I was trying to read a proof on the Banach Tarski Paradox (https://jfr.unibo.it/article/view/6927) and have a question on the rotation of set (3) in Section 3.7.
Section 3.3 previously describes each string of rotations composed of the unit rotations x, x-1, z, z-1, as a path or a sequence of unit rotations (where the unit is the angle of acos(1/3)), which can be visualized as steps on the surface of the sphere.
My question arises where the author takes the set S(x-1), i.e. the set of points whose path from the initial point ends with x-1, and somehow cancels it out by rotating by x.
This strikes me as an incorrect operation. Since the string is a path of steps, shouldn't a rotation by x be performed not just on the final x-1, but to all the steps in the path? To give an example:
Take the path x-1zx. According to the proof, rotation by x should give the result xx-1zx, which then reduces to zx.
What should have been done is that every step of the path should be subject to the rotation, including the origin. Due to the normalization rule introduced in Section 3.5 of the proof, there is no such thing as a positional placeholder in the system, which may be part of the confusion. If we define 1 as the positional placeholder, then the example is as follows:
Take the path x-1,z,x,1, where the 1 denotes the origin (commas are used to separate the steps for clarity). A rotation by x yields the following: xx-1,xz,xx,x1, which reduces to 1,xz,xx,x relative to the original position of the origin.
This can be generalized for a path of any length in S(x-1), which can be described as x-1...1 (there are an infinite number of leading 1s before and after the string of course, but only the one denoting the origin is needed, just as only one leading zero is required for decimals).
To me, the paradox arises from the cancellation of the final x-1 due to inaccurate application of the normalization rule, which hides the final step. The second-last step then is seen as the last step, which is analogous to multiplying the set by three (if we consider there are only four permissible steps, then it is equivalent to multiplication by the base (4) less one).
Is something incorrect in my understanding?