarski now, I have to admit that these are probably completely unrelated to Banach-Tarski.

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    $\begingroup$ Equiconsistency (or even consistency strength) is not the issue here. The axiom of choice, on the other hand, is the reason why the theorem holds. This is not a red herring: If you remove choice from the axioms of set theory, even if you add dependent choice, you can no longer prove the Banach-Tarski theorem. You may want to explain better what you mean, since as stated this is misleading. $\endgroup$ – Andrés E. Caicedo Jan 15 '14 at 1:11
  • $\begingroup$ @AndresCaicedo After reading the 70+ page proof of the consistency of Peano arithmetic by Gerhard Gentzen, I realized that he really had proved the consistency by finitary means. What he actually needed where not the axioms of ZFC or any other foundational theory of this sort, but the principle that predicative definitions and constructions are OK. John Horton Conway made a remark pointing in this very direction when he explained how his surreal numbers can be constructed inside ZFC, and why this construction just obscures the real principles why the surreal numbers are a valid construction. $\endgroup$ – Thomas Klimpel Jan 15 '14 at 8:28
  • $\begingroup$ @AndresCaicedo I may extend the answer when I have time, but not with the content of the previous comment. The reason why Choice is a red herring is that there is nothing wrong with Choice, but people talk about Banach-Tarski when they want to express their uneasiness with ZFC. There may be reasons to be uneasy with ZFC, but Choice should not be one of these reasons. It expresses an important property that we want the universe of sets to have. Perhaps it is slightly too strong, but people should discuss the real contentious points of ZFC before focusing on such minor details. $\endgroup$ – Thomas Klimpel Jan 15 '14 at 8:35
  • $\begingroup$ I don't understand how this is even an answer to the question. Sure, it's a point for a discussion on this topic, but not an answer to the question. The fact that choice holding or failing is equiconsistent has nothing to do with this. The question is about implications rather than equiconsistency. I agree that if this was a discussion over lunch, coffee or drinks, this would be a good point to raise. But unfortunately, MSE is not a place for discussions of this sort. $\endgroup$ – Asaf Karagila Jan 15 '14 at 11:04
  • $\begingroup$ @AsafKaragila The funny thing is that "being equiconsistent" is a red herring too. The important point of my answer is that there is nothing wrong with the Axiom of Choice. This is different from being equiconsistent, i.e. the axiom "every subset of the reals is measurable" would also be equiconsistent, but there would nevertheless be something wrong with it if used in the foundations of set theory. But I think I understand now why my answer is easy to misinterpret, and I will try to improve it later. $\endgroup$ – Thomas Klimpel Jan 15 '14 at 12:41

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