Regularity on the circle

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

1. If A≤B and B≤C, then A≤C. (Transitivity)
2. A≤A. (Reflexivity)
3. Either A≤B or B≤A (or both). (Totality)
4. If A is a proper subset of B, then A<B. (Regularity)

Moreover, we have a very plausible invariance condition:

5. If r is any reflection in a line going through the center of the circle T and A≤B, then rA≤rB,

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 A≤B and B≤A. 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 rA≤A or A≤rA by totality. If rA≤A then A=r²A≤rA (the square of a reflection is the identity). If A≤rA then rA≤r²A=A. In both cases, thus, A~rA.

Therefore, if A≤B, then rA~A≤B and so rA≤B. 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 A≤B, then we have rA≤B and rA≤rB. 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.