Some Closure Operators and Topologies on a Hyper BCK-algebra

EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 1, 2020, 1-8 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Some Closure Operators and Topologies on a Hyper BCK-algebra Rachel M. Patangan1 , Sergio R. Canoy, Jr.2,∗ 1 Department of Applied Mathematics, College of Arts and Sciences, Agusan del Sur State College of Agriculture and Technology, Bunawan, Agusan del Sur, Philippines 2 Department of Mathematics and Statistics, College of Science and Mathematics, Center for Graph Theory, Algebra, and Analysis-PRISM, Mindanao State University Iligan Institute of Technology, 9200, Iligan City, Philippines Abstract. Given a hyper BCK-algebra (H, ∗, 0), we introduce some subsets of H and use them to generate two closure operators on H. In this paper, we show that each of the two closure operators on H can be utilized to form a base for some topology on H. Moreover, we show that each of the induced topologies coincides with a previously known topology on a hyper BCK-algebra. 2020 Mathematics Subject Classifications: 06F35, 03G25 Key Words and Phrases: Hyper BCK-algebra, hyper order, closure operator 1. Introduction The study of BCK-algebras was initiated by Y. Imai and K. Iséki [4] in 1966 as a generalization of the concept of set theoretic difference and propositional calculi. The hyperstructure theory (or multialgebras) was introduced in 1934 by F. Marty [6] at the 8th Congress of Scandinavian Mathematicians in 1934. In [5], Y.B. Jun et al. applied the hyperstructures to BCK-algebras and introduced the notion of a hyper BCK-algebra which is a generalization of a BCK-algebra. By using the sets LH (A) and RH (A), we introduce the bases BL (H) and BR (H) and the induced topologies τL (H) and τR (H) by these sets, respectively, and investigate their related properties [7, 8]. In this paper, we present two closure operators on a hyper BCK-algebra and consider the respective topologies they generate. It is shown that these topologies coincide, respectively, with the topologies generated by BL (H) and BR (H). 2. Preliminaries A hyper BCK-algebra is a nonempty set H endowed with a hyperoperation “ ∗ ” and a constant 0 satisfying the following axioms: for all x, y, z ∈ H, Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i1.3632 Email addresses: [email protected] (R. Patangan), [email protected] (S. Canoy, Jr.) ∗ http://www.ejpam.com 1 c 2020 EJPAM All rights reserved. R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 (H1) (x ∗ z) ∗ (y ∗ z) ≪ x ∗ y, (H3) x ∗ H ≪ x, (H2) (x ∗ y) ∗ z = (x ∗ z) ∗ y, (H4) x ≪ y and y ≪ x imply x = y, 2 where for every A, B ⊆ H, A ≪ B if and only if for each a ∈ A, there exists b ∈ B such that 0 ∈ a ∗ b. In particular, for every x, y ∈ H, x ≪ y if and only if 0 ∈ x ∗ y. In such case, we call “ ≪ ” the hyper order in H. Throughout this study, (H1 , ∗1 , 01 ) (or simply H1 ) and (H2 , ∗2 , 02 ) (or simply H2 ) are hyper BCK-algebras. Let H be a hyper BCK-algebra and A ⊆ H. The sets LH (A) and RH (A) are given as follows: LH (A) := {x ∈ H | x ≪ a ∀a ∈ A} = {x ∈ H | 0 ∈ x ∗ a ∀a ∈ A} and RH (A) := {x ∈ H | a ≪ x ∀a ∈ A} = {x ∈ H | 0 ∈ a ∗ x ∀a ∈ A}. If A = {a}, we write LH ({a}) = LH (a) and RH ({a}) = RH (a). Let X be a nonempty set and let P(X) denote the power set of X. A mapping φ : P(X) → P(X) is called a closure operator on X, if for all A, B ∈ P(X), the following properties hold [2]: (i) A ⊆ φ(A) (ii) φ2 (A) = φ(A) (iii) A ⊆ B ⇒ φ(A) ⊆ φ(B). 3. Known Results Proposition 1. [7] Let H be a hyper BCK-algebra and A, B ⊆ H. Then the following hold: (i) RH (∅) = H. T (ii) RH (A) = RH (a). a∈A (iii) For any ∅ 6= A ⊆ H such that A 6= {0}, 0 ∈ / RH (A). (iv) If A ⊆ B, then RH (B) ⊆ RH (A). The next result is generated by Albaracin and Vilela. Proposition 2. [1] Let A and B be subsets of a hyper BCK-algebra H. Then the following hold: (i) LH (∅) = H. (ii) If A ⊆ B, then LH (B) ⊆ LH (A). R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 (iii) LH (A) = T 3 LH (a). a∈A (iv) For any A ⊆ H, 0 ∈ LH (A). If 0 ∈ A, then LH (A) = {0}. Theorem 1. [8] Let H be a hyper BCK-algebra. Then BL (H) = {LH (A) : A ⊆ H} is a basis for some topology on H. Denote by τL (H) the topology generated by BL (H). Theorem 2. [7] Let H be a hyper BCK-algebra. Then BR (H) = {RH (A) : ∅ 6= A ⊆ H} is a basis for some topology on H. Denote by τR (H) the topology generated by BR (H). 4. Topology Induced by RH LH Let H be any hyper BCK-algebra with H 6= {0}. By Proposition 1(iii), 0 ∈ / RH (H). By definition of a closure operator, we have the following remark. Remark 1. RH : P(H) → P(H) is not a closure operator for every hyper BCK-algebra H 6= {0}. Theorem 3. Let H be a hyper BCK-algebra and A, B ⊆ H. Then the following properties hold: (i) A ⊆ RH (LH (A)). (ii) If A ⊆ B, then RH (LH (A)) ⊆ RH (LH (B)). (iii) [RH LH ]2 (A) = RH LH [RH (LH (A))] = RH (LH (A)). Proof. (i) Let A ⊆ H and y ∈ LH (A). Then y ≪ a for all a ∈ A. Pick x ∈ A. Then y ≪ x for every y ∈ LH (A). This T means that x ∈ RH (y) for every y ∈ LH (A). Thus, by Proposition 1(ii), x ∈ RH (y) = RH LH (A). Therefore, A ⊆ RH (LH (A)). y∈LH (A) (ii) Let A, B be subsets of H. Suppose A ⊆ B. By Proposition 2(ii), LH (B) ⊆ LH (A). Thus, by Proposition 1(iv), RH LH (A) ⊆ RH LH (B). (iii) By (i) and (ii), RH LH (A) ⊆ [RH LH ]2 (A). We are left to prove that [RH LH ]2 (A) ⊆ (RH LH )(A). First, we need to show that LH (A) ⊆ LH [RH (LH (A))]. Let x ∈ RH (LH (A)). Then y ≪ x for all y ∈ LH (A). Choose z ∈ LH (A). Then z ≪ x for every x ∈ RH (LH (A)). This impliesTthat z ∈ LH (x) for every x ∈ RH (LH (A)). Hence, by Proposition 2(iii), z ∈ LH (x) = LH [RH (LH (A))]. Consequently, x∈RH (LH (A)) LH (A) ⊆ LH [RH (LH (A))]. Therefore, by Proposition 1(iv), RH LH [RH (LH (A))] ⊆ RH (LH (A)). R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 4 Theorem 4. Let H be a hyper BCK-algebra. The function RH LH : P(H) → P(H) is a closure operator on H. Proof. Since H is a hyper BCK-algebra, H 6= ∅. Hence, by the definition of a closure operator and Theorem 3, RH LH is a closure operator. Theorem 5. The family BRL (H) = {A : RH (LH (A)) = A, A ⊆ H} is a basis for some topology on H. Proof. Since RH (LH (H)) ⊆ H and by Theorem 3(i), RH LH (H) = H, it follows that H ∈ BRL (H). Thus, BRL (H) 6= ∅. Let A, B ∈ BRL (H) and x ∈ A ∩ B. Then by Theorem 3(i), we have A ∩ B ⊆ RH (LH (A ∩ B)). Also, since A ∩ B ⊆ A and A ∩ B ⊆ B, RH (LH (A ∩ B)) ⊆ RH (LH (A)) and RH (LH (A ∩ B)) ⊆ RH (LH (B)), by Theorem 3(ii). Since A and B are elements of BRL (H), we have RH (LH (A)) = A and RH (LH (B)) = B. Consequently, RH (LH (A ∩ B)) ⊆ A ∩ B. Combining the two set inclusions, we have RH (LH (A ∩ B)) = A ∩ B, that is, x ∈ A ∩ B ∈ BRL (H). Therefore, BRL (H) is a basis for some topology on H. Denote by τRL (H) the topology generated by BRL (H). Example 1. Consider the infinite hyper BCK-algebra (H, ∗, 0) given by Harizavi in [3], where H = {0, 1, 2, . . .} and “ ∗ ” is defined as follows:  {0, x} if x ≤ y x∗y = {x} if x > y for all x, y ∈ H. Now, let r ∈ H. Then RH (LH (r)) = RH ({0, 1, 2, . . . , r}) = {r, r + 1, r + 2, . . .} 6= {r}. Hence, for any s ∈ H, {s} ∈ / BRL (H). Let ∅ 6= A ⊆ H be a finite set and let u = min A. Since RH (LH (A)) = RH LH (u) = RH ({0, 1, 2, . . . , u}) = {u, u + 1, u + 2, . . .} = RH (u) 6= A, it follows that for any finite set A ⊆ H, A ∈ / BRL (H). Next, suppose that A ⊆ H is an infinite set and let p = min A. Then RH (LH (A)) = RH (p) = {p, p + 1, p + 2, . . .}. Thus, A ∈ BRL (H) if and only if A = {p, p + 1, p + 2, . . .}. Therefore, τRL (H) = {∅, H} ∪ {{p, p + 1, p + 2, . . .} : p ∈ H}. { n1 In the next example, we extend the set H in Example 1 by adjoining to it the set : n = 2, 3, ...}. Example 2. Consider now the infinite hyper BCK-algebra (H, ◦, 0) given also by Harizavi in [3], where H = {0, 1, 2, . . .} ∪ { n1 : n = 2, 3, ...} and “ ◦ ” is defined as follows:  {0, x} if x ≤ y x◦y = {x} if x > y 1 : for all x, y ∈ H. For convenience, let N = {1, 2, ...}, N0 = N ∪ {0}, and let Jk = { m m is a positive integer and m ≥ k}. Let p ∈ H. If p = 0, then LH (p) = {0} and RH (LH (p)) = H. If p ∈ N, then LH (p) = J2 ∪ {0, 1, 2, ..., p}. Hence, RH (LH (p)) = RH (J2 ∪ {0, 1, 2, . . . , p}) = {p, p + 1, p + 2, . . .}. If p = n1 for some n ∈ {2, 3, ...}, then R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 5 1 1 , n+2 , ...}. It follows that RH (LH (p)) = RH (Jn ∪ {0}) = LH (p) = Jn ∪ {0} = {0, n1 , n+1 {x ∈ H : a ≤ x for all a ∈ (Jn ∪ {0})} = N ∪ { 12 , 31 , ..., n1 }. Now let ∅ 6= A ⊆ H. If 0 ∈ A, then LH (A) = {0} and RH (LH (A)) = H. Thus, A ∈ BRL (H) if and only if A = H. Next, suppose that 0 ∈ / A. Suppose first that A ∩ J2 6= ∅. Suppose that A∩J2 is infinite and suppose further that there exists z ∈ LH (A). Then z ≤ a for all a ∈ A. It follows that z ≤ b for all b ∈ A∩J2 , contrary to the assumption that A∩J2 is infinite. Thus, LH (A) = ∅. Next, suppose that A ∩ J2 is finite and let 1q = min(A ∩ J2 ). Then LH (A) = Jq ∪ {0}. Hence, RH (LH (A)) = N ∪ { 21 , 31 , ..., 1q }. Therefore, A ∈ BRL (H) if and only if A = N ∪ { 21 , 13 , ..., 1q }. Suppose now that A ∩ J2 = ∅. Then LH (A) = J2 ∪ {0, 1, 2, ..., r}, where r = min A. Thus, RH (LH (A)) = RH (J2 ∪ {0, 1, 2, . . . , r}) = {r, r + 1, r + 2, . . .}. Thus, A ∈ BRL (H) if and only if A = {r, r + 1, r + 2, . . .}. Therefore, τRL (H) = {∅, H} ∪ {{p, p + 1, p + 2, . . .} : p ∈ N} ∪ {N ∪ { 21 , 31 , ..., k1 } : k = 2, 3, ...}. It can be observed that the topology τRL (H) in Example 1 coincides with the topology τR (H) in Example 2.5 [7]. In fact, the next result shows that this equality is true for any hyper BCK-algebra H. Theorem 6. Let H be a hyper BCK-algebra. Then the topology τRL (H) coincides with the topology τR (H). Proof. By Theorem 5, a basis for τRL (H) is the family BRL (H) = {A : RH LH (A) = A, A ⊆ H} while a basis for τR (H) is the family BR (H) = {RH (A) : ∅ 6= A ⊆ H}. Let A ∈ BRL (H). Then RH LH (A) = A. Set D = LH (A). Then RH (D) = A ∈ BR (H), showing that BRL (H) ⊆ BR (H). Next, let U ∈ BR (H). Then there exists B ⊆ H such that RH (B) = U . This implies that for every u ∈ U , b ≪ u for all b ∈ B. Hence, B ⊆ LH (U ). By Proposition 1(iv), RH LH (U ) ⊆ RH (B) = U . Also, by Theorem 3(i), we have U ⊆ RH LH (U ). Thus, U = RH LH (U ) ∈ BRL (H), it shows that BR (H) ⊆ BRL (H). Therefore, BRL (H) = BR (H). Accordingly, τRL (H) = τR (H). 5. Topology Induced by LH RH Let H be any hyper BCK-algebra with H 6= {0}. By Proposition 2(iv), LH (H) = {0}. By definition of a closure operator, we obtain the following remark. Remark 2. LH : P(H) → P(H) is not a closure operator for every hyper BCK-algebra H 6= {0}. Theorem 7. Let A, B be subsets of a hyper BCK-algebra H. Then the following properties hold: (i) A ⊆ LH RH (A). (ii) If A ⊆ B, then LH (RH (A)) ⊆ LH (RH (B)). (iii) [LH RH ]2 (A) = LH RH [LH (RH (A))] = LH (RH (A)). R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 6 Proof. (i) Let A ⊆ H and x ∈ RH (A). Then a ≪ x for all a ∈ A. Select b ∈ A. Then b ≪ x for every x ∈ RH (A). This means that b ∈ LH (x) for every x ∈ RH (A). Hence, b ∈ LH (RH (A)). Therefore, A ⊆ LH (RH (A)). (ii) Let A, B ⊆ H such that A ⊆ B. By Proposition 1(iv), RH (B) ⊆ RH (A). Thus, by Proposition 2(ii), LH (RH (A)) ⊆ LH (RH (B)). (iii) By (i) and (ii), LH (RH (A)) ⊆ [LH RH ]2 (A). We are left to prove that [LH RH ]2 (A) ⊆ LH (RH (A)). First, we need to show that RH (A) ⊆ RH [LH (RH (A))]. Let x ∈ LH (RH (A)). Then x ≪ y for all y ∈ RH (A). Take z ∈ RH (A). It follows that x ≪ z for every x ∈ LH (RH (A)). This implies that z ∈ RH (x) for each x ∈ LH (RH (A)). Hence, z ∈ RH [LH (RH (A))]. This shows that RH (A) ⊆ RH [LH (RH (A))]. Thus, by Proposition 2(ii), LH RH [LH (RH (A))] ⊆ LH (RH (A)). By combining the two set inclusions, we have LH RH [LH (RH (A))] = LH (RH (A)). Theorem 8. Let H be a hyper BCK-algebra. The function LH RH : P(H) → P(H) is a closure operator on H. Proof. Since H is a hyper BCK-algebra, H 6= ∅. Then by the definition of a closure operator and Theorem 7, LH RH is a closure operator. Theorem 9. The family BLR (H) = {A : LH RH (A) = A, ∅ 6= A ⊆ H} is a basis for some topology on H. Proof. Since LH RH (H) ⊆ H and by Theorem 7(i), we have LH RH (H) = H and so, H ∈ BLR (H). Thus, BLR (H) 6= ∅. Next, let A, B ∈ BLR (H) and x ∈ A ∩ B. Then by Theorem 7(i), we have A ∩ B ⊆ LH RH (A ∩ B). Also, since A ∩ B ⊆ A and A ∩ B ⊆ B, LH RH (A ∩ B) ⊆ LH RH (A) and LH RH (A ∩ B) ⊆ LH RH (B) by Theorem 7(ii). Since A, B ∈ BLR (H), we have LH RH (A) = A and LH RH (B) = B. Hence, LH RH (A ∩ B) ⊆ A ∩ B. Accordingly, LH RH (A ∩ B) = A ∩ B ∈ BLR (H). This shows that BLR (H) is a basis for some topology on H. Denote by τLR (H) the topology generated by BLR (H). Example 3. Consider the hyper BCK-algebra (H, ∗, 0) in Example 1. Let n ∈ H. Then LH RH (n) = LH ({n, n + 1, n + 2, . . .}) = {0, 1, 2, . . . , n} = 6 {n}. Hence, for any m ∈ H, {m} ∈ / BLR (H). Let A be any infinite subset of H. Then LH RH (A) = LH (∅) = H 6= A. It follows that A ∈ / BLR (H). Now, suppose that A is any finite set of H such that A = {0, 1, 2, . . . , r}. Then LH RH (A) = LH ({r, r + 1, r + 2, . . .}) = {0, 1, 2, . . . , r} = A. Thus, {0, 1, 2, . . . , r} ∈ BLR (H). Now, for any finite set B 6= A, LH RH (B) 6= B. Therefore, τLR (H) = {∅, H} ∪ {{0, 1, 2, . . . , r} : r ∈ H}. Example 4. Consider the infinite hyper BCK-algebra (H, ◦, 0) given in Example 2. Again, 1 : m is a positive integer for convenience, let N = {1, 2, ...}, N0 = N ∪ {0}, and let Jk = { m R. Patangan, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 13 (1) (2020), 1-8 7 and m ≥ k}. Let p ∈ H. If p = 0, then RH (0) = {x ∈ H : 0 ≤ x} = H and LH (RH (0)) = LH (H) = {0}. It follows that {0} ∈ BLR (H). Let p ∈ N. Then RH (p) = {x ∈ H : p ≤ x} = {p, p + 1, p + 2, . . .}. It follows that LH (RH (p)) = LH ({p, p + 1, p + 2, . . .}) = J2 ∪ {0, 1, 2, . . . , p}. If p = n1 for some n ∈ {2, 3, ...}, then RH (p) = {x ∈ H : n1 ≤ x} = N ∪ 1 1 { 21 , 31 , ..., n1 }. Hence, LH (RH (p)) = LH (N ∪ { 12 , 13 , ..., n1 }) = Jn ∪ {0} = {0, n1 , n+1 , n+2 , ...}. 0 Next, let ∅ 6= A ⊆ H with A 6= {0}. Suppose first that A ∩ N 6= ∅. If A ∩ N is infinite, then RH (A) = ∅ and LH (RH (A)) = H. Suppose A ∩ N is finite and let s = max(A ∩ N). Then RH (A) = RH (s) and LH (RH (A)) = J2 ∪ {0, 1, 2, . . . , s}. Hence, A ∈ BLR (H) if and only if A = H or A = J2 ∪ {0, 1, 2, . . . , s}. Suppose now that A ∩ N = ∅. Since A 6= {0}, A ∩ J2 6= ∅. Let d = min{k ∈ {2, 3, ...} : 1 1 1 1 1 k ∈ A}. Then RH (A) = RH ( d ) = N ∪ { 2 , 3 , ..., d } and LH (RH (A)) = Jd ∪ {0} = 1 1 1 1 1 {0, d , d+1 , d+2 , ...}. Consequently, A ∈ BLR (H) if and only if A = {0, d1 , d+1 , d+2 , ...}. 1 1 1 , ...} : Therefore, τLR (H) = {∅, {0}, H}∪{J2 ∪{0, 1, 2, . . . , p} : p ∈ H}∪{{0, k , k+1 , k+2 k = 2, 3, ...}. It can be seen that the topology τLR (H) generated in Example 3 is equal to the topology τL (H) generated in Example 2.5 [8]. The next result says that the equality of these topologies holds for any hyper BCK-algebra H. Theorem 10. Let H be a hyper BCK-algebra. The the topology τLR (H) coincides with the topology τL (H). Proof. By Theorem 9, a basis for τLR (H) is given by BLR (H) = {A : LH RH (A) = A, ∅ 6= A ⊆ H} while a basis for τL (H) is the family BL (H) = {LH (A) : ∅ 6= A ⊆ H}. Let A ∈ BLR (H). Then LH RH (A) = A. Put B = RH (A) ⊆ H. Then LH (B) = A ∈ BL (H), showing that BLR (H) ⊆ BL (H). Next, let V ∈ BL (H). Then there exists D ⊆ H such that LH (D) = V . Now, for every v ∈ V , v ≪ d for all d ∈ D. Thus, D ⊆ RH (V ). By Proposition 2(ii), LH RH (V ) ⊆ LH (D) = V . By Theorem 7(i), we have V ⊆ LH RH (V ). By combining the two set inclusions, we obtain V = LH RH (V ) ∈ BLR (H). Accordingly, BLR (H) = BL (H), showing that τLR (H) = τL (H). Conclusion Right and left applications of a hyper order associated with a hyper BCKalgebra can give rise to two closure operators which , in turn, can be used to generate bases for some topologies on the given hyper structure. The two induced topologies turn out to coincide, respectively, with some previously known topologies on this hyper BCK-algebra. Acknowledgements The authors would like to thank the referees for their invaluable comments and corrections which led to the improvement of the manuscript. This research is funded by ASSCAT and MSU-Iligan Institute of Technology. REFERENCES 8 References [1] J Albaracin and J Vilela. Zero Divisor Graph of Finite Hyper BCK-algebra involving hyperatoms. Far East Journal Mathematical Sciences, 103(4):743–755, 2018. [2] S Burris and H Sankappanavar. A course in Universal Algebra. Springer-Verlag, New York., Heidelbery, Berlin, 1981. [3] H Harizavi. On Direct Sum of Branches in Hyper BCK-algebras. Iranian Journal of Mathematical Sciences and Informatics, 11(2):43–55, 2016. [4] Y Imai and K Iséki. On Axiom systems of Propositional Calculi XIV. Proc. Japan Academy, 42:19–22, 1966. [5] Y Jun M Zahedi, X Xin and R Borzooei. On Hyper BCK-algebras. Italian Journal of Pure and Applied Mathematics, 8:127–136, 2000. [6] F Marty. Sun une Generalization da la Notion de Group. Stockholm: 8th Congress Math. Scandinaves, pages 45–49, 1934. [7] R Patangan and S Canoy Jr. A Topology on a Hyper BCK-algebra. JP Journal of Algebra, Number Theory and Applications, 40:787–797, 2018. [8] R Patangan and S Canoy Jr. A Topology on a Hyper BCK-algebra via Left Application of a Hyper Order. JP Journal of Algebra, Number Theory and Applications, 40:321– 332, 2018.