1. Introduction

A triplet (*X*, τ_{1}, τ_{2}), where *X* is a non-empty set and τ_{1}, τ_{2} 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*, τ_{1}, τ_{2}) 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 τ_{i} and τ_{j} respectively, where *i, j* ∈ {1, 2}, *i* ≠ *j*.

**Definition 2.1.** A subset *A* of (*X*, τ) is called *b*-open, if *A* ⊂ *int(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*, τ_{1}, τ_{2}) is said to be *(i, j)*-*b*-open if *A* ⊂ *i-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 (τ_{i}, τ_{j}).

**Definition 2.3.** Let *A* be a subset of a bitopological space (*X*,τ_{1}, τ_{2}). 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*, τ_{1}, τ_{2}) is said to be *(i, j)-*generalized *b*-closed (in short, *(i, j)-gb-*closed) set if *( j, i)-bcl(A)* ⊂ *U* whenever *A* ⊂ *U* and *U* is τ_{i}-open in *X*.

We denote the family of all *(i, j)*-*gb*-closed sets and *(i, j)*-*gb*-open sets in(*X*,τ_{1}, τ_{2}) 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*, τ_{1}, τ_{2}) 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*, τ_{1}, τ_{2}) is *(i, j)*-*gb*-open if and only if *U* ⊂ *(j, i)*-*bint(A)*, whenever *U* is τ_{i}-closed and *U* ⊂ *A*.

**Lemma 2.2.** For any subset *A* of a bitopological space (*X*, τ_{1}, τ_{2}), A ⊂ (*i, j*)m-m*gbcl(A)*.

**Lemma 2.3.** Let (*X*, τ_{1}, τ_{2}) 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 *x* ∉ *V*, there exist a (*j, i)-gb*-open set *U* such that *x* ∉ *U* and *V* ⊂ *U*, for *i, j* = 1, 2, *i* ≠ *j*.

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

**Proof.** (a)⇒(b) Let *V* be an *(i, j)*-*gb*-closed set and *x* ∉ *V*. 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 *V* ⊂ *X* \ (*j, i)-gbcl*({x}) = *U*. Thus *V* ⊂ *U* and *x* ∉ *U*.

(b) ⇒ (c) Let *V* be a (*i, j*)-*gb*-closed set and *x* ∉ *V*. By hypothesis, there exists a (*j, i*)-*gb*-open set *U* such that *x* ∉ *U* and *V* ⊂ *U*. 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 *x* ∈ *U*. Now, *X*\*U* is (*i, j*)-*gb*-closed and *x* ∉ *X*\*U*. By (c), (*j, i*)-*gbcl*({x})⋂ (*X*\*U*) = ∅. Which implies (*j, i*)*-gbcl*({x}) ⊂ *U*. Hence (*X*, τ_{1}, τ_{2}) is pairwise _{
gb-Ro
} space.

**Theorem 3.2.** Let (*X*, τ_{1}, τ_{2}) 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 *x* ∈ *X* \ (*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*, τ_{1}, τ_{2}) is not pairwise *gb*-_{
Ro
} space.

Conversely, Let *U* be a (i, j)-*gb*-open set such that *x* ∈ *U*. Suppose (*j, i*)-*gbcl*({x}) is not contained in *U*. Then there exists a *y* ∈ (*j, i*)-*gbcl*({x}) such that *y* ∉ *U* and (*i, j*)-*gbcl*({y})⋂ *U* = ∅, since *X*\*U* is (*i, j*)-*gb*-closed and *y* ∈ *X*\*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 *A* ⊂ *X*. 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 *x* ∈ *X* is the set

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

= {*y : x* ∈ (*i, j*)-*gbcl*({y})}.

**Theorem 3.3.** Let (*X*, τ_{1}, τ_{2}) be a bitopological space and *A* be a subset of *X*. Then (*i, j*)-*gb-*ker(*A*) = {*x* ∈ *X*: (*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 *A* ⊂ *X* \ (*i, j*)-*gbcl*({x}). But *x* ∉ *X* \ (*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 *U* ∈ *GBO*(*i, j*) such that *x* ∉ *U* and *A* ⊂ *U*. Let *y* ∈ (*i, j*)-*gbcl*({x})⋂ *A*. Then *y* ∈ (*i, j*)-*gbcl*({x}) and *y* ∈ *A* ⊂ *U*. Hence *U* ∈ *GBO*(*i, j*) such that *y* ∈ *U* and *x* ∉ *U*, 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*) = { *x* ∈ *X*: (*i, j*)-*gbcl*({x})⋂ A≠ ∅}.

**Theorem 3.4.** Let (*X*, τ_{1}, τ_{2}) be a bitopological space. Then ⋂ {(*i, j*)-*gbcl*({x}) : x ∈ X} = ∅ if and only if (*i, j*)-*gb*-ker({x})≠ *X*, for every *x* ∈ *X*.

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

Conversely assume that (*i, j*)-*gb*-ker({x})≠ *X*, for every *x* ∈ *X*. Let ⋂ {(*i, j*)-*gbcl*({x}) : *x* ∈ *X*} ≠ ∅. If there is some *y* ∈ *X* such that *y* ∈ ⋂{(*i, j*)-*gbcl*({x}) : *x* ∈ *X*}, 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}) : *x* ∈ *X* } = ∅.

**Theorem 3.5.** Let (*X*, τ_{1}, τ_{2}) be a bitopological space. Then the following statements are equivalent :

(a) (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
Ro
} space.

(b) For any *x* ∈ *X*, (*i, j*)-*gbcl*({x}) = (*j, i*)-*gb*-ker({x}) , for *i, j* =1, 2 and *i* ≠ *j*.

(c) For any *x* ∈ *X*, (*i, j*)-*gbcl*({x}) ⊂ (*j, i*)-*gb*-ker({x}) , for *i, j* =1, 2 and *i* ≠ *j*.

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

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

(f) For any (*i, j*)-*gb*-closed set *V* and *x* ∉ *V*, there exist a (*j, i*)-*gb*-open set *U* such that *x* ∉ *U* and *V* ⊂ *U*, for *i, j* =1, 2 and *i* ≠ *j*.

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

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

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

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

**Proof.** (a)⇒ (b) Let *x, y* ∈ *X*. 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, y* ∈ *X* 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, y* ∈ *X* 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 *x* ∉ *V*. Then for any *y* ∈ *V*, 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 *x* ∉ *U* and *V* ⊂ *U*.

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

(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 *A* ⋂ *U* ≠ ∅. Let *x* ∈ *A*⋂ *U*. By (h), *U* = ⋃ {*V*: *V* is (*j, i*)-*gb*-closed and *V* ⊂ *U*}.Then there is a (*j, i*)-*gb*-closed *V* such that *x* ∈ *V* ⊂ *U*. Therefore *x* ∈ *A*⋂ *V* and so *A*⋂ *V* ≠ ∅.

(i)⇒ (j) Let *V* be a (*i, j*)-*gb*-closed set such that *x* ∉ *V*. 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 *W* ⊂ *X* \*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*, τ_{1}, τ_{2}), for any *x* ∈ *X*, (*i, j*)-*gbcl*({x})⋂ (*j, i*)-*gb*-ker({x}) = {x} holds for *i, j* =1, 2 and *i* ≠ *j*, then (*i, j*)-*gbcl*({x}) = {x}.

**Proof.** Since (*X,* τ_{1}, τ_{2}) 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*, τ_{1}, τ_{2}) is a pairwise *gb*-_{
Ro
} space, then for any *x, y* ∈ *X*, 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*, τ_{1}, τ_{2}) 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*, τ_{1}, τ_{2}) is a pairwise *gb*-_{
Ro
} space, then for any *x, y* ∈ *X*, 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*, τ_{1}, τ_{2}) 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 *U* ⋂ *V* =∅ and (*i, j*)-*gbcl*({x}) ⊂ *U*, (*j, i*)-*gbcl*({y}) ⊂ *V* , for *i, j* =1, 2 and *i* ≠ *j*.

**Theorem 4.1.** If (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space, then it is pairwise *gb*-_{
Ro
} space.

**Proof.** Suppose that (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space. Let *U* be an (*i, j*)-*gb*-open set containing *x*. Then for each *y* ∈ *X*\*U*, (j, i) -*gbcl*({x}) ≠ (*i, j*)-*gbcl*({y}). Since (*X*, τ_{1}, τ_{2}) 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
}: *y* ∈ *X*\*U*}. Then *X*\*U* ⊂ *A*, *x* ∉ *A* and *A* is (*j, i*)-*gb*-open set. Therefore (*j, i*)-*gbcl*({x}) ⊂ *X*\*A* ⊂ *U* and hence (*X*, τ_{1}, τ_{2}) 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, y* ∈ *X* 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 *x* ∈ *V* , *y* ∈ *U* and *U* ⋂ *V* = ∅, for *i, j* =1, 2 and *i* ≠ *j*.

**Proof.** Suppose that (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space. Let *x, y* ∈ *X* 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 *x* ∈ *V* , *y* ∈ *U* and *U* ⋂ *V* =∅. Therefore (*i, j*)-*gbcl*({x})⋂ (*j, i*)-*gbcl*({y}) = ∅. So by Theorem 3.2, (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
Ro
} space. Then (*i, j*)-*gbcl*({x}) ⊂ *V* and (*j, i*)-*gbcl*({y}) ⊂ *U*. Hence (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space.

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

(a) (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space.

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

**Proof.** (a)⇒ (b) Suppose that (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
R1
} space. Let *x, y* ∈ *X* 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 *x* ∈ *U*, *y* ∈ *V* and *U* ⋂ *V* = ∅. Then G_{1} = *X*\*V* is (*i, j*)-*gb*-closed and G_{2} = *X*\*U* is (*j, i*)-*gb*-closed set such that *x* ∈ G_{1}, y ∉ G_{1}, *y* ∈ G_{2}, *x* ∉ G_{2} and *X* = G_{1} ⋃ G_{2}.

(b)⇒ (a) Let *x, y* ∈ *X* such that (*i, j*)-*gbcl*({x}) ∉ (*j, i*)-*gbcl*({y}). Therefore for any *x, y* ∈ *X*, *x* ∉ *y* , we have (*i, j*)-*gbcl*({x})⋂ (*j, i*)-*gbcl*({y}) = ∅. Then by Theorem 3.2, (*X*, τ_{1}, τ_{2}) is pairwise *gb*-_{
Ro
} space. By (b), there is an (*i, j*)-*gb*-closed set G_{1} and a (*j, i*)-*gb*-closed set G_{2} such that *x* ∈ G_{1} , *y* ∉ G_{1} , *y* ∈ G_{2} , *x* ∉ G_{2} and *X* = G_{1} ⋃ G_{2}. Therefore *x* ∈ *X*\G_{2} = *U*, which is (*j, i*)-*gb*-open and *y* ∈ *X*\G_{1} = *V*, which is (*i, j*)-*gb*-open. Which implies that (*i, j*)-*gbcl*({x})⊂ *U*, (*j, i*)-*gbcl*({y}) ⊂ *V* and *U* ⋂ *V* =∅ . Hence the result.