α (β,β) -Topological Abelian Groups

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1 Global Journal of Pure and Applied Mathematics. ISSN Volume 13, Number 6 (2017), pp Research India Publications α (β,β) -Topological Abelian Groups Alias B. Khalaf Department of Mathematics, College of Science, University of Duhok, Kurdistan-Region, Iraq. Hariwan Z. Ibrahim Department of Mathematics, Faculty of Science, University of Zakho, Kurdistan-Region, Iraq. Abstract The purpose of this paper is to introduce and study the concept of α (β,β) -topological abelian group. Also the subgroups in α (β,β) -topological group are studied and several basic theorems are introduced. The factor group space of α (β,β) -topological groups is defined and some basic results are presented. AMS subject classification: Primary: 22A05, 22A10; Secondary: 54C05. Keywords: Operations, α β -open set, abelian groups, α (β,β) -topological abelian group. 1. Introduction Topological groups are objects that combine two separate structures-the structure of a topological space and the algebraic structure of a group-linked by the requirement that the group operations are continuous with respect to the underlying topology. Topological groups have the algebraic structure of a group and the topological structure of a topological space and they are linked by the requirement that multiplication and inversion are continuous functions. In the literature the most papers are related to the case of topological groups where group operations are continuous mappings. However, there are also many nice papers and interesting, deep results on other classes of topologized

2 2292 Alias B. Khalaf and Hariwan Z. Ibrahim groups. In all these cases continuity play the crucial role. We refer the reader to the monograph [1] and the survey paper [10] where detail information about mentioned classes of topologized groups can be found, as well as references related to them. In 1965, Njastad [9] initiated and explored a new class of generalized open sets in a topological space called α-open sets and proved that the collection of all α-open sets in (X, τ) is a topology on X, finer that τ. H. Z. Ibrahim [2] introduced and discussed an operation of a topology αo(g) into the power set P (G) of a space G and also he introduced the concept of α β -open sets and investigated the related topological properties of the associated topology αo(g, τ) β and αo(g, τ) by using operation β. These α β - open sets were used to define four new separation axioms called α β T 0, α β T 1 and α β T Preliminaries Through out this paper, we mean by a triple (G, +,τ) an abelian group G equipped with the topology τ. Let A be a subset of a topological space (G, τ). We denote the interior and the closure of a set A by Int(A) and Cl(A) respectively. A subset A of a topological space (G, τ) is called α-open [9] if A Int(Cl(Int(A))). ByαO(G, τ), we denote the family of all α-open sets of G. An operation β : αo(g, τ) P (G) [2] is a mapping satisfying the condition, V V β for each V αo(g, τ). We call the mapping β an operation on αo(g, τ). A subset A of G is called an α β -open set [2] if for each point x A, there exists an α-open set U of G containing x such that U β A. The complement of an α β -open set is said to be α β -closed. We denote the set of all α β -open sets of (G, τ) by αo(g, τ) β. The α β -closure [2] of a subset A of G with an operation β on αo(g) is denoted by α β Cl(A) and is defined to be the intersection of all α β -closed sets containing A. An operation β on αo(g, τ) is said to be α-regular if for every α-open sets U and V of each x G, there exists an α-open set W of x such that W β U β V β. Definition 2.1. [5] An operation ρ : αo(g G) P(G G) is said to be α-associated with β and β,if(u V) ρ = U β V β holds for each sets U,V αo(g). Definition 2.2. [4] Let (G, τ) be a topological space and x G, then a subset N of G is said to be α β -neighbourhood of x, if there exists an α β -open set U in G such that x U N. Definition 2.3. [6] Two subsets A and B of a topological space (G, τ) are called α β - separated if (α β Cl(A) B) (A α β Cl(B)) = φ. Definition 2.4. [6] A subset C of a space G is said to be α β -disconnected if there are nonempty α β -separated subsets A and B of G such that C = A B, otherwise C is called α β -connected. Definition 2.5. [6] A set C is called maximal α β -connected set if it is α β -connected and if C D G where D is α β -connected, then C = D. A maximal α β -connected

3 α (β,β) -Topological Abelian Groups 2293 subset C of a space G is called an α β -component of G. Definition 2.6. [2] A topological space (G, τ) with an operation β on αo(g) is said to be: 1. α β T 0 if for any two distinct points x,y X, there exists an α β -open set U such that either x U and y/ U or y U and x/ U. 2. α β T 1 if for any two distinct points x,y X, there exist two α β -open sets U and V containing x and y, respectively, such that y/ U and x/ V. 3. α β T 2 if for any two distinct points x,y X, there exist two α γ -open sets U and V containing x and y, respectively, such that U V = φ. Definition 2.7. [5] A function f : (G, τ) (G,τ is said to be α (β,β -open if for any α β -open set A of (G, τ), f (A) is α β -open in (G,τ. Definition 2.8. [2] A mapping f : (G, τ) (G,τ is said to be α (β,β -continuous if for each x of G and each α β -open set V containing f(x), there exists an α β -open set U such that x U and f(u) V. Definition 2.9. [2] A mapping f : (G, τ) (G, τ) is said to be α (β,β) -homeomorphism, if f is bijective, α (β,β) -continuous and f 1 is α (β,β) -continuous. Corollary [6] A function f : G G is α (β,β -continuous if and only if f 1 (V ) is α β -open in G, for every α β -open set V in G. We recall some of the well known definitions and results which can be found in most of text books of abstract algebra we refer to [3] and [8]. Definition A group G is an algebraic structure consisting of a non-empty set equipped with an operation on its elements that satisfies four conditions, namely closure, associativity, identity and invertibility. Moreover, if the operation is abelian then G is called an abelian group. Definition Let G and G be abelian groups, then a mapping f : G G is called a homomorphism if f(a+ b) = f(a)+ f(b)for all a,b G. A homomorphism, being also: a bijection, is called an isomorphism. Let f : G G and g : G G be homomorphisms of abelian groups, then g f : G G is a homomorphism of abelian groups. The kernel of the homomorphism f is denoted by ker(f ) and ker(f ) ={a G : f(a)= 0}. Definition Let G be an abelian group and B G. Then B is called a subgroup, if B is a group with respect to the existing operations.

4 2294 Alias B. Khalaf and Hariwan Z. Ibrahim A subset C of an abelian group G is called symmetric if C = C. Let D be a subset of G, then D ( D) and D ( D) are symmetric subsets of G. D = n(d ( D)) = ((D ( D)) + +(D ( D))) }{{} n=1 n=1 n times is the smallest subgroup in G containing D (it is called a subgroup generated by the subset D). Definition Let B,C be subgroups of an abelian group G, then G/B is denoted to be the factor group of the group G by B, that is, the set {a + B : a A} of cosets with the following operation of addition: (a + B) + (c + B) = a + c + B. The mapping π : G G/B is denotes to be the natural homomorphism: π(a) = a + B for all a G. It is clear that π 1 (π(s)) = S + B for any subset S G. In particular, if S = S + B, then π 1 (π(s)) = S. Theorem Let f : A B and A i A for i I. If f 1 (f (A i )) = A i for all i I except, possibly, one i 0 I, then ( ) f A i = f(a i ). i I i I Theorem Let (G, +) be a group and τ be a topology on G. Then, (G, +,τ)is a topological group if and only if for any elements a,b of G and open set U with a b U, there exist open sets V and W containing a and b respectively such that V W U. Recall the following definitions from [7]. Definition Let (G, ) be a group and τ be a topology on G. The inversion map is β-α-continuous if given a G and O αo(g, τ) such that a 1 O, then there is U αo(g, τ) with a U and (U β ) 1 O β, where (U β ) 1 ={x 1 : x U β }. Definition Let (G, ) be a group and τ be a topology on G. The multiplication is jointly β-α-continuous in both variables if given a,b G and O αo(g, τ) such that a b O, then there exist U,V αo(g, τ) with a U, b V and U β V β O β. Definition Let (G, ) be a group and (G, τ) be a topological space. A triple (G,,τ) is called a β-α-topological group if the inversion map is β-α-continuous and the multiplication map is jointly β-α-continuous in both variables. 3. α (β,β) -Topological Group Definition 3.1. Let (G, +) be abelian group and τ be a topology on G. A triple (G, +,τ) is said to be an α (β,β) -topological group if the following conditions are satisfied:

5 α (β,β) -Topological Abelian Groups For any two elements a,b G and U αo(g, τ) β such that a + b U, there exist V,W αo(g, τ) β with a V, b W and V + W U. 2. For any element a G and U αo(g, τ) β such that a U, there exists V αo(g, τ) β with a V and V U. The following example shows that the α (β,β) -topological group need not be a β-αtopological group. Example 3.2. Let (Z 4, + 4 ) be a group and τ be the discrete topology on Z 4. For each A αo(z 4,τ), we define β on αo(z 4,τ)by { A β {1, 2, 3} if A ={2}, = Z 4 if A = {2}. Then, (Z 4, + 4,τ)is an α (β,β) -topological group, but it is not a β-α-topological group. Theorem 3.3. Let (G, +) be a group and τ be a topology on G. Then, (G, +,τ)is an α (β,β) -topological group if and only if for any elements a,b of G and α β -open set U with a b U, there exist α β -open sets V and W containing a and b respectively such that V W U. Proof. Let a,b G, hence a b G. Let U be an α β -open set containing a b, then there are α β -open sets V containing a and R containing b such that V + R U, and also there is an α β -open set W containing b such that W R. Hence, V W U. Conversely, let a G and U be an α β -open set containing a. Then, U contains the element 0 a and hence there exist α β -open sets V containing 0 and W containing a such that V W U and W = 0 W V W U. Now let a,b G and U αo(g, τ) β such that a + b U. By hypothesis there exist α β -open sets U and R containing a and b respectively such that V R U and there is an α β -open set W containing b such that W R. Now,V + W = V + ( ( W)) V + ( R) U. Therefore, (G, +,τ)is an α (β,β) -topological group. Proposition 3.4. Let G be an α (β,β) -topological abelian group, a G, B and C be subsets of G. Then, the following statements are true: 1. The mappings f : G G and f a : G G, where f(x) = x and f a (x) = x + a, are both α (β,β) -homeomorphisms from the topological space G onto itself. 2. The following conditions are equivalent: (a) B is α β -open (α β -closed). (b) B is α β -open (α β -closed). (c) B + a is α β -open (α β -closed). 3. If the subset B is α β -open, then B + C is also an α β -open.

6 2296 Alias B. Khalaf and Hariwan Z. Ibrahim Proof. 1. It is clear that the mappings f and f a are bijective. Let U be an α β -open set in G containing f(x) = x. Then, by G is an α (β,β) - topological abelian group, there is V αo(g, τ) β with x V and f(v) = V U, then f is α (β,β) -continuous, so if U is α β -open, then f 1 (U) = U is α β -open. Let U be an α β -open set in G containing f a (x) = x + a. Then, by G is an α (β,β) -topological abelian group, there is V,W αo(g, τ) β with x V, a W and f a (V ) = V + a V + W U, then f a is α (β,β) -continuous. As for the inverse mappings, their α (β,β) -continuity is obvious by virtue of f 1 = f and f 1 a = f a. 2. (a) (b) is obvious in view of the fact that B = f(b). Since B + a = f a (f ( B)), (b) (c) is true. Finally, (c) (a) follows from the fact that B = (f a ) 1 (B + a). 3. Since B + C = c C(B + c), it follows that B + C is a union of α β -open subsets and hence, it is α β -open. Corollary 3.5. Let G be an α (β,β) -topological abelian group and a G. A subset U G is an α β -neighborhood of the element a if and only if U a is an α β -neighborhood of 0. Proof. The proof follows from Proposition 3.4. Proposition 3.6. Let B and C be subsets of an α (β,β) -topological abelian group G. Then the following statements are true: 1. α β Cl(B + C) α β Cl(B) + α β Cl(C). 2. α β Cl( B) = α β Cl(B). 3. α β Cl(B C) α β Cl(B) α β Cl(C). Proof. 1. Let x α β Cl(B)+α β Cl(C) and let U be an α β -open set containing the element x. Then, x = b+c, where b α β Cl(B) and c α β Cl(C), and, hence, there exist α β - open sets V and W in G of the elements b and c respectively, such that V +W U. By virtue of the fact that V B = φ and W C = φ, elements b 1 V B and c 1 W C can be found. Thus, b 1 + c 1 B + C and b 1 + c 1 V + W U, that is (B + C) U = φ. Consequently, α β Cl(B + C) α β Cl(B) + α β Cl(C).

7 α (β,β) -Topological Abelian Groups The validity of α β Cl( B) = α β Cl(B) results from the fact that the mapping x x is an α (β,β) -homeomorphism of the topological space G onto itself (see Proposition 3.4 (1)). 3. Inclusion α β Cl(B C) α β Cl(B) α β Cl(C) results from items (1) and (2). Proposition 3.7. Let G be an α (β,β) -topological abelian group and ρ : αo(g G) P(G G) be an α-associated operation with β and β. Then, the mapping m of topological space G G onto a topological space G, where m(a, b) = a + b for all a,b G, is α (ρ,β) -continuous. Proof. Let U be an α β -open set in G containing m(x, y) = x + y. Since G is α (β,β) - topological abelian group, then there are α β -open sets V containing x and W containing y such that V + W U, and also there are α-open sets V 1 containing x and W 1 containing y such that V β 1 V, W β 1 W and (V 1 W 1 ) ρ = V β 1 W β 1 V W, implies that m(v W) = V + W U and (x, y) V W αo(g G) ρ, hence m is α (ρ,β) -continuous. Definition 3.8. A family B x of subsets of an α (β,β) -topological abelian group G is called a basis of α β -neighborhoods of x G if any subset of B x is an α β -neighborhood of x and any α β -neighborhood of the element x contains some subset from B x. Proposition 3.9. Let a family B 0 of subsets of an α (β,β) -topological abelian group G be a basis of α β -neighborhoods of zero in G and β be an α-regular operation on αo(g). Then, the following conditions are satisfied: 1. 0 V. 2. For any subsets U and V from B 0, there exists a subset W B 0 such that W U V. 3. For any subset U B 0, there exists a subset V B 0 such that V + V U. 4. For any subset U B 0, there exists a subset V B 0 such that V U. Besides, if a G, then B a ={a + V V B 0 } is a basis of α β -neighborhoods of the element a. Proof. The fulfillment of conditions (1) and (2) results from the definition of a basis of α β -neighborhoods of an element in a topological space. (3) Let U be any α β -neighborhood of 0 = Since G is an α (β,β) -topological abelian group, then there are α β -neighborhoods W 1 and W 2 of 0 and W 1 + W 2 U. Let V = W 1 W 2, then V is α β -neighborhood of 0 and V + V W 1 + W 2 U.

8 2298 Alias B. Khalaf and Hariwan Z. Ibrahim The fulfillment of condition (4) result from the definition of an α (β,β) -topological abelian group. If a G, then the mapping f a : G G is an α (β,β) -homeomorphism in view of Proposition 3.4, and, hence, B a ={a + V V B 0 }={f a (V ) V B 0 } is a basis of α β -neighborhoods of the element a. Proposition Let S be a subset of an α (β,β) -topological abelian group G with a basis B 0 of α β -neighborhoods of zero. Then, α β Cl(S) = (S + V). Proof. Let x α β Cl(S) and V B 0. Let also V B 0 and V V, then (x + V S = φ, because x α β Cl(S). Then, x S V S+V. Therefore, α β Cl(S) S+V, and, hence, α β Cl(S) (S + V). Conversely, let y (S + V), and let U be an α β -neighborhood of zero in G. Let s choose an α β -neighborhood V B 0 of zero such that V U. Since y S + V, then S (y + U) S (y V) = φ, that is (S + V) α β Cl(S). Proposition Let B 0 be a basis of α β -neighborhoods of zero of an α (β,β) -topological abelian group G. Then, V is an α β -closed set. Proof. By Proposition 3.10, we have V is an α β -closure of a one element subset {0} in G. Corollary Let U and V be α β -neighborhoods of zero of α (β,β) -topological abelian group G such that V + V U, then α β Cl(V ) U. Proof. It is obvious that the family B 0 of all α β -neighborhoods of zero in G is a basis of α β -neighborhoods of zero of G. In the view of Proposition 3.10, α β Cl(V ) = W B 0 (V + W) V + V U. Proposition Let G be an α (β,β) -topological abelian group and β be an α-regular operation on αo(g). Then, the following statements are true: 1. G has a basis of α β -neighborhoods of zero consisting of symmetric α β -open neighborhoods.

9 α (β,β) -Topological Abelian Groups G has a basis of α β -neighborhoods of zero consisting of symmetric α β -closed neighborhoods. Proof. 1. Let B 0 be a basis of α β -neighborhoods of zero in G. Then for every V B 0, there exists an α β -open subset W v in G such that 0 W v V. In view of Proposition 3.4, the subset W v is α β -open, besides, 0 W v. Hence, subset W v ( W v ) is an α β -open symmetric neighborhood of zero, being contained in V. Then from condition (2) of Proposition 3.9, follows that the family B 0 ={W v ( W v ) V B 0 } is a basis of symmetric α β -open neighborhoods of zero in G. Thus, statement (1) is proved. 2. Let B 0 ={α βcl(u) U B 0 }. From Proposition 3.6 and from the symmetry of every α β -neighborhood U B 0 follows that α βcl(u) = α β Cl( U) = α β Cl(U). Hence, B 0 is a family of symmetric α β-closed neighborhoods of zero of G. Let s check that B 0 is a basis of α β-neighborhoods of zero in G. Indeed, if U B 0, then, in view of condition (3) of Proposition 3.9, there exists an α β- neighborhood U B 0 such that U +U U. Consequently, in view of Corollary 3.12, α β Cl(U U, and, hence, B 0 is a basis of α β-neighborhoods of zero in G. Corollary Let G be an α (β,β) -topological abelian group, a G and β be an α-regular operation on αo(g). Then, the following statements are true: 1. The element a has a basis of α β -neighborhoods consisting of α β -open neighborhoods. 2. The element a has a basis of α β -neighborhoods consisting of α β -closed neighborhoods. Proof. Follows from Proposition 3.13 and Proposition 3.4. Theorem For any α (β,β) -topological abelian group G and β an α-regular operation on αo(g), the following conditions are equivalent: 1. G is an α β T 2 -space. 2. {0} is α β -closed subset in G. 3. If B 0 is a basis of α β -neighborhoods of zero of G, then 4. G is an α β T 0 -space. V ={0}.

10 2300 Alias B. Khalaf and Hariwan Z. Ibrahim 5. G is an α β T 1 -space. Proof. (1) (2): If0 = a α β Cl({0}), then due to item (1), there exist α β -open sets U and V containing a and 0 respectively such that U V = φ. In particular, 0 / U, that is U {0} =φ, that contradicts the fact that a α β Cl({0}). (2) (3): Let {0} be α β -closed subset in G and B 0 be a basis of α β -neighborhoods of zero in G. Then, due to Proposition 3.10, we get {0} =α β Cl({0}) = ({0}+V)= V. (3) (4): Let B 0 be a basis of α β -neighborhoods of zero in G, and V ={0}. Let x,y G and x = y, then x y = 0, hence, there exists an α β -neighborhood V 0 B 0 such that x y/ V 0. Therefore, x/ y + V 0. Thus, G is α β T 0 -space. (4) (5): Let G be an α β T 0 -space and x,y be distinct elements of G. Then there exists an α β -open subset V such that: Case 1. 0 V, x + y / V. Let U be a symmetric α β -neighborhood of 0 such that U + U V. Then x + y/ U + U, hence x + U y + U = φ. Case 2. 0 / V, x+y V, then 0 y+x+v, and y+x / y+x+v = W. Let W 1 be a symmetric α β -neighborhood of 0 such that W 1 +W 1 W. Then y+x / W 1 +W 1, hence x + W 1 y + W 1 = φ. Thus, G is α β T 1 -space. (5) (1): Let x,y X such that x y = 0. Then there exists an α β -open set U containing 0 such that x y / U. Then, there exists symmetric α β -open set W containing 0 such that W + W U. Let V 1 = x + W and V 2 = y + W and note that V 1 αo(g, x) β, V 2 αo(g, y) β and V 1 V 2 = φ, since if a V 1 V 2, then (a x) W and a y W. It follows that x y = (a y)+( (a x)) W +W U, which is a contradiction. So, we have V 1 V 2 = φ. Therefore, G is α-β-t 2. Definition Let (G, +,τ)be an α (β,β) -topological abelian group. A subset B of G is called a subgroup of α (β,β) -topological group (G, τ) if B is a subgroup of G and B is endowed with family αo(g) β B induced by the αo(g) β. Theorem Let B be a subgroup of an α (β,β) -topological group (G, +,τ). Then (B, +, αo(g) β B) is a topological group. Proof. Let b 1,b 2 B and U αo(g) β B with b 1 b 2 U. Then U = U 1 B, where U αo(g) β and b 1 b 2 U 1. Let V 1 and V 1 be α β-open sets containing b 1 and b 2 respectively such that V 1 V 1 U 1. Then V = V 1 B and V = V 1 B are in αo(g) β B containing b 1 and b 2 respectively. Besides, V V (V 1 V 1 ) B

11 α (β,β) -Topological Abelian Groups 2301 U 1 B = U. Hence, by Theorem 2.16, (B, +, αo(g) β B) is a topological group. Proposition Let B be a subgroup of an α (β,β) -topological abelian group G. Then α β Cl(B) is a subgroup of the α (β,β) -topological group G. Proof. From Proposition 3.6, we have α β Cl(B) α β Cl(B) α β Cl(B B) = α β Cl(B), hence, α β Cl(B) is a subgroup of the α (β,β) -topological group G. Proposition Let B be a subgroup of an α (β,β) -topological abelian group G, then 1. If B is α β -open subgroup of the group G, then B is α β -closed in G. 2. B is α β -open if and only if it is an α β -neighborhood of at least one of its elements. Proof. 1. Let, B 0 be a basis of α β -neighborhoods of zero of G, then, taking into account that B is an α β -neighborhood of zero of G, and applying Proposition 3.10, we obtain that α β Cl(B) = (V + B) B + B = B. Therefore, B is an α β -closed subgroup of the α (β,β) -topological group G. 2. Let B be an α β -open subgroup of an α (β,β) -topological abelian group G, then B is an α β -neighborhood of each of its elements. Conversely, let b 0 B and B be an α β -neighborhood of the element b 0, then for any element b B (due to Proposition 3.4), B = b + (B b 0 )isanα β -neighborhood of the element b. Therefore, B is an α β -open subgroup of the α (β,β) -topological abelian group G, being an α β -neighborhood of each of its elements. Corollary Let B 0 be a basis of α β -neighborhoods of zero of an α (β,β) -topological abelian group, then V is the smallest α β -closed subgroup of G. Proof. The proof results from Propositions 3.11 and Proposition Let B be the smallest α β -closed subgroup of an α (β,β) -topological abelian group G and U be an α β -neighborhood of zero in G. Then there exists an α β -neighborhood V of zero in G such that: 1. V U. 2. V + B = V.

12 2302 Alias B. Khalaf and Hariwan Z. Ibrahim Proof. Let W be an α β -neighborhood of zero in G such that W + W U. Let V = W + B. Then V is an α β -neighborhood of zero in G, and, due to Corollary 3.20, V = W + B W + W U, that is, (1) is proved. Since V + B = W + B + B = W + B = V, then (2) is proved. The proof of the following result is obvious. Corollary An α β -connected α (β,β) -topological abelian group G has no α β -open subgroups different from G. Corollary An α β -connected α (β,β) -topological abelian group G is generated as a group by any α β -neighborhood of each of its elements. Proof. Let a G and U be an α β -neighborhood of a, then the subgroup < U > generated by the set U is an α β -neighborhood of the element a. Due to Proposition 3.19, <U>is an α β -open subgroup of G, and, according to Corollary 3.22, <U>= A. Proposition For an α (β,β) -topological abelian group G, the following statements are true: 1. If a G, and C(G) is an α β -component containing zero, then C(G) + a is an α β -component of a. 2. If C(G) is an α β -component containing zero, then C(G) is an α β -closed subgroup. Proof. 1. Let a G. The mapping f a : G G, where f a (x) = x + a for x G, is due to Proposition 3.4, an α (β,β) -homeomorphism mapping of the topological space G to itself and, besides, f a (0) = a. Then, C(G) + a is an α β -component of a. 2. Let c C(G), then C(G) c is an α β -component of c, besides, 0 C(G) c and, hence, C(G) c C(G). Then C(G) C(G) = c C(G) (C(G) c) C(G), that is, C(G) is a subgroup of the group G. Since C(G) is an α β -closed subset in G. Proposition Let G be α (β,β) -topological abelian group, G be α (β,β -topological abelian group and f : G G be a homomorphic mapping of G to G. Then:

13 α (β,β) -Topological Abelian Groups f is an α (β,β -continuous if and only if f 1 (U is an α β -neighborhood of zero in G for any α β -neighborhood U of zero in G. 2. f is an α (β,β -open if and only if f(u)is an α β -neighborhood of zero in G for any α β -neighborhood U of zero in G. Proof. 1. If f is an α (β,β ) -continuous, then it is, in particular, α (β,β ) -continuous at the point 0 G, and f(0) = 0. Then the necessity of the condition in the statement (1) results from Corollary Conversely, let a G and V be an α β -neighborhood of the element f(a). Then, due to Proposition 3.4, V f(a)is an α β -neighborhood of zero in G. Therefore, U = f 1 (V f(a))is the α β -neighborhood of zero in G. Then U + a is the α β -neighborhood of a in G, besides, f(u+a) = f(u)+f(a)= f(f 1 (V f (a)))+f(a) V f(a)+f(a)= V, that means that the mapping f is α (β,β -continuous. 2. If f is an α (β,β ) -open mapping, and U is an α β -neighborhood of zero in G, then, there exists an α β -open V in G containing zero such that V U. Then f(v) is an α β -open subset in G, besides, 0 f(v) f(u), that is, f(u) is an α β -neighborhood of zero in G. Conversely, let B be an α β -open subset in G, and b f(b). Then there exists an element b B such that f(b) = b. Since B is an α β -neighborhood of the element b in G, then B b is an α β -neighborhood of zero in G. Hence, f(b b) is an α β -neighborhood of zero in G, besides, f(b b) = f(b) f(b)= f(b) b. Therefore, f(b)= f(b b)+b is an α β -neighborhood of b in G. Thus, f(b) is an α β -neighborhood of each of its elements, that is, f(b)is an α β -open subset in G. Proposition Let G be α (β,β) -topological abelian group, B be a subgroup of G, B 0 (G) be a basis of α β -neighborhoods of zero in G, β be an α-regular operation on αo(g) and π : G G/B be a natural homomorphism of the group G to its factor group G/B. Then the system B 0 (G/B) ={π(u) U B 0 (G)} of the subsets of factor group G/B satisfies conditions (1)-(4) in Proposition 3.9.

14 2304 Alias B. Khalaf and Hariwan Z. Ibrahim Proof. It is clear that 0 = π(0) π(u) for any α β -neighborhood U B 0 (G), that is, condition (l) is fulfilled. Let U,V B 0 (G). There exists W B 0 (G) such that W U V. Then π(w) B 0 (G/B), besides, π(w) π(u) π(v), that is, condition (2) is fulfilled. Let U B 0 (G), then there exists V B 0 (G) such that V + V U. Then π(v) B 0 (G/B), besides, π(v) + π(v) = π(v + V) π(u), that is, condition (3) is fulfilled. Since π( V)= π(v) for each V B 0 (G), then condition (4) is fulfilled. Definition Let G be an α (β,β) -topological abelian group with a basis B 0 (G) of α β -neighborhoods of zero, B be its subgroup, π be the natural homomorphism of the group G to its factor group G/B. The group G/B endowed with the family, whose basis of α β -neighborhoods of zero is the system B 0 (G/B) ={π(u) U B 0 (G)}, is called a factor group of the α (β,β) -topological group G by the subgroup B. Proposition Let G/B be the factor group of an α (β,β) -topological abelian group G by a subgroup B. Then the natural homomorphism π : G G/B is α (β,β) -open and α (β,β) -continuous. Proof. Let B 0 (G) be a basis of α β -neighborhoods of zero of G. Then, due to Definition 3.27, B 0 (G/B) ={π(u) U B 0 (G)}, is a basis of α β -neighborhoods of zero of the group G/B. Let V be an α β -neighborhood of zero of G, then there exists an α β -neighborhood U B 0 (G) such that U V. Then π(u) π(v). Since π(u) B 0 (G/B), then π(v) is an α β -neighborhood of zero in G/B. Therefore, due to Proposition 3.25, π is α (β,β) -open. Let W be an α β -neighborhood of zero in G/B, then there exists an α β -neighborhood T B 0 (G) such that π(t) W. Then π 1 (W) T, hence, π 1 (W) is an α β - neighborhood of zero in G. From Proposition 3.25, follows that π is α (β,β) -continuous. Theorem Let G be an α (β,β) -topological abelian group, G be an α (β,β -topological abelian group, f : G G be a homomorphism from G onto G, and B = ker(f ). Let π : G G/B be the natural homomorphism and π 1 : G/B G be the natural isomorphism, then: 1. If f is an α (β,β -continuous, then π 1 is an α (β,β -continuous of the topological group G/B onto the topological group G. 2. If f is an α (β,β -open, then π 1 is an α (β,β -open isomorphism of the topological group G/B onto the topological group G.

15 α (β,β) -Topological Abelian Groups 2305 Proof. 1. Let f be an α (β,β -continuous and U be an α β -neighborhood of zero in G. Then f 1 (U) is an α β -neighborhood of zero in G, and π(f 1 (U)) is an α β - neighborhood of zero in G/B. From the definition of the mappings π and π 1 follows that π 1 (π(f 1 (U))) = U. Since π 1 is an isomorphism, then π(f 1 (U)) = π 1 1 (U). Since π(f 1 (U)) is an α β -neighborhood of zero in G/B, we obtain that π 1 is an α (β,β ) -continuous isomorphism. 2. Let f be an α (β,β -open and V be an α β -neighborhood of zero in G/B. Then π 1 (V ) is an α β -neighborhood of zero in G, and f(π 1 (V )) is an α β -neighborhood of zero in G. Since f(π 1 (V )) = π 1 (V ), then π 1 is an α (β,β ) -open isomorphism. Proposition Let G be an α (β,β) -topological abelian group, B be its subgroup, and π : G G/B be the natural homomorphism from G to the factor group G/B. If C G and C + B = C, then π(α β Cl(C)) = α β Cl(π(C)). Proof. Let B 0 (G) be a basis of α β -neighborhoods of zero of G. According to the Definition 3.27, B 0 (G/B) ={π(u) U B 0 (G)} is a basis of α β -neighborhoods of zero of G/B. Due to Proposition 3.10, Since α β Cl(π(C)) = then, due to Theorem 2.15, U B 0 (G) U B 0 (G) (π(c) + π(u)) = U B 0 (G) (C + U)+ B = (C + B) + U = C + U, π(c + U) = π U B 0 (G) π(c + U). (C + U) = π(α β Cl(C)). Proposition Let B be an α β -connected subgroup of an α (β,β) -topological abelian group G. If the group G/B is α (β,β) -connected, then the group G is α (β,β) -connected. Proof. Assume the contrary hypothesis, that is, that S is a non-empty α β -open and α β - closed subset of the α (β,β) -topological group G and S = G. Since S B is an α β -open and α β -closed subset of the α β -connected group B, then either S B = φ (in that case B G \ S), or S B = B (in that case B S). Hence, in any case B is contained in some α β -open and α β -closed subset T, different from G. Let s show that T + B = T.

16 2306 Alias B. Khalaf and Hariwan Z. Ibrahim Indeed, let t T, then t (t + B) T, that is, (t + B) T is a non-empty α β -open and α β -closed subset of the α (β,β) -connected space t + B. Hence, (t + B) T = t + B. Therefore, t + B T for any t T. Thus, T + B = t T (t + B) T. Since T + B T, then T + B = T. Due to Propositions 3.28 and 3.30, π(t) is a non-empty α β -open and α β -closed subset of the α β -connected group G/B. Then π(t) = G/B. Since T + B = T, then T = G. This is a contradiction to the assumption. This completes the proof. References [1] A. V. Arhangel skii, M. G. Tkachenko, Topological Groups and Related Structures, Atlantis Press/World Scientific, Amsterdam Paris, (2008). [2] H. Z. Ibrahim, On a class of α γ -open sets in a topological space, Acta Scientiarum. Technology, 35(3) (2013), [3] L. Kaplansky, Topological algebra, Notes Mat. Inst. Mat. Pure Apl., (16) (1959). [4] A. B. Khalaf and H. Z. Ibrahim, Some properties of operations on αo(x), International Journal of Mathematics and Soft Computing, 6 (1) (2016), [5] A. B. Khalaf and H. Z. Ibrahim, Some new functions via operations defined on α-open sets, Journal of Garmian University, no. 12 (2017). [6] A. B. Khalaf and H. Z. Ibrahim, α γ -connectedness and some properties of α (γ,β) - continuous functions, Accepted in The First International Conference of Natural Science (ICNS) from 11th-12th July 2016, Charmo University. [7] A. B. Khalaf and H. Z. Ibrahim, Topological group via operations defined on α-open sets, submitted. [8] A. G. Kurosh, Lectures on General Algebra, Nauka, (1962). [9] O. Njastad, On some classes of nearly open sets, Pacific J. Math. 15 (1965), [10] M. Tkachenko, Paratopological and semitopological groups vs topological groups, In: K.P. Hart, J. van Mill, P. Simon (eds.), Recent Progress in General Topology III, Atlantis Press, (2014),