Exercise 20 solution:

(a) By Exercise 19, H(x + k) = H(x) + k for all integers k (see the solution to Exercise 6(b)). Suppose H(J) = {y_{0}} for some interval J ⊂ R. Let i, j ∈ Z. By Proposition 3, H(F^{i}(J) + j) = HF^{i}(J) + j = R_{α}^{i}H(J) + j = R_{α}^{i}(y_{0}) + j = y_{0} + iα + j, a real number. Hence H is constant on F^{i}(J) + j for all integers i and j.

(b) Let K equal the union of all open intervals on which H is constant. By definition K is open. If K is not dense in R, there exist x ∈ R and ϵ > 0 such that K ∩ (x − ϵ, x + ϵ) = ∅. Since H is increasing, if x_{1}, x_{2} ∈ (x − ϵ, x + ϵ), x_{1} < x_{2}, then H(x_{1}) < H(x_{2}). That is, H((x − ϵ, x + ϵ)) is an open interval V. If, for some integers i and j, F^{i}(J) + j ∩ (x − ϵ, x + ϵ) ≠ ∅, then H would be constant on an open interval in (x − ϵ, x + ϵ), a contradiction. Hence F^{i}(J) + j ∩ (x − ϵ, x + ϵ) = ∅ for all i, j ∈ Z. This implies R_{α}^{i}(H(J)) + j ∩ V = ∅ for all i and j, contradicting the fact { R_{α}^{i}(H(J)) + j : i, j ∈ Z } is dense in R. Thus K is an open, dense subset of R.