Case I: z = (n, −m) with n, m ∈ N. Then τ (z) = (−m, s) with s = m − ⌊(n − m)x⌋. Case I.1: s ≤ 0. Then τ (z) ∈ (−N) 2 and we are done by (iii). Case I.2: s > 0. Case I.2.1: s ≥ m. Then τ (z) ∈ M and we are done by (ii). Case I.2.2: s < m. Then n > m, τ (z) ∈ L and τ (z) 1 = m + s < z 1 , hence we are done by induction hypothesis. Case II: z = (−n, m) with n, m ∈ N. Case II.1: m ≥ n. Then z ∈ M and we are done by (ii). Case II.2: m < n. Case III: z = (n, m) with n, m ∈ N. Then τ (z) = (m, −p) with p ≥ m and τ 2 (z) = (−p, q) with q ≥ p. Thus τ 2 (z) ∈ M and our assertion follows from (ii).(v) By (iii) and (iv) we finally see Z 2 ⊆ S(x, x + 1), thereby completing the proof of the lemma.Proof. If m > n > 0 then τ (x,−x−1) (n, m) = (m, p) with p > m. Thus the sequence (τ k (x,−x−1) (1, 2)) k∈N is strictly increasing with respect to the norm · 1 .Proof of Theorem 2.1. (i) By the Schur-Cohn criterion ( 6 ) and ([4, Lemmas 4.1 and 4.2]) we know that E 2 ⊆ D 2 ⊆ E 2 .(ii) Let (x, y) ∈ L. We are going to show that (x, y) belongs to D 2 .Case I: x < 1. Case I.1: |y| < 1 + x. Then we are done by (i). Case I.2: |y| = 1 + x. Case I.2.1: y < 0. Then −y = 1 + x, x ≤ 0, and we are done by Lemma 2.2. Case I.2.2: y ≥ 0. Thus y = 1 + x. Case I.2.2.1: x ≤ 0. We are done by Lemma 2.2. Case I.2.2.2: x > 0. Our assertion drops out of Lemma 2.3. Case II: x = 1. Then y ∈ {−1, 0, 1} and the assertion can easily be checked.
