Hacker News Reader

Comment by abstractbill | Hacker News Reader

The most interesting thing about the BT paradox, for me, is its interplay with the Axiom Of Choice. Most of the mathematics I did as part of my PhD would fall apart without AC, but I still remember being very unsettled by BT.

The existence of nonmeasurable sets, such as those in the Banach–Tarski paradox, has been used as an argument against the axiom of choice. Nevertheless, most mathematicians are willing to tolerate the existence of nonmeasurable sets, given that the axiom of choice has many other mathematically useful consequences.

View on HN

The funny things is that the same result for an orbit on a sphere doesn't need the Axiom Of Choice at all. It just uses the concept of actual infinity.

"the axiom of choice has many other mathematically useful consequences." is an understatement. It's like a life in Space. You can live in Space (in a specially design isolated constructions e.g., a space station); you even can go outside (not for long and only if you are wearing a space-suit) but It is much much easier to live on the Earth.

I'd be interested in seeing a reference for the result on the sphere, if you have one.

So would I. I think it's wrong; I'm pretty sure that the Hausdorff paradox, which is kinda-sorta the equivalent on the sphere, requires some Choice. But maybe I'm confused.

On the other hand, you don't need any Choice for the Sierpinski-Mazuriewicz paradox, which says that there's a subset A of the plane, which can be written as the disjoint union of two "smaller" subsets B and C, but where A, B, and C are all congruent. (But A is countable, so this isn't nearly as startling as Banach-Tarski.) Specifically, let T be translation by (1,0), and let R be rotation about the origin by 1 radian; then let A be the set of points you can get from (0,0) by applying some sequence of T and R, let B consist of the points where the last operation in the sequence was T rather than R, and let C be everything else. Then B = T(A) and C = R(A). (The key point here is that it turns out that the sequence of Ts and Rs is unique, so that B is well-defined.)

It is the theorem from a dead-tree math-textbook where it is used as a step in a Banach-Tarski Paradox proving.

The textbook is not in English. Here's a translation of the theorem:

An orbit O [1] can be decomposed into 4 sets: A, B, C, D. Using rotation these sets can be combined into 2 orbits:

  A ∪ aB = O; C ∪ bD = O

Proof:

  A = H(a)x; B = H(a')x; C = H(b)x; D = H(b')x

The theorem statement follows from the fact that the free group H can be decomposed into 4 parts:

  H(a),H(a'),H(b),H(b')

and doubled by rotations:

  H = aH(a') ∪ H(a); H = bH(b') ∪ H(b)

∎

__ [1]: The term `orbit` is used in the same sense as in [2]. `H` is a free group similar to the one from the step 3 in [2], and `a`, `b` are rotations defined similar to Step 2 in [2] i.e., they are generators of H:

  H = {e}∪H(a)∪H(a')∪H(b)∪H(b')

, where `e` is the unit of the group H:

  aa' = e; bb' = e

__ [2]: http://en.wikipedia.org/wiki/Banach–Tarski_paradox

I can't follow either of the proofs yet, but one question does pose itself: given a decomposition of the sphere, don't you get a decomposition of the ball, just by projecting towards the origin?

Of course what happens at the origin itself is not clear. Perhaps there's no way of getting around that difficulty. If you could deal with that, however, a proof on the sphere would be equivalent to a proof on the ball.

An orbit is not a sphere. To get the result for a sphere we need AC in some form.

The result for the sphere are easily generalized on a ball (first without a center, then with the center).

The funny things is that the same result for an orbit on a sphere doesn't need the Axiom Of Choice at all. It just uses the concept of actual infinity.

I had to read this several times before I understood what you were saying. But you are right. The point is that you can get what wikipedia calls a paradoxical decomposition of a groups of rotations, and that doesn't require choice. Converting that into a proof of B-T requires taking a member of each equivalence class, and that does require choice.

The proof on WikiPedia is pretty good, actually:

http://en.wikipedia.org/wiki/Banach-Tarski_paradox#A_sketch_...

The point about the orbit on a sphere not needing AC is really step 1 of the proof there, except using rotations on a sphere as the "location" of the free group.

As a side issue, B-T is proof that there isn't a finitely additive, isometry-invariant measure on R^3 that's defined for all subsets. Interestingly, there is such a measure on R^2, although it's now 26 years since I proved it.

Dependent choice is often enough to work with, and BT does not follow from DC.

Reply on news.ycombinator.com