EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 4, 2016, 367-382 ISSN 1307-5543 – www.ejpam.com Hypergroups Associated With Ternary Relations And The Associated Join Space Of The Spherical Geometry Sakarapani Govindarajan PG and Research Department of Mathematics, A.V.C. College (Autonomous), (affiliated to Bharathi- dasan University, Trichy) Mannampandal, Tamil Nadu, India Abstract. The paper deals with hypergroupoids obtained from ternary relations.The correspondence between ternary relation and hypergroups, studied by the author, especially in the case when the rela- tionρ is symmetric and reflexive, is analyzed here in the most general context.A necessary and sufficient condition on a ternary relation ρ is obtained for the associated hypergroupoid to be a hypergroup or a join space. Extension of a hyperoperation associated with a ternary relation being employed to obtain a associated join space of the spherical geometry. 2010 Mathematics Subject Classifications: 20N20; 04A05 Key Words and Phrases: Ternary relation, Hypergroups, Join Spaces, Spherical Geometry, Associated join space 1. Introduction Hyper structure theory was born during the 8th Congress of Scandinavian Mathematicians in 1934, when F. Marty [17] defined hypergroups, a natural generalization of the concept of group, and began to analyze their properties and applied to them to non commutative groups, rational fractions and algebraic functions etc,. Since then various connection between hyper- groups and other subjects of theoretical and applied mathematics have been established. The most important applications to geometry, topology, cryptography and code theory, graphs and hypergraphs, probability theory, binary relations, theory of fuzzy sets and rough sets, automata theory are found in [8]. The first association between binary relation and hyperstructures ap- peared in J. Nieminen [18], who studied hypergroups related to connected simple graphs. In the same direction P. Corsini [3] worked, considering different hyperoperations associated with graphs. J. Chvalina [1] used ordered structures for the construction of semihypergroups and hyper- groups. Later on, I. Rosenberg [20] introduced a hyperoperation obtained by a binary relation; the new hypergroupoid has been investigated by P. Corsini [5, 6], P. Corsini and V. Leoreanu [7] Email address: govindavc@gmail.com http://www.ejpam.com 367 c© 2016 EJPAM All rights reserved. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 368 and recently by I. Cristea and M. Ştefanescu [10]. P.Corsini introduced a new hyperoperation obtained by a binary relation [4]. Another approach to the connection between hypergroups and ordered set is initiated by M. Ştefanescu [12] and recently hypergroupoids associated with n-ary (n ≥ 3) relations are studied by I. Cristea and M. Ştefanescu [11] and I. Cristea [9] alone. B. Davvaz and T. Vougioklis[13] introduced the concept of n-ary hypergroups as a generalization of hyper- groups in the sense of Marty. I. Cristea [9] introduced a new hypergroupoid associated with n-ary relations and obtained necessary conditions for the new hypergropoid to be a hypergroup or a join space; the new hypergroupoid has been investigated by S. Govindarajan (the author) and G. Ramesh [14] to find sufficient condition if any so that the hypergroupoid introduced by I. Cristea to be a hypergroup or a join space. In the same direction, albeit with different hyperoperations went the paper by the author and G. Ramesh [15]. Next, the author and G. Ramesh established a new correspondence between n-ary relations and hypergroupoids and found conditions on a ternary relationρ, such that Hρ is a hypergroup or a join space [16]. In the same paper, the author and G. Ramesh studied these hypergroups under the union and intersection of relations. In the papers [14] and [16], we obtained a necessary and sufficient condition on the n(n ≥ 3)-ary relation ρ for Hρ to be a hypergroup, especially in the case when ρ is reflexive and symmetric relation. In this paper, the analysis of connection between hyperpgroups and the n(n= 3)-ary relations is continued. We confine with ternary relations, by remarking that any results obtained with regards to n(n = 3)-ary relation can be obtained analogously to the case of n(n ≥ 4)-ary relation. Here, we consider the ternary relation in a most general context and, we find a necessary and sufficient con- ditions such that Hρ is a hypergroup or a join space. Several extensions of hyperoperations being employed, among them, one for which coincides with W. Prenowitz [19] extension of hyperoperation associated with spherical geometry is remarkable, but in a little different form. That is, W. Prenowitz extended the hyperoperation by adjoining an ideal element e and e 6∈ S, but we set e ∈ S. The n-ary relations were studied for their applications in theory of dependence space. Moreover, they are used in database Theory, providing a convenient tool for database mod- eling. The operations such as union, intersection, difference and Cartesian product on n-ary relations are useful to describe information manipulations in databases. These operations are useful in the construction of new database. First we recall some necessary definitions: For a non empty set H, we denote by P∗(H) the set of all non empty subsets of H. • A non empty set H, endowed with a mapping, called hyperoperation • ◦ : H ×H −→ P∗(H) is called a hypergroupoid. • A hypergroupoid which verifies the following conditions: (1) (x ◦ y) ◦ z = x ◦ (y ◦ z), for all x , y, z ∈ H, (2) x ◦H = H = H ◦ x , for all x ∈ H, (reproduction axiom) is called a hypergroup. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 369 • If, for any x , y ∈ H, x ◦ y = H, then (H,◦) is called the total hypergroup. • If A and B are nonempty subsets of H, then we denote the set A◦ B = ⋃b∈B a∈A a ◦ b. • If A and B are nonempty subsets of H, then we denote A/B = ⋃b∈B a∈A a/b. • A commutative hypergroupoid (H;◦) is called a join space if the following implication holds: for any (a, b, c, d) ∈ H4, a/b∩ c/d 6= ;=⇒ a◦d∩ b◦ c 6= ; (transposition axiom). For more details on hypergroup theory, see [2, 8, 21]. 2. Properties of the n-ary Relations In this section we present some basic notions about the n-ary relations defined on a non- empty set H, n ∈ N a natural number such that n ≥ 3, and ρ ⊆ Hn is an n-ary relation on H. Definition 1 ([9, 11]). The relation ρ is said to be : (1) reflexive if for any x ∈ H, the n-tuple (x , x , . . . , x) ∈ ρ; (2) n-transitive if it has the following property: if (x 1 , . . . , x n ) ∈ ρ, (y 1 , . . . , y n ) ∈ ρ hold and if there exist natural numbers i 0 > j 0 such that 1 < i 0 ≤ n, 1 ≤ j 0 < n, x i0 = y j0 , then the n-tuple (x i1 , . . . , x i k , y j k+1 , . . . , y jn ) ∈ ρ for any natural number 1 ≤ k < n and i 1 , . . . , i k , jk+1, . . . , j n such that 1≤ i1 < . . .< ik < i0, j0 < jk+1 < . . .< jn ≤ n; (3) symmetric if (x1, x2, . . . , xn) ∈ ρ implies (xn, xn−1, . . . , x1) ∈ nρ (4) strongly symmetric if (x1, . . . , xn) ∈ ρ implies (xσ(1), . . . , xσ(n)) ∈ ρ for any permutation σ of the set {1,. . . ,n}; (5) n-ary preordering on H if it is reflexive and n-transitive; (6) an n -equivalence on H if it is reflexive,strongly symmetric and n-transitive; (7) diagonal n-ary relation on H if ρ = {(x , x , . . . , x) | x ∈ H}. 2.1. Projections and Join Relations Definition 2 ([9]). Let ρ be an n-ary relation on a nonempty set H and k < n. The (i1, . . . , ik)- projection of ρ, denoted by ρi1,ik , is a k- ary relation on H defined by: if (a1, a2, a3, an) ∈ ρ, then (ai1 , . . . , aik) ∈ ρi1,...,ik Definition 3 ([9]). Let ρ be an n-ary relation on a nonempty set H λ an m-ary relation on the same set H. The join relation of ρ and λ, denoted by Jp(ρ,λ) where 1< p < n, 1< p < m is an m+n−p relation on H that consists of m+n−p-tuples (a1, . . . , am−p, c1, . . . , cp, b1, . . . , bn−p) ∈ ρ and (c1, c2, . . . , cp, b1, . . . , bn−p) ∈ λ. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 370 Let ρ be a ternary relation on H. The join relation J2(ρ,ρ) denoted by α is a 4-ary relation such that (x , y, z) ∈ J2(ρ,ρ)⇐⇒ (x , y, z), (y, z, t) ∈ ρ. If we denote the join relation J2(ρ,ρ) by the symbol α then its projection relations are denoted by α 1,2,4 and α 1,3,4 . 2.2. Some Results on Hypergroupoids Associated with n - ary Relations To each ternary relation ρ on H, the hyperproduct is defined in [9] as follows: (β) for any x , y ∈ H, x ⊗ρ y = {z ∈ H | (x , z, y) ∈ ρ}. The hyperproduct defined in (β) associated to ternary relation ρ is generalized to the case of an n-ary relation ρ(n≥ 3), using projection as in the following: (δ1) For any i ∈ {2, . . . , n− 1}, x ⊗i y = {z ∈ H | (x , z, y) ∈ ρ 1,i,n } (δ2) x ⊗ρ y = {z ∈ H | (x , z, y) ∈ ⋃n−1 i=2 ρ1,i,n }= ⋃n−1 i=2 x ⊗i y . This (H,⊗ρ) is a hypergroupoid if and only if the projection ρ1,n is the total relation, that is ρ1,n = H × H. The necessary and sufficient condition on ρ for the hypergroupoid Hρ to be a quasi hypergroup and necessary condition on ρ for Hρ to be a semihypergroup are obtained by I. Cristea [9]. They are useful in the succeeding sections and so stated below. Proposition 1 ([9, Proposition 11]). Let ρ be an n-ary relation on H. Then (H,⊗ρ) is a quasi- hypergroup if and only if ρ1,n = H × H, and there exists i, j with 2 ≤ i, j ≤ n − 1, such that ρ1,i = ρ j,n = H ×H. Using the projection of J2(ρ,ρ), necessary condition for a hypergroupoid (H,⊗ρ) to be a semihypergroup is obtained in [9] as follows: Proposition 2 ([9, Proposition 16]). Let ρ be a reflexive and symmetric ternary relation H. If ρ 6⊂ α1,2,4 or ρ 6⊂ α1,3,4, then the hyperoperation ⊗ρ is not associative. Corollary 1 ([9, Corollary 17]). Let ρ be a reflexive and symmetric ternary relation H. If (H,⊗ρ) is a semihypergroup, then ρ ⊂ α1,2,4 ∩α1,3,4. Proposition 3. Let ρ be a reflexive and symmetric n-ary relation H which satisfies the condition (S) (x , a1, . . . an−2, y) ∈ ρ ⇐⇒ (x , aσ(1), . . . aσ(n−2), y) ∈ ρ for any permutation σ of the set {1,2,. . . ,n-1}. If, there exists j ∈ {2, . . . , n − 1} such that ρ1, j,n 6⊂ α1,n,2n−2 or ρ1, j,n 6⊂ α1,n−1,2n−2, then the hyperoperation ⊗ρ is not associative. Proposition 4 ([9, Proposition 14]). Let ρ be a ternary relation on H such that ρ1,3 = ρ1,2 = H ×H or ρ1,3 = ρ2,3 If ρ is 3-transitive, then (H;⊗ρ) is the total hypergroup. Example 1. A ternary relation ρ is 3-transitive if and only if it satisfies the following conditions: (i) If (x , y, z) ∈ ρ, (y, u, v) ∈ ρ, then (x , u, v) ∈ ρ (ii) If (x , y, z) ∈ ρ, (z, u, v) ∈ ρ, then (x , y, u) ∈ ρ, (x , y, v) ∈ ρ, (x , u, v) ∈ ρ, (y, u, v) ∈ ρ (iii) If (x , y, z) ∈ ρ, (u, z, v) ∈ ρ, then (x , y, v) ∈ ρ. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 371 3. Hypergroups Associated with Ternary Relations We begin by remarking that the hypergroupoid (H,⊗ρ) is a semihypergroup if and only if: the following bi-implication holds: (β1) ¨ ∀a, b, c, z ∈ H, there exists x ∈ H, such that (a, x , b), (x , z, c) ∈ ρ ⇐⇒ there exits y ∈ H, such that (a, z, y), (b, y, c) ∈ ρ Now, let ρ be a ternary relation on H such that ρ1,3 = ρ1,2 = ρ2,3 = H×H. The expression (β1) together with ρ1,3 = ρ1,2 = ρ2,3 = H ×H and Example 1 shows that ρ is 3-transitive. That is, if (H,⊗ρ) is a quasi hypergroup and ⊗ρ satisfies the associative axiom, then ρ is 3-transitive. Therefore, from Proposition 4 and from the above paragraph it derives that: Theorem 1. Let ρ be a ternary relation on H such that ρ1,3 = ρ1,2 = ρ2,3 = H × H. If (H,⊗ρ) is a hypergroup, then it is the total hypergroup. However, we can prove the existence of a hypergroup (H,⊗ρ) which is different from the total hypergroup by restricting elements of ρ. Proposition 5. Let ρ be a ternary relation on H such that ρ 1,3 = H ×H. Consider the following condition: (τ1)      (x , x , y), (x , y, y) ∈ ρ ∀(x , y) ∈ H2 (x , y, z) 6∈ ρ if x , y, z are distinct (x , y, x) 6∈ ρ if x 6= y (τ2) ¨ if a, b, c, d are distinct, then (a, b, c) ∈ ρ, (b, c, d) ∈ ρ ⇐⇒ (a, b, d) ∈ ρ, (a, c, d) ∈ ρ. Then (H,⊗ρ) is a hypergroup if and only if: either (τ1) alone holds or both (τ1) and (τ2) are simultaneously holds. Proof. Case(1): Suppose (τ1) alone holds. Then, ∀(x , y) ∈ H2 (x , x , y), (x , y, y) ∈ ρ. Hence, ∀(x , y) ∈ H2, {x , y} ⊂ x ⊗ρ y . Therefore, (H,⊗ρ) is a quasi-hypergroup. We clearly have ∀(x , y, z) ∈ H3, x ⊗ρ (y ⊗ρ z) = ⋃ a∈y⊗ρz x ⊗ρ a = {x , y, z}= (x ⊗ρ y)⊗ρ z. So, ⊗ρ is associative, whence. (H,⊗ρ) is a hypergroup. Case(2): Suppose both (τ1) and (τ2) simultaneously hold. It follows from Case (1) that (H,⊗ρ) is a quasi-hypergroup. All that remains to be proved is that ⊗ρ associative, and we dispose of this by showing that ∀(x , y, z) ∈ H3, (x ⊗ρ y)⊗ρ z ⊂ x ⊗ρ (y ⊗ρ z), and conversely. Let a ∈ (x ⊗ρ y)⊗ρ z. Thus =⇒ there exists u ∈ x ⊗ρ y such that a ∈ u⊗ρ z. Hence, for any a ∈ (x ⊗ρ y)⊗ρ z, we have (x , u, y) ∈ ρ and (u, a, z) ∈ ρ (η). We show that a ∈ x ⊗ρ (y ⊗ρ z) to get that (x ⊗ρ y)⊗ρ z ⊂ x ⊗ρ (y ⊗ρ z). We consider the following situations: S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 372 (i) Let (x , u, y) ∈ ρ with (u, a, z) ∈ ρ. Using the (u, a, z) ∈ ρ in (τ2), we obtain that (u, y, a), (y, a, z) ∈ ρ. By the (τ1) and from (y, a, z) ∈ ρ, it follows that (x , a, a) ∈ ρ with (y, a, z) ∈ ρ. Thus, there exist a = v ∈ y ⊗ρ z with a ∈ x ⊗ρ a. So, in this case a ∈ x ⊗ρ (y ⊗ρ z). (ii) Now, we consider the (η)with x = y . From (τ1), we obtain, for any x ∈ H, (x , x , x) ∈ ρ. From (η), we obtain that u = x and (x , a, z) ∈ ρ it follows a ∈ (x ⊗ρ x) ⊗ρ z. Thus a ∈ x ⊗ρ (x ⊗ρ z). We have, (x , a, z) ∈ ρ and (by the (τ1)) (x , a, a) ∈ ρ. Thus, there exist v = a ∈ H such that (x , a, z) ∈ ρ with (x , a, a) ∈ ρ, so, in this case, a ∈ x ⊗ρ (x ⊗ρ z). (iii) Finally, we consider the (η) with y = z. Put y = z in (η). Then, it follows that (x , u, z) ∈ ρ and (u, a, z) ∈ ρ. By the (τ2), we obtain that (x , u, a) ∈ ρ. Now, we have (x , u, a) ∈ ρ and (u, a, z) ∈ ρ. Using (x , u, a) ∈ ρ and (u, a, z) ∈ ρ in (τ2), we obtain that (x , a, z) ∈ ρ. Now, (τ1) yields that (z, z, z) ∈ ρ. Hence, (x , a, z) ∈ ρ with (z, z, z) ∈ ρ and so a ∈ x ⊗ρ (z ⊗ρ z). We have proved that, for any x , y, z, a ∈ H, a ∈ x ⊗ρ (y ⊗ρ z) =⇒ a ∈ x ⊗ρ (y ⊗ρ z); i.e. (x ⊗ρ y)⊗ρ z ⊂ x ⊗ρ (y ⊗ρ z). Similar argument establishes the other inclusion x ⊗ρ (y ⊗ρ z) ⊂ (x ⊗ρ y)⊗ρ z. Conversely, we suppose that Hρ is a semihypergroup and (x , y, z) 6∈ ρ with x , y, z are all distinct. Assume to the contrary that (x , x , z), (x , z, z) 6∈ ρ, for some x , z ∈ H. Then, it follows that z 6∈x ⊗ρ (y ⊗ρ z) = x ⊗ρ {y, z}= {x , y} ∪ x ⊗ρ z z ∈(x ⊗ρ y)⊗ρ z = {x , y} ⊗ρ z = {y, z} ∪ x ⊗ρ z which is a contradiction to that Hρ is a semihypergroup. Thus (τ1) holds. Next, we prove that (τ2) holds. Suppose to the contrary that (τ2) does not holds. Let (x , u, y), (u, y, z) ∈ ρ with x , u, y, z are all distinct. Suppose that (x , y, z) 6∈ ρ. We have from (τ1) that (z, z, z) ∈ ρ as (z, z) ∈ H2. It follows that y 6∈ x ⊗ρ z ⊂ x ⊗ρ (z ⊗ρ z) but y ∈ u⊗ρ z ⊂ (x ⊗ρ y)⊗ρ z which is a contradiction to that Hρ is a semihypergroup. Thus (τ2) holds. Remark 1. If Hρ is a hypergroup and (τ1) alone holds, then ρ is symmetric ternary relation on H. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 373 Proof. It follows by (τ1) of Proposition 5. Remark 2. If Hρ is a hypergroup, then ρ is reflexive relation on H. Proof. According to Proposition 5, we have ∀(x , y) ∈ H2, (x , x , y) ∈ ρ, so, for (x , x) ∈ H2, (x , x , x) ∈ ρ, whence ρ is reflexive. Remark 3. (H,⊗ρ) is a quasihypergroup does not imply the condition (τ1) of Proposition 5, as we can see in the following example. Example 2. Let H = {x , y, z} and ρ ={(x , x , x), (y, y, y), (z, z, z), (x , y, z), (z, y, x), (x , x , y), (y, x , x), (x , x , z), (x , z, z), (y, z, z), (z, x , x), (z, z, x), (z, z, y)}. We have clearly, ∀x ∈ H, x ⊗ρ x = {x}, x ⊗ρ y = {x}= y ⊗ρ x , x ⊗ρ z = {x , y, z}= z ⊗ρ x , y ⊗ρ z = {z}= z ⊗ρ y. So, we obtain the hypergroupoid: Table 1: Example 2 Hypergroupoid ⊗ρ x y z x x x x,y,z y x y z z x,y, z z z We have ∀x ∈ H, x ⊗ρ H = H = H ⊗ρ x. So, (H;⊗ρ) is a quasi hypergroup, but we find also (x , y, y), (y, y, x) 6∈ ρ. Notice that (x , y, z) ∈ ρ and x 6= y 6= z. Moreover, (H,⊗ρ) is a hypergroup. Remark 4. If (x , y, z) 6∈ ρ with x , y, z are all distinct, then the condition (τ2) is neither necessary nor sufficient for (H,⊗ρ) to be a hypergroup as we see in the following example. Example 3. Let us consider the ternary relation ρ on a non empty set H defined as follows: ρ = {(x , x , y), (x , y, y) | (x , y) ∈ H2}. We find ρ1,3 = ρ1,2 = ρ2,3 = H ×H and ∀(x , y) ∈ H2, x ⊗ρ y = {x , y}= x ⊗ρ y. We clearly have ∀(x , y, z) ∈ H3, x ⊗ρ (y ⊗ρ z) = {x , y, z}= (x ⊗ρ y)⊗rho z. Moreover (H,⊗ρ) is a quasi-hypergroup, since ρ1,2 = ρ2,3 = H ×H. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 374 Now, we give an example to show that (τ2) is a sufficient condition for (H,⊗ρ) to be a semihypergroup though (H,⊗ρ) is a quasi-hypergroup. Example 4. On the set H = {1, 2,3, 4}, we consider the ternary relation ρ defined as follows: ρ ={(x , x , y), (x , y, y) | (x , y) ∈ H2} ∪ {(1, 2,1), (1,2, 3), (1,3, 2), (1,2, 4), (2, 3,2), (2,3, 1), (2, 3,4), (3, 2,1), (4,2, 1), (4,3, 2)}. We obtain the hypergroupoid: Table 2: Example 4 Hypergroupoid ⊗ρ 1 2 3 4 1 1,2 1,2,3 1,2,3 1,2,4 2 1,2,3 2,3 2,3 2,3,4 3 1,2,3 2, 3 3 3,4 4 1,2,4 2,3,4 3,4 4 Clearly, we have ρ1,3 = ρ1,2 = ρ2,3 = H ×H. So, (H,⊗ρ) is a quasi-hypergroup, but ⊗ρ is not associative. Indeed, we have (1,2, 3) ∈ ρ and (2,3, 4) ∈ ρ, but (1,3, 4) 6∈ ρ while (1, 2,4) ∈ ρ. Therefore, 1⊗ρ (4⊗ρ 4) = {1, 2,4} 6= {1,2, 3,4}= (1⊗ρ 4)⊗ρ 4. If we suppose (1,3, 4) ∈ ρ then (τ2) is satisfied, and then ⊗ρ is associative. Proposition 6. Let ρ be a reflexive and symmetric ternary relation H such that ρ1,3 = H × H. Then (H,⊗ρ) is a hypergroup, which is different from the total hypergroup if and only if : (1) ρ1,2 = ρ2,3 = H ×H (2) ρ ⊂ α1,2,4 ∩α1,3,4 (3) either (τ1) alone holds or both (τ1) and (τ2) are simultaneously holds. Proof. (=⇒) It results by Proposition 5 and Corollary 1. (⇐=) The conditions of Proposition 5 are verified. Further, (3) =⇒ (2). 3.1. Hypergroups Associated with Reflexive Symmetric Ternary Relations Recall now what a spherical geometry is (see [19], [8, defn. 25, p.37 in]). This definition is useful in order to characterize the hypergroup Hρ Definition 4. An (abstract) spherical geometry is a system (S, R), where S is a set of elements called points and R is a ternary relation on S called betweenness, which satisfies the following postulates: S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 375 (i) If (x , y, z) ∈ R, then x , y, z are distinct; (ii) if (x , y, z) ∈ R, then (z, y, x) ∈ R; (iii) for any x, there exists a unique x ′ such that x 6= x ′ and the following implication holds: (x , u, v) ∈ R=⇒ (u, v, x ′) ∈ R; (iv) if y 6= x and y 6= x ′, then there exists u such that (x , u, y) ∈ R. W. Prenowitz [19] defined the following hyperoperation on S: ∀(x , y) ∈ S2, y 6= x , y 6= x ′, x ◦ y = {t | (x , t, y) ∈ R}, x ◦ x = {x}. Now, let us set: R = ρ. Again, recall that the join relation J2(ρ,ρ) denoted by α is a 4-ary relation such that (x , y, z) ∈ J2(ρ,ρ)⇐⇒ (x , y, z), (y, z, t) ∈ ρ. Suppose that ρ is betweenness relation. Then the postulates (iii) and (iv) of Definition 4 together with the definition of join relation and projection relations shows that, for any (x , y, z) ∈ ρ, there exists a unique x ′ such that x 6= x ′ and (x , y, x ′) ∈ α1,2,4, (x , z, x ′) ∈ α1,3,4. We observe that the product ⊗ ρ and ◦ are identical and the relation ρ = R if and only if ρ is reflexive and symmetric ternary relation on H. Hence, we obtain the following proposition: Proposition 7. Let ρ be a reflexive and symmetric ternary relation on H with |H| ≥ 3 such that ρ1,3 = H ×H. If ρ satisfies the following the postulates, then ⊗ ρ is not associative (i) If (x , y, z) ∈ ρ, then x , y, z are distinct; (ii) if x 6= y, then (x , y, x) 6∈ ρ; (iii) for any (x , y, z) ∈ ρ, there exists a unique x ′ such that x 6= x ′ and (x , y, x ′) ∈ α1,2,4, (x , z, x ′) ∈ α1,3,4. Proof. The ρ verifies the postulates of the betweenness relation on H. Now, the proof follows from [8, Theorem 27, p.39]. Let ρ be a reflexive ternary relation H such that ρ1,3 = H×H. Let (x , a, y) ∈ ρ be arbitrary and x , a, y are distinct. Let (H;◦ρ) be the hypergroupoid defined as follows. ∀(x , y) ∈H2, x 6= y, x ◦ρ y = {a | (x , a, y) ∈ ρ} ∪ {x , y} ∀x ∈H, x ◦ρ a = {x}. Indeed, ◦ρ is an extension of ⊗ρ. If we suppose, ∀ (x , y) ∈ H2, (x , y, x) 6∈ ρ when x 6= y; (1) then, we have clearly: ∀x ∈ H, x ◦ρ x = {x}= x ⊗ ρ x ⇐⇒ (x , x , x) ∈ ρ S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 376 ∀(x , y) ∈ H2, x 6= y, x ◦ρ y = {x , a, y}= x ⊗ ρ y ∪ {x , y} ⇐⇒ ∀x 6= y, (x , a, y) ∈ ρ ∀x ∈ H, x ◦ρ a = {x}= x ⊗ ρ a⇐⇒ (x , x , a) ∈ ρ If (x ′, a′, y ′) ∈ ρ and x ′, a′, z′, a are distinct, then we extend the product as follows: a ◦ρ a′ = {a, a′} whenever 6 ∃ u ∈ H such that (a, u, a′) ∈ ρ. Again, if (a, u, a′) ∈ ρ, then we define the product recursively as above. Proposition 8. Let ρ be a reflexive and symmetric ternary relation on H with |H| ≥ 3 such that ρ1,3 = H ×H. If ρ satisfies the following conditions: (i) any (a, b, c) ∈ ρ with a, b, c distinct =⇒ (x , b, y) ∈ ρ, ∀(x , y) ∈ H2 and x 6= y (ii) ∀(x , y) ∈ H2, x 6= y, (x , y, x) 6∈ ρ (iii) ∀x ∈ H, (x , x , b) ∈ ρ then the extension (H;◦ρ) of (H; ⊗ ρ ) defined by setting ∀(x , y) ∈ H2, x 6= y, x ◦ρ y = {z | (x , z, y) ∈ ρ} ∪ {x , y} is a join space. Proof. Suppose (x , z, y) ∈ ρ. Then, we clearly have ∀(x , y) ∈ H2, x 6= y, x ◦ρ y = {z | (x , z, y) ∈ ρ} ∪ {x , y} ∀x ∈ H, x ◦ρ x = {x} ∀x ∈ H, x ◦ρ z = {x} Since, for any x , y ∈ H, {x , y} ⊂ x ◦ρ y; it follows that (H;◦ρ) is a quasihypergroup. Moreover, since ρ is symmetric, it follows that x ◦ρ y = y ◦ρ x , for any x , y ∈ H, and therefore (H;◦ρ) is commutative. Now, we prove that the hyper operation < ◦ρ > is associative. We have clearly, x ◦ρ (y ◦ρ z) =x ◦ρ {y, z′, z | (y, z′, z) ∈ ρ} =x ◦ρ y ∪ x ◦ρ z′ ∪ x ◦ρ {z′ | (y, z′, z) ∈ ρ} =x ◦ρ y ∪ x ◦ρ z, since x ◦ρ z′ = {x} ⊂ x ◦ρ y =x ◦ρ y ∪ x ◦ρ {x , z′′, z | (x , z′′, z) ∈ ρ} , since x ◦ρ z = {z′′ | {x , z′′, z) ∈ ρ} ∪ {x , z} =(x ◦ρ y) ◦ρ z, since {x , z′′} ⊂ (x ◦ρ y) It remains to check the condition of the join space. Set a, b, c, d ∈ H such that a/b ∩ c/d 6= ;; then there exists x ∈ a/b ∩ c/d, that is a ∈ x ◦ρ b = {z | (x , z, b) ∈ ρ} ∪ {x , b} and c ∈ x ◦ρ d = {z | (x , z, d) ∈ ρ} ∪ {x , d}. We consider the following situations: S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 377 (i) Let a, b, c, d, z are distinct and z ∈ a ◦ρ d. Recall that, for any x 6= y 6= z, (x , z, y) ∈ ρ =⇒ (x ′, z, y ′) ∈ ρ, for all x ′ 6= y ′. Then, it follows that z ∈ b ◦ρ c and therefore z ∈ a ◦ρ d ∩ b ◦ρ c. Similarly, if z ∈ b ◦ρ c then z ∈ a ◦ρ d ∩ b ◦ρ c. (ii) If a, b, c, d, z are distinct and a = x then c ∈ a ◦ρ d, and since c ∈ b ◦ρ c, it follows that c ∈ a ◦ρ d ∩ b ◦ρ c. (iii) If a, b, c, d, z are distinct and c = x then a ∈ c ◦ρ b. By the symmetry of ρ, it follows that a ∈ b ◦ρ c, and since a ∈ a ◦ρ d, it follows that a ∈ a ◦ρ d ∩ b ◦ρ c. By the same way, we establish that a ∈ a ◦ρ d ∩ a ◦ρ c and c ∈ a ◦ρ c ∩ a ◦ρ c in the case a = b and c = d respectively. We can conclude that a ◦ρ d ∩ b ◦ρ c 6= ;, so (H;◦ρ) is a join space. Let < ◦ρ > be the extension of < ⊗ ρ > as it is defined in Proposition 8 and ρ be a ternary relation on H. Let α be the join relation J2(ρ,ρ). Using the projections of α, we give the following proposition. Proposition 9. Let ρ be a reflexive and symmetric ternary relation on H with |H| ≥ 3 such that ρ1,3 = ρ1,2 = ρ2,3 = H ×H. If ρ satisfies the following condition: (1) If x,y,z are distinct and (x , y, z) ∈ α1,2,4 ∩α1,3,4, then (x , y, z) ∈ ρ and conversely; then (H;◦ρ) is a hypergroup. Proof. Since ρ1,2 = ρ2,3 = H ×H, it follows, by Proposition 1, that (H;◦ρ) is a quasihyper- group. It remains to check the associative axiom. First of all, we shall check the following equality: ∀(x , y, z) ∈ H3, x ◦ρ (y ◦ρ z) = x ◦ρ y ∪ y ◦ρ y ∪ y ◦ρ z. (2) We suppose that (x , y, z) ∈ α1,2,4 ∩α1,3,4 and x , y, z are distinct, then (x , y, z) ∈ ρ and x , y, z are distinct. Then (1) =⇒ (x , y, z) ∈ ρ and x , y, z are distinct. Now, we verify: u ∈ x ◦ρ (y ◦ρ z)⇐⇒ u ∈ x ◦ρ y ∪ y ◦ρ y ∪ y ◦ρ z. =⇒ There exists v ∈ y ◦ρ z such that u ∈ x ◦ρ v. Hence (x , u, v) ∈ ρ with (y, v, z) ∈ ρ. Recall that, (x , y, z) ∈ α1,2,4 ∩α1,3,4⇐⇒ (x , y, t, z) ∈ J2(ρ,ρ)⇐⇒ (x , y, t), (y, t, z) ∈ ρ. (3) Set t = v. Then, we have (x , y, v), (y, v, z) ∈ ρ, whence (x , u, v) ∈ ρ and (x , y, v) ∈ ρ. By the condition specified and by the (3), if (a, t, b) ∈ α1,2,4 ∩α1,3,4 and (a, s, b) ∈ α1,2,4 ∩α1,3,4 ∈ ρ, t 6= s then (a, t, b) ∈ ρ and (a, s, b) ∈ ρ; S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 378 it follows that (a, t, s) ∈ ρ or (a, s, t) ∈ ρ. Therefore, from (x , u, v) ∈ ρ, (x , y, v) ∈ ρ and u 6= y it results (x , u, y) ∈ ρ or (x , y, u) ∈ ρ. If (x , u, y) ∈ ρ, then u ∈ x ◦ρ y . If (x , y, u) ∈ ρ, then (x , y, z) ∈ α1,2,4∩α1,3,4 =⇒ (y, u, z) ∈ ρ, whence u ∈ y ◦ρ z. If u= y , then u= y ∈ y ◦ρ y (by the reflexivity of ρ). Therefore, u ∈ x ◦ρ y ∪ y ◦ρ y ∪ y ◦ρ z. ⇐= Suppose u ∈ x ◦ρ y . Then (x , u, y) ∈ ρ. Let (x , y, z) ∈ ρ ⊂ α1,2,4 ∩α1,3,4. Since (x , u, y) ∈ ρ ⊂ α1,2,4 ∩α1,3,4, it follows that there exists a v ∈ H such that (x , u, v) ∈ ρ, (u, v, z) ∈ ρ. Thus u ∈ x ◦ρ v. Again, (x , y, z) ∈ ρ ⊂ α1,2,4 ∩α1,3,4 =⇒ (y, v, z) ∈ ρ. Hence, from (x , u, v) ∈ ρ and (y, v, z) ∈ ρ, it follows that u ∈ x ◦ρ v with v ∈ y ◦ρ z. Therefore u ∈ x ◦ρ (y ◦ρ z). Now, suppose u = y . Then we obtain (x , y, v), (y, v, z) ∈ ρ since (x , y, z) ∈ α1,2,4 ∩α1,3,4. Hence, from (x , y = u, v) ∈ ρ and (y, v, z) ∈ ρ, it follows u ∈ x ◦ρ v with v ∈ y ◦ρ z. Therefore u ∈ x ◦ρ (y ◦ρ z). Finally, suppose u ∈ y ◦ρ z. By the definition of the product ◦ρ, u ∈ x ◦ρ u, whence (x , u, u) ∈ ρ and so u ∈ x ◦ρ (y ◦ρ z). Thus, we have established the following result: u ∈ x ◦ρ (y ◦ρ z)⇐⇒ u ∈ x ◦ρ y ∪ y ◦ρ y ∪ y ◦ρ z. By the same way, we prove the following result: u ∈ (x ◦ρ y) ◦ρ z⇐⇒ u ∈ x ◦ρ y ∪ y ◦ρ y ∪ y ◦ρ z. We find x ◦ρ (y ◦ρ z) = (x ◦ρ y) ◦ρ z. This completes the proof. 4. The Associated Join Space Of The Spherical Geometry Proposition 10. Let ρ be a reflexive and symmetric ternary relation on H with |H| ≥ 3 such that ρ1,2 = ρ2,3 = ρ1,3 = H ×H. Let ρ satisfies the following postulates: (i) If (x , y, z) ∈ ρ, then x , y, z are distinct; (ii) for any x, there exists a unique x ′ such that (x , y, x ′) ∈ α1,3,4; (iii) if x 6= y, then (x , y, x) 6∈ ρ. Then the extension (H;e◦ρ) of (H; ⊗ ρ ) defined by the following setting is a join space. ∀x 6= y, choose t ∈ H such that x 6= t 6= x ′. Set ∀x 6= y, xe◦ρ y = {x , t, y}∀x ∈ H, xe◦ρ t = {x} = xe◦ρe = {x} and for any t, t ′ ∈ xe◦ρ y, set te◦ρ t ′ = {t, t ′} S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 379 Proof. Suppose ρ be a reflexive and symmetric ternary relation on H. Then, from the hypothesis, we obtain the following implications: (i) If (x , y, z) ∈ ρ, then x , y, z are distinct; (ii) if (x , y, z) ∈ ρ, then (z, y, x) ∈ ρ are distinct. (iii) For any x , there exists a unique x ′ such that (x , y, x ′) ∈ α1,3,4 =⇒ for any x , there exists a unique x ′ and an element u ∈ H such that (x , u, y) ∈ ρ and (u, y, x ′) ∈ ρ, (iv) if x 6= y , then (x , y, x) 6∈ ρ =⇒∀x ∈ H, x e◦ρ x = {x}. Hence, ρ is the betweenness relation on H. Since ρ1,2 = ρ2,3 = H ×H, it follows, by Proposition 1, that (H;e◦ρ) is a quasihypergroup. Moreover, since ρ is symmetric, it follows that x e◦ρ y = y e◦ρ x , for any x , y ∈ H, and therefore (H;e◦ρ) is commutative. Now, we prove that the hyper operation 〈e◦ρ〉 is associative. We shall check the following inclusion: ∀(x , y, z) ∈ H3, (x e◦ρ y) e◦ρ z ⊂ x e◦ρ (y e◦ρ z). (4) Let u ∈ (x e◦ρ y) e◦ρ z. Then, there exists v ∈ x e◦ρ y such that u ∈ x e◦ρ v. Hence, (x , v, y) ∈ ρ with (v, u, z) ∈ ρ. We distinguish the following cases: (i) If x 6= y and y 6= x ′ then (v, y, x ′) ∈ ρ; it follows (x , y, x ′) ∈ α1,3,4 ⊂ ρ. Set: y = e. Hence, from x e◦ρ e = {x} and v ∈ x e◦ρ y , it follows v = x . Thus, (x , u, z) ∈ ρ. On the other hand, from z e◦ρ y = z and from the symmetry, it follows (y, z, z) ∈ ρ. Therefore, u ∈ x e◦ρ (y e◦ρ z). By the same way, we check the following other inclusion: ∀(x , y, z) ∈ H3, (x e◦ρ y) e◦ρ z ⊇ x e◦ρ (y e◦ρ z) (5) (ii) If y = x ′, then (x e◦ρ y) e◦ρ z ={x , t, x ′} e◦ρ z ={x , e, x ′} e◦ρ z =x e◦ρ z, ∪ {z} ∪ x ′ e◦ρ z ={x , e, z} ∪ {z} ∪ x ′ e◦ρ z ={x , e, z} ∪ {x ′, e, z, }= {x , e, x ′, z} =x e◦ρ (x ′ e◦ρ z) so (H;e◦ρ) is a commutative hypergroup. Now, let us check the following implication: a/b ∩ c/d 6= ;=⇒ a e◦ρ d ∩ b e◦ρ c 6= ;. Notice that ∀a ∈ H, a has a unique inverse a′ and ∀(a, b) ∈ H2, a/b = a e◦ρ b′. S. Govindarajan / Eur. J. Pure Appl. Math, 9 (2016), 367-382 380 So, a/b ∩ c/d 6= ;=⇒ a e◦ρ b′ ∩ c e◦ρ d ′ 6= ; whence {a} ∩ b e◦ρ (c e◦ρ d ′) 6= ;, hence a/d ′ ∩ b e◦ρ c 6= ;, that is, a e◦ρ d ∩ b e◦ρ c 6= ; (By [8, Theorem (64, 2), p.12]). Therefore, (H;e◦ρ) is a join space. Thus, we obtain a join space with identity e′ in H. To illustrate the application of this proposition, let us consider a ternary relation defined on the set H = {1, 2,3, 4} with ρ ⊃ {(1, 2,3), (2,3, 4), (1,3, 2), (3, 2,4), (3, 2,1), (4, 3,2), (2,3, 1), (4,2, 3)} and ρ1,3 = ρ1,2 = ρ2,3 = H ×H. Let x = 1, x ′ = 4. Choose t = 2 and t ′ = 3 and set: ∀x ∈ H, x ◦ρ x = {x}. Applying the hyperoperation defined in the proposition, we obtain the hypergroupoid: Table 3: Join Space Associated with Spherical Geometry ⊗ ρ 1 2 3 4 1 1 1,2,3 1,2,3 1,2, 3,4 2 1,2,3 2 2,3 2,3,4 3 1,2,3 2, 3 3 2,3,4 4 1,2,3,4 2,3,4 2,3,4 4 Clearly, we have (H;◦ρ) a join space. We have consider only one element x = 1 and its inverse x ′ = 4. For the remaining ele- ments 2, 3, we can easily find their inverses in infinite case which is necessary for the spherical geometry. 5. Conclusion We have characterized all the ternary relation ρ such that the hypergroupoid Hρ is a hy- pergroup or a join space. We have stated some connections between a general ternary relation ρ and hypergroups and then proved them. Moreover, a correspondence between this hyper- groupoid and the join space obtained by W. Prenowitz from betweenness (ternary) relation has been established. 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