By Golasinski M., et al. (eds.)

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**Example text**

Suppose that B (c, ), B (c , ) are hyperbolic balls satisfying and B (c, ) ⊆ B (c , ). 18) . Proof. 18), c = c and > 0. a, a motion µ such that µ (c) = 0, µ (c ) = λj, λ > 0. e. e. 1 + x2 = cosh implies 1 + λ2 1 + x2 − λjx = cosh . Applying this implication twice, namely for x = j sinh i ∈ X, i2 = 1, ij = 0 we obtain 1 + λ2 cosh − λ sinh = cosh = and for x = i sinh with 1 + λ2 cosh , a contradiction, since λ > 0 and > 0. Thus c = c . 18), hyp (c, x) = . Hence = . 5. Balls, hyperplanes, subspaces 49 a euclidean hyperplane of X.

If x + y = x + y holds true for x, y ∈ X, then x, y are linearly dependent. Proof. Squaring both sides, we obtain xy = chapter 1. x y . 1) in the case of (X eucl). Let x be a solution. 2. M. , by Lemma 2, x (γ) − x (β), x (β) − x (α) must be linearly dependent. Put p := x (0), q := x (1) − x (0). 1), q = 1. If 0 < 1 < ξ, we obtain x (ξ) − x (1) = for a suitable · x (1) − x (0) = q ∈ R. Thus ξ − 1 = x (ξ) − x (1) = q = | |. Moreover, ξ − 0 = x (ξ) − p = x (1) + q − p = |1 + |. Hence = ξ − 1 and thus x (ξ) = x (1) + q = p + ξq for ξ > 1, a formula which holds also true for ξ = 1, ξ = 0, but also in the cases 0 < ξ < 1, ξ < 0 < 1 as similar arguments show.

E. e. l = p + Ra, by applying Proposition 12. On the other hand, p + Ra ⊥ H (a, 0). The point of intersection is r = p − pa a2 a. It remains to consider p ∈ H in the hyperbolic case. Put H = H (a, 0), a2 = 1. If p − (pa) a = 0, we deﬁne p − (pa) a . j := p − (pa) a Take ω ∈ O (X) with ω (e) = j, where e is the axis of our underlying translation group, and t ∈ R with ωTt ω −1 (p) = (pa) a, in view of (T2) for j. Because of ωTt ω −1 (x) = x + [(xj)(cosh t − 1) + 1 + x2 sinh t] j for x ∈ X, we obtain ωTt ω −1 (H) = H on account of j ∈ a⊥ .