Sierpinski-mazurkiewicz paradox: One plus one equals one

You may know the paradox of the Real-Time-Tarski-dividing a three-dimensional ball into a limited number of parts and then assembling two balls that are exactly the same as the original. This paradox tells us how difficult we seem to be to use the right of choice. logic. Today I have heard of another similar paradox called Sierpinski-mazurkiewicz. Its conclusions are intuitive and unacceptable, and the derivation does not rely on the principle of choice.
Sierpinski-mazurkiewicz paradox: There is a point set S on the plane. We can divide it into two subsets A and B, so that after a rotates 1 radian, it completely overlaps with S, B is also the same as s after translating a unit. In other words, if such a point set exists, we can divide it into two identical subsets! This sounds incredible, but the structure is extremely simple.

Consider all polynomials with non-negative integers

F = {A0 + A1 * x + A2 * x ^ 2 + A3 * x ^ 3 +... + an * x ^ n | N, A0, A1 ,..., an ε n}

So that X in each polynomial is equal to e ^ I, we can obtain several points on the complex plane. Note that because E ^ I is a super number, it is not the root of any Integer Polynomial, so for any two different integer polynomials, the results of X = e ^ I cannot be the same (otherwise, the two formula minus e ^ I is the root of the integer equation ). In this case, the points in the above complex plane correspond one to one with the polynomials in F. Set this vertex set to S. So that subset A is the point corresponding to all polynomials without constant terms in F, so that subset B is the point corresponding to all polynomials whose constant items are not 0 in F. Obviously, the intersection of A and B is an empty set and the complete set is a division of S.
After a point set a is rotated 1 radian clockwise, it completely overlaps with S. Why? Because the clockwise rotation of 1 radian is equivalent to multiplying a plural number by E ^ (-I), and E ^ (-I) is equal to 1/X. After all polynomials without any constant terms are multiplied by 1/X, it is obviously one-to-one correspondence with all polynomials in F.
After a Point Set B is moved to the left, it completely overlaps with S. Why? Because moving one unit left is equivalent to subtracting 1 from a plural number, and according to the previous definition, the constant items in the B set are all positive integers, and the value of the constant item after subtracting 1 is all non-negative integers, it is exactly the whole Polynomial in F.
So the strange thing is that a equals S, B equals S, and A plus B equals S.

 
Compared with the paradox mentioned here, Sierpinski-mazurkiewicz has only a handful of points and is divided into finite portions. Unfortunately, this point set is unbounded. Although the latter is bounded, however, the entire point set is unmeasurable and needs to be divided into an infinite number of parts, and the selection principle is used. Is there a similar paradox when the vertex set is bounded and can only be divided into finite portions? We still do not know.

Reference: http://www.math.hmc.edu/funfacts/ffiles/30001.1-2-8.shtml