International Journal of Analysis and Applications ISSN 2291-8639 Volume 11, Number 1 (2016), 54-60 http://www.etamaths.com THE ROLE OF COMPLETE PARTS IN TOPOLOGICAL POLYGROUPS M. SALEHI SHADKAMI, M.R. AHMADI ZAND AND B. DAVVAZ∗ Abstract. A topological polygroup is a polygroup P together with a topology on P such that the polygroup’s binary hyperoperation and the polygroup’s inverse function are continuous with respect to the topology. In this paper, we present some facts about complete parts in polygroups and we use these facts to obtain some new results in topological polygroups. We define the concept of cp-resolvable topological polygroups. A non-empty subset X of a topological polygroup is called cp- resolvable if there exist disjoint dense subsets A and B such that at least one of them is a complete part. Then, we investigate a few properties of cp-resolvable topological polygroups. 1. Introduction Hypergroup that is based on the notion of hyperoperation was introduced by Marty in [13] and studied extensively by many mathematicians. Applications of hypergroups have mainly appeared in special subclasses. For example, polygroups that form an important subclass of hypergroups are studied by Comer [3, 4]. Polygroups have been applied in many area, such as geometry, lattices, combinatorics and color scheme. There exists a rich bibliography on polygroups [6]. This book contains the principal definitions endowed with examples and the basic results of the theory. Aghabozorgi et al. [1] defined perfect and solvable polygroups. Also see [11]. Till now, only a few papers treated the notion of topological hyperstructures, for example see [2, 8, 10]. Heidari et al. [9] defined the notion of topological polygroups. By considering the relative topology on subpolygroups they proved some properties of them. In particular, they proved the topological isomorphism theorems of topological polygroups. The purpose of this paper is as stated in the abstract. 2. Basic definitions and results Let H be a non-empty set and P∗(H) be the set of all non-empty subsets of H. Then, the mapping ◦ : H × H → P∗(H) is called a hyperoperation and (H,◦) is called a hypergroupoid. If A and B are two non-empty subsets of H and x ∈ H, then we define A◦B = ⋃ a∈A b∈B a◦ b, A◦x = A◦{x} and x◦A = {x}◦A. If a◦ (b◦c) = (a◦b) ◦c for all a,b,c ∈ H (associativity axiom), then (H,◦) is called a semihypergroup and it is called a quasihypergroup if for every x ∈ H, we have x◦H = H = H◦x (reproduction axiom). The couple (H,◦) is called a hypergroup if it is a semihypergroup and a quasihypergroup. Thus, a hy- pergroup is a generalization of a group. Let (H,◦) be a semihypergroup and A be a non-empty subset of H. We say that A is a complete part of H if for any natural number n and for all a1,a2, . . . ,an ∈ H , the following implication holds: A∩ n∏ i=1 ai 6= ∅⇒ n∏ i=1 ai ⊆ A. The complete parts are introduced for the first time by Koskas[12] and are studied by Corsini [5] and many others. A special subclass of hypergroups is the class of polygroups. We recall the following 2010 Mathematics Subject Classification. 20N20, 22A30. Key words and phrases. hyperstructure; polygroup; topological polygroup; complete part; cp-resolvable polygroup; natural mapping. c©2016 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 54 THE ROLE OF COMPLETE PARTS IN TOPOLOGICAL POLYGROUPS 55 definition from [3]. A polygroup is a system P =< P,◦,e,−1 >, where ◦ : P ×P → P∗(P), e ∈ P, −1 is a unitary operation on P and the following axioms hold for all x,y,z ∈ P : (1) (x◦y) ◦z = x◦ (y ◦z); (2) e◦x = x◦e = x; (3) x ∈ y ◦z implies y ∈ x◦z−1 and z ∈ y−1 ◦x. The following elementary facts about polygroups follow easily from the axioms: e ∈ x◦x−1 ∩x−1 ◦x , e−1 = e , (x−1)−1 = x, and (x ◦ y)−1 = y−1 ◦ x−1. A non-empty subset K of a polygroup P is a subpolygroup of P if and only if a,b ∈ K implies a ◦ b ⊆ K and a ∈ K implies a−1 ∈ K. The subpolygroup N of P is normal in P if and only if a−1 ◦N ◦a ⊆ N for every a ∈ P . Proposition 2.1. Let A and B be non-empty subsets of a polygroup P =< P,◦,e,−1 > such that A is a complete part and x ∈ P . Then, (1) x−1 ◦x◦A = x◦x−1 ◦A = A; (2) A−1 is a complete part; (3) x◦A and A◦x are complete parts; (4) B ⊆ x−1 ◦A if and only if x◦B ⊆ A. Remark 1. Suppose that A is a complete part of a polygroup P , n,m ∈ N and (x1,x2, . . . ,xn) ∈ Pn, (y1,y2, . . . ,ym) ∈ Pm. By Proposition 2.1, n∏ i=1 xi ◦A◦ m∏ j=1 yj and n∏ i=1 xi ◦A−1 ◦ m∏ j=1 yi are complete parts. Let {As}s∈S be a family of complete parts of P. Then, we observe that ⋃ s∈S As is a complete part and if I = ⋂ s∈S As 6= ∅, then I is a complete part. Proposition 2.2. If A and B are two non-empty subsets of a polygroup P such that A is a complete part, then A◦B and B ◦A are complete parts. Proof. Suppose that n ∈ N and n∏ i=1 ai∩(A◦B) 6= ∅ where (a1,a2, . . . ,an) ∈ Pn. Then, there exists b ∈ B such that n∏ i=1 ai∩(A◦b) 6= ∅. By Proposition 2.1, A◦b is a complete part and so n∏ i=1 ai ⊆ A◦b ⊆ A◦B, i.e., A◦B is a complete part. Similarly, B ◦A is a complete part. � We observe that if A is a complete part and Bi,Cj are non-empty subsets of P for i = 1, 2, ...,n, j = 1, ...,m, where n,m ∈ N, then C1 ◦ ...◦Cm ◦A◦B1 ◦B2...◦Bn is a complete part. Proposition 2.3. Let P be a polygroup and Bα be a complete part for every α ∈ I. Then, a◦ ⋂ α∈I Bα =⋂ α∈I (a◦Bα) and ( ⋂ α∈I Bα) ◦a = ⋂ α∈I (Bα ◦a). Proof. Clearly, a◦ ⋂ α∈I Bα ⊆ ⋂ α∈I a◦Bα. Conversely, suppose that t ∈ ⋂ α∈I a◦Bα. So, for every α ∈ I, t ∈ a◦Bα. Now, if s ∈ I, then t ∈ a◦Bs. Hence, there exists bs ∈ Bs such that t ∈ a◦ bs. Thus, for every α ∈ I, t ∈ a◦bs∩a◦Bα. So, for every α ∈ I , a◦bs ⊆ a◦Bα. Therefore, by Proposition 2.1, for every α ∈ I , bs ∈ a−1 ◦a◦ bs ⊆ a−1 ◦a◦Bα = Bα, i.e., bs ∈ ⋂ α∈I Bα. Thus, t ∈ a◦ bs ⊆ a◦ ( ⋂ α∈I Bα), and so a◦ ⋂ α∈I Bα = ⋂ α∈I (a◦Bα). Similarly the equality ( ⋂ α∈I Bα) ◦a = ⋂ α∈I (Bα ◦a) holds. � Remark 2. Being complete part is necessary in Proposition 2.3 as it is illustrated in the following example. Example 1. Consider the set of integer numbers Z and define the hyperoperation ◦ on it as follows: for every m ∈ Z, m◦ 0 = m and if m,n ∈ Z\{0}, then m◦n = { E, m + n ∈ E Ec, m + n ∈ Ec where E = 2Z. If A = {1, 2} and B = {1, 4}, then A and B are not complete parts and 1◦A = 1◦B = Z. But the equality 1 ◦ (A∩B) = 2Z holds. 56 SHADKAMI, ZAND AND DAVVAZ The following result is a direct consequence of Proposition 2.3. Corollary 2.4. Let A and C be non-empty subsets of a polygroup P . If Bα is a complete part for every α ∈ I, then A◦ ( ⋂ α∈I Bα) ◦C = ⋂ α∈I (A◦Bα ◦C). 3. On topological polygroups and cp-resolvable topological polygroups By using complete parts in topological polygroups, some interesting results were obtained by a few authors, in the classical case, see [2, 8, 9, 10, 14]. In this section, we study subpolygroups and dense subsets of a topological polygroup. Let (H,τ) be a topological space. Then, the family B = {SV | V ∈ τ}, where SV = {U ∈ P∗(H) | U ⊆ V} is a base for a topology on P∗(H). This topology is denoted by τ∗ [10]. Let (H,τ) be a topological space. We consider the product topology on H × H and the topology τ∗ on P∗(H). Let H be a topological space and A ⊂ Y ⊂ H, intY A denotes the interior of A in the subspace Y , and the closure of A in the subspace Y is denoted by clY A; the interior of A in H is denoted by A ◦, and the closure of A in H is denoted by A. Definition 3.1. [9] Let P =< P,◦,e,−1 > be a polygroup and (P,τ) be a topological space. Then, the system P = (P,◦,e,−1 ,τ) is called a topological polygroup if the mappings µ : P × P → P∗(P) and ι : P → P defined by µ(x,y) = x◦y and ι(x) = x−1 are continuous. Let U be an open subset of a topological polygroup P such that U is a complete part. Then, a ◦ U and U ◦ a are open subsets of P for every a ∈ P [9]. Let P be a topological polygroup such that every open subset of P is a complete part. Let U be an open base at e. Then, the families {x◦U | x ∈ P ,U ∈U} and {U ◦x | x ∈ P ,U ∈U} are open bases for P [9]. Theorem 3.2. [9] Let P = (P,◦,e,−1 ,τ) be a topological polygroup and U be a base at e. Then, the following assertions hold. (1) For every U ∈U and x ∈ U there exists V ∈U such that x◦V ⊆ U. (2) For every U ∈U there exists V ∈U such that V ◦V ⊆ U. (3) For every U ∈U there exists V ∈U such that V −1 ⊆ U. Let P = (P,◦,e,−1 ,τ) be a topological polygroup. We denote by νP (e) the set of all neighborhoods of e. Theorem 3.3. [9] Let P be a topological polygroup such that every open subset of P is a complete part. Then, for every U ∈ νP (e) there exists V ∈ νP (e) such that V ⊆ U. Let every open subset of a topological polygroup P be a complete part. If F is a compact subset of P , then for every a ∈ P , a◦F and F ◦a are compact [14]. Theorem 3.4. [14] Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part, F be a compact subset of P and G be a closed subset of P . Then, the sets F ◦G and G◦F are closed in P . Lemma 3.5. [9] If H is a subpolygroup of a topological polygroup P = (P,◦,e,−1 ,τ) and every open subset of P is a complete part, then H is a subpolygroup of P . Theorem 3.6. [9] Let P be topological polygroup such that every open subset of P is a complete part. Then, a subpolygroup K of P is open if and only if its interior is non-empty. Theorem 3.7. [9] Let P be a topological polygroup such that every open subset of P is a complete part. Then, every open subpolygroup is closed. Theorem 3.8. [14] Let H be a non-empty subset of a topological polygroup P = (P,◦,e,−1 ,τ) and every open subset of P is a complete part. Then, we have (1) H = ⋂ U∈ νP (e) U ◦H = ⋂ U∈νP (e) H ◦U = ⋂ U∈νP (e),V∈νP (e) U ◦H ◦V ; (2) If H is a normal subpolygroup, then H is a normal subpolygroup. THE ROLE OF COMPLETE PARTS IN TOPOLOGICAL POLYGROUPS 57 Lemma 3.9. If P is a topological polygroup such that every open subset of P is a complete part, then a◦A = a◦A. Proof. By Theorem 3.8 and Proposition 2.3, a◦A = ⋂ U∈νP (e) (a ◦ A) ◦ U = ⋂ U∈νP (e) a ◦ (A ◦ U) = a◦ ⋂ U∈νP (e) (A◦U) = a◦A. � Remark 3. Begin complete part is necessary in Lemma 3.9 as it is illustrated in the following example. Example 2. Suppose that the multiplication table for a polygroup P =< P,◦, 1,−1 >, where P = {1, 2}, 1−1 = 1 and 2−1 = 2 is ◦ 1 2 1 {1} {2} 2 {2} {1, 2} If τ = {∅,{1},{1, 2}}, then P = (P,◦,−1 ,τ) is a topological polygroup. If A = {1}, then A = {1, 2} , 2 ◦A = {2} and 2 ◦A = {2}. But the equality 2 ◦A = {1, 2} holds. Let X be a topological space and x ∈ X. Recall that a family β of open subsets of X is called a base for X at the point x if all elements of β contain x and, for every neighborhood O of x, there exists U ∈ β such that x ∈ U ⊆ O, and that the character of X at the point x is denoted by χ(x,X) where χ(x,X) = min{ |β| | β is a base for X at the point x} and that if X has a countable base at each point x ∈ X, then X is called first-countable [7]. Proposition 3.10. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and Y is a dense subspace of P . Then, χ(y,Y ) = χ(y,P) for each y ∈ Y . Proof. Suppose that β is a base for P at a point y ∈ Y such that χ(y,P) = |β|. Then, the family βY = {U ∩Y | U ∈ β } is a base for Y at y. Thus, χ(y,Y ) = min{ |β1| | β1 is a base for Y at y}≤ |βY | ≤ |β| = χ(y,P). Conversely, let βY be a base for Y at a point y such that χ(y,Y ) = |βY |. For every U ∈ βY , choose an open set VU in P such that U = VU ∩Y . We claim that the family β = { VU| U ∈ βY} is a base for P at the point y. Indeed, we have (1) For every VU ∈ β, y ∈ VU . (2) Let O be an open subset of P containing y. Then, by Theorem 3.2, there exists W1 ∈ νP (e) such that y◦W1 ⊆ O. So, by Theorem 3.3, there exists W ∈ νP (e) such that W ⊆ W1. Hence, y ◦W ⊆ y ◦W1 ⊆ O. Since W2 = y ◦W ∩Y is an open subset of Y and y ∈ W2, there exists U ∈ βY such that U ⊆ W2. Since U = VU ∩Y and Y = P , it follows that VU = U ⊆ y ◦W = y ◦W ⊆ O by Lemma 3.9. Thus, y ∈ VU ⊆ O and so χ(y,P) = min{ |β1| | β1 is a base for P at y}≤ |β| ≤ |βY | = χ(y,Y ). � Lemma 3.11. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and let H be a subpolygroup of P . Then, every open subset of H is a complete part. Proof. Suppose that U ∈P∗(H) is an open subest of H and for n ∈ N, n∏ i=1 ai ∩U 6= ∅, where ai ∈ H. Thus, there exists an open subset V of P such that U = V ∩H. Hence, n∏ i=1 ai∩V 6= ∅. So, n∏ i=1 ai ⊆ V . On the other hand n∏ i=1 ai ⊆ H and hence, n∏ i=1 ai ⊆ V ∩H = U. � Lemma 3.12. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part. If Y is a first-countable subpolygroup of P , then Y is also first-countable. Proof. Suppose that K = Y . Then, by Lemma 3.5, K is a subpolygroup of P and so by Proposition 3.10, χ(e,K) = χ(e,Y ) ≤ℵ0. � Definition 3.13. Let P = (P,◦,e,−1 ,τ) be a topological polygroup and U ∈ νP (e). A subset A of P is called U-disjoint, if b /∈ a◦U, for any distinct a,b ∈ A. 58 SHADKAMI, ZAND AND DAVVAZ Example 3. Suppose that the multiplication table for a topological polygroup P = (P,τ,◦, 1,−1 ) where P = {e,a,b}, e−1 = e, a−1 = a and b−1 = b is ◦ e a b e {e} {a} {b} a {a} {e} {b} b {b} {b} {e,a} and τ = {∅,{e,a},{e,a,b}}. Let A = {a,b} , U = {e,a}. Then, a ◦ U = {a,e} , b ◦ U = {b} and b /∈ a◦U , a /∈ b◦U. Thus, A is U-disjoint but A∩U 6= ∅. We recall that a family {As}s∈S of subsets of a topological space X is called discrete if every point x ∈ X has a neighborhood that intersects at most one set of the given family [7]. Lemma 3.14. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and U,V ∈ νP (e) such that V 4 ⊆ U and V −1 = V . If a subset A of P is U-disjoint, then the family of open sets {a◦V | a ∈ A} is discrete in P . Proof. Suppose that x ∈ P , we claim that x◦V intersects at most one element of the family {a◦V | a ∈ A}. Suppose to the contrary that, there exist distinct elements a,b ∈ A such that x◦V ∩a◦V 6= ∅ and x◦V ∩b◦V 6= ∅. Let t ∈ x◦V ∩a◦V , hence there exists v1,v2 ∈ V such that t ∈ x◦v1 , t ∈ a◦v2. Thus, x ∈ t◦v−11 , a ∈ t◦v −1 2 that is x −1◦a ⊆ v1◦t−1◦t◦v−12 ⊆ V ◦t −1◦t◦V −1 = V ◦V since V is a complete part. Similarly, we can prove that b−1 ◦x ⊆ V ◦V . So, b−1 ◦a ⊆ b−1 ◦x◦x−1 ◦a ⊆ V 4 ⊆ U. Then, a ∈ b◦ b−1 ◦a ⊆ b◦U which is a contradiction. � Let (P,◦,e,−1 ,τ) be a topological polygroup and N be a normal subpolygroup of P . Let π : P → P/N be the natural mapping defined by π(x) = N ◦ x. Recall that (P/N,τ) is a topological space, where τ is the quotient topology induced by π, that is τ = {U ⊆ P/N | π−1(U) ⊆ P is open}. Lemma 3.15. [9] Let P = (P,◦,e,−1 ,τ) be a topological polygroup and N be a normal subpolygroup of P . Let π : P → P/N be the natural mapping. Then, (1) π−1({N ◦x | x ∈ X}) = N ◦X for every nonempty subset X of P and π−1(π(X)) = N ◦X; (2) {N ◦x | x ∈ X} = {N ◦y | y ∈ N ◦X} for every nonempty subset X of P ; (3) If every open subset of P is a complete part, then the natural mapping π is open. Theorem 3.16. [9] Let N be a normal subpolygroup of a topological polygroup P = (P,◦,e,−1 ,τ) and every open subset of P is a complete part. Then, < P/N,�,N,−I ,τ > is a topological polygroup, where N ◦x�N ◦y = {N ◦z|z ∈ x◦y} and (N ◦x)−I = N ◦x−1. Theorem 3.17. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and N is a normal subpolygroup of P . Let π : P → P/N be the natural mapping. Then, the family {π(U ◦x) | U ∈ νP (e)} is a local base of the space P/N at the point N ◦x ∈ P/N. Proof. Since U ∈ νP (e) is a complete part, it follows that U ◦ x is an open subset of P. Then, for every y ∈ N, y ◦ (U ◦x) is open. So, π−1(π(U ◦x)) = N ◦ (U ◦x) = ⋃ y∈N y ◦ (U ◦x) is an open subset of P . Thus, π(U ◦ x) is an open subset of P/N by Lemma 3.15. Now, suppose that W is an open neighborhood of N ◦x in P/N and O = π−1(W). Clearly O ⊆ P is open. Since N ◦x ⊆ W , it follows that x ∈ π−1(N ◦ x) ⊆ π−1(W) = O. So, there exists U ∈ νP (e) such that U ◦ x ⊆ O. Therefore, N ◦x ∈ π(U ◦x) ⊆ π(O) = W and this completes the proof. � Lemma 3.18. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part. Let N be a normal subpolygroup of P and π : P → P/N be the natural mapping. If U ,V ∈ νP (e) and V ◦V −1 ⊆ U, then π(V ) ⊆ π(U). Proof. Suppose that π(x) ∈ π(V ), where x ∈ P . We prove that π(x) ∈ π(U). Since V is a complete part, it follows that x◦V ⊆ P is open. By Lemma 3.15, since π is open, it follows that π(x◦V ) is open in P/N. Hence, π(x◦V ) ∩π(V ) 6= ∅. Thus, there exists a y ∈ P such that π(y) ∈ π(x◦V ) ∩π(V ). Therefore, there exist t ∈ x◦V and z ∈ V such that π(y) = π(t) = N ◦ t and π(y) = π(z) = N ◦ z. Since t ∈ x◦V , there exists b ∈ V such that t ∈ x◦ b. So, x ∈ t◦ b−1 and x ∈ N ◦x ⊆ N ◦ t◦ b−1 = N ◦z◦b−1 ⊆ N ◦V ◦V −1 ⊆ N ◦U. Then, π(x) ∈ π(N ◦U) = π(U) and this completes the proof. � THE ROLE OF COMPLETE PARTS IN TOPOLOGICAL POLYGROUPS 59 Theorem 3.19. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part , N is a normal subpolygroup of P and W ∈ νP/N (N). Then, there exists V ∈ νP/N (N) such that V ⊆ W . Proof. Suppose that π : P → P/N is the natural mapping. Since π is continuous, it follows that U = π−1(W) ⊆ P is open and e ∈ U. Then, U ∈ νP (e). Thus, by Theorem 3.2, there exists V1 ∈ νP (e) such that V1 ◦V −11 ⊆ U. Hence, by Lemma 3.18, π(V1) ⊆ π(U) = W . Since π is open and V1 ∈ νP (e), V = π(V1) ∈ νP/N (N) and this completes the proof. � Recall that a continuous mapping f : X → Y is perfect if f is a closed mapping and all fibers f−1(y) are compact subsets of X. Theorem 3.20. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and let N be a compact normal subpolygroup of P . Then, the natural mapping π : P → P/N is perfect. Proof. Suppose that H ⊆ P is a closed subset. Then, by Theorem 3.4, N ◦ H is closed in P. Since π−1(π(H)) = N ◦H, so π−1(π(H)) ⊆ P is closed. Then, π(H) is closed in P/N. Hence, π is closed. Now, suppose that y ∈ P/N. Then, there exists x ∈ P such that π(x) = y. Since N is compact, it follows that π−1(y) = π−1(π(x)) = N ◦x is compact. Therefore, π−1(y) is compact. � Lemma 3.21. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and N is a normal subpolygroup of P . (1) If P is compact, then P/N is compact. (2) If N and P/N are compact, then P is compact. Proof. (1) Since π is continuous and P is compact, it follows that π(P) = P/N is compact. (2) Since N is compact, by Theorem 3.20, the natural mapping π : P → P/N is perfect. Then, π−1(P/N) = P is compact [[7], Theorem 3.7.2]. � Definition 3.22. Let X be a non-empty subset of a topological polygroup P = (P,◦,e,−1 ,τ). Then, X is called cp-resolvable if there exist two disjoint dense subsets of X such that at least one of them is a complete part of P . Lemma 3.23. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and let H be a cp-resolvable subpolygroup of P . If x ∈ P , then x◦H is cp-resolvable. Proof. Suppose that A and B are disjoint dense subspaces of H such that one of them is a complete part of P . Then, by Lemma 3.9, clx◦H(x ◦ A) = x◦A ∩ x ◦ H = x ◦ A ∩ x ◦ H = x ◦ H since H = clH(A) = A∩H. Similarly, clx◦H(x◦B) = x◦H. Without lose of generality we can assume that A is a complete part of P . Thus, by Proposition 2.1, x ◦ A is a complete part and we observe that x◦A∩x◦B = ∅. Thus, x◦H is cp-resolvable. � We recall that if N is a normal subpolygroup of P , then the relation x,y ∈ P, xNPy if and only if x−1 ◦y ∩N 6= ∅ is an equivalence relation. The equivalence class of the element x ∈ P is denoted by NP (x) [6]. Lemma 3.24. [6] If N is a normal subpolygroup of P , then x◦N = NP (x). Theorem 3.25. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part and let {Pα|α ∈ I} be a pairwise disjoint family of non-empty subsets of P such that every Pα is cp-resolvable. If P = ⋃ α∈I Pα, then P is cp-resolvable. Proof. Since each Pα is cp-resolvable, there exist disjoint dense subsets Aα and Bα of Pα such that Aα is a complete part of P and so Aα ∩Pα ∩ Pα = clPαAα = Pα. If A = ⋃ α∈I Aα and B = ⋃ α∈I Bα, then A is a complete part by Remark 1 and A and B are disjoint dense subsets of P . Thus, P is cp-resolvable. � Theorem 3.26. Let P = (P,◦,e,−1 ,τ) be a topological polygroup such that every open subset of P is a complete part. Then, the following conditions hold. 60 SHADKAMI, ZAND AND DAVVAZ (1) If H is a proper dense subpolygroup of P and H is a complete part, then P is cp-resolvable. (2) If a normal subpolygruop N of P is cp-resolvable, then P is cp-resolvable. (3) If P contains a non-closed normal subpolygroup N which is a complete part, then P is cp- resolvable. Proof. (1) Suppose that A = H and B = P \ H. Since the subpolygroup A is not closed, then by Theorem 3.7, A is not open. Thus, int(A) = ∅. Therefore, A = B = P . (2) By Lemma 3.24, P is equal to the disjoint unions ( ⋃ x/∈N x◦N) ∪N. Thus, P is cp-resolvable by Lemma 3.23 and Theorem 3.25. (3) By Theorem 3.8, B = N is a normal subpolygroup of P , and so by (1); B is cp-resolvable. Therefore, by (2), P is cp-resolvable. � References [1] H. Aghabozorgi, B. Davvaz, M. Jafarpour, Solvable polygroups and derived subpolygroups, Comm. 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