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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 transl
by d0mine 18y ago
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 http://en.wikipedia.org/wiki/Banach–Tarski_paradox
- tome 18y agoI 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.
- d0mine 18y agoAn 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).