By Smirnov V. A.

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**Sample text**

5. If one of the ends of (AB) is on the ray OC , the other end is either on OC or coincides with O. Proof. Let, say, B ∈ OC . 3 OB = OC . Assuming the contrary to the statement of the lemma, we have c A ∈ OB ⇒ [AOB] ⇒ O ∈ (AB), which contradicts the hypothesis. 6. If (AB) has common points with the ray OC , either both ends of (AB) lie on OC , or one of them coincides with O. 3 Proof. By hypothesis ∃M M ∈ (AB) ∩ OC . M ∈ OC =⇒ OM = OC . 2 c the lemma and let, say, A ∈ OM . Then [AOM ] & [AM B] =⇒ [AOB] ⇒ O ∈ (AB) - a contradiction.

N − 1, with the ray AcnAk , complementary to the ray AnAk , where k ∈ {1, 2, . . e. n−1 A1An = i=1 (Ai Ai+1 ] ∪ AcnAk . Proof. (See Fig. 4. ✷ Point Sets on Rays Given a point O on a line a, a nonempty point set B ⊂ Pa is said to lie on line a on the same side (on the opposite side) of the point O as (from) a nonempty set A ⊂ Pa iff for all A ∈ A and all B ∈ B the point B lies on the same side (on the opposite side) of the point O as (from) the point A ∈ A. If the set A (the set B) consists of a single element, we say that the set B (the point B) lies on line a on the same side of the point O as the point A (the set A).

19. Indeed, by hypothesis, C ∈ Int∠AOB = (aOA )B ∩(aOB )A . 19 OC ⊂ Int∠AOB = (aOA )B ∩ (aOB )A . 5. If a point C lies outside an angle ∠AOB, the ray OC lies completely outside ∠AOB: OC ⊂ Ext∠AOB. 76 73 Our use of the notation αAOB is in agreement with the definition on p. 3. 19 makes this notion well defined in its ”any of the points” part. 3 it follows that this lemma can be reformulated as: If one of the points of a ray OC lies outside an angle ∠AOB, the whole ray OC lies outside the angle ∠AOB.

### A-Simplicial Objects and A-Topological Groups by Smirnov V. A.

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