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Proyecciones (Antofagasta)

versión impresa ISSN 0716-0917

Proyecciones (Antofagasta) vol.36 no.4 Antofagasta dic. 2017

http://dx.doi.org/10.4067/S0716-09172017000400589 

Articles

Pairwise Generalized b-Ro Spaces in Bitopological Spaces

Diganta Jyoti Sarma1 

Binod Chandra Tripathy2 

1Central Institute of Technology, Kokrajhar, BTAD, Assam, India, e-mail: dj.sarma@cit.ac.in

2 Tripura University, Department of Mathematics, Suryamaninagar, Tripura, India, e-mail: tripathybc@rediffmail.com

Abstract:

The main purpose of this paper is to introduce pairwise generalized b-Ro spaces in bitopological spaces with the help of generalized b-open sets in bitopological spaces and give several characterizations of this spaces. We also introduce generalized b-kernel of a set and investigate some properties of it and study the relationship between this space and other bitopological spaces.

Key Words: Bitopological spaces; pairwise gb-Ro spaces; pairwise gb-R1 spaces; (i, j)-gb-kernel; (i, j)-gb-open sets

1. Introduction

A triplet (X, τ1, τ2), where X is a non-empty set and τ1, τ2 are topologies on X is called a bitopological space. Kelly (9) initiated the study of bitopological spaces. Later on several authors turned their attention towards generalizations of various concepts of topology by considering bitopological spaces. Andrijevic (6) introduced a new class of generalized open sets called b-open sets in the field of topology and studied several fundamental and interesting properties. Later on Khadra and Nasef (1) and Al-Hawary and Al-Omari (4) defined the notions of b-open sets in bitopological spaces. Al-Hawary (2), (3) studied about the pre-open sets. Tripathy and Sarma (18), (19), (20), (21), (22) have done some works on bitopological spaces using this notion. Ganster and Steiner (8) introduced the concept of generalized b-closed sets in topological spaces. There after Tripathy and Sarma (22) have extended this notion to bitopological spaces. It is observed from literature that there has been a considerable work on different retatively weak form of seperation axiom like Ro axiom. For instance, semi- Ro , pre- Ro , b- Ro are some of the varient form of Ro property that have been investigated by different researchers as seperate entities. Khaleefa (10) has introduced and studied new types of separation axioms termed by generalized b Ro and generalized b- R1 by using generalized b-open sets due to Ganster and Steiner (8). Tripathy and Acharjee (20) have done some works on bitopological spaces.

In this paper, we introduce the notion of pairwise gb- Ro spaces and pairwise gb- R1 spaces in bitopological spaces and investigate some of their properties. In particular, the notion of (i, j)-gb-kernel of a set is also defined in bitopological spaces.

2. Preliminaries

Throughout this paper, (X, τ1, τ2) denotes a bitopological space on which no separation axioms are assumed. For a subset A of X, i-int(A) and j-cl(A) denotes the i-interior and j-closure of A with respect to the topology τi and τj respectively, where i, j ∈ {1, 2}, ij.

Definition 2.1. A subset A of (X, τ) is called b-open, if Aint(cl(A))cl(int(A)) and called b-closed if X\A is b-open.

One may refer to Andrijevic (6) for the above definition.

The following definitions are due to Al-Hawary and Al-Omari (4).

Definition 2.2. A subset A of a bitopological space τ (X, τ1, τ2) is said to be (i, j)-b-open if Ai-int(j-cl(A))j-cl (i-int(A)). The complement of an (i, j)-b-open set is (i, j)-b-closed.

By (i, j) we mean the pair of topologies (τi, τj).

Definition 2.3. Let A be a subset of a bitopological space (X1, τ2). Then

(i) the (i, j)- b-closure of A denoted by (i, j)- b cl(A), is defined by the intersection of all (i, j)- b-closed sets containing A.

(ii) the (i, j)- b-interior of A denoted by (i, j)- b int(A), is defined by the union of all (i, j)- b-open sets contained in A.

The following Definitions and results are due to Tripathy and Sarma (19).

Definition 2.4. A subset A of a bitopological space (X, τ1, τ2) is said to be (i, j)-generalized b-closed (in short, (i, j)-gb-closed) set if ( j, i)-bcl(A)U whenever AU and U is τi-open in X.

We denote the family of all (i, j)-gb-closed sets and (i, j)-gb-open sets in(X1, τ2) by GBC(i, j) and GBO(i, j) respectively.

Definition 2.5. The (i, j)-generalized b-closure of a subset A of a bitopological space (X, τ1, τ2) is the intersection of all (i, j)-gb-closed sets containing A and is denoted by (i, j)-gbcl(A).

Lemma 2.1. A subset A of a bitopological space (X, τ1, τ2) is (i, j)-gb-open if and only if U(j, i)-bint(A), whenever U is τi-closed and UA.

Lemma 2.2. For any subset A of a bitopological space (X, τ1, τ2), A ⊂ (i, j)m-mgbcl(A).

Lemma 2.3. Let (X, τ1, τ2) be a bitopological space. If A is (i, j)- gb-closed subset of X, then A = (i, j)-gbcl(A)

Lemma 2.4. A point x(i, j)- gbcl(A) if and only if for every (i, j)-gb-open set U containing x, U ⋂ A ≠ ∅.

3. Pairwise gb - Ro Spaces

Definition 3.1. A bitopological space (X, τ1, τ2) is said to be pairwise generalized b- Ro (in short, pairwise gb- Ro ) spaces if ( j, i)-gbcl({x}) ⊂ U , for every (i, j)-gb-open set U containing x and i, j =1, 2, i ≠ j.

Theorem 3.1. Let (X, τ1, τ2) be a bitopological space. Then the following statements are equivalent :

(a) (X, τ1, τ2) is pairwise gb-Ro space.

(b) For any (i, j)-gb-closed set V and xV, there exist a (j, i)-gb-open set U such that xU and VU, for i, j = 1, 2, ij.

(c) For any (i, j)-gb-closed set V and xV, (j, i)-gbcl ({x})⋂ V = ∅ for i, j =1, 2, ij.

Proof. (a)⇒(b) Let V be an (i, j)-gb-closed set and xV. By (a), we have (j, i)-gbcl({x}) ⊂ X \ V. Put U = X \ (j, i)-gbcl({x}). Then U is (j, i)-gb-open and VX \ (j, i)-gbcl({x}) = U. Thus VU and xU.

(b) ⇒ (c) Let V be a (i, j)-gb-closed set and xV. By hypothesis, there exists a (j, i)-gb-open set U such that xU and VU. Which implies U ⋂ (j, i)-gbcl({x}) = ∅ since U is (j, i)-gb-open. Hence V ⋂ (j, i)-gbcl({x}) = ∅.

(c)⇒(d) Let U be a (i, j)-gb-open set such that xU. Now, X\U is (i, j)-gb-closed and xX\U. By (c), (j, i)-gbcl({x})⋂ (X\U) = ∅. Which implies (j, i)-gbcl({x}) ⊂ U. Hence (X, τ1, τ2) is pairwise gb-Ro space.

Theorem 3.2. Let (X, τ1, τ2) be a bitopological space. Then X is pairwise gb- Ro space if and only if for any two distinct points x and y of X, either (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}) = ∅ or {x,y} ⊂ (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}).

Proof. Let (i, j)-gbcl({x})⋂(j, i)-gbcl({y})≠ ∅ and {x, y} is not contained in (i, j)-gbcl({x})⋂(j, i)-gbcl({y}). Let z ∈ (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}) and x ∉ (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}). Now, x ∉ (j, i)-gbcl({y}) implies xX \ (j, i)-gbcl({y}), which is a (j, i)-gb-open set containing x. Since z ∈ (j, i)-gbcl({y}), so (i, j)-gbcl({x}) is not contained in X \ (j, i)-gbcl({y}). Hence (X, τ1, τ2) is not pairwise gb- Ro space.

Conversely, Let U be a (i, j)-gb-open set such that xU. Suppose (j, i)-gbcl({x}) is not contained in U. Then there exists a y ∈ (j, i)-gbcl({x}) such that yU and (i, j)-gbcl({y})⋂ U = ∅, since X\U is (i, j)-gb-closed and yX\U. Hence {x, y} is not contained in (i, j)-gbcl({y})⋂(j, i)-gbcl({x}) and (i, j)-gbcl({y})⋂ (j, i)-gbcl({x})≠ ∅.

Now, we introduce the concept of (i, j)-gb-kernel of a set and utilizing it to characterize the notion of pairwise gb- Ro space.

Definition 3.2. Let (X, τ1, τ2) be a bitopological space and AX. The intersection of all (i, j)-gb-open sets containing A is called the (i, j)-gb-kernel of A and is denoted by (i, j)-gb-ker(A).

The (i, j)-gb-kernel of a point xX is the set

(i, j)-gb-ker({x}) = ⋂{U : U is (i, j)-gb-open and xU}.

= {y : x ∈ (i, j)-gbcl({y})}.

Theorem 3.3. Let (X, τ1, τ2) be a bitopological space and A be a subset of X. Then (i, j)-gb-ker(A) = {xX: (i, j)-gbcl({x})⋂ A ≠ ∅}.

Proof. Let x ∈ (i, j)-gb-ker(A) and (i, j)-gbcl({x})⋂ A = ∅. Therefore (i, j)-gbcl({x})⊂ X \A and so AX \ (i, j)-gbcl({x}). But xX \ (i, j)-gbcl({x}), which is a (i, j)-gb-open sets containing A. Thus x ∉ (i, j)-gb-ker(A), a contradiction. Consequently, (i, j)-gbcl({x})⋂ A≠ ∅.

Conversely, let (i, j)-gbcl({x})⋂ A ≠ ∅. If possible, let x ∉ (i, j)-gb-ker(A). Then there exists UGBO(i, j) such that xU and AU. Let y ∈ (i, j)-gbcl({x})⋂ A. Then y ∈ (i, j)-gbcl({x}) and yAU. Hence UGBO(i, j) such that yU and xU, which is a contradiction, since y ∈ (i, j)-gbcl({x}) ∈ GBC(i, j). Therefore x ∈ (i, j)-gb-ker(A). Hence (i, j)-gb-ker(A) = { xX: (i, j)-gbcl({x})⋂ A≠ ∅}.

Theorem 3.4. Let (X, τ1, τ2) be a bitopological space. Then ⋂ {(i, j)-gbcl({x}) : x ∈ X} = ∅ if and only if (i, j)-gb-ker({x})≠ X, for every xX.

Proof. Assume that ⋂ {(i, j)-gbcl({x}) : xX } = ∅. Let (i, j)-gb-ker({x}) = X. If there is some yX, then X is the only (i, j)-gb-open set containing y. Which shows y ∈ (i, j)-gbcl({x}), for every xX. Therefore ⋂ {(i, j)-gbcl({x}) : xX} ≠ ∅, a contradiction. Hence (i, j)-gb-ker({x})≠ X, for every xX.

Conversely assume that (i, j)-gb-ker({x})≠ X, for every xX. Let ⋂ {(i, j)-gbcl({x}) : xX} ≠ ∅. If there is some yX such that y ∈ ⋂{(i, j)-gbcl({x}) : xX}, then every (i, j)-gb-open set containing y must contain every point of X. This shows that X is the only (i, j)-gb-open set containing y. Therefore (i, j)-gb-ker({x}) = X, a contradiction. Hence ⋂ {(i, j)-gbcl({x}) : xX } = ∅.

Theorem 3.5. Let (X, τ1, τ2) be a bitopological space. Then the following statements are equivalent :

(a) (X, τ1, τ2) is pairwise gb- Ro space.

(b) For any xX, (i, j)-gbcl({x}) = (j, i)-gb-ker({x}) , for i, j =1, 2 and ij.

(c) For any xX, (i, j)-gbcl({x}) ⊂ (j, i)-gb-ker({x}) , for i, j =1, 2 and ij.

(d) For any x, yX, y ∈ (i, j)-gb-ker({x}) if and only if x ∈ (j, i)-gb-ker({y}), for i, j =1, 2 and ij.

(e) For any x, yX, y ∈ (i, j)-gbcl({x}) if and only if x ∈ (j, i)-gbcl({y}), for i, j =1, 2 and ij.

(f) For any (i, j)-gb-closed set V and xV, there exist a (j, i)-gb-open set U such that xU and VU, for i, j =1, 2 and ij.

(g) For each (i, j)-gb-closed set V, V = ⋂ {U: U is (j, i)-gb-open and VU}, for i, j =1, 2 and ij.

(h) For each (i, j)-gb-open set U, U = ⋃ {V : V is (j, i)-gb-closed and VU}, for i, j =1, 2 and ij.

(i) For every non-empty subset A of X and for any (i, j)-gb-open set U such that AU ≠ ∅, there exists a (j, i)-gb-closed V such that AV ≠ ∅ and VU, for i, j =1, 2 and ij.

(j) For any (j, i)-gb-closed set V and xV, (j, i)-gbcl({x}) ⋂ V =∅, for i, j =1, 2 and ij.

Proof. (a)⇒ (b) Let x, yX. Then by Definition 3.2, y ∈ (j, i)-gb-ker({x}) ⇔ x ∈ (j, i)-gbcl({y}). Since X is pairwise gb-Ro space, therefore by Theorem 3.2, we have x ∈ (j, i)-gbcl({y}) ⇔ y ∈ (i, j)-gbcl({x}). Thus y ∈ (j, i)-gb-ker({x}) ⇔ x ∈ (j, i)-gbcl({y}) ⇔ y ∈ (i, j)-gbcl({x}). Hence (i, j)-gbcl({x}) = (j, i)-gb-ker({x})

(b)⇒ (c) It is obvious.

(c)⇒ (d) Let x, yX and y ∈ (i, j)-gb-ker({x}). Then by Definition 3.2, x ∈ (i, j)-gbcl({y}). Therefore by (c), x ∈ (i, j)-gbcl({y}) ⊂ (j, i)-gb-ker({y}). Thus x ∈ (j, i)-gb-ker({y}). Similarly, we can prove the other part also.

(d)⇒ (e) Let x, yX and y ∈ (i, j)-gbcl({x}). Then by Definition 3.2, x ∈ (i, j)-gb-ker({y}). Therefore by (d), y ∈ (j, i)-gb-ker({x}) and so x ∈ (j, i)-gbcl({y}). Similarly, we can prove the other part also.

(e)⇒ (f) Let V be a (i, j)-gb-closed set and xV. Then for any yV, we have (i, j)-gbcl({y}) ⊂ V and x ∉ (i, j)-gbcl({y}). Therefore by (e), y ∉ (j, i)-gbcl({x}). That is there exists a (j, i)-gb-open set Uy such that y Uy and x ∉ Uy . Let U = ⋃y∈V { Uy : Uy is (j, i)-gb-open, y Uy and x Uy }. Hence U is (j, i)-gb-open set such that xU and VU.

(f)⇒ (g) Let V be an (i, j)-gb-closed set in X and W = ⋂ {U: U is (j, i)-gb-open and VU}. Clearly, VW. Suppose that, xV. Therefore by (f), there is a (j, i)-gb-open set U such that xU and VU. So xW and thus WV. Hence V = W = ⋂{U: U is (j, i)-gb-open and VU}.

(g)⇒ (h) It is obvious.

(h)⇒ (i) Let A be a non-empty subset of X and U be a (i, j)-gb-open set in X such that AU ≠ ∅. Let xAU. By (h), U = ⋃ {V: V is (j, i)-gb-closed and VU}.Then there is a (j, i)-gb-closed V such that xVU. Therefore xAV and so AV ≠ ∅.

(i)⇒ (j) Let V be a (i, j)-gb-closed set such that xV. Then X \V is (i, j)-gb-open set containing x and {x}⋂ (X \V) ≠ ∅. Therefore by (i), there is a (j, i)-gb-closed set W such that WX \V and {x}⋂ W ≠ ∅. Hence (j, i)-gbcl({x}) ⊂ X \V and so (j, i)-gbcl({x}) ⋂ V =∅.

(j)⇒ (a) Follows from Theorem 3.1.

Theorem 3.6. In a pairwise gb- Ro space (X, τ1, τ2), for any xX, (i, j)-gbcl({x})⋂ (j, i)-gb-ker({x}) = {x} holds for i, j =1, 2 and ij, then (i, j)-gbcl({x}) = {x}.

Proof. Since (X, τ1, τ2) is pairwise gb-Ro space, therefore by Theorem 3.5 (b), we have (i, j)-gbcl({x}) =(j, i)-gb-ker({x}). Hence the result follows.

Theorem 3.7. If (X, τ1, τ2) is a pairwise gb- Ro space, then for any x, yX, either (i, j)-gbcl({x})⋂ (j, i)-gbcl({x}) = (i, j)-gbcl({y})⋂ (j, i)-gbcl({y}) or {(i, j)-gbcl({x})⋂ (j, i)-gbcl({x})}⋂ {(i, j)-gbcl({y})⋂ (j, i)-gbcl({y})} = ∅.

Proof. Let (X, τ1, τ2) is a pairwise gb- Ro space. Suppose that {(i, j)-gbcl({x})⋂ (j, i)-gbcl({x})}⋂ {(i, j)-gbcl({y})⋂ (j, i)-gbcl({y})} ≠ ∅. Let z ∈{(i, j)-gbcl({x})⋂ (j, i)-gbcl({x})}⋂ {(i, j)-gbcl({y})⋂ (j, i)-gbcl({y})}. Then (i, j)-gbcl({z})⊂ (i, j)-gbcl({x})⋂ (i, j)-gbcl({y}) and (j, i)-gbcl({z})⊂ (j, i)-gbcl({x})⋂ (j, i)-gbcl({y}). Since z ∈ (i, j)-gbcl({x}), we have by (e) of Theorem 3.5, x ∈ (j, i)-gbcl({z}). Therefore (j, i)-gbcl({x}) ⊂ (j, i)-gbcl({z})⊂ (j, i)-gbcl({y}). Similarly, z ∈ (j, i)-gbcl({x}) implies (i, j)-gbcl({x}) ⊂ (i, j)-gbcl({y}), z ∈ (i, j)-gbcl({y}) implies (j, i)-gbcl({y}) ⊂ (j, i)-gbcl({x}) and also from z ∈ (j, i)-gbcl({y}) implies (i, j)-gbcl({y}) ⊂ (i, j)-gbcl({x}). Thus (i, j)-gbcl({x}) = (i, j)-gbcl({y}) and (j, i)-gbcl({x}) = (j, i)-gbcl({y}). Hence the result follows.

Theorem 3.8. If (X, τ1, τ2) is a pairwise gb- Ro space, then for any x, yX, either (i, j)-gb-ker({x})⋂ (j, i)-gb-ker({x}) = (i, j)-gb-ker({y})⋂ (j, i)-gb-ker({y})or {(i, j)-gb-ker({x})⋂ (j, i)-gb-ker({x})} ⋂ {(i, j)-gb-ker({y})⋂ (j, i)-gb-ker({y})} = ∅.

Proof. The proof is similar to that of Theorem 3.7 which follows from Definition of (i, j)-gb-ker({x}) and Theorem 3.5.

4. Pairwise gb - R1 Spaces

Definition 4.1. A bitopological space (X, τ1, τ2) is said to be pairwise generalized b- R1 (in short, pairwise gb- R1 ) if for every pair of distinct points x and y of X such that (i, j)-gbcl({x}) ≠ (j, i)-gbcl({y}), there exists a (j, i)-gb-open set U and an (i, j)-gb-open set V such that UV =∅ and (i, j)-gbcl({x}) ⊂ U, (j, i)-gbcl({y}) ⊂ V , for i, j =1, 2 and ij.

Theorem 4.1. If (X, τ1, τ2) is pairwise gb- R1 space, then it is pairwise gb- Ro space.

Proof. Suppose that (X, τ1, τ2) is pairwise gb- R1 space. Let U be an (i, j)-gb-open set containing x. Then for each yX\U, (j, i) -gbcl({x}) ≠ (i, j)-gbcl({y}). Since (X, τ1, τ2) is pairwise gb- R1 , there exists an (i, j)-gb-open set Uy and a (j, i)-gb-open set Vy such that Uy Vy = ∅ and (i, j)-gbcl({y}) ⊂ Vy , (j, i)-gbcl({x}) ⊂ U y. Let A = ⋃ { Vy : yX\U}. Then X\UA, xA and A is (j, i)-gb-open set. Therefore (j, i)-gbcl({x}) ⊂ X\AU and hence (X, τ1, τ2) is pairwise gb- Ro space.

Theorem 4.2. A bitopological space (X, τ1, τ2) is pairwise gb- R1 if and only if for every x, yX such that (i, j)-gbcl({x}) ≠ (j, i)-gbcl({y}), there exists an (i, j)-gb-open set U and a (j, i)-gb-open set V such that xV , yU and UV = ∅, for i, j =1, 2 and ij.

Proof. Suppose that (X, τ1, τ2) is pairwise gb- R1 space. Let x, yX such that (i, j)-gbcl({x}) ≠ (j, i)-gbcl({y}). Then there exists an (i, j)-gb-open set U and a (j, i)-gb-openset V such that x ∈ (i, j)-gbcl({x}) ⊂ V and y ∈ (j, i)-gbcl({y}) ⊂ U.

Conversely, suppose that there exists an (i, j)-gb-open set U and a (j, i)-gb-open set V such that xV , yU and UV =∅. Therefore (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}) = ∅. So by Theorem 3.2, (X, τ1, τ2) is pairwise gb- Ro space. Then (i, j)-gbcl({x}) ⊂ V and (j, i)-gbcl({y}) ⊂ U. Hence (X, τ1, τ2) is pairwise gb- R1 space.

Theorem 4.3. Let (X, τ1, τ2) be a bitopological space. Then the following are equivalent:

(a) (X, τ1, τ2) is pairwise gb- R1 space.

(b) For any x, yX, xy and (i, j)-gbcl({x}) ≠ (j, i)-gbcl({y}) implies that there exists an (i, j)-gb-closed set G1 and a (j, i)-gb-closed set G2 such that x ∈ G1 , y ∉ G1 , y ∈ G2 , x ∉ G2 and X = G1 ⋃ G2, for i, j =1, 2 and ij.

Proof. (a)⇒ (b) Suppose that (X, τ1, τ2) is pairwise gb- R1 space. Let x, yX such that (i, j)-gbcl({x}) ∉ (j, i)-gbcl({y}). Therefore by Theorem 4.2, there exists an (i, j)-gb-open set V and a (j, i)-gb-open set U such that xU, yV and UV = ∅. Then G1 = X\V is (i, j)-gb-closed and G2 = X\U is (j, i)-gb-closed set such that x ∈ G1, y ∉ G1, y ∈ G2, x ∉ G2 and X = G1 ⋃ G2.

(b)⇒ (a) Let x, yX such that (i, j)-gbcl({x}) ∉ (j, i)-gbcl({y}). Therefore for any x, yX, xy , we have (i, j)-gbcl({x})⋂ (j, i)-gbcl({y}) = ∅. Then by Theorem 3.2, (X, τ1, τ2) is pairwise gb- Ro space. By (b), there is an (i, j)-gb-closed set G1 and a (j, i)-gb-closed set G2 such that x ∈ G1 , y ∉ G1 , y ∈ G2 , x ∉ G2 and X = G1 ⋃ G2. Therefore xX\G2 = U, which is (j, i)-gb-open and yX\G1 = V, which is (i, j)-gb-open. Which implies that (i, j)-gbcl({x})⊂ U, (j, i)-gbcl({y}) ⊂ V and UV =∅ . Hence the result.

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Received: January 2017; Accepted: May 2017

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