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Was reading through this question and the answer given by @triple_sec lists some mindboggling results that are implied by the axiom of choice.

Specifically:
[Geometry] Banach–Tarski paradox. (The axiom of choice makes it possible to cut an object into a finite number of pieces in such a weird way that you can reassemble two copies of the same object of the same size!)

[Measure theory] Existence of sets that are not Lebesgue measurable.

How do we make sense of the Banach-Tarski problem? The reason I pointed out the existence of non-measurable sets made me think perhaps the paradox from Banach-Tarski theorem arises because we are "incorrectly/inconsistently" [I am not entirely sure how to formalize this notion] measuring these pieces of the ball.

We can't take a ball of chocolate and cut it up and re-assemble it into two balls of chocolate, both as large as the ball of chocolate was to start with, can we? Of course not!

Question: Is AC necessarely the culprit? Is the paradox invariant w.r.t the definition of measure (area/volume?) of a set?

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    $\begingroup$ What are you asking exactly ? $\endgroup$ – user171326 Jul 20 '17 at 13:39
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    $\begingroup$ AC is the culprit because we need AC to show that there are non-measurable sets, which are in turn the culprit of the paradox. The paradox is not invariant with respect to the definition of measure, but Lebesgue measure is an analog of physical volume/area/whatever, so that's why the paradox can be phrased in terms of a physical object and size as a sort of loose description. $\endgroup$ – Jonathan Hebert Jul 20 '17 at 13:47
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    $\begingroup$ I don't think you should try to relate Banach-Tarski and the real world, as the first one is a purely mathematical phenomenon without any connection to the real world. If you want to get a feeling for Banach Tarski, the big point is that this is true for the free group with two generators. Once you know it, you can prove that there is a group of rotations which generate $F_2$ and finally apply axiom of choice for "cover" the sphere by lot of copies of $F_2$ and picking a representant in every copies. $\endgroup$ – user171326 Jul 20 '17 at 13:50
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    $\begingroup$ We can't cut a ball of chocolate to five pieces and assemble two. But we are not mathematical objects nor a chocolate ball is a perfect continuous ball, nor reality as we experience it is continuous, but rather very much discrete and finite. $\endgroup$ – Asaf Karagila Jul 20 '17 at 13:53
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    $\begingroup$ The problem is that you cannot really measure all sets in space in many models of set theory. AC does not allow it, nor does CH, etc. it's not the inconsistency of any measure, we can make consistent measures, but those paradoxical sets from B-T, or Vitali sets, or sets from free ultrafilters etc. will not be measurable for any "sensible" measure. Maybe wanting to measure all sets is too much to ask for. $\endgroup$ – Henno Brandsma Jul 20 '17 at 14:02
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Do you expect all functions $f\colon\Bbb R^3\to\Bbb R^3$ to be linear? Or at least polynomial? Or at least analytic? Or at least smooth? Or at least continuously differentiable? Or at least continuous? Or at least Baire measurable?

Well, no. We know that amongst the functions, the continuous ones make very few of them. We know that amongst the continuous functions those which are differentiable make a negligible set, let alone polynomials or linear transformations.

We know that most of the objects we are interested in are themselves the pathological ones, and those things we call "pathologies" are in fact the common law of the mathematical land.

So why do we expect that all the sets of reals, or the plane, or the Euclidean space should be Lebesgue measurable? We shouldn't. Because it does not sit with the rest of the mathematical results that we have so far. Only with our wishful intuition, which itself gets rebuilt over time in order to accommodate these sort of facts. (To paraphrase von Neumann, you don't understand this, you get used to it.)

Choice is certainly not the culprit. Rather it is the mixture of Infinity and Power Set which allow us to prove the existence of all manners of odd sets. Sure, it is consistent that the axiom of choice fails and the Banach–Tarski theorem fails too. So to some extent, you could argue this is choice's fault. But in all these models we can partition the real numbers into more sets than numbers. So we can also partition $\Bbb R^3$ into more sets than points. And if that is not a paradoxical decomposition, I don't know what is.

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The Banach-Tarski paradox shows that (assuming AC) there can be no finitely additive full (i.e. defined for all subsets) measure (so weaker than Lebesgue measure, which is countably additive) on $\mathbb{R}^n$ for $n \ge 3$ that is preserved by translation and rotations. The "paradoxical sets" cannot be measurable for any such finitely additive measure that agrees with Lebesgue measure where possible (so that spheres have their usual volume/measure).

In fact Banach himself had shown earlier that there is a finitely additive measure on $\mathbb{R}^2$ (that agrees with Lebesgue measure when both are defined) defined on all subsets of the plane, and which is preserved by translations and rotations. (And also on $\mathbb{R}$ for translations) This shows that the Banach-Tarski paradox is impossible in the plane or the line.

So the problem really is that you cannot measure everything at the same time (only nice sets like the Lebesgue measurable ones), in a nice invariant way. Vitali showed (using AC) that we cannot have a countably additive translation invariant full measure, and even CH shows this without AC (as Ulam showed). The existence of a full countably additive measure on the reals that extends Lebesgue measure implies the existence of a so-called real-valued measurable cardinal $\le \mathfrak{c}$ and cannot be shown from ZFC alone.

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  • $\begingroup$ I do not think the first sentence says what it is supposed to say. $\endgroup$ – bertram Jul 20 '17 at 14:12
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    $\begingroup$ Thanks for editing. I still don't think it says what it is supposed to say: somewhere should be the words "defined for all subsets". (Sorry to be nitpicky; the current formulation seems likely to confuse beginners.) $\endgroup$ – bertram Jul 20 '17 at 14:18
  • $\begingroup$ @bertram You're right; I do mention it later. $\endgroup$ – Henno Brandsma Jul 20 '17 at 14:20
  • $\begingroup$ Isn't the counting measure on $\Bbb{R}^3$ full, finitely additive and invariant by translations & rotations? $\endgroup$ – mattecapu Feb 21 '18 at 19:23
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    $\begingroup$ @GabrielRomon wikipedia mentions it, I think I read about it in the Stan Wagon book on the Banach-Tarski paradox. According to his own answer it's indeed in his book. $\endgroup$ – Henno Brandsma Jul 19 '18 at 16:46

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