Exercise 1 solution: Let r_{ω} : T → T, r_{ω}(θ) = θ + ω. Note that r_{ω}^{n}(θ) = θ + nω.

(a) Let p/q be rational. Then r_{p/q}^{q}(θ) = θ + q · p/q = θ + p. Since p is an integer, θ and θ + p are the same point on T, and r_{p/q}^{q}(θ) = θ. Thus all points are periodic under r_{p/q}^{q} with period q.

(b) Suppose ω is irrational. Let θ ∈ T. Suppose r_{ω}^{n}(θ) = r_{ω}^{m}(θ). Then θ + nω = θ + mω + k for some integer k, implying (n − m)ω = k. Since ω is irrational, n = m. Hence o(θ, r_{ω}) is an infinite subset of T.

Let ϵ > 0 be given. Since T is bounded, the Bolzano-Weierstrass Theorem implies o(θ, r_{ω}) has a limit point in T. Hence there exist integers n > m with |r_{ω}^{n}(θ)−r_{ω}^{m}(θ)| < ϵ. Letting k = n − m, we have |r_{ω}^{k}(θ) − θ| < ϵ since r_{ω} preserves lengths. Now r_{ω}^{k}(θ) maps the arc connecting θ and r_{ω}^{k}(θ) to the arc connecting r_{ω}^{k}(θ) and r_{ω}^{2k}(θ), and this arc has length less than ϵ. The set { θ, r_{ω}^{k}(θ), r_{ω}^{2k}(θ), …} thus partitions T into arcs of length less than ϵ, so that o(θ, r_{ω}) is dense in T. (This argument is from An Introduction To Chaotic Dynamical Systems (2nd edition) by R.L. Devaney, p. 22.)