International Journal of Analysis and Applications

Volume 16, Number 4 (2018), 556-568

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-16-2018-556

ON g-β-IRRESOLUTE FUNCTIONS ON GENERALIZED TOPOLOGICAL SPACES

M K GHOSH∗

Department of Mathematics, Kalyani Mahavidyalaya, Kalyani-741235, Nadia, West Bengal, India

∗Corresponding author: manabghosh@gmail.com

Abstract. In this paper, we introduce and investigate a new kind of function namely g-β-irresolute function

along with its two weak and strong forms in generalized topological spaces. Several characterizations and

interesting properties of these functions are discussed.

1. Introduction

Concepts of generalized topological spaces (GTS), generalized open sets and generalized continuity (=

(g,g′)-continuous functions) were introduced by A. Császár [8, 11, 14]. Since then, several research works de-

voted to generalize the existing notions of topological spaces to generalized topological spaces have appeared.

In [22], [23], W. K. Min introduced the notions of weak (g,g′)-continuity and almost (g,g′)-continuity on

generalized topological spaces. The concept of g-α-irresolute functions on generalized topological spaces was

introduced by Bai and Zuo [4]. In 2013, Bayhan et al. [7] investigated some functions between generalized

topological spaces. Recently, Acikgoz et al. [2] also studied some functions between GTS’s.

On the other hand, Abd El-Monsef et al. [1] introduced the notions of β-open sets and β-continuity

in topological spaces early in 1983. Andrijevic [3] introduced the notion of semi-preopen sets which are

equivalent to β-open sets. Since then, β-open sets [1] played a significant role in the theory of generalized

open sets in topological spaces. In [21], Mahmoud and El-Monsef defined and studied β-irresolute functions.

Received 2018-02-08; accepted 2018-04-06; published 2018-07-02.

2010 Mathematics Subject Classification. 54A05, 54C05, 54C08.

Key words and phrases. generalized topology; g-β-open; g-β-regular; g-β-θ-open; g-β-irresolute; weakly g-β-irresolute;

strongly g-β-irresolute.

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.

556

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-556


Int. J. Anal. Appl. 16 (4) (2018) 557

T. Noiri [26] studied some weak and strong forms of β-irresolute functions in 2003. This work is concerned

with the extension various forms of β-irresolute functions to generalized topological spaces.

2. Preliminaries

A collection g of subsets of X is called a generalized topology (briefly GT) on X [11] if and only if ∅∈ g

and Gi ∈ g for i ∈ I 6= ∅ implies G =
⋃
i∈I Gi ∈ g. A set X with a GT g on X is called a generalized

topological space (GTS) and is denoted by (X,g). By a space X or (X,g), we will always mean a GTS. A

GT g on X is called a strong GT [13] if X ∈ g. For a space (X,g), the elements of g are called g-open sets

and the complements of g-open sets are called g-closed sets.

For A ⊂ X, the g-closure of A, denoted by cA is the intersection of all g-closed sets containing A and

the g-interior of A, denoted by iA is the union of all g-open sets contained in A. It was pointed out in [14]

that each of the operations iA and cA are monotonic i.e. if A ⊂ B ⊂ X, then iA ⊂ iB and cA ⊂ cB,

idempotent [16], i.e. if A ⊂ X, then i(iA) = iA and c(cA) = cA, iA is restricting [16], i.e. if A ⊂ X, then

iA ⊂ A, cA is enlarging [16], i.e., if A ⊂ X, then A ⊂ cA. In a space (X,g), for A ⊂ X, x ∈ iA if and only

if there exists an g-open set V containing x such that V ⊂ A and x ∈ cA if and only if V ∩A 6= ∅ for every

g-open set V containing x [9]. In a space (X,g), A ⊂ X is g-open if and only if A = iA and is g-closed if

and only if A = cA [8] and cA = X \ i(X \A).

A subset A of a topological space is called β-open [1] if A ⊂ cl(int(cl(A))). The complement of a β-open

set is called β-closed. For a subset A of a topological space (X,τ), the β-closure of A, denoted by βcl(A) is

the intersection of all β-open sets containing A and the β-interior of A, denoted by βint(A) is the union of

all β-open sets contained in A.

In a GTS (X,g), a subset A of X is said to be g-β-open (resp. g-α-open, g-preopen, g-semiopen) [14]

if A ⊂ cicA (resp. A ⊂ iciA, A ⊂ icA, A ⊂ ciA). We denote by β(gX) (resp. α(gX), π(gX), σ(gX)) the

class of all g-β-open (resp. resp. g-α-open, g-preopen, g-semiopen) sets of (X,g). From [14], it is clear that,

g ⊂ α(gX) ⊂ σ(gX) ⊂ β(gX), α(gX) ⊂ π(gX) ⊂ β(gX) and each of β(gX) (resp. α(gX), π(gX), σ(gX))

forms a GT on X. The complements of a g-β-open sets (resp. g-α-open, g-preopen, g-semiopen) is called

g-β-closed (resp. g-α-closed, g-preclosed, g-semiclosed) set. We denote by β(gX,x), the set of all g-β-open

sets of (X,g) containing x ∈ X and by βc(gX) the class of all g-β-closed sets of (X,g). For A ⊂ X, we

denote by βcA the intersection of all g-β-closed sets containing A and by βiA the union of all g-β-open sets

contained in A.

3. g-β-regular sets and g-β-θ-open sets

We first state a lemma which will be used in the sequel. Proofs can be checked easily and therefore

omitted.



Int. J. Anal. Appl. 16 (4) (2018) 558

Lemma 3.1. The following hold for a subset A of GTS (X,g):

(i) Arbitrary union of g-β-open sets is g-β-open.

(ii) Arbitrary intersection of g-β-closed sets is g-β-closed.

(iii) βiA = A∩ cicA.

(iv) βcA = A∪ iciA.

(v) x ∈ βcA if A∩U 6= ∅ for every g-β-open set U of X containing x.

(vi) βc(X \A) = X \βiA.

(vii) A is g-β-closed if and only if A = βcA.

(viii) βiA is g-β-open and βcA is g-β-closed.

Lemma 3.2. [22] In a GTS (X,g), X is both g-semiopen and g-β-open.

Definition 3.1. A subset A of a space X is said to be g-β-regular if it is both g-β-open and g-β-closed. The

family of all g-β-regular sets of a space X is denoted by βr(X) and those of containing a point x of X by

βr(X,x).

Theorem 3.1. For a subset A of a GTS (X,g),

(i) A ∈ β(gX) if and only if βcA ∈ βr(X).

(ii) A ∈ βc(gX) if and only if βiA ∈ βr(X).

Proof: (i) First suppose, A ∈ β(gX). Then A ⊂ cicA and therefore, βc(cA) ⊂ βc(cicA) = cicA ⊂ cic(βcA)

i.e. βcA is g-β-open. Since βcA is g-β-open and g-β-closed, βcA ∈ βr(X). Next suppose, βcA ∈ βr(X).

Then A ⊂ βcA ⊂ cic(βcA) ⊂ cic(cA) = cicA. Hence A ∈ β(gX).

(ii) Follows from (i) and Lemma 3.1 (vi).

Theorem 3.2. The following are equivalent for a subset A of a GTS (X,g).

(i) A ∈ βr(X);

(ii) A = βiβcA;

(iii) A = βcβiA.

Proof: Proofs of (i) ⇒ (ii) and (i) ⇒ (iii) are obvious.

Proofs of (ii) ⇒ (i) and (iii) ⇒ (i) follow from Lemma 3.1 and Theorem 3.1.

Definition 3.2. In a GTS (X,g), a point x ∈ X is said to be in the g-β-θ-closure of A, denoted by β-θ-cA,

if A∩βcV 6= ∅ for every g-β-open set V of X containing x.

If β-θ-cA = A, then A is said to be g-β-θ-closed. The complement of a g-β-θ-closed set is said to be

g-β-θ-open. For a subset A of X, union of all g-β-θ-open sets contained in A is said to be g-β-θ-interior of

A, denoted by β-θ-iA.



Int. J. Anal. Appl. 16 (4) (2018) 559

Lemma 3.3. For a subset A of a space (X,g),

β-θ-cA = ∩{V : A ⊂ V and V is g-β-θ-closed } = ∩{V : A ⊂ V and V ∈ βr(X)}

Proof: We give a proof of the first equality, because that of the other is quite similar. Suppose that, x 6∈ β-

θ-cA. Then there exists, g-β-open set V containing x such that βcV ∩ A = ∅. Therefore by Theorem 3.1,

X\βcV is g-β-regular and so g-β-θ-closed set containing A such that x 6∈ X\βcV . Hence, x 6∈ ∩{V : A ⊂ V

and V is g-β-θ-closed set}.

Conversely, suppose that, x 6∈ ∩{V : A ⊂ V and V is g-β-θ-closed set}. Then there exist, a g-β-θ-closed

set V containing A and x 6∈ V . Also, there exists a U ∈ β(gX) such that x ∈ U ⊂ βcU ⊂ X \V . Then we

have, βcU ∩A ⊂ βcU ∩V = ∅ and so x 6∈ β-θ-cA.

Lemma 3.4. Let A and B be any subset of a GTS (X,g). Then the following properties hold:

(i) x ∈ β-θ-cA if and only if A∩V 6= ∅ for every V ∈ βr(X,x).

(ii) If A ⊂ B then β-θ-cA ⊂ β-θ-cB.

(iii) β-θ-c(β-θ-cA) = β-θ-cA.

(iv) intersection of an arbitrary family of g-β-θ-closed sets in X is g-β-θ-closed in X.

(v) A is g-β-θ-open if and only if for each x ∈ A, there exists V ∈ βr(X,x), such that x ∈ V ⊂ A.

(vi) If A ∈ β(g) then βcA = β-θ-cA.

(vii) If A ∈ βr(X) then A is g-β-θ-closed.

(viii) A ∈ βr(X) if and only if A is g-β-θ-open and g-β-θ-closed.

Proof: We give only the proofs of (iii) and (iv). Others proofs are obvious.

(iii) We have β-θ-cA ⊂ β-θ-c(β-θ-cA). Now, if x /∈ β-θ-cA, there exits V ∈ βr(X,x) such that A∩V = ∅.

Since V ∈ βr(X,x), we have β-θ-cA∩V = ∅. This implies x 6∈ β-θc(β-θcA) and so β-θ-c(β-θ-cA) ⊂ β-θ-cA.

(iv) Let Aα be a g-β-θ-closed for each α ∈ ∆. Then for each α ∈ ∆, we have Aα = β-θ-cAα. Therefore β-

θ-c(∩α∈∆Aα) ⊂ ∩α∈∆β-θ-cAα=∩α∈∆Aα ⊂ β-θ-c(∩α∈∆Aα). Hence, β-θ-c(∩α∈∆Aα) = ∩α∈∆Aα. Therefore,

∩α∈∆Aα is g-β-θ-closed.

Corollary 3.1. For a subset A of a GTS (X,g), the following properties hold:

(i) A is g-β-θ-open in X if and only if for each x ∈ A there exists V ∈ βr(X,x) such that x ∈ V ⊂ A.

(ii) β-θ-cA is g-β-θ-closed and β-θ-iA is g-β-θ-open.

(iii) Arbitrary union of g-β-θ-open sets is g-β-θ-open.

Theorem 3.3. For a subset A of a GTS (X,g), the following properties hold:

(i) If A ∈ β(gX) then βcA = β-θ-cA.

(ii) A ∈ βr(X) if and only if A is g-β-θ-open and g-β-θ-closed.



Int. J. Anal. Appl. 16 (4) (2018) 560

Proof: (i) Let A ∈ β(gX) and x /∈ βcA. Then, there exists V ∈ β(gX,x) such that A∩V = ∅. Now, since

A ∈ β(gX) we have A∩βcV = ∅. This implies x /∈ β-θ-cA and so β-θ-cA ⊂ β-cA. Also, for every subset A

of X, we have β-cA ⊂ β-θ-cA. Hence, βcA = β-θ-cA.

(ii) Suppose A ∈ βr(X). Then A = βcA=β-θ-cA. Hence A is g-β-θ-closed. Again, since X \A ∈ βr(X), we

get X \A is g-β-θ-closed and so A is β-θ-open. The converse part is obvious.

Remark 3.1. It is clear that in a GTS (X,g), g-β-regular ⇒ g-β-θ-open ⇒ g-β-open. But the converses

are not necessarily true.

Example 3.1. Let X = {a,b,c,d} and g = {∅,{a},{a,b},{b,c},{a,b,c}} be a GT on X. Then the subsets

{a,b}, {a,b,c} and {a,b,d} of X are g-β-θ-open but not g-β-regular.

Example 3.2. Let X = {a,b,c,d} and g = {∅,{a,c},{b,c},{a,b,c}} be a GT on X. Then the subset {b,c}

of X is g-β-open but not g-β-θ-open.

4. g-β-irresolute functions

Definition 4.1. Let gX and gY be generalized topologies on X and Y respectively. Then a function

f : (X,gX) → (Y,gY ) is defined to be generalized continuous or more properly (gX,gY )-continuous [11]

if f−1(V ) ∈ gX for each V ∈ gY .

Definition 4.2. A function f : (X,gX) → (Y,gY ) is called (β,gY )-continuous [24] if f−1(V ) ∈ β(gX) for

each V ∈ gY .

Definition 4.3. [4] A function f : (X,gX) → (Y,gY ) is called g-α-irresolute if f−1(V ) is g-α-open in X

for every g-α-open set V of Y .

Definition 4.4. A function f : (X,gX) → (Y,gY ) is called g-β-irresolute if the inverse image of each

g-β-open set of Y is g-β-open in X.

Definition 4.5. A function f : (X,gX) → (Y,gY ) is said to be g-β-irresolute at x ∈ X if for each V ∈

β(gY ,f(x)), there exists U ∈ β(gX,x) such that f(U) ⊂ V .

Definition 4.6. A function f : (X,gX) → (Y,gY ) is called weakly g-β-irresolute (resp. strongly g-β-

irresolute) if for each point x ∈ X and each g-β-open set V of Y containing f(x), there exists a g-β-open

set U of X containing x such that f(U) ⊂ βcV (resp. f(βcU) ⊂ V ).

Remark 4.1. From the above definitions we have the following implications:

Strongly g-β-irresolute ⇒ g-β-irresolute ⇒ weakly g-β-irresolute and g-α-irresolute ⇒ g-β-irresolute ⇒

(β,gY )-continuous.



Int. J. Anal. Appl. 16 (4) (2018) 561

We now state basic properties of a g-β-irresolute function. Some results of the following Theorem may be

analogous to Theorem 3.18 of [24] in terms of other terminologies.

Theorem 4.1. Let f : (X,gX) → (Y,gY ) be a function. Then the following are equivalent:

(i) f is g-β-irresolute;

(ii) f−1(F) is g-β-closed in X for every g-β-closed subset F of Y ;

(iii) f(βcA) ⊂ βc(f(A)) for every subset A of X;

(iv) βc(f−1(B)) ⊂ f−1(βcB) for every subset B of Y ;

(v) f−1(βiV ) ⊂ βi(f−1(V )) for every subset V of Y ;

(vi) for every x ∈ X and for every g-β-open set V containing f(x), there exists a g-β-open set U of X

containing x such that f(U) ⊂ V ;

Proof: (i) ⇒ (ii): Obvious.

(ii) ⇒ (iii): Let A be any subset of X. Then since f−1(βc(f(A))) is a g-β-closed set we get βcA ⊂

βc(f−1(f(A))) ⊂ βc(f−1(βc(f(A)))) = f−1(βc(f(A))). Hence f(βcA) ⊂ βc(f(A)).

(iii) ⇒ (iv): For any subset V of Y , using (iii) we get f(βc(f−1(V ))) ⊂ βc(ff−1(V )) ⊂ βcV . Therefore,

βc(f−1(V )) ⊂ f−1f(βc(f−1(V )) ⊂ f−1(βcV ).

(iv) ⇒ (v): For any subset V of Y , using (iv) we get, f−1(βc(Y\V )) ⊃ βc(f−1(Y\V )) = βc(X\f−1(V )). Now

by Lemma 3.1, f−1(βiV ) = f−1(Y \βc(Y \V )) = X\f−1(βc(Y \V )) ⊂ X\βc(X\f−1(V )) = βi(f−1(V )).

(v) ⇒ (i): Let V be any g-β-open subset of Y . Then f−1(V ) = f−1(βiV ) ⊂ βi(f−1(V )) ⊂ f−1(V ). This

implies f−1(V ) = βi(f−1(V )) i.e. f−1(V ) is g-β-open set of X. Hence f is g-β-irresolute.

(i) ⇒ (vi): Let f be g-β-irresolute. Also let x ∈ X and V ∈ β(gY ,f(x)). Then x ∈ f−1(V ) = βi(f−1(V )).

If we set U = f−1(V ), then U ∈ β(gX) and f(U) ⊂ V . Hence f is g-β-irresolute for each x ∈ X.

(vi) ⇒ (i): Let V ∈ β(gY ) and x ∈ f−1(V ). Then f(x) ∈ V . So, there exists U ∈ β(gX,x) such that f(U) ⊂

V . Then x ∈ U ⊂ f−1f(U) ⊂ f−1(V ) and x ∈ U = βiU ⊂ βi(f−1(V )). Therefore f−1(V ) ⊂ βi(f−1(V ))

and so f−1(V ) = βi(f−1(V )). Hence f is g-β-irresolute.

Theorem 4.2. Let f : (X,gX) → (Y,gY ) be a bijective function. Then f is g-β-irresolute if and only if

βi(f(U)) ⊂ f(βiU) for every subset U of X.

Proof: Let V be any subset of X. Then by above Theorem 4.1, f−1(βi(f(V ))) ⊂ βi(f−1f(V )) = βiV .

Therefore, ff−1(βi(f(V ))) ⊂ f(βiV ) so βi(f(V )) ⊂ f(βiV ).

Conversely, let V be any g-β-open set of Y . Then V = βiV = βi(ff−1(V )) ⊂ f(βi(f−1(V ))) i.e.

f−1(V ) ⊂ f−1f(βi(f−1(V )). Since f is bijective, this implies f−1(V ) ⊂ f−1f(βi(f−1(V ))) = βi(f−1(V ))

i.e f−1(V ) = βi(f−1(V )). Therefore f−1(V ) is g-β-open set of X and so f is g-β-irresolute.



Int. J. Anal. Appl. 16 (4) (2018) 562

Definition 4.7. [10] A GTS (X,g) is called β-compact if each cover of X by g-β-open sets of X, has a

finite subcover.

Theorem 4.3. Let f : (X,gX) → (Y,gY ) be a g-β-irresolute function. If (X,gX) is β-compact then so is

(Y,gY ).

Proof: Let {Uα : α ∈ Λ} be a g-β-open cover of Y . Then since f is g-β-irresolute, {f−1(Uα) : α ∈ Λ} is a

g-β-open cover of X. Now, since (X,gX) is β-compact, there exists a finite subcover, say

{f−1(Uα1 ),f−1(Uα2 ), ...,f−1(Uαn )} such that {ff−1(Uα1 ),ff−1(Uα2 ), ...,ff−1(Uαn )}⊂{Uα1,Uα2, ...,Uαn}

is a finite subcover of (Y,gY ). Hence (Y,gY ) is β-compact.

5. Properties of weakly g-β-irresolute functions

Theorem 5.1. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is weakly g-β-irresolute;

(ii) f−1(V ) ⊂ βi(f−1(βcV )) for every g-β-open set V of Y ;

(iii) βc(f−1(V )) ⊂ f−1(βcV ) for every g-β-open set V of Y .

Proof: (i) ⇒ (ii): Let V be any g-β-open set of Y and x ∈ f−1(V ). Then f(U) ⊂ βcV for some U ∈ β(gX,x).

This implies U ⊂ f−1(βcV ) and x ∈ U ⊂ βi(f−1(βcV )). Hence f−1(V ) ⊂ βi(f−1(βcV )).

(ii) ⇒ (iii): Let V ∈ β(gY ) and x 6∈ f−1(βcV ). Then f(x) /∈ βcV . So, there exists W ∈ β(gY ,f(x)) such

that V ∩W = ∅. Now, since V is g-β-open, we have V ∩βcW = ∅ and so f−1(V ) ∩βi(f−1(βcW)) = ∅. As

x ∈ f−1(W) ⊂ βi(f−1(βcW)) ∈ β(gX), we have x /∈ βc(f−1(V )). Hence, βc(f−1(V )) ⊂ f−1(βcV ).

(iii) ⇒ (i): For x ∈ X, suppose V ∈ β(gY ,f(x)). Then by Lemma 3.1, βcV ∈ βr(Y ) and x /∈ f−1(βc(Y\βcV ).

Since Y \βcV is a g-β-open set of Y , we get x /∈ βc(f−1(Y \βcV )). So there exists U ∈ β(gX,x) such that

f−1(Y \βcV )∩U = ∅. This implies f(U)∩(Y \βcV ) = ∅ and so f(U) ⊂ βcV i.e. f is weakly g-β-irresolute.

Theorem 5.2. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is weakly g-β-irresolute;

(ii) βc(f−1(V )) ⊂ f−1(β-θ-cV ) for every subset V of Y ;

(iii) f(βcU) ⊂ β-θ-c(f(U)) for every subset U of X;

(iv) f−1(F) ∈ βc(gX) for every g-β-θ-closed set F of Y ;

(v) f−1(G) ∈ β(gX) for every g-β-θ-open set G of Y .

Proof: (i) ⇒ (ii): Let V be any subset of Y and x /∈ f−1(β-θ-cV ). Then f(x) /∈ β-θ-cV and so there exists

Q ∈ β(gY ,f(x)) such that V ∩βcQ = ∅. Since f is weakly g-β-irresolute, there exists P ∈ β(gX,x) such that

f(P) ⊂ βcQ. Hence f(P) ∩ V = ∅ and so P ∩ f−1(V ) = ∅. Therefore, x /∈ βc(f−1(V )) and consequently

βc(f−1(V )) ⊂ f−1(β-θ-cV ).



Int. J. Anal. Appl. 16 (4) (2018) 563

(ii) ⇒ (iii) : For any subset U of X, we have βcU ⊂ βc(f−1(f(U))) ⊂ f−1(β-θ-f(U)) and so f(βcU) ⊂ β-θ-

c(f(U)).

(iii) ⇒ (iv): For any g-β-θ-closed set F of Y , f(βc(f−1(F))) ⊂ β-θ-c(f(f−1(F))) ⊂ β-θ-cF = F. This

implies βc(f−1(F)) ⊂ f−1(F) and hence βc(f−1(F)) = f−1(F). Therefore f−1(F) ∈ βc(gX).

(iv) ⇒ (v): Obvious.

(v) ⇒ (i): For any x ∈ X, let Q ∈ β(gY ,f(x)). Then by Theorem 3.1 (i) and Theorem 3.3 (ii), we get

βcQ is g-β-θ-open in Y . If we set P = f−1(βcQ), then P ∈ β(gX,x) and f(P) ⊂ βcQ. Hence f is weakly

g-β-irresolute.

Theorem 5.3. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is weakly g-β-irresolute;

(ii) for each x ∈ X and V ∈ β(gY ,f(x)), there exists U ∈ β(gX,x) such that f(βcU) ⊂ βcV ;

(iii) f−1(R) ∈ βr(X) for every R ∈ βr(Y ).

Proof: (i) ⇒ (ii): For any x ∈ X, let V ∈ β(gY ,f(x)). Then by Theorem 3.1 and 3.3, βcV is g-β-θ-open and

g-β-θ-closed in Y . If we set U = f−1(βcV ), then by Theorem 5.2, we have U ∈ βr(X) and so U ∈ β(gX,x).

Also we have f(βcU) ⊂ βcV .

(ii) ⇒ (iii): Let R ∈ βr(Y ) and x ∈ f−1(R). Then we have f(x) ∈ R and there exists U ∈ β(gX,x) such

that f(βcU) ⊂ R. This implies x ∈ U ⊂ βcU ⊂ f−1(R) and so f−1(R) ∈ β(gX). Again since Y \R ∈ βr(Y )

f−1(Y \R) = X \f−1(R) ∈ β(gX). Thus f−1(R) ∈ βc(gX) and consequently f−1(R) ∈ βr(X).

(iii) ⇒ (i): For any x ∈ X, suppose V ∈ β(gY ,f(x)). Then by Theorem 3.1, we get βcV ∈ βr(Y,f(x)) and

f−1(βcV ) ∈ βr(X,x). If we take, U = f−1(βcV ), then U ∈ β(gX,x) and f(U) ⊂ βcV . Hence f is weakly

g-β-irresolute.

Theorem 5.4. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is weakly g-β-irresolute;

(ii) f−1(V ) ⊂ β-θ-i(f−1(β-θ-cV )) for every g-β-open set V of Y ;

(iii) β-θ-c(f−1(V )) ⊂ f−1(β-θ-cV ) for every g-β-open set V of Y .

Proof: The proof is quite similar to the Proof of Theorem 5.1, if we observe that every g-β-closed set is

g-β-θ-closed.

Theorem 5.5. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is weakly g-β-irresolute;

(ii) β-θ-c(f−1(V )) ⊂ f−1(β-θ-cV ) for every subset V of Y ;

(iii) f(β-θ-cU) ⊂ β-θ-c(f(U)) for every subset U of X;



Int. J. Anal. Appl. 16 (4) (2018) 564

(iv) f−1(F) is g-β-θ-closed in X for every g-β-θ-closed set F of Y ;

(v) f−1(G) is g-β-θ-open set in X for every g-β-θ-open set G of Y .

Proof: The proof is quite similar to Proof of Theorem 5.2 and hence omitted.

Definition 5.1. A GTS (X,g) is said to be g-β-regular if for each F ∈ βc(gX) and each x /∈ F , there exist

disjoint g-β-open sets U and V such that x ∈ U and F ⊂ V .

Lemma 5.1. The following properties are equivalent in a GTS (X,g):

(i) X is g-β-regular;

(ii) For each U ∈ β(gX) and each x ∈ U, there exists V ∈ β(gX) such that x ∈ V ⊂ βcV ⊂ U;

(iii) For each U ∈ β(gX) and each x ∈ U, there exists V ∈ βr(X) such that x ∈ V ⊂ U

Proof: Follows from Theorem 3.1.

Theorem 5.6. A function f : (X,gX) → (Y,gY ) is g-β-irresolute if and only if it is weakly g-β-irresolute

and (Y,gY ) is g-β-regular.

Proof: Suppose that f is weakly g-β-irresolute. Let V be any g-β-open set of Y and x ∈ f−1(V ), then

f(x) ∈ V . Now, since Y is g-β-regular, by above Lemma 5.1, there exists W ∈ β(gX) such that f(x) ∈ W ⊂

βcW ⊂ V . Again since f is weakly g-β-irresolute, there exists U ∈ β(gX,x) such that f(U) ⊂ βcW. This

implies x ∈ U ⊂ f−1(V ) and f−1(V ) ∈ β(gX). Hence f is g-β-irresolute. The converse part is obvious.

Proposition 5.1. [28] Let (X,gX) and (Y,gY ) be generalized topological spaces and let U = {U × V :

U ∈ gX,V ∈ gY}. Then U generates a generalized topology gX×Y on X ×Y , called the generalized product

topology on X ×Y , i.e. gX×Y = {all possible union of members of U}

Proposition 5.2. [28] Let (X,gX) and (Y,gY ) be generalized topological spaces, gX×Y be the generalized

topology on X ×Y , A ⊂ X, B ⊂ Y and K ⊂ X ×Y . Then the following hold:

(i) K is gX×Y -open if and only if for each (x,y) ∈ K, there exist Ux ∈ gX and Vy ∈ gY such that (x,y) ∈

Ux ×Vy ⊂ K.

(ii) c(A×B) = cA× cB.

(iii) i(A×B) = iA× iB.

Proposition 5.3. Let (X,gX) and (Y,gY ) be generalized topological spaces, gX×Y be the generalized topology

on X ×Y , A ⊂ X, B ⊂ Y and K ⊂ X ×Y . Then the following hold:

(ii) βc(A×B) = βcA×βcB.

(iii) βi(A×B) = βiA×βiB.



Int. J. Anal. Appl. 16 (4) (2018) 565

Theorem 5.7. A function f : (X,gX) → (Y,gY ) is weakly g-β-irresolute if the graph function, defined by

G(f) = (x,f(x)) for each x ∈ X, is weakly g-β-irresolute.

Proof: Let x ∈ X and V ∈ β(gX). Then using Lemma 3.2, we get X×V is g-β-open set of X×Y containing

G(f). Since G is weakly g-β-irresolute, there exists U ∈ β(gX,x) such that G(U) ⊂ βc(X ×V ) ⊂ X ×βcV .

Hence f(U) ⊂ βcV i.e. f is weakly g-β-irresolute.

Definition 5.2. [27] A GTS (X,g) is said to be g-β-T2 if and only if for each pair of distinct points

x,y ∈ X, there exits disjoint g-β-open sets containing x and y respectively.

Lemma 5.2. A GTS (X,g) is g-β-T2 if and only if for each pair of distinct points x,y ∈ X, there exist

U ∈ β(gX,x) and V ∈ β(gX,y) such that βcU ∩βcV = ∅.

Proof: Follows from Theorem 3.1.

Theorem 5.8. If Y is a g-β-T2 space and f : (X,gX) → (Y,gY ) is a weakly g-β-irresolute injection, then

X is g-β-T2.

Proof: Let x,y be any two distinct points of X, then since f is an injection, we have f(x) 6= f(y). Now Y

being g-β-T2, by Lemma 5.1, there exists U ∈ β(gY ,f(x)) and V ∈ β(gY ,f(y)) such that βcU ∩βcV = ∅.

Again since f is weakly g-β-irresolute, there exist P ∈ β(gX,x) and Q ∈ β(gX,y) such that f(P) ⊂ βcU

and f(Q) ⊂ βcV . This implies P ∩Q = ∅. Therefore X is g-β-T2.

Definition 5.3. A function f : (X,gX) → (Y,gY ) is said to have strongly g-β-closed graph if for each

(x,y) ∈ (X ×Y ) \G(f), there exist U ∈ β(gX,x) and V ∈ β(gY ,y) such that (βcU ×βcV ) ∩G(f) = ∅.

Theorem 5.9. If a function f : (X,gX) → (Y,gY ) is weakly g-β-irresolute, where Y is g-β-T2, then G(f)

is strongly g-β-closed.

Proof: Let (x×y) ∈ (X×Y )\G(f). Then since y 6= f(x), by Lemma 5.1, there exists, U ∈ β(gX,f(x)) and

V ∈ β(gY ,y) such that βcU ∩βcV = ∅. Again since f is weakly g-β-irresolute, by Theorem 5.3, there exists

W ∈ β(gX,x) such that f(βcW) ⊂ βcU. This implies f(βcW) ∩βcV = ∅ and so (βcW ×βcV ) ∩G(f) = ∅.

Hence G(f) is strongly g-β-closed in X ×Y .

Theorem 5.10. If a function f : (X,gX) → (Y,gY ) is weakly g-β-irresolute injection and G(f) is strongly

g-β-closed, then X is g-β-T2.

Proof: Let x,y ∈ X and x 6= y. Since f is an injection f(x) 6= f(y) and (x,f(y)) /∈ G(f). Again since G(f)

is strongly g-β-closed, there exists U ∈ β(gX,x) and V ∈ β(gY ,f(y)) such that f(βcU) ∩βcV = ∅. Also, f

being weakly g-β-irresolute, there exists W ∈ β(gX,y) such that f(W) ⊂ βcV . Hence f(βcU) ∩f(W) = ∅

and so U ∩W = ∅. Therefore X is g-β-T2.



Int. J. Anal. Appl. 16 (4) (2018) 566

Definition 5.4. A GTS (X,g) is said to be connected [29] if there are no nonempty disjoint sets A,B ∈ g

such that A∪B = X. A GTS (X,g) is said to be β-connected [29] if (X,β(gX)) is connected.

Theorem 5.11. If a function f : (X,gX) → (Y,gY ) is a weakly g-β-irresolute surjection and X is β-

connected, then Y is β-connected.

Proof: If possible, suppose that Y is not β-connected. Then there exists nonempty disjoint sets A,B ∈ β(gY )

such that Y = A ∪ B. This implies A,B ∈ βr(Y ) by Lemma 3.2. Now, since f is weakly g-β-irresolute,

by Lemma 3.2 and Theorem 5.3, we get f−1(A),f−1(B) ∈ βr(X). Moreover f being a surjection, X =

f−1(A) ∪f−1(B) where f−1(A) and f−1(B) are disjoint nonempty sets. Therefore X is not β-connected.

6. Properties of strongly g-β-irresolute functions

Theorem 6.1. For a function f : (X,gX) → (Y,gY ), the following are equivalent:

(i) f is strongly g-β-irresolute;

(ii) for each x ∈ X, and each V ∈ β(gY ,f(x)), there exists U ∈ β(gX,x) such that f(β-θ-cU) ⊂ V ;

(iii) for each x ∈ X and each V ∈ β(gY ,f(x)), there exists U ∈ βr(X,x) such that f(U) ⊂ V ;

(iv) for each x ∈ X and for each V ∈ β(gY ,f(x)), there exist an g-β-open set U in X containing x such

that f(U) ⊂ V ;

(v) f−1(G) is g-β-θ-open in X for every G ∈ β(gY );

(vi) f−1(F) is g-β-θ-closed in X for every F ∈ βc(gY );

(vii) f(β-θ-cA) ⊂ βc(f(A)) for every subset A of X;

(viii) β-θ-c(f−1(B)) ⊂ f−1(βcB) for every subset B of Y .

Proof: We first observe that (i) to (iv) are equivalent from Theorem 3.1 and Theorem 3.3.

(iv) ⇒ (v): Let G ∈ β(gY ) and x ∈ f−1(G). Then we have f(x) ∈ G and there exists a g-β-θ-open set U in

X containing x such that f(U) ⊂ G. Therefore, x ∈ U ⊂ f−1(G). Hence by using Corollary 3.1, f−1(G) is

g-β-θ-open in X.

(v) ⇒ (vi): Obvious.

(vi) ⇒ (vii): Let A be any subset of X. Then f−1(βc(f(A))) is g-β-θ-closed in X and so we get β-θ-cA ⊂ β-

θ-c(f−1(f(A))) ⊂ β-θ-c(f−1(βc(f(A)))) = f−1(βc(f(A))). Hence f(β-θ-cA) ⊂ β-c(f(A)).

(vii) ⇒ (viii): Let B be any subset of Y . Then we have f(β-θ-c(f−1(B))) ⊂ βc(f(f−1(B))) ⊂ βcB. Hence

β-θ-c(f−1(B)) ⊂ f−1(βcB).

(viii) ⇒ (i): Let x ∈ X and V ∈ β(gY ,f(x)). Since Y \ V ∈ βc(gY ), we have β-θ-c(f−1(Y \ V )) ⊂

f−1(βc(Y \V )) = f−1(Y \V ). This implies f−1(Y \V ) is g-β-θ-closed in X and so f−1(V ) is a β-θ-open

set containing x. Then there exists U ∈ β(gX,x) such that βcU ⊂ f−1(V ) i.e. f(βcU) ⊂ V . Therefore f is

strongly g-β-irresolute.



Int. J. Anal. Appl. 16 (4) (2018) 567

Theorem 6.2. A g-β-irresolute function f : (X,gX) → (Y,gY ) is strongly g-β-irresolute if and only if X is

a g-β-regular space.

Proof: First let, every g-β-irresolute function be strongly g-β-irresolute. The identity function id : (X,gX) →

(X,gX) is g-β-irresolute and therefore strongly g-β-irresolute. Therefore, for any P ∈ β(gX) and any point

x = id(x) ∈ P , there exists Q ∈ β(gX,x) such that id(βcQ) ⊂ P . This implies x ∈ Q ⊂ βcQ ⊂ P . Hence by

Lemma 5.1, X is g-β-regular.

Conversely, let f : (X,gX) → (Y,gY ) be g-β-irresolute and X be g-β-regular. Then for any x ∈ X and

any Q ∈ β(gY ,f(x)), we get f−1(Q) ∈ β(gX,x). Now, since X is g-β-regular, there exists P ∈ β(gX,x) such

that x ∈ P ⊂ βcP ⊂ f−1(Q) i.e. f(βcP) ⊂ Q. Hence f is strongly g-β-irresolute.

Corollary 6.1. Let X be a g-β-regular space. Then a function f : (X,gX) → (Y,gY ) is strongly g-β-

irresolute if and only if it is g-β-irresolute.

Theorem 6.3. Let f : (X,gX) → (Y,gY ) be a function and G(f) : X → X ×Y be the graph of f. If G(f)

is strongly g-β-irresolute, then f is strongly g-β-irresolute and X is g-β-regular.

Proof: Suppose G(f) is strongly g-β-irresolute. To show, f is strongly g-β-irresolute, let x ∈ X and

Q ∈ β(gY ,f(x)). Now by Lemma 3.2, we have X × Q is a g-β-open set of X × Y containing G(f).

Since G(f) is strongly g-β-irresolute, there exists P ∈ β(gX,x) such that G(βcP) ⊂ X × Q. This implies

f(βcP) ⊂ Q and so f is strongly g-β-irresolute. To show X is g-β-regular, let P ∈ β(gX,x). Then since

G(f) ∈ P ×Y , using Lemma 3.2 we get, P ×Y is g-β-open set in X ×Y . Hence there exists S ∈ β(gX,x)

such that G(βcS) ⊂ P ×Y . Therefore we obtain, x ∈ S ⊂ βcS ⊂ P . So by Lemma 5.1, X is g-β-regular.

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	1. Introduction
	2. Preliminaries
	3. g–regular sets and g—open sets
	4. g–irresolute functions
	5. Properties of weakly g–irresolute functions
	6. Properties of strongly g–irresolute functions
	References