Abstract: In this paper a new class of sets called πgβ_closed sets is introduced and its properties are studied. Further the notion of πgβ-T½ space and πgβ-continuity are introduced.

Andrijevic [3] introduced a new class of generalized open sets in a topological space, the so-called β-open sets. This type of sets was discussed by Ekici and Caldas [11] under the name of γ-open sets. The class of β-open sets is contained in the class of semi-pre-open sets and contains all semi-open sets and preopen sets. The class of β-open sets generates the same topology as the class of preopen sets. Since the advent of these notations, several research paper with interesting results in different respects came to existence ([1,3,6,11,12,21,22,23]).Levine[16] introduced the concept of generalized closed sets in topological space and a class of topological spaces called T½spaces. Extensive research on generalizing closedness was done in recent years as the notions of a generalized closed, generalized semi-closed α-generalized closed, generalized semi -pre-open closed sets were investigated in [2,7,16,18,19].The finite union of regular open sets is said to be π-open. The complement of a π-open set is said to be π-closed.

The aim of this paper is to study the notion of πgβ-closed sets and its various characterizations are given in this paper. In Section 3, the basic properties of πgβ-closed sets are studied. In section 4, the characterize πgβ-open sets is given. Finally in section 5, πgβ-continuous and πgβ-irresolute functions are discussed.

2. Preliminaries

Throughout this paper (X, τ) and (Y, σ) represent non-empty topological spaces on which no separation axioms are assumed unless otherwise mentioned. For a subset A of a space (X, τ) cl (A) and int (A) denote the closure of A and the interior of A respectively. (X, τ) will be replaced by X if there is no chance of confusion.

Let us recall the following definitions which shall be required later.

Definition 2.1: A subset A of a space (X, τ) is called

(1) a pre-open set [17] if A⊂int (cl (A)) and a preclosed set if cl(int(A))⊂A;

(2) a semi-open set [15] if A⊂cl (int(A)) and a semi-closed set if int(cl(A))⊂A

(3) a α-open set [20] if A⊂int(cl(int(A))) and a α-closed set if cl(int(cl(A)))⊂A

(4) a semi-preopen set[1] A⊂cl(intcl(A)) and a semi-pre-closed set if int(cl(int(A)))⊂A

(5) a regular open set if A=int(cl(A)) and a regular closed set if A=cl(int(A));

(6) b-open [3] or sp-open[8],γ-open[11]if A⊂cl(int(A))∪int(cl(A)).

The complement of a b-open set is said to be b-closed [3].The intersection of all b-closed sets of X containing A is called the b-closure of A and is denoted by bCl(A).The union of all b-open sets of X contained in A is called b-interior of A and is denoted by bInt(A).The family of all b-open (resp α-open, semi-open,preopen,β-open, b-closed,preclosed)subsets of a space X is denoted by bO(X)(resp αO(X),SO(X),PO(X),βO(X),bC(X),PC(X)) and the collection of all b-open subsets of X containing a fixed point x is denoted by bO(X,x).

The sets SO(X, x), α(X, x), PO(X, x), βO(X, x) are defined analogously.

Definition 2.3: A subset A of a space (X, τ) is called

(1) a generalized closed (briefly g-closed) [16] if cl(A)⊂U whenever A⊂U and U is open.

(2) a generalized b-closed (briefly gb-closed) [13] if bcl(A)⊂U whenever A⊂U and U is open.

(3) πg-closed [10] if cl (A) ⊂U whenever A⊂U and U is π-open.

(4) πgp-closed [24] if pcl (A) ⊂U whenever A⊂U and U is π-open.

(5) πgα-closed [14] if αcl (A) ⊂U whenever A⊂U and U is π-open.

(6) πgsp-closed [25] if spcl (A) ⊂U whenever A⊂U and U is π-open.

Definition 2.4: A function f: (X, τ) ⟶ (Y, σ) is called

(1) π-irresolute [4] if f‾ˡ (V) is π -closed in (X, τ) for every π -closed of (Y, σ);

(2) b- irresolute: [11] if for each b-open set V in Y, f ‾ ˡ (V) is b-open in X:

(3) b-continuous: [11] if for each open set V in Y, f ‾ ˡ (V) is b-open in X.

3. πgβ-closed sets

Definition 3.1: A is a subset of (X, τ) is called πgβ-closed if βcl (A) ⊂U whenever A⊂U and U is π-open in (X, τ). BY πGBC (τ) we mean the family of all πgβ-closed subsets of the space (X, τ).

Theorem 3.2:

1. Every closed set is πgβ-closed

2. Every g-closed is πgβ-closed

3. Every α-closed set is πgβ-closed

4. Every pre-closed set is πgβ-closed.

5. Every gβ-closed set is πgβ-closed.

6. Every πg-closed set is πgβ-closed.

7. Every πgp-closed set is πgβ-closed.

8. Every πgα–closed set is πgβ-closed.

9. Every πgs-closed set is πgβ-closed.

10. Every πgβ-closed set is πgsp-closed

Proof: Straight forward converse of the above need not be true as seen in the following examples.

Example3.3:Consider X={a,b,c,d},τ = {ɸ,{a},{d},{a,d},{c,d},{a,c,d},X}.Let A= {c}.Then A is πgβ-closed but not closed, g-closed, α-closed,pre-closed,gβ-closed, πg-closed.

Example 3.4:Let X ={a,b,c,d,e} and τ ={ɸ ,{a,b},{c,d},{a,b,c,d},X}.Let A ={a}.Therefore A is πgβ-closed but not πgα-closed.

Example3.5: Let X={a,b,c,d,e} and τ = {ɸ ,{a,b},{c,d},{a,b,c,d},X}.

Let A = {a, b}.Therefore A is πgβ-closed but not πgp-closed.

Rewmark 3.6: The above discussions are summarized in the following diagram.

Semi closed set sg-closed set gsp-closed set πgβ-closed set πgsp-closed

semi pre closed set pre closed β-closed gβ-closed

Theorem3.7: If A is π-open and πgβ-closed, then A is β-closed.

Proof: Let A is π-open and πgβ-closed.Let A⊂A where A is π-open. Since A is πgβ-closed, βcl (A) ⊂A.Then

A=βcl (A).Hence A is β-closed.

Theorem 3.8: Let A be a πgβ-closed in (X, τ).Then βcl (A)-A does not contain any non empty π-closed set.

Proof:Let F be a non empty π-closed set such that F⊂βcl(A)-A.since A is πgβ closed,A⊂X-F where X-F is π-open implies βcl(A)⊂X-F.Hence F⊂X- βcl(A).Now F⊂βcl(A)∩(X-βcl(A)) implies F=ɸ which is a contradiction. Therefore βcl (A) does not contain any non empty π–closed set.

Corollary3.9: Let A be πgβ-closed in (X, τ).Then A is β-closed iff βcl (A)-A is π- closed.

Proof: Let A be β-closed. Then βcl (A) =A. This implies βcl (A)-A=ɸ which is π-closed. Assume βcl (A)-A is π-closed. Then βcl (A)-A=ɸ. Hence βcl (A) =A.

Remark 3.10: Finite union of πgβ-closed sets need not be πgβ-closed.

Example 3.11: Consider X={a,b,c}, τ ={ ɸ ,{a},{b},{a,b},X}.Let A = {a},B={b}.Here A and B are πgβ-closed but A∪B ={a,b} is not πgβ-closed.

Remark 3.12: Finite intersection of πgβ-closed sets need not be πgβ-closed.

Example3.13:Consider X ={a,b,c,d},τ ={ ɸ,{a},{b},{a,b},{a,b.c},{a,b,d},X}.Let A= {a,b,c},B={a,b,d}.Here A and B are πgβ-closed but A∩B ={a,b} is not πgβ-closed.

Definition3.14 [5]: Let (X, τ) be a topological space A⊂X and xϵX is said to be β-limit point of A iff every β-open set containing x contains a point of A different from x.

Definition 3.15[5]: Let (X,τ ) be a topological space,A⊂X.The set of all β-limit points of A is said to be β-derived set of A and is denoted by Db [A]

Lemma 3.16 [5]: If D (A) = Db [A], then we have cl (A) = βcl (A)

Lemma3.17 [5]: If D (A) ⊂Db [A] for every subset A of X.Then for any subsets F and B of X, we have

Βcl (F∪B) = βcl(F)∪βcl(B)

Theorem3.18: Let A and B be πgβ-closed sets in (X, τ) such that D [A] ⊂Db [A] and D [B] ⊂Db [B]

Then A∪B is πgβ-closed.

Proof: Let U be π–open set such that A∪B⊂U.Since A and B are πgβ-closed sets we have β cl(A)⊂U and βcl(B)⊂U.Since D[A]⊂Db [A] and D[B]⊂Db [B] ,by lemma 3.16,cl(A)=βcl(A) and cl(B)=βcl(B).Thus

βcl(A∪B)⊂cl (A∪B)=cl(A)∪cl(B)=βcl(A)∪βcl(B)⊂U.This implies A∪B is πgβ-closed.

Theorem3.19: If A is πgβ-closed set and B is any set such that A⊂B⊂βcl (A), then B is πgβ-closed set.

Proof: Let B⊂U and U be π-open. Given A⊂B.Then A⊂USince A is πgβ-closed, A⊂U implies βcl (A) ⊂U.By assumption it follows that βcl(B)⊂βcl(A)⊂U.Hence B is a πgβ-closed set.

4. πgβ-open sets

Definition 4.1: A set A⊂X is called πgβ-open if and only if its complement is πgβ-closed.

Remark4.2: βcl(X-A) =X-βint (A)

By πGBO (τ) we mean the family of all πgβ-open subsets of the space (X, τ).

Theorem4.3: If A⊂X is πgβ-open iff F⊂βint (A) whenever F is π-closed and F⊂A

Proof: Necessity: Let A be πgβ-open.Let F be π-closed and F⊂A.Then X-A⊂X-F where X-F is π-open. By assumption, βcl(X-A) ⊂X-F.By remark 4.2, X- β int (A) ⊂X-F.Thus F⊂βint (A).

Sufficiency: Suppose F is π-closed and F⊂A such that F⊂βint (A).Let X-A⊂U where U is π -open. Then X-U⊂A when X-U is π-closed. By hypothesis, X-U⊂β int (A).X-β int (A) ⊂U. βcl(X-A) ⊂U.Thus X-A is πgβ-closed and A is πgβ-open.

Theorem4.4: If βint (A) ⊂B⊂A and A is πgβ-open then B is πgβ-open.

Proof: Let βint (A) ⊂B⊂A.Thus X-A⊂X-B⊂βcl(X-A).Since X-A is πgβ-closed, by theorem 3.19, (X-A) ⊂(X-B) ⊂βcl(X-A) implies (X-B) is πgβ-closed.

Remark 4.5: For any A⊂X, βint (β cl (A)-A) = ɸ

Theorem 4.6: If A⊂X is πgβ-closed, then βcl (A)-A is πgβ-open.

Proof: Let A be πgβ-closed let F be π–closed set.F⊂βcl (A)-A. By theorem3.8, F=ɸ .By remark4.5,

Lemma4.7 [24]: Let A⊂X.If A is open or dense, then πO (A, τ /A) =V ∩ A such that V ϵ π O(X, τ).

Theorem 4.8: Let B ⊂ A ⊂ X where A is πgβ-closed and π-open set. Then B is πgβ-closed relative to A iff B is

Πgβ-closed in X.

Proof: Let B⊂A ⊂ X.Where A is πgβ-closed and π-open set. Let B be πgβ-closed in A.Let B⊂U where U is π-open in X.Since B⊂A, B=B∩A⊂U∩A, this implies βcl(B) =βclA (B)⊂U∩A⊂U.Hence, B is πgβ -closed in X.

Let B be πgβ-closed in X.Let B⊂O where O is π-open in A.Then O=U∩A where U is π-open in X.This implies B⊂ O=U ∩ A ⊂ U.Since B is π g β –closed in X, βcl (B) ⊂U.Thus βclA (B) =A∩β cl(B)⊂U A =O.Hence,B is πgβ-closed relative to A.

Corollary 4.9: Let A be π-open, πgβ-closed set. Then A∩F is πgβ-closed whenever FϵβC(X).

Proof: Since A is πgβ-closed and π-open, then βcl (A) ⊂A and thus A is β-closed. Hence A∩F is β-closed in X which implies A∩F is πgβ-closed in X.

Definition 4.10: A space (X,τ ) is called a πgβ-T1/2 space if every πgβ-closed set is β-closed.

Theorem 4.11:

(i)BO (τ) ⊂πGBO (τ)

(ii) A space (X, τ) is πgβ-T1/2 iff BO (τ) =πGBO (τ).

Proof :( i) Let A be β-open, then X-A is β-closed so X-A is πgβ-closed.Thus A is πgβ-open.Hence

BO (τ) ⊂πGBO (τ)

(ii)Necessity: Let (X, τ) be πgβ-T1/2 space. Let AϵπGBO (τ).Then X-A is πgβ-closed.By hypothesis-A is β-closed thus AϵBO (τ).Thus πGBO (τ) =BO (τ).

Suffiency: Let BO (Τ) =πGBO (τ).Let A be πgβ-closed.Then X-A is πgβ-open X-AϵπGBO (τ).X-AϵBO (τ).Hence A is β-closed. This implies (X, τ) is πgβ-T1/2 space.

Lemma 4.12: Let A be a subset of (X, τ) and xϵX.Then xϵβcl (A) iff V∩ɸ for every β-open set V containing x.

Theorem 4.13: For a topological space (X, τ) the following are equivalent

(i)X is πgβ-T1/2 space.

(ii) Every singleton set is either π-closed or β-open.

Proof: To prove (i) ⇒ (ii): Let X be a πgβ-T1/2 space. Let xϵX and assuming that {x} is not π-closed. Then clearly X-{x} is not π-open. Hence X-{x} is trivially a πgβ-closed.Since X is πgβ-T1/2 space-{x} is β-closed. Therefore {x} is β –open.

(ii) ⇒ (i): Assume every singleton of X is either π-closed or β-open. Let A be a πgβ-closed set. Let {x} ϵ β cl (A).

Case (i): Let {x} be π closed. Suppose {x} does not belong to A. Then {x} ϵ β cl (A)-A.By theorem3.8, {x} ϵ A. Hence β cl (A) ⊂ A.

Case (ii): Let {x} be β-open. Since {x} ϵ β cl (A), we have {x} ∩ A ≠ɸ ⇒ {x} ϵ A. Therefore β cl (A) ⊂ A. Therefore A is β-closed.

5. πgβ-continuous and πgβ-irresolute functions

Definition 5.1:A function f:(X,τ )→(Y,σ ) is called πgβ-continuous if every f ‾ 1 (V) is πgβ-closed in (X,τ ) for every closed set V of (Y,σ ).

Definition 5.2:A function f(X,τ)⟶(y,σ )is called πgβ-irresolute if f ‾ 1 (V) is πgβ-closed in (X,τ ) for every πgβ- closed set V in (Y,σ )

Proposition 5.3: Every πgβ-irresolute function is πgβ-continuous.

Remark 5.4: Converse of the above need not be true.

Proof: Let f :( X, τ) ⟶ (Y, σ) be a πgβ-continuous function and V be a πgβ-closed in (Y, σ).

But every πgβ-closed set need not be closed in (Y, σ).So there exists some sets which is not closed in (Y, σ).By definition there exits some sets which are not πgβ-closed in (X, τ), which implies f is not πgβ-irresolute.

Remark 5.6: Composition of two πgβ-continuous functions need not be πgβ-continuous.

Proof: Let f :( X τ) ⟶ (X, σ) and g :( X, σ) ⟶ (X, η) be two πgβ-continuous functions.

Let V be a closed set in (X, η).Since g is a πgβ-continuous function, g‾ 1 (V) is πgβ-closed set in (X, σ).But every πgβ-closed is not closed. Therefore there exist some sets in (X, σ) which is not πgβ-closed in (X, τ).

Hence g₀f is not πgβ-continuous.

Definition5.8: A function f: X ⟶ Y is said to be pre β-closed if f (U) is β-closed in Y for each β-closed set in X.

Proposition 5.9: Let f :(X,τ )→(Y,σ ) be π-irresolute and pre β-closed map. Then f (A) is πgβ-closed in Y for every πgβ-closed set A of X.

Proof: Let A be πgβ–closed in X.Let f(A)⊂V where V is π-open in Y.Then A⊂ f ‾ 1 (V) and A is πgβ–closed in X implies βcl(A)⊂f ‾ 1 (V).Hence f(β cl(A))⊂V.Since f is pre β-closed, βcl(A)⊂βcl(f(βcl(A)))=f(βcl(A))⊂V.Hence f(A) is πgβ-closed in Y.

Definition 5.10: A topological space X is a πgβ-space if every πgβ-closed set is closed.

Proposition 5.11: Every πgβ space is πgβ-T1/2 space.

Theorem 5.12: Let f :(X,τ ) ⟶ (Y,σ ) be a function.

(1)If is πgβ-irresolute and X is πgβ- T1/2 space, then f is β-irresolute.

(2)If f is πgβ-continuous and X is πgβ-T1/2 space, then f is β-continuous.

Proof :( 1) Let V be β-closed in Y.Since f is πgβ-irresolute, f ‾ 1 (V) is πgβ-closed in X.Since X is πgβ-T1/2 space, f ‾ 1 (V) is β-closed in X.Hence f is β-irresolute.

(2)Let V be closed in Y.Since f is πgβ continuous, f ‾ 1 (V) is πgβ-closed in X.By assumption, it is β-closed. Therefore f is β-continuous.

Definition5.13 [14]: A function f: (X,τ ) ⟶ (Y,σ ) is π-open map in Y for every π-open in X

Theorem5.14: If the bijective f: (X,τ ) ⟶ (Y,σ ) is β-irresolute and π-open map, then f is πgβ-irresolute.

Proof: Let V be πgβ-closed in Y.Let f ‾ 1 (V)⊂U where U is π-open in X.Then V ⊂ f(U)and f(U)is π-open implies βcl(V)⊂f(U).Since f is β-irresolute,(f‾1(βcl(V))) is β-closed. Hence βcl(f‾1 (V))⊂βcl(f‾1(βcl(V)))=f‾1(βcl(V))⊂U. Therefore is πgβ-irresolute.

Theorem 5.15: If f: X⟶Y is π-open, β-irresolute, pre β-closed subjective function. If X is πgβ-T1/2 space, then Y is πgβ-T1/2 space.

Proof: Let V be πgβ-closed in Y.Let f ‾ 1 (V)⊂U where U is π-open in X.Then F⊂f(U) and F is a πgβ–closed set in Y implies β cl(F)⊂f(U).Since f is β-irresolute, β cl( f ‾ 1 (F))⊂β cl(f ‾ 1β cl(F)))= f ‾ 1(β cl(F)) ⊂ U.Therefore f ‾ 1 (F) = F is β-closed in Y.Hence Y is πgβ-T1/2 space.