International Journal of Analysis and Applications Volume 16, Number 4 (2018), 518-527 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-16-2018-518 A NEW TYPE OF CONNECTED SETS VIA BIOPERATIONS HARIWAN Z. IBRAHIM∗ Department of Mathematics, Faculty of Education, University of Zakho, Kurdistan-Region, Iraq ∗Corresponding author: hariwan math@yahoo.com Abstract. The purpose of this paper is to introduce the notion of α (γ,γ ′ ) -separated sets and study their properties in topological spaces, then we introduce the notions of α (γ,γ ′ ) -connected and α (γ,γ ′ ) -disconnected sets. We discuss the characterizations and properties of α (γ,γ ′ ) -connected sets and then properties under (α (γ,γ ′ ) , α (β,β ′ ) )-continuous functions. The α (γ,γ ′ ) -components in a space X is also introduced. 1. Introduction Njastad [5] introduced α-open sets in a topological space and studied some of their properties. Ibrahim [1] introduced and discussed an operation of a topology αO(X) into the power set P(X) of a space X and also in [2] he introduced the notion of αO(X,τ)(γ,γ′ ), which is the collection of all α(γ,γ′ )-open sets in a topological space (X,τ). In addition, Ibrahim [3] introduced the concept of (α(γ,γ′ ), α(β,β′ ))-closed and (α(γ,γ′ ), α(β,β′ ))-continuous functions and investigated some of their basic properties. Mishra [4] introduced α-τ-disconnectedness and α-τ-connectedness in topological spaces. In this paper, the author introduce and study the characterizations and properties of α(γ,γ′ )-connected and α(γ,γ′ )-disconnected spaces and then properties under (α(γ,γ′ ), α(β,β′ ))-continuous functions. Received 2018-01-18; accepted 2018-03-19; published 2018-07-02. 2010 Mathematics Subject Classification. Primary 22A05, 22A10, Secondary 54C05. Key words and phrases. α-open; bioperations; α (γ,γ ′ ) -connected set; α (γ,γ ′ ) -disconnected set. c©2018 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 518 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-16-2018-518 Int. J. Anal. Appl. 16 (4) (2018) 519 2. Preliminaries Throughout the present paper, (X,τ) and (Y,σ) (or simply X and Y ) denotes a topological spaces on which no separation axioms is assumed unless explicitly stated. For a subset A of a space X, Cl(A) and Int(A) represent the closure of A and the interior of A, respectively. Definition 2.1. [5] A subset A of a topological space (X,τ) is said to be α-open if A ⊆ Int(Cl(Int(A))). The complement of an α-open set is said to be α-closed. The family of all α-open (resp., α-closed) sets in a topological space (X,τ) is denoted by αO(X,τ) (resp., αC(X,τ)). The intersection of all α-closed sets containing A is called the α-closure of A and is denoted by αCl(A). Definition 2.2. [4] The subsets A and B of a topological space (X,τ) are called α-τ-separated sets if (αCl(A) ∩B) ∪ (A∩αCl(B)) = φ. Definition 2.3. [1] An operation γ : αO(X,τ) → P(X) is a mapping satisfying the condition, V ⊆ V γ for each V ∈ αO(X,τ). We call the mapping γ an operation on αO(X,τ). The operation id : αO(X,τ) → P(X) is defined by V id = V for any set V ∈ αO(X,τ). This operation is called the identity operation on αO(X,τ). Definition 2.4. [2] A nonempty subset A of (X,τ) is said to be α(γ,γ′ )-open if for each x ∈ A, there exist α-open sets U and V of X containing x such that Uγ ∪V γ ′ ⊆ A. A subset F of (X,τ) is said to be α(γ,γ′ )-closed if its complement X \F is α(γ,γ′ )-open. The set of all α(γ,γ′ )-open sets of (X,τ) is denoted by αO(X,τ)(γ,γ′ ). Definition 2.5. [2] Let A be a subset of a topological space (X,τ). (1) The union of all α(γ,γ′ )-open sets contained in A is called the α(γ,γ′ )-interior of A and is denoted by α(γ,γ′ )-Int(A). (2) The intersection of all α(γ,γ′ )-closed sets containing A is called the α(γ,γ′ )-closure of A and denoted by α(γ,γ′ )-Cl(A). Proposition 2.1. [2] Let A and B be subsets of (X,τ). Then the following hold: (1) A ⊆ α(γ,γ′ )-Cl(A). (2) If A ⊆ B, then α(γ,γ′ )-Cl(A) ⊆ α(γ,γ′ )-Cl(B). (3) A is α(γ,γ′ )-closed if and only if α(γ,γ′ )-Cl(A) = A. (4) α(γ,γ′ )-Cl(A) is α(γ,γ′ )-closed. Proposition 2.2. [2] For a point x ∈ X, x ∈ α(γ,γ′ )-Cl(A) if and only if V ∩A 6= φ for every α(γ,γ′ )-open set V containing x. Definition 2.6. [3] A function f : (X,τ) → (Y,σ) is said to be (α(γ,γ′ ), α(β,β′ ))-closed if for α(γ,γ′ )-closed set A of X, f(A) is α(β,β′ )-closed in Y . Int. J. Anal. Appl. 16 (4) (2018) 520 Proposition 2.3. [3] Let f : (X,τ) → (Y,σ) be a function. Then, f is (α(γ,γ′ ), α(β,β′ ))-closed if and only if α(β,β′ )-Cl(f(A)) ⊆ f(α(γ,γ′ )-Cl(A)) for every subset A of X. Theorem 2.1. [3] Suppose that f : (X,τ) → (Y,σ) is (α(γ,γ′ ), α(β,β′ ))-continuous. Then, (1) f−1(V ) is α(γ,γ′ )-open for every α(β,β′ )-open set V of (Y,σ). (2) For each point x ∈ X and each α(β,β′ )-open W of (Y,σ) containing f(x), there exist α(γ,γ′ )-open U of (X,τ) containing x such that f(U) ⊆ W . 3. α(γ,γ′ )-Connected and α(γ,γ′ )-Disconnected Sets Throughout this section, let γ,γ ′ : αO(X,τ) → P(X) be operations on αO(X,τ) and β,β ′ : αO(Y,σ) → P(Y ) be operations on αO(Y,σ). Definition 3.1. Two subsets A and B of a topological space (X,τ) are called α(γ,γ′ )-separated if (α(γ,γ′ )- Cl(A) ∩B) ∪ (A∩α(γ,γ′ )-Cl(B)) = φ. Remark 3.1. Each two α(γ,γ′ )-separated sets are always disjoint, since A ∩ B ⊆ A ∩ α(γ,γ′ )-Cl(B) = φ. The converse may not be true in general, as it is shown in the following example. Example 3.1. Let X = {1, 2, 3} and τ = {φ,X,{2}}. For each A ∈ αO(X), we define two operations γ and γ ′ , respectively, by Aγ = Aγ ′ =   A if 3 ∈ AX if 3 /∈ A. Since αO(X,τ)(γ,γ′ ) = {φ,X,{2, 3}}, then {2} and {3} are disjoint subsets of X, but not α(γ,γ′ )-separated. From the fact that αCl(A) ⊆ α(γ,γ′ )-Cl(A), for every subset A of X. Then every α(γ,γ′ )-separated set is α-τ-separated. But the converse may not be true as shown in the following example. Example 3.2. Let X = {1, 2, 3, 4} and τ = {φ,X,{1},{2},{1, 2}}. For each A ∈ αO(X), we define two operations γ and γ ′ , respectively, by Aγ = Aγ ′ =   A if 4 ∈ AX if 4 /∈ A. Since αO(X,τ)(γ,γ′ ) = {φ,X,{1, 2, 4}}, then the subsets {3} and {4} are α-τ-separated, but not α(γ,γ′ )- separated. Theorem 3.1. If A and B are any two nonempty subsets in a space X, then the following statements are true: (1) If A and B are α(γ,γ′ )-separated, A1 ⊆ A and B1 ⊆ B, then A1 and B1 are also α(γ,γ′ )-separated. Int. J. Anal. Appl. 16 (4) (2018) 521 (2) If A ∩ B = φ such that each of A and B are both α(γ,γ′ )-closed (α(γ,γ′ )-open), then A and B are α(γ,γ′ )-separated. (3) If each of A and B is α(γ,γ′ )-closed (α(γ,γ′ )-open) and if H = A ∩ (X \ B) and G = B ∩ (X \ A), then H and G are α(γ,γ′ )-separated. Proof. (1) Since A1 ⊆ A, then α(γ,γ′ )-Cl(A1) ⊆ α(γ,γ′ )-Cl(A). Then, B ∩ α(γ,γ′ )-Cl(A) = φ implies B1 ∩α(γ,γ′ )-Cl(A) = φ and B1 ∩α(γ,γ′ )-Cl(A1) = φ. Similarly A1 ∩α(γ,γ′ )-Cl(B1) = φ. Hence, A1 and B1 are α(γ,γ′ )-separated. (2) Since A = α(γ,γ′ )-Cl(A), B = α(γ,γ′ )-Cl(B) and A ∩ B = φ, then α(γ,γ′ )-Cl(A) ∩ B = φ and α(γ,γ′ )-Cl(B) ∩ A = φ. Hence, A and B are α(γ,γ′ )-separated. If A and B are α(γ,γ′ )-open, then their complements are α(γ,γ′ )-closed. Hence, α(γ,γ′ )-Cl(A) ⊆ X \ B and α(γ,γ′ )-Cl(B) ⊆ X \ A. Therefore, A and B are α(γ,γ′ )-separated. (3) If A and B are α(γ,γ′ )-open, then X \ A and X \ B are α(γ,γ′ )-closed. Since H ⊆ X \ B, α(γ,γ′ )- Cl(H) ⊆ α(γ,γ′ )-Cl(X \ B) = X \ B and so α(γ,γ′ )-Cl(H) ∩ B = φ. Thus G ∩ α(γ,γ′ )-Cl(H) = φ. Similarly, H∩α(γ,γ′ )-Cl(G) = φ. Hence H and G are α(γ,γ′ )-separated. If A and B are α(γ,γ′ )-closed, then α(γ,γ′ )-Cl(H) ⊆ A and α(γ,γ′ )-Cl(G) ⊆ B. Thus, H and G are α(γ,γ′ )-separated. � Theorem 3.2. The sets A and B of a space X are α(γ,γ′ )-separated if and only if there exist U and V in αO(X,τ)(γ,γ′ ) such that A ⊆ U, B ⊆ V and A∩V = φ and B ∩U = φ. Proof. Let A and B be α(γ,γ′ )-separated sets. Set V = X \ α(γ,γ′ )-Cl(A) and U = X \ α(γ,γ′ )-Cl(B). Then U,V ∈ αO(X,τ)(γ,γ′ ) such that A ⊆ U, B ⊆ V and A∩V = φ, B ∩U = φ. On the other hand, let U,V ∈ αO(X,τ)(γ,γ′ ) such that A ⊆ U, B ⊆ V and A∩V = φ, B∩U = φ. Since X\V and X\U are α(γ,γ′ )- closed, then α(γ,γ′ )-Cl(A) ⊆ X\V ⊆ X\B and α(γ,γ′ )-Cl(B) ⊆ X\U ⊆ X\A. Thus α(γ,γ′ )-Cl(A)∩B = φ and α(γ,γ′ )-Cl(B) ∩A = φ. � Theorem 3.3. In any topological space (X,τ), the following statements are equivalent: (1) φ and X are the only α(γ,γ′ )-open and α(γ,γ′ )-closed sets in X. (2) X is not the union of two disjoint nonempty α(γ,γ′ )-open sets. (3) X is not the union of two disjoint nonempty α(γ,γ′ )-closed sets. (4) X is not the union of two nonempty α(γ,γ′ )-separated sets. Proof. (1) ⇒ (2): Suppose (2) is false and that X = A∪B, where A,B are disjoint nonempty α(γ,γ′ )-open sets. Since X \A = B is α(γ,γ′ )-open and nonempty, we have that A is a nonempty proper α(γ,γ′ )-open and α(γ,γ′ )-closed set in X, which shows that (1) is false. (2) ⇔ (3): This is clear. Int. J. Anal. Appl. 16 (4) (2018) 522 (3) ⇒ (4): If (4) is false, then X = A∪B, where A,B are nonempty and α(γ,γ′ )-separated. Since α(γ,γ′ )- Cl(B)∩A = φ, we conclude that α(γ,γ′ )-Cl(B) ⊆ B, so B is α(γ,γ′ )-closed. Similarly, A must be α(γ,γ′ )-closed. Therefore, (3) is false. (4) ⇒ (1): Suppose (1) is false and that A is a nonempty proper α(γ,γ′ )-open and α(γ,γ′ )-closed subset of X. Then, B = X \ A is nonempty, α(γ,γ′ )-open and α(γ,γ′ )-closed, so A and B are α(γ,γ′ )-separated and X = A∪B, so (4) is false. � Definition 3.2. A subset C of a space X is said to be α(γ,γ′ )-disconnected if there are nonempty α(γ,γ′ )- separated subsets A and B of X such that C = A ∪ B, otherwise C is called α(γ,γ′ )-connected. If C is α(γ,γ′ )-disconnected, such a pair of sets A,B will be called an α(γ,γ′ )-disconnection of C. Example 3.3. Let X = {1, 2, 3} and τ = {φ,X,{1},{2},{1, 2},{2, 3}}. For each A ∈ αO(X), we define two operations γ and γ ′ , respectively, by Aγ = Aγ ′ =   A if 3 ∈ ACl(A) if 3 /∈ A. Then, X is α(γ,γ′ )-disconnected because there exist a pair {1},{2, 3} subsets of X such that {1}∪{2, 3} = X, and (α(γ,γ′ )-Cl({1}) ∩{2, 3}) ∪ ({1}∩α(γ,γ′ )-Cl({2, 3})) = ({1}∩{2, 3}) ∪ ({1}∩{2, 3}) = φ. Example 3.4. Let X = {1, 2, 3} and τ = {φ,X,{1},{3},{1, 3}}. For each A ∈ αO(X), we define two operations γ and γ ′ , respectively, by Aγ = Aγ ′ =   A if 2 ∈ AX if 2 /∈ A. Then, X is α(γ,γ′ )-connected, since there does not exist a pair A,B of nonempty α(γ,γ′ )-separated subsets of X such that X = A∪B. Remark 3.2. Every indiscrete space is α(γ,γ′ )-connected. Remark 3.3. Every discrete space contains more than one element is α(id,id′ )-disconnected. Remark 3.4. A space X is α(γ,γ′ )-connected if any (therefore all) of the conditions (1) − (4) in Theorem 3.3 hold. Remark 3.5. According to the Definition 3.2 and Remark 3.4, a space X is α(γ,γ′ )-disconnected if we can write X = A∪B, where the following (equivalent) statements are true: (1) A and B are disjoint, nonempty and α(γ,γ′ )-open. (2) A and B are disjoint, nonempty and α(γ,γ′ )-closed. (3) A and B are nonempty and α(γ,γ′ )-separated. Int. J. Anal. Appl. 16 (4) (2018) 523 Theorem 3.4. A space X is α(γ,γ′ )-disconnected if and only if there exists a nonempty proper subset A of X which is both α(γ,γ′ )-open and α(γ,γ′ )-closed in X. Proof. Follows from Remark 3.5. � Definition 3.3. Let A be a subset of a space X, then the α(γ,γ′ )-boundary of A is defined as α(γ,γ′ )- Cl(A) \α(γ,γ′ )-Int(A) and is denoted by α(γ,γ′ )-Bd(A). Proposition 3.1. Let A be any subset of a topological space (X,τ). Then, the following statements are hold: (1) α(γ,γ′ )-Cl(A) = α(γ,γ′ )-Int(A) ∪α(γ,γ′ )-Bd(A). (2) α(γ,γ′ )-Bd(A) = α(γ,γ′ )-Cl(A) ∩α(γ,γ′ )-Cl(X \A). Proof. Obvious. � Theorem 3.5. A space X is α(γ,γ′ )-connected if and only if every nonempty proper subset of X has a nonempty α(γ,γ′ )-boundary. Proof. Suppose that a nonempty proper subset A of an α(γ,γ′ )-connected space X has empty α(γ,γ′ )- boundary. Since α(γ,γ′ )-Cl(A) = α(γ,γ′ )-Int(A) ∪α(γ,γ′ )-Bd(A). Thus, A is both α(γ,γ′ )-closed and α(γ,γ′ )- open. By Theorem 3.4, X is α(γ,γ′ )-disconnected. This contradiction, hence proves that A has a nonempty α(γ,γ′ )-boundary. Conversely, suppose X is α(γ,γ′ )-disconnected. Then by Theorem 3.4, X has a nonempty proper subset A which is both α(γ,γ′ )-closed and α(γ,γ′ )-open. Then, α(γ,γ′ )-Cl(A) = A, α(γ,γ′ )-Cl(X \ A) = X \ A and α(γ,γ′ )-Cl(A) ∩α(γ,γ′ )-Cl(X \A) = φ. So A has empty α(γ,γ′ )-boundary, this is a contradiction. Hence, X is α(γ,γ′ )-connected. � Lemma 3.1. Suppose M,N are α(γ,γ′ )-separated subsets of X. If C ⊆ M ∪N and C is α(γ,γ′ )-connected, then C ⊆ M or C ⊆ N. Proof. Since C∩M ⊆ M and C∩N ⊆ N, then C∩M and C∩N are α(γ,γ′ )-separated and C = C∩(M∪N) = (C∩M)∪(C∩N). But C is α(γ,γ′ )-connected so (C∩M) and (C∩N) can not form an α(γ,γ′ )-disconnection of C. Therefore, either C ∩M = φ, so C ⊆ N or C ∩N = φ, so C ⊆ M. � Theorem 3.6. Suppose C and Ci (i ∈ I) are α(γ,γ′ )-connected subsets of X and that for each i, Ci and C are not α(γ,γ′ )-separated. Then, S = C ∪Ci is α(γ,γ′ )-connected. Proof. Suppose that S = M ∪N, where M and N are α(γ,γ′ )-separated. By Lemma 3.1, either C ⊆ M or C ⊆ N. Without loss of generality, assume C ⊆ M. By the same reasoning we conclude that for each i, Int. J. Anal. Appl. 16 (4) (2018) 524 either Ci ⊆ M or Ci ⊆ N. But if some Ci ⊆ N, then C and Ci would be α(γ,γ′ )-separated. Hence every Ci ⊆ M. Therefore, N = φ and the pair M,N is not an α(γ,γ′ )-disconnection of S. � Corollary 3.1. Suppose that for each i ∈ I, Ci is an α(γ,γ′ )-connected subset of X and that for all i 6= j, Ci ∩Cj 6= φ. Then, ∪{Ci : i ∈ I} is α(γ,γ′ )-connected. Proof. If I = φ, then ∪{Ci : i ∈ I} = φ is α(γ,γ′ )-connected. If I 6= φ, pick i0 ∈ I and let Ci0 be the central set C in Theorem 3.6. For all i ∈ I, Ci ∩Ci0 6= φ, so Ci and Ci0 are not α(γ,γ′ )-separated. By Theorem 3.6, ∪{Ci : i ∈ I} is α(γ,γ′ )-connected. � Corollary 3.2. Suppose that for all x,y ∈ X, there exists an α(γ,γ′ )-connected set Cxy ⊆ X with x,y ∈ Cxy. Then, X is α(γ,γ′ )-connected. Proof. Certainly X = φ is α(γ,γ′ )-connected. If X 6= φ, choose a ∈ X. By hypothesis there is, for each y ∈ X, an α(γ,γ′ )-connected set Cay containing both a and y. By Corollary 3.1, X = ∪{Cay : y ∈ X} is α(γ,γ′ )-connected. � Corollary 3.3. Suppose C is an α(γ,γ′ )-connected subset of X and A ⊆ X. If C ⊆ A ⊆ α(γ,γ′ )-Cl(C), then A is α(γ,γ′ )-connected. Proof. For each a ∈ A, {a} and C are not α(γ,γ′ )-separated. By Theorem 3.6, A = C ∪ ⋃ {{a} : a ∈ A} is α(γ,γ′ )-connected. � Remark 3.6. In particular, the α(γ,γ′ )-closure of an α(γ,γ′ )-connected set is α(γ,γ′ )-connected. Theorem 3.7. Let f : (X,τ) → (Y,σ) be a function. Consider the following statements. (1) f is (α(γ,γ′ ), α(β,β′ ))-continuous. (2) f−1(V ) ⊆ α(γ,γ′ )-Int(f −1(V )) for every α(β,β′ )-open set V of Y . (3) f(α(γ,γ′ )-Cl(A)) ⊆ α(β,β′ )-Cl(f(A)) for every subset A of X. (4) α(γ,γ′ )-Cl(f −1(B)) ⊆ f−1(α(β,β′ )-Cl(B)) for every subset B of Y . Then, the following implications are true: (1) ⇒ (2) ⇒ (3) ⇒ (4). Proof. (1) ⇒ (2). Let V be any α(β,β′ )-open set of Y and x ∈ f −1(V ). Then, f(x) ∈ V . Since f is (α(γ,γ′ ), α(β,β′ ))-continuous, there exists an α(γ,γ′ )-open set U of X containing x such that f(U) ⊆ V and hence U ⊆ f−1(V ), this implies that x ∈ α(γ,γ′ )-Int(f −1(V )). Thus, it follows that f−1(V ) ⊆ α(γ,γ′ )- Int(f−1(V )). (2) ⇒ (3). Let A be any subset of X and f(x) /∈ α(β,β′ )-Cl(f(A)). Then, by Proposition 2.2, there exists an α(β,β′ )-open set V of Y containing f(x) such that V ∩ f(A) = φ and hence f −1(V ) ∩ A = φ. Also f(x) ∈ V implies x ∈ f−1(V ). Then by (2) we obtain that x ∈ α(γ,γ′ )-Int(f −1(V )). Hence, there exists an Int. J. Anal. Appl. 16 (4) (2018) 525 α(γ,γ′ )-open set U of X containing x such that U ⊆ f −1(V ). Then U ∩ A = φ and so x /∈ α(γ,γ′ )-Cl(A). This implies f(x) /∈ f(α(γ,γ′ )-Cl(A)). Thus, f(α(γ,γ′ )-Cl(A)) ⊆ α(β,β′ )-Cl(f(A)). (3) ⇒ (4). Let B be any subset of Y . Since f(f−1(B)) ⊆ B, so, we have α(β,β′ )-Cl(f(f −1(B))) ⊆ α(β,β′ )- Cl(B). Also, f−1(B) ⊆ X, then by (3), we have f(α(γ,γ′ )-Cl(f −1(B))) ⊆ α(β,β′ )-Cl(f(f −1(B))) ⊆ α(β,β′ )- Cl(B). Thus, α(γ,γ′ )-Cl(f −1(B)) ⊆ f−1(α(β,β′ )-Cl(B)). � Corollary 3.4. Let f : X → Y be an (α(γ,γ′ ), α(β,β′ ))-continuous and injective function. If K is α(γ,γ′ )- connected in X, then f(K) is α(β,β′ )-connected in Y. Proof. Suppose that f(K) is α(β,β′ )-disconnected in Y . There exist two α(β,β′ )-separated sets P and Q of Y such that f(K) = P∪Q. Set A = K∩f−1(P) and B = K∩f−1(Q). Since f(K)∩P 6= φ, then K∩f−1(P) 6= φ and so A 6= φ. Similarly B 6= φ. Now, A∪B = (K ∩f−1(P)) ∪ (K ∩f−1(Q)) = K ∩ (f−1(P) ∪f−1(Q)) = K ∩ f−1(P ∪ Q) = K ∩ f−1(f(K)) = K. Since f is (α(γ,γ′ ), α(β,β′ ))-continuous, then by Theorem 3.7 , α(γ,γ′ )-Cl(f −1(Q)) ⊆ f−1(α(β,β′ )-Cl(Q)) and B ⊆ f −1(Q), then α(γ,γ′ )-Cl(B) ⊆ f −1(α(β,β′ )-Cl(Q)). Since P ∩α(β,β′ )-Cl(Q) = φ, then A∩α(γ,γ′ )-Cl(B) ⊆ A∩f −1(α(β,β′ )-Cl(Q)) ⊆ f −1(P)∩f−1(α(β,β′ )-Cl(Q)) = φ and then A ∩ α(γ,γ′ )-Cl(B) = φ. Similarly, B ∩ α(γ,γ′ )-Cl(A) = φ. Thus, A and B are α(γ,γ′ )-separated. Therefore, K is α(γ,γ′ )-disconnected, this is contradiction. Hence, f(K) is α(β,β′ )-connected. � Theorem 3.8. If f : (X,τ) → (Y,σ) is an onto (α(γ,γ′ ), α(β,β′ ))-continuous function and X is α(γ,γ′ )- connected, then Y is α(β,β′ )-connected. Proof. Suppose that Y is α(β,β′ )-disconnected and A,B is an α(β,β′ )-disconnection of Y . By Remark 3.5, A and B are both α(β,β′ )-open sets. Since f is (α(γ,γ′ ), α(β,β′ ))-continuous, so by Theorem 2.1, f −1(A) and f−1(B) are both nonempty α(γ,γ′ )-open sets in X. Now, f −1(A) ∪f−1(B) = f−1(A∪B) = f−1(Y ) = X, and f−1(A) ∩ f−1(B) = f−1(A ∩ B) = f−1(φ) = φ. Then by Remark 3.5, f−1(A),f−1(B) is a pair of α(γ,γ′ )-disconnection of X. This contradiction shows that Y is α(β,β′ )-connected. � Corollary 3.5. For a bijective (α(γ,γ′ ), α(β,β′ ))-closed function f : X → Y , if C is α(β,β′ )-connected in Y , then f−1(C) is α(γ,γ′ )-connected in X. Proof. Suppose that f−1(C) is α(γ,γ′ )-disconnected in X. There exist two α(γ,γ′ )-separated sets M and N of X such that f−1(C) = M ∪ N. Set K = C ∩ f(M) and L = C ∩ f(N). Since C = f(M) ∪ f(N), then C ∩ f(M) 6= φ and so K 6= φ. Similarly L 6= φ. Now, K ∪ L = (C ∩ f(M)) ∪ (C ∩ f(N)) = C ∩ (f(M) ∪ f(N)) = C ∩ f(M ∪ N) = C ∩ f(f−1(C)) = C. Since f is (α(γ,γ′ ), α(β,β′ ))-closed, then by Proposition 2.3, α(β,β′ )-Cl(f(N)) ⊆ f(α(γ,γ′ )-Cl(N)) and L ⊆ f(N), then α(β,β′ )-Cl(L) ⊆ f(α(γ,γ′ )-Cl(N)). Since M∩α(γ,γ′ )-Cl(N) = φ, then K∩α(β,β′ )-Cl(L) ⊆ K∩f(α(γ,γ′ )-Cl(N)) ⊆ f(M)∩f(α(γ,γ′ )-Cl(N)) = φ and then K ∩ α(β,β′ )-Cl(L) = φ. Similarly, L ∩ α(β,β′ )-Cl(K) = φ. Thus, K and L are α(β,β′ )-separated. Therefore, C is α(β,β′ )-disconnected, this is contradiction. Hence, f −1(C) is α(γ,γ′ )-connected. � Int. J. Anal. Appl. 16 (4) (2018) 526 Definition 3.4. A set C is called a maximal α(γ,γ′ )-connected set if it is α(γ,γ′ )-connected and if C ⊆ D ⊆ X where D is α(γ,γ′ )-connected, then C = D. A maximal α(γ,γ′ )-connected subset C of a space X is called an α(γ,γ′ )-component of X. If X is itself α(γ,γ′ )-connected, then X is the only α(γ,γ′ )-component of X. Theorem 3.9. For each x ∈ X, there is exactly one α(γ,γ′ )-component of X containing x. Proof. For any x ∈ X, let Cx = ⋃ {A : x ∈ A ⊆ X and A is α(γ,γ′ )-connected}. Then, {x}∈ Cx, since Cx is a union of α(γ,γ′ )-connected sets each containing x, Cx is α(γ,γ′ )-connected by Corollary 3.1. If Cx ⊆ D and D is α(γ,γ′ )-connected, then D was one of the sets A in the collection whose union defines Cx, so D ⊆ Cx and therefore Cx = D. Therefore, Cx is an α(γ,γ′ )-component of X that contains x. � Corollary 3.6. A space X is the union of its α(γ,γ′ )-components. Proof. Follows from Theorem 3.9. � Corollary 3.7. Two α(γ,γ′ )-components are either disjoint or coincide. Proof. Let Cx and Cy be α(γ,γ′ )-components and Cx 6= Cy. If p ∈ Cx ∩Cy, then by Corollary 3.1, Cx ∪Cy would be an α(γ,γ′ )-connected set strictly larger than Cx. Therefore, Cx ∩Cy = φ. � Theorem 3.10. Each α(γ,γ′ )-connected subset of X is contained in exactly one α(γ,γ′ )-component of X. Proof. Let A be an α(γ,γ′ )-connected subset of X which is not in exactly one α(γ,γ′ )-component of X. Suppose that C1 and C2 are α(γ,γ′ )-components of X such that A ⊆ C1 and A ⊆ C2. Since C1 ∩ C2 6= φ and by Corollary 3.1, C1∪C2 is another α(γ,γ′ )-connected set which contains C1 as well as C2, a contradiction to the fact that C1 and C2 are α(γ,γ′ )-components. This proves that A is contained in exactly one α(γ,γ′ )-component of X. � Theorem 3.11. A nonempty α(γ,γ′ )-connected subset of X which is both α(γ,γ′ )-open and α(γ,γ′ )-closed is α(γ,γ′ )-component. Proof. Suppose that A is α(γ,γ′ )-connected subset of X which is both α(γ,γ′ )-open and α(γ,γ′ )-closed. By Theorem 3.10, A is contained in exactly one α(γ,γ′ )-component C of X. If A is a proper subset of C, then C = (C∩A)∪(C∩(X\A)) and (C∩A), (C∩(X\A)) is an α(γ,γ′ )-disconnection of C, which is a contradiction. Thus, A = C. � Theorem 3.12. Every α(γ,γ′ )-component of X is α(γ,γ′ )-closed. Proof. Suppose that C is an α(γ,γ′ )-component of X. Then, by Remark 3.6, α(γ,γ′ )-Cl(C) is α(γ,γ′ )-connected containing α(γ,γ′ )-component C of X. This implies that C = α(γ,γ′ )-Cl(C) and hence C is α(γ,γ′ )-closed. � Int. J. Anal. Appl. 16 (4) (2018) 527 References [1] H. Z. Ibrahim, On a class of α (γ,γ ′ ) -open sets in a topological space, Acta Sci., Technol., 35 (3) (2013), 539-545. [2] H. Z. Ibrahim, On α (γ,γ ′ ) -open sets in topological spaces, (submitted). [3] H. Z. Ibrahim, On (α (γ,γ ′ ) , α (β,β ′ ) )-functions, (submitted). [4] S. Mishra, On α-τ-disconnectedness and α-τ-connectedness in topological spaces, Acta Sci., Technol., 37 (3) (2015), 395- 399. [5] O. Njastad, On some classes of nearly open sets, Pac. J. Math. 15 (1965), 961-970. 1. Introduction 2. Preliminaries 3. (, ')-Connected and (, ')-Disconnected Sets References