WU Yao-qiang

(College of Arts and Science,Suqian college,Suqian,223800,P.R.China)

§1. Introduction and Preliminaries

O.Njastad[1]defined an operator α on topological spaces and studied α-open sets in 1965.E.Ahmad and F.U.Rehman[2]defined and investigated the properties of(α,β)-continuous in 1993.′A.Cs′asz′ar[3,4]further investigated common generalization concept of γ-open sets in 1997,and studied separation axioms for generalized topologies in 2004,respectively.E.Rosas et al[5,6]introduced the(α,β)-semi open sets and studied the(α,β)-semi Tispaces for i=0,1,2 in 2005,and obtained some results of new generalized separation axioms in 2007,respectively.In 2015 the notion of(α,β)-γ-open sets was introduced[7].In this paper,We introduce the concept of γ-(α,β)-open sets and characterize γ-(α,β)-Tispaces(i=0,1,2,3,4)in topological space.

We recall some definitions of the most essential concepts and some important results needed in the following.

Definition 1.1 Let(X,T)be a topological space,and expX denote the collection of the power set of X.Then we say a mapping α is an operator associated to T on X,if α:expX→expX satisfies the following conditions:

(1)Possessing the property of monotony(i.e,A ? B implies α(A)? α(B));

(2)A?α(A)for every subset A?X;

(3)α(?)= ?,α(X)=X;

(4)G∩α(A)?α(G∩A)for every subset A?X and an open set G∈T.

Definition 1.2 Let(X,T)be a topological space,and α be an operator associated to T on X.(1)A subset A ? X is said to be α-open if for every x ∈ A there exists an open set U such that x ∈ Uand α(U) ? A.The complement of an α-open set is α-closed.(2)The union of all α-open sets of a set A ? X is called the α-interior of A,the intersection of all α-closed sets containing A is called the α-closure of A.We denote the α-interior and the α-closure of A by iα(A)and cα(A).

Theorem 1.1[3]Let(X,T)be a topological space,and α be operator associated to T on X.Then following holds:

(1) ?,X is α-open;

(2)If G is an open set,then G is an α-open set;

(3)Any union of α-open sets is α-open;

Definition 1.3 Let(X,T)be a topological space,and α,β be operators associated to T on X.A subset A ? X is said to be an(α,β)-open set if for every x ∈ A,there exist open sets U,V such that x ∈ U,x ∈ V and α(U)∪ β(V) ? A.The complement of an(α,β)-open set is(α,β)-closed.

Theorem 1.2[7]Let(X,T)be a topological space,and α,β be operators associated to T on X.Then the following statements hold:

(1)If G is an(α,β)-open set,then G is an α-open set(β-open set);

(2)Any union of(α,β)-open sets is(α,β)-open.

Lemma 1.1[7]Let(X,T)be a topological space,and α be operator associated to T on X.Then the following statements hold:

(1)The set iα(A)is the largest α-open subset of A;

(2)The set cα(A)is the smallest α-closed set containing A;

(3)iα(A)? A ? α(A)? cα(A).

Theorem 1.3 Let(X,T)be a topological space,and α,β be operators associated to T on X.Then the following holds:

If α =id,then the notion of(α,β)-open set is exactly to the notion of β-open set.

§2. γ-(α,β)-Open Sets

Definition 2.1 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.A subset A ? X is said to be a γ-(α,β)-open set if for each x ∈ A,there exists an(α,β)-open set V such that x ∈ V and γ(V)? A.The complement of a γ-(α,β)-open set is a γ-(α,β)-closed set.

Theorem 2.1 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.If A is a γ-(α,β)-open set,then A is(α,β)-open.

Proof In fact,if γ =id,then the notion of γ-(α,β)-open set is exactly to the notion of(α,β)-open set.

The following example shows that there exist(α,β)-open sets are not γ-(α,β)-open.

Example 2.1 Let X={a,b,c}and T={?,{a},,{a,b},X},consider the operators α,β,γ defined as follows:α(A)=A,if A={a}and otherwise α(A)=c(A);β(A)=id(A)and γ(A)=c(A),where c(A)denotes the closure of set A.Then the(α,β)-open set of(X,T)are{?,{a},X}and the γ-(α,β)-open set are{?,X}.Therefore{a}is an(α,β)-open set,but that is not γ-(α,β)-open.

Theorem 2.2 Let(X,T)be a topological space,and α,β,γ be operators associated to T on Xthen the following statements hold:

(1)Any union of γ-(α,β)-open sets is γ-(α,β)-open;

(2)If A and B are γ-(α,β)-open sets,then A ∩ B is γ-(α,β)-open.Where γ-(α,β)is a regular operation.i.e for each x ∈ X and any(α,β)-open sets U and V containing x,there exists a(α,β)-open set W containing x such that α(W)? α(U)∩α(V).

Proof (1)Suppose Aiis γ-(α,β)-open where i∈ I.Given x ∈ Si∈IAi,then x ∈ Aifor some i.There exists an(α,β)-open set Visuch that x ∈ Viand γ(Vi) ? Ai? Si∈IAi.Then

(2)Suppose A and B are γ-(α,β)-open sets.Let x ∈ A ∩ B.By Definition 2.1,there exist(α,β)-open sets U and V such that x ∈ U,x ∈ V,γ(U) ? A and γ(V)? B,respectively.Furthermore,γ(U)∩γ(V)? A ∩B,Since γ-(α,β)is a regular operation,there exists a(α,β)-open set W containing x such that γ(W) ? γ(U)∩ γ(V).Hence γ(W) ? A ∩ B,this implies A ∩ B is γ-(α,β)-open.

Corollary 2.1 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X,then the following statements hold:

(1)Any intersection of γ-(α,β)-closed sets is γ-(α,β)-closed;

(2)If A and B are γ-(α,β)-closed sets,then A ∪ B is γ-(α,β)-closed,where γ-(α,β)is a regular operation.

Theorem 2.3 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements are equivalent:

(1)A is γ-(α,β)-open;

(2)A=D ∩γ(A)for some subset D ? X where cγ(D)=X.

Proof(1)?(2)Let D=(X ? γ(A))∪ A.Then cγ(D)=cγ(X ? γ(A))∪ cγ(A)? (X ?cγ(A))∪ cγ(A)=X,by Lemma 1.1.Then cγ(D) ? X,thus cγ(D)=X.By the other wise D∩γ(A)=((X ?γ(A))∪A)∩γ(A)=A∩γ(A)=A.

(2)?(1)is clear!

Definition 2.2 The union of all γ-(α,β)-open sets of a set A ? X is called the γ-(α,β)-interior of A.The intersection of all γ-(α,β)-closed sets containing A is called the γ-(α,β)-closure of A.we denote the γ-(α,β)-interior and the γ-(α,β)-closure of A by iγ?(α,β)(A)and cγ?(α,β)(A).

Theorem 2.4 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements hold:

(1)iγ?(α,β)(A) ? iγ?(α,β)(B)if A ? B;

(2)cγ?(α,β)(A)? cγ?(α,β)(B)if A ? B;

(3)A is a γ-(α,β)-open set ? iγ?(α,β)(A)=A;

(4)A is a γ-(α,β)-closed set ? cγ?(α,β)(A)=A.

(5)x ∈ iγ?(α,β)(A) ? there exists a γ-(α,β)-open set U such that x ∈ U ? A;

(6)x ∈ cγ?(α,β)(A) ? if x ∈ V for all γ-(α,β)-open set V such that V ∩ A 6= ?.

Proof The proof method of Statements(1)-(6)which is similar to the theorem 1.3[7].

§3. γ-(α,β)-TiSpaces(i=0,1,2,3,4)

Definition 3.1 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.

(1)(X,T)is said to be a γ-(α,β)-T0space if for every pair of distinct points x,y ∈ X,there exists a γ-(α,β)-open set P such that x ∈ P,y 6∈ P or x 6∈ P,y ∈ P;

(2)(X,T)is said to be a γ-(α,β)-T1space if for every pair of distinct points x,y ∈ X,there exist a pair of γ-(α,β)-open sets P1and P2such that x ∈ P1,y 6∈ P1and x 6∈ P2,y ∈ P2;

(3)(X,T)is said to be a γ-(α,β)-T2space if for every pair of distinct points x,y ∈ X,there exist a pair of disjoint γ-(α,β)-open sets P1and P2such that x ∈ P1,y ∈ P2;

(4)(X,T)is said to be a γ-(α,β)-regular space if for any γ-(α,β)-closed set F and x 6∈F,there exist a pair of disjoint γ-(α,β)-open sets P1and P2such that x ∈ P1,F ? P2;Furthermore,(X,T)is said to be a γ-(α,β)-T3space if X is γ-(α,β)-regular and γ-(α,β)-T1;

(5)(X,T)is said to be a γ-(α,β)-normal space if for every pair of disjoint γ-(α,β)-closed sets F1and F2,there exist a pair of disjoint γ-(α,β)-open sets P1and P2such that F1? P1,F2? P2;Furthermore,(X,T)is said to be a γ-(α,β)-T4space if X is γ-(α,β)-normal and γ-(α,β)-T1.

From the above definitions,the following relations hold:

Remark 3.1 γ-(α,β)-T4? γ-(α,β)-T3? γ-(α,β)-T2? γ-(α,β)-T1? γ-(α,β)-T0

Theorem 3.1 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements are equivalent:

(1)X is a γ-(α,β)-T0space;

(2)for every pair of distinct points x,y ∈ X,cγ?(α,β)({x})6=cγ?(α,β)({y}).

Proof Sufficiency:By hypothesis,for every pair of distinct points x,y∈X,there exists a γ-(α,β)-open set P such that x ∈ P,y 6∈ P or x 6∈ P,y ∈ P.In the first case,if x ∈ P,y 6∈ P,it shows that x 6∈ cγ?(α,β)({y}).In fact,suppose x ∈ cγ?(α,β)({y}),then easily implies P∩{y}6= ?by Theorem 2.3.Thus y ∈ P,that is a contradiction.Hence cγ?(α,β)({x})6=cγ?(α,β)({y}).Similarly,in the second one,we also have the same result.

Necessity:It is clear that cγ?(α,β)({x})cγ?(α,β)({y})6= ? or cγ?(α,β)({y})cγ?(α,β)({x})6=? by hypothesis.In the first case,it shows that x 6∈ cγ?(α,β)({y}).In fact,suppose that x ∈cγ?(α,β)({y}),then{x} ? cγ?(α,β)({y}).Thus cγ?(α,β)({x}) ? cγ?(α,β)({y}).So cγ?(α,β)({x})cγ?(α,β)({y})= ?.It is a contradiction.On the other hand,it is clear that Xcγ?(α,β)({y})is γ-(α,β)-open.Set P=Xcγ?(α,β)({y}),then x ∈ P and y 6∈ P by definition 2.2.This shows that X is a γ-(α,β)-T0space.

Theorem 3.2 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements are equivalent:

(1)X is a γ-(α,β)-T1space;

(2)For every singleton set{x}is γ-(α,β)-closed where x ∈ X;

(3)Each finite subset A ? X is γ-(α,β)-closed,where γ-(α,β)is a regular operation;

(4)For any subset A ? X is the intersection of all super sets γ-(α,β)-open containing it.i.e A=T{S:A ? S and S is a γ-(α,β)-open set}.

Proof (1) ?(2)Let X is a γ-(α,β)-T1space,then for every pair of distinct points x,y ∈ X,there exist a pair of γ-(α,β)-open sets P1and P2such that x ∈ P1,y 6∈ P1and x 6∈ P2,y ∈ P2,then P2∩{x}= ?.Thus imply that y ∈ P2? X{x}.Then y 6∈ cγ?(α,β)({x})by theorem 2.4,therefore{x}=cγ?(α,β)({x}).This shows that{x}is γ-(α,β)-closed by theorem 2.4.

(2) ?(3)Set G={xi:i=1,2,···,n} ? X,i.e.G={x1} ∪ {x2} ∪ ···∪ {xn}.Because{xi}is γ-(α,β)-closed for every i by theorem 3.2(2),then G is γ-(α,β)-closed by Corollary 2.1.

(3) ?(4)Set GA=T{S:A ? S and S is a γ-(α,β)-open set}.It is clear that A ? GA.Then we will proof A ? GA.Let x 6∈ A,then A ? X{x},therefore{x}is γ-(α,β)-closed by theorem 3.2(3).Hence x 6∈GA,this shows that A?GA.Thus A=GA.

(4)?(1)Set Px={S:x ∈ S and S is γ-(α,β)-open}.It is clear that{x}=TS∈PxS,then y 6∈T

S∈PxS for each y ∈ X and x 6=y.Thus there exists a γ-(α,β)-open set S such that x ∈ S and y 6∈S.Similarly,Given{y}=S where Q={S:y ∈ S and S is γ-(α,β)-openy}.Then there exists a γ-(α,β)-open set S such that y ∈ S and x 6∈ S.This implies X is a γ-(α,β)-T1space.

Theorem 3.3 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.If X is a γ-(α,β)-T2space,then for each pair of distinct points x,y ∈ X,there exists a γ-(α,β)-open set P such that x ∈ P and y 6∈ P.

Proof By hypothesis,suppose X is a γ-(α,β)-T2space,then for every pair of distinct points x,y ∈ X,there exist a pair of disjoint γ-(α,β)-open sets P and P2such that x ∈ P,y ∈ P2.It is clear that P ? XP2and P2is γ-(α,β)-closed.Then cγ?(α,β)(P) ? XP2by theorem 2.4.Hence y 6∈P.

Theorem 3.4 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements are equivalent:

(1)X is a γ-(α,β)-regular space;

(2)If U is a γ-(α,β)-open set and x ∈ U,there exist a pair of sets P and Q,such that x ∈ P ? Q ? U where P is γ-(α,β)-open and Q is γ-(α,β)-closed.

(3)If U is a γ-(α,β)-open set and x ∈ U,there exists a γ-(α,β)-open set P such that x ∈ P ? cγ?(α,β)(P) ? U.

Proof (1) ?(2)since U is a γ-(α,β)-open set and x ∈ U,then XU is γ-(α,β)-closed and x 6∈ XU.Thus there exist a pair of disjoint γ-(α,β)-open sets P1and P2such that x ∈ P1,XU ? P2by definition3.1.Therefore P1? XP2? U where XP2is γ-(α,β)-closed.Set P=P1,Q=XP2,thus x∈P?Q?U.

(2) ?(3)By hypothesis,let x ∈ P ? Q ? U where P is γ-(α,β)-open and Q is γ-(α,β)-closed.Then P ? cγ?(α,β)(P)? cγ?(α,β)(Q)=Q.Therefore,x ∈ P ? cγ?(α,β)(P)? U.

(3) ?(1)For any x ∈ X,let F be a γ-(α,β)-closed set and x 6∈ F.Then x ∈ XF where XF is γ-(α,β)-open.Hence there exists a γ-(α,β)-open set P such that x ∈ P ?cγ?(α,β)(P) ? XF by hypothesis.Set U=Xcγ?(α,β)(P),It is clear that U is γ-(α,β)-open.Hence X is a γ-(α,β)-regular space.

Theorem 3.5 Let(X,T)be a topological space,and α,β,γ be operators associated to T on X.Then the following statements are equivalent:

(1)X is a γ-(α,β)-normal space;

(2)If U is a γ-(α,β)-open set and F ? U for any γ-(α,β)-closed F,there exist a pair of sets P and Q,such that F ? P ? Q ? U where P is γ-(α,β)-open and Q is γ-(α,β)-closed;

(3)If U is a γ-(α,β)-open set and F ? U for any γ-(α,β)-closed F,there exists a γ-(α,β)-open set P such that F ? P ? cγ?(α,β)(P)? U.

Proof (1) ?(2)Let U is a γ-(α,β)-open set and F ? U for any γ-(α,β)-closed F.It is clear that F ∩ (XU)= ? where XU is γ-(α,β)-closed.Since X is a γ-(α,β)-normal space,then there exists a pair of disjoint γ-(α,β)-open sets P1and P2with F ? P1and XU ? P2.Hence P1?XP2?U.Set P=P1,Q=XP2.Thus F?P?Q?U.

(2) ?(3)By hypothesis,let F ? P ? Q ? U,where P is γ-(α,β)-open and Q is γ-(α,β)-closed.Then P ? cγ?(α,β)(P)? cγ?(α,β)(Q)=Q.Thus F ? P ? cγ?(α,β)(P)? U.

(3) ?(1)Let F1,F2be γ-(α,β)-closed where F1∩ F2= ?.Then F1? XF2where XF2is γ-(α,β)-open.Therefore,there exists a γ-(α,β)-open set P such that F1? P ?cγ?(α,β)(P) ? XF2.So F2? Xcγ?(α,β)(P),where Xcγ?(α,β)(P)is a γ-(α,β)-open and P ∩ Xcγ?(α,β)(P)= ?.It shows that X is γ-(α,β)-normal.

From the above conclusions,the following corollaries are easily obtained:

Corollary 3.1 Let(X,T)be a γ-(α,β)-T1space,and α,β,γ be operators associated to T on X.Then X be a γ-(α,β)-T3space if and only if there exists a γ-(α,β)-open set P such that x ∈ P ? cγ?(α,β)(P) ? U,where U is a γ-(α,β)-open set and x ∈ U.

Corollary 3.2 Let(X,T)be a γ-(α,β)-T1space,and α,β,γ be operators associated to T on X.Then X be a γ-(α,β)-T4space if and only if there exists a γ-(α,β)-open set P such that F ? P ? cγ?(α,β)(P) ? U,where U is a γ-(α,β)-open set and F ? U for any γ-(α,β)-closed F.