Exercise 24 solution:

(a) Let ϵ = 0. Then F_{ω,0}(x) = x+ω, and ρ(f_{ω,0}) = ω. Hence A_{r} = { (ω, 0) : ρ(f_{ω,0}) = r } = {(r, 0)}.

(b) Let ϵ ≤ 1 and (ω, ϵ) ∈ A_{r_{1}} ∩ A_{r_{2}}. Then ρ(f_{ω,ϵ}) = r_{1} and ρ(f_{ω,ϵ}) = r_{2}. Since ϵ ≤ 1, F_{ω,ϵ} is a homeomorphism, and r_{1} = r_{2} by Proposition 1. Hence A_{r_{1}} ∩ A_{r_{2}} = ∅ for ϵ ≤ 1 and r_{1} ≠ r_{2}.