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Example text
Observe d (x, y) ≥ 0 for all x, y ∈ S, since (i), (ii), (iii) imply 0 = d (x, x) ≤ d (x, y) + d (y, x) = 2d (x, y). 38 Chapter 2. Euclidean and Hyperbolic Geometry (i) is called the axiom of coincidence, (ii) the symmetry axiom and (iii) the triangle inequality. Proposition 1. (X, eucl), (X, hyp) are metric spaces, called the euclidean, hyperbolic metric space, respectively, over X. Proof. d of step D of the proof of Theorem 7. e. eucl (x, y) ≤ eucl (x, z)+ eucl (z, y). It remains to prove (iii) for (X, hyp).
Then [a, b] = {x (ξ) | α ≤ ξ ≤ β} and l (a, b) = {x (ξ) | ξ ∈ R}. 9) 44 Chapter 2. Euclidean and Hyperbolic Geometry Proof. 9) is a subset of [a, b]. This follows from α ≤ ξ ≤ β and hyp x (α), x (β) = |α − β| = β − α, hyp x (α), x (ξ) = ξ − α, hyp x (ξ), x (β) = β − ξ. e. with β − α = hyp x (α) x (β) = hyp x (α), z + hyp z, x (β) . Define ξ := α + hyp x (α), z . e. α ≤ ξ ≤ β. 9). 10) hyp z, x (β) = β − ξ = hyp x (ξ), x (β) . 11) We take a motion f with f (a) = 0 and f (b) = λe, λ > 0. e. that f (a) = e sinh η1 , f x (ξ) = e sinh η2 , f (b) = e sinh η3 with η3 = |η2 | + |η3 − η2 | and λ = sinh η3 .
That images of lines under motions are lines follows immediately from the definition of lines. In fact! 1), d f x (ξ) , f x (η) = d x (ξ), x (η) = |ξ − η| for all ξ, η ∈ R. This holds true in euclidean as well as in hyperbolic geometry. In both geometries also holds true the Proposition 5. e. with l a, b. Proof. 3) we know that there exists a motion f such that f (a) = 0 and f (b) = λe, λ > 0, e a fixed element of X with e2 = 1. In the euclidean case there is exactly one line {(1 − α) p + αq | α ∈ R}, p = q, through 0, λe, namely {βe | β ∈ R}.