|
Proof of Lemma 2. Observe first that all the members of the sequence c(n) are different. Indeed, if c(k)=c(m) for k > m, then (k-m)x = [kx]-[mx]. This is a contradiction, since the product of a non-zero integer (k-m) and an irrational number x cannot be an integer. Take now a positive integer n such that 1/n < b-a. The numbers c(1), c(2), ..., c(n+1) being all different and belonging to the interval [0,1], we infer by Dirichlet's pigeon-hole principle that for some i and s such that i and (i+s) both lie between 1 and (n+1) the following inequality holds: In the sequel we shall draw from the following useful representation. Wrap the real axis into a circumference T of perimeter 1 with a distinguished point 0. For two numbers a and b in [0,1] we denote by (a,b) the arc of T which corresponds to the interval (a,b) on the real axis. Let be an anticlockwise revolution by an angle of is equal to c(n) Hence, due to (1) we know that the length of an arc between the points b(i) and b(i+s) is smaller than b-a. This means that the s-th iterate of f is a revolution by an angle of This obviously implies that infinitely many of the points b(s), b(2s), b(3s), ... belong to the arc (a,b). Indeed, if we start from the fixed point 0 and walk for an infinitely long time along the circumference T in one and the same direction making steps of length |
radians. Instead of watching the numbers c(n) in the interval [0,1] we shall look at the images of the distinguished point 0 under iterations of f on T. After a moment of reflection we note that the length of an arc (0,b(n)), where 
radians; the direction of this revolution is of no importance to us. 
, then infinitely many times we shall step into the arc (a,b), since its length b-a is greater than the length of our step. This completes the proof. 