Hence, we have something akin to a "paradoxical decomposition" of the free group. You may be unimpressed by this, since the free group isn't really a geometric object. If you recall, we said earlier that the free group was important for two reasons, and one of them was that it had a paradoxical decomposition. We now discuss the second reason. Consider the unit sphere in xyz space. Pick a point on the sphere P , and let mean rotation of this point counterclockwise around the x -axis by , and mean rotation around the z -axis counterclockwise by that same angle. If we define so that , it turns out that this is precisely the free group if we define to mean first doing one rotation, then the other, and to be clockwise rotations respectively.

The isomorphism between this group, with and so on, is not easy, and requires some hard linear algebra.

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It will not be given here. From now on, we will use to refer to this group of rotations on the sphere. Before continuing to a complete proof, we must discuss the Axiom of Choice. The axiom states than given any collection of sets possibly infinitely many sets - possibly uncountably many sets! We begin by removing the center of the unit ball by choosing some circle in the ball which contains the center, and using the method by which we removed a point from the circle earlier.

Now that we have done this, we can work on the surface of the ball, , by understanding that everything we do on the surface can extend downward to the interior of the ball in a ray towards the center. Now we define the orbit of a point x of the sphere under the operations of the free group to be.

For almost all the points on the surface, this will yield a copy of the free group, except a couple of bad points: for example, the "north pole" is fixed under the b operation. So define , that is, the set of all the points on the sphere which are held still by some rotation in. Since B is countable, and we can remove any countable set of points from the sphere, remove this set:.

Now we define an orbit equivalence relation: for two points , define whenever and are in the same orbit, i. We state without proof that this is an equivalence relation and thus partitions into equivalence classes, which will be disjoint. Note there will be uncountably many orbits, since each only has countably many points and the union of all orbits covers the uncountable set. Using the axiom of choice, we pick a single point from each orbit and call this set.

Note that.

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Make sure you understand the notation: is a set of rotations, is a set of points, and is the set of points resulting from applying every rotation in to every point in. Similarly we will be defining as sets of rotations and etc. Let's define. But we remember. This immediately yields:.

Since we also had , we have. Hence and , a paradoxical decomposition of. Since we've already shown that is rigid-body equivalent to , this proves the Banach-Tarski paradox. Some view the Banach-Tarski paradox as an absurd result, and evidence that the Axiom of Choice is false. First of all we mention descriptive combinatorics which provides tools for a solution of the circle-squaring problem with Lebesgue measurable pieces or a radical simplification of the construction of paradoxical decomposition of the unit ball in R3 with the pieces having Property of Baire.

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## Banach-Tarski Paradox -- Math Fun Facts

This thread leads also to some applications of paradoxical decomposition to the theory of games with incomplete information which are quite close to physical reality. There is still also an intensive feedback between amenability and paradoxical decomposition. However only recently it appeared that transversality theory and theory of o-minimal structures have some significant meaning for paradoxical decompositions. Grzegorz Tomkowicz is a self-educated Polish mathematician who has made several important contributions to the theory of paradoxical decompositions and invariant measures Keep up with the latest from Cambridge University Press on our social media accounts.

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## No. 3200: THE BANACH-TARSKI PARADOX

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