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

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**Extra resources for Algebraic topology - old and new: M.M.Postnikov memorial conf.**

**Example text**

We can put some of the results above (as well as some of the results we encounter in their particular cases below) into a broader context as follows. A point set A is called convex if A ∈ A & B ∈ A implies (AB) ⊂ A for all points A, BA. 16. Consider a ray OA , a point B ∈ OA , and a convex set A of points of the line aOA . If B ∈ A but O∈ / A then A ⊂ OA . 51 c Proof. Suppose that there exists C ∈ OA ∩ A. Then O ∈ A in view of convexity, contrary to hypothesis. Since c A ⊂ Pa and OA ∩ A = ∅, O ∈ / A, we conclude that A ⊂ OA .

1. If points C, D lie respectively on the sides h = OA and k = OB of the angle ∠(h, k) then ∠COD = ∠(h, k). Proof. (See Fig. 3. 2. Given an angle ∠AOB, we have B ∈ / aOA , A ∈ / aOB , and the points A, O, B are not collinear. 3 c c , Proof. Otherwise, we would have B ∈ aOA & B = O =⇒ B ∈ OA ∨ B ∈ OA =⇒ OB = OA ∨ OB = OA contrary to hypothesis that OA , OB form an angle. 3 ¬∃b (A ∈ b & O ∈ b & B ∈ b) and A ∈ / aOB . ✷ 71 In practice the letter used to denote the vertex of an angle is usually omitted from its ray-pair notation, so we can write simply ∠(h, k) 72 Thus, the angle ∠AOB exists if and only if the points A, O, B do not colline.

2, B ∈ aOA contradicts the fact that the rays OA , OB form an angle. 2. In the future the reader will encounter many such analogies. 88 Obviously, this means that given an angle ∠(h, k), none of the interior points of an angle ∠(k, m) adjacent to it, lies inside ∠(h, k). 15 is applied here to every point of the ray O . C 90 If O ⊂ Int∠AOB we have nothing more to prove. 47: Given an angle ∠(h, k), all points inside any angle ∠(k, m) adjacent to it, lie outside ∠(h, k). 48: If points B, C lie on one side of aOA , and OB = OC , either OB lies inside ∠AOC, or OC lies inside ∠AOB.