| After reading the approach it seems like too much work! Anyone willing to critique my suggested simpler proof (which didn’t occur to me until after reading the article). TLDR; It is basically the same as proofs that all countable sets have the same cardinality. (TLDR of that: map set of positive integers x to the even numbers by doubling, and the odd numbers by doubling and subtracting 1. Take the union of even and odd and you end up with the set you started with, the positive integers x). For a circle: We can identify all the points on a circle as the points p associated with the [x,y] coordinates of the complex numbers p = e^(2.c.i.pi), where 0 <= c < 1. (And . is multiply.) If we take each of those points p and rotate it by doubling its c, we now have the same points represented by the expression p = e^(2.c.i.pi), where x <= 0 < 2. So the same number of points, but two passes around the circle, 0 <= c < 1 and 1 <= c < 2. We can move the second set of points in the x positive direction by 2 or more to avoid the overlap. We have now rearranged points of one circle into two. For the surface of a sphere: We simply divide a sphere up into points defined by a stack of circles at real-valued vertical z positions, z <= -1 <= 1. And their real [x,y] points are the real and imaginary parts of each circle e^(2.c.i.pi).circumference(z), where 0 <= c < 1, and circumference(z) is the cos(z). Again, rotate the points by doubling c, so that they are now located at c, where 0 <= c < 2. There are now two overlapping sphere surfaces. We can move the second in any direction by 2 to avoid the overlap. Similar generalizations work for including the volume. Anyone understand why this simpler proof is wrong, or why the more complex proof in the article does something better? |