Exercise 8 solution:

(a) Let y = F^{i}(x)+j ∈ eo(x). Then π(y) = π(F^{i}(x)+j) = πF^{i}(x) = f^{i}π(x), implying π(y) ∈ o(π(x), f), or y ∈ π^{−1}(o(π(x), f)). We have eo(x) ⊂ π^{−1}(o(π(x), f)).

Suppose π(y) ∈ o(π(x), f). Then π(y) = f^{i}π(x) for some integer i. Hence π(y) = πF^{i}(x), and y = F^{i}(x) + j for some integer j. We have that π^{−1}(o(π(x), f)) ⊂ eo(x), and thus eo(x) = π^{−1}(o(π(x), f)).

(b) Let x ∈ R and I ⊂ R be an interval. Since o(π(x), r_{ω}) is dense in T, r_{ω}^{i}π(x) = πR_{ω}^{i}(x) ∈ π(I) for some integer i. This implies R_{ω}^{i}(x) + j ∈ I for some integer j, so that eo(x) ∩ I ≠ ∅. Thus eo(x) is dense in R.