Saturday, August 17, 2013

Regularity on the circle

Suppose that a point is uniformly chosen on the circumference of the circle T. Write AB for "the point is at least as likely to be in B as in A" and say A<B when AB but not BA. Here are some very plausible axioms:

  1. If AB and BC, then AC. (Transitivity)
  2. AA. (Reflexivity)
  3. Either AB or BA (or both). (Totality)
  4. If A is a proper subset of B, then A<B. (Regularity)
Moreover, we have a very plausible invariance condition:
  1. If r is any reflection in a line going through the center of the circle T and AB, then rArB,
i.e., the probability comparison holds between A and B if and only if it holds between their reflections.

Proposition. There is no relation ≤ satisfying (1)-(5) for all countable subsets A, B and C of T.

I do not as yet know if the Proposition is true if we replace reflections by rotations in (5).

Totality and/or Regularity should go. Other cases suggest to me that both should go.

Proof of Proposition: Suppose ≤ satisfies (1)-(5). Say that A~B if and only if AB and BA. It is easy to see that ~ is transitive since ≤ is transitive and if A~B then rA~rB. Now observe that rA~A. For either rAA or ArA by totality. If rAA then A=r2ArA (the square of a reflection is the identity). If ArA then rAr2A=A. In both cases, thus, A~rA.

Therefore, if AB, then rA~AB and so rAB. Now, any rotation can be written as the composition of a pair of reflections (a rotation by angle θ equals the composition of reflections in lines subtending angle θ/2). Thus, for every every rotation r, if we have AB, then we have rAB and rArB. It follows easily that A<B if and only if rA<B.

Now, let r be a rotation by an angle which is an irrational number of degrees and let x0 be any point on the circle. Let A be the set {x0,rx0,r2x0,r3x0,...}. Observe that rA={rx0,r2x0,r3x0,r4x0,...} is a proper subset of A (x0 is not equal to rnx0 for any positive integer n as r was a rotation by an irrational number of degrees). Thus, rA<A by Regularity. Thus, A<A, which is a contradiction.

1 comment:

Alexander R Pruss said...

The answer to the question after the Proposition is negative: