Volume 2007, Article ID 16292,11pages doi:10.1155/2007/16292
Research Article
Regular Generalized ω-Closed Sets
Ahmad Al-Omari and Mohd Salmi Md NooraniReceived 12 September 2006; Revised 1 December 2006; Accepted 16 January 2007 Recommended by Lokenath Debnath
In 1982 and 1970, Hdeib and Levine introduced the notions ofω-closed set and gener- alized closed set, respectively. The aim of this paper is to provide a relatively new notion of generalized closed set, namely, regular generalizedω-closed, regular generalizedω- continuous,a-ω-continuous, and regular generalizedω-irresolute maps and to study its fundamental properties.
Copyright © 2007 A. Al-Omari and M. S. M. Noorani. This is an open access article dis- tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop- erly cited.
1. Introduction
All through this paper (X,τ) and (Y,σ) stand for topological spaces with no separation axioms assumed, unless otherwise stated. LetA⊆X, the closure ofAand the interior ofA will be denoted by Cl(A) and Int(A), respectively.Ais regular open ifA=Int(Cl(A)) and Ais regular closed if its complement is regular open; equivalentlyAis regular closed ifA= Cl(Int(A)), see [1]. Let (X,τ) be a space and letAbe a subset ofX. A pointx∈Xis called a condensation point ofAif for eachU∈τ withx∈U, the setU∩Ais uncountable.
Ais calledω-closed [2] if it contains all its condensation points. The complement of an ω-closed set is called ω-open. It is well known that a subsetW of a space (X,τ) isω- open if and only if for eachx∈W, there existsU∈τ such that x∈U andU−W is countable. The family of allω-open subsets of a space (X,τ), denoted byτωorωO(X), forms a topology onX finer thanτ. Theω-closure andω-interior, that can be defined in a manner similar to Cl(A) and Int(A), respectively, will be denoted by Clω(A) and Intω(A), respectively. Several characterizations ofω-closed subsets were provided in [3, 2,4]. Levine [5] introduced the notion of generalized closed sets and a class of topological spaces calledT1/2-spaces. He defined a subsetAof a space (X,τ) to be generalized closed
set (brieflyg-closed) if Cl(A)⊆U wheneverU∈τandA⊆U. Generalized semiclosed [6] (resp.,α-generalized closed [7],θ-generalized closed [8], generalized semi-preclosed [9],δ-generalized closed [10],ω-generalized closed [3,11]) sets are defined by replacing the closure operator in Levine’s original definition by the semiclosure (resp.,α-closure, θ-closure, semi-preclosure,δ-closure,ω-closure) operator.
2. Regular generalizedω-closed sets
A subsetAof (X,τ) is called regular generalized closed (simply,rg-closed) (see [12]) if Cl(A)⊂UwheneverA⊂UandUis regular open. Analogously, we begin this section by introducing the class of regular generalizedω-closed sets.
Definition 2.1. A subsetAof (X,τ) is called regular generalizedω-closed (simply,rgω- closed) if Clω(A)⊂UwheneverA⊂UandUis regular open. A subsetBof (X,τ) is called regular generalizedω-open (simply,rgω-open) if the complement ofBisrgω-closed sets.
We have the following relation forrgω-closed with the other known sets:
ω-c-closed
closed g-closed rg-closed
ω-closed gω-closed rgω-closed
(2.1)
Example 2.2. LetRbe the set of all real numbers, letQbe the set of all rational numbers, with the topologyτ= {R,φ,R−Q}. ThenA=R−Qis notgω-closed, sinceAis open, thusω-open andA⊆A, Clω(A)A(becauseAis notω-closed). Also the only regular open set containingAisX. ThusAisrgω-closed.
Example 2.3. LetX= {a,b,c,d}, with the topologyτ= {φ,X,{a},{b},{a,b},{a,b,c}}. Then the set{a}is notrg-closed, see [13]. But{a}isrgω-closed set, sinceXis finite and τωis discrete topology.
It is clear that if (X,τ) is a countable space, thenrgω(X,τ)=ᏼ(X), wherergω(X,τ) is the set of allrgω-closed subsets ofXandᏼ(X) is the power set ofX.
Since every closed set isω-closed we have the following.
Lemma 2.4. For every subsetAof (X,τ), Clω(A)⊂Cl(A).
The proof of the following result follows from the fact that every regular open set is an open set together withLemma 2.4.
Theorem 2.5. Everygω-closed set andrg-closed set arergω-closed.
Theorem 2.6. LetAbe anrgω-closed subset of (X,τ). Then Clω(A)−Adoes not contain any nonempty regular closed set.
Proof. LetFbe a regular closed subset of (X,τ) such thatF⊆Clω(A)−A. ThenF⊆X− Aand henceA⊆X−F. SinceAisrgω-closed set andX−Fis a regular open subset of (X,τ), Clω(A)⊆X−Fand soF⊆X−Clω(A). ThereforeF⊆Clω(A)∩(X−Clω(A))=
φ.
Theorem 2.7. A subsetAof (X,τ) isrgω-open if and only ifF⊆Intω(A) wheneverFis a regular closed subset such thatF⊆A.
Proof. LetAbe anrgω-open subset ofX and letFbe a regular closed subset ofXsuch thatF⊆A. ThenX−Ais anrgω-closed set andX−A⊆X−F. SinceX−A isrgω- closed,X−Intω(A)=Clω(X−A)⊆X−F. ThusF⊆Intω(A). Conversely, ifF⊆Intω(A) whereFis a regular closed subset of (X,τ) such thatF⊆A, then for any regular open subsetUsuch thatX−A⊆U, we haveX−U⊆Aand thusX−U⊆Intω(A). That is, X−Intω(A)=Clω(X−A)⊆U. ThereforeX-Aisrgω-closed.
Lemma 2.8 [14]. For every openUin a topological spaceXand everyA⊆X, Cl(U∩A)= Cl(U∩Cl(A)).
Recall that two nonempty setsAandBofXare said to be separated if Cl(A)∩B=φ= A∩Cl(B).
Theorem 2.9. IfAandBare open,rgω-open, and separated sets, thenA∪Bisrgω-open.
Proof. LetFbe a regular closed subset ofA∪B. ThenF∩Cl(A)⊆A, sinceAis open and byLemma 2.8we haveF∩Cl(A) is regular closed hence by Theorem 2.7F∩Cl(A)⊆ Intω(A). Similarly,F∩Cl(B)⊆Intω(B). Then we haveF⊆Intω(A∪B) and henceA∪B
isrgω-open.
The following example shows that the union ofrgω-open sets need not bergω-open.
Example 2.10. LetXbe an uncountable set and letA,B,C,Dbe subsets ofX, such that each of them is uncountable set and the family{A,B,C,D}is a partition ofX. We defined the topologyτ= {φ,X,{A},{B},{A,B},{A,B,C}}. Choosex,y /∈Aandx=y. ThenH= A∪ {x}andG=A∪ {y}arergω-closed, since only regular open set containingH,Gis X. ButH∩G= {A}and{A}is regular open inXand Clω(A)A, since{A}is notω- closed. ThusH∩Gis notrgω-closed. Therefore the union ofrgω-open sets need not be rgω-open.
The proof of the following result is straightforward sinceτω is a topology onX and thus omitted.
Theorem 2.11. IfAandBarergω-closed sets, thenA∪Bisrgω-closed.
Theorem 2.12. LetAbe argω-closed subset of (X,τ). IfB⊆Xsuch thatA⊆B⊆Clω(A), thenBis alsorgω-closed. LetBbe a subset of (X,τ) and letAbe anrgω-open subset such that Intω(A)⊆B⊆A. ThenBis alsorgω-open.
The proof is obvious.
Theorem 2.13. IfAbe anrgω-closed subset of (X,τ), then Clω(A)−Aisrgω-open set.
Proof. LetAbe anrgω-closed subset of (X,τ) and letF be a regular closed subset such thatF⊆Clω(A)−A. ByTheorem 2.6,F=φand thusF⊆Intω(Clω(A)−A). ByTheorem
2.7, Clω(A)−Aisrgω-open set.
We first recall the following lemmas to obtain further results forrgω-closed sets.
Lemma 2.14 [3]. If Y is an open subspace of a space X and A is a subset of Y, then Clω|Y(A)=Clω(A)∩(Y).
Lemma 2.15. IfAis a regular open andrgω-closed subset of a spaceX, thenAisω-closed inX.
The proof is obvious.
Theorem 2.16. LetY be an open subspace of a spaceXandA⊆Y. IfAisrgω-closed inX, thenAisrgω-closed inY.
Proof. LetUbe a regular open set ofY such thatA⊆U. ThenU=V∩Yfor some regu- lar open setVofX. SinceAisrgω-closed inX, we have Clω(A)⊆Uand byLemma 2.14, Clω|Y(A)=Clω(A)∩(Y)⊆V∩Y=U. HenceAisrgω-closed inX.
Corollary 2.17. IfAis anrgω-closed regular open set andBis anω-closed set of a space X, thenA∩Bisrgω-closed.
Theorem 2.18. LetAbe anrgω-closed set. ThenA=Clω(Intω(A)) if and only if Clω(Intω
(A))−Ais regular closed.
Proof. If A=Clω(Intω(A)), then Clω(Intω(A))−A=φand hence Clω(Intω(A))−A is regular closed. Conversely, let Clω(Intω(A))−Abe regular closed, since Clω(A)−Acon- tains the regular closed set Clω(Intω(A))−A. ByTheorem 2.6Clω(Intω(A))−A=φand
henceA=Clω(Intω(A)).
Lemma 2.19 [3]. Let (A,τA) be an antilocally countable subspace of a space (X,τ). Then Cl(A)=Clω(A).
We call (X,τ) an antilocally countable space if each nonempty open set is an uncount- able set.
Corollary 2.20. In an antilocally countable subspace of a space (X,τ), the concepts of rgω-closed set andrg-closed set coincide.
Lemma 2.21 [3]. Let (X,τ) and (Y,σ) be two topological spaces. Then (τ×σ)ω⊆τω×σω. Theorem 2.22. IfA×Bisrgω-open subset of (X×Y,τ×σ), thenAisrgω-open subset in (X,τ) andBisrgω-open subset in (Y,σ).
Proof. LetFA be a regular closed subset of (X,τ) and letFB be a regular closed subset of (Y,σ) such thatFA⊆AandFB⊆B. ThenFA×FBis regular closed in (X×Y,τ×σ) such thatFA×FB⊆A×B. By assumptionA×Bisrgω-open in (X×Y,τ×σ) and so
FA×FB⊆Intω(A×B)⊆Intω(A)×Intω(B) byLemma 2.21. ThereforeFA⊆Intω,FB⊆
Intω(B). HenceA,Barergω-open.
The converse of the above need not be true in general.
Example 2.23. LetX=Y =Rwith the usual topologyτ. LetA= {{R−Q} ∪[√2, 5]} andB=(1, 7). ThenAandBarergω-open (ω-open) subsets of (R,τ), whileA×B is notrgω-open in (R×R,τ×τ), since the setF=[√2, 3]×[3, 5] is regular closed set con- tained inA×BandFIntω(A×B). The point (√2, 4)∈F and (√2, 4)∈/ Intω(A×B), because if (√2, 4)∈Intω(A×B), then there exist open setU containing√2 and open setV containing 4 such that (U×V)−(A×B) is countable but (U×V)−(A×B) is uncountable for any open setUcontaining√2 and open setV containing 4.
3. Regular generalizedω-T1/2space
Recall that a space (X,τ) is calledT1/2 [5] if everyg-closed set is closed or equivalently if every singleton is open or closed, Dunham [15]. We introduce the following relatively new definition.
Definition 3.1. A space (X,τ) is a regular generalizedω-T1/2(simply,rgω-T1/2) if every rgω-closed set in (X,τ) isω-closed.
Theorem 3.2. For a space (X,τ), the following are equivalent.
(1)Xis argω-T1/2.
(2) Every singleton is either regular closed orω-open.
Proof. (1)⇒(2) Suppose{x}is not a regular closed subset for somex∈X. ThenX− {x} is not regular open and henceXis the only regular open set containingX− {x}. Therefore X− {x}isrgω-closed. Since (X,τ) isrgω-T1/2space,X− {x}isω-closed and thus{x}is ω-open.
(2)⇒(1) LetAbe anrgω-closed subset of (X,τ) andx∈Clω(A). We show thatx∈A.
If{x}is regular closed andx /∈A, thenx∈(Clω(A)−A). Thus Clω(A)−Acontains a nonempty regular closed set{x}, a contradiction toTheorem 2.6. Sox∈A. If{x}isω- open, sincex∈Clω(A), then for everyω-open setUcontainingx, we haveU∩A=φ. But {x}isω-open then{x} ∩A=φ. Hencex∈A. So in both cases we havex∈A. Therefore
Aisω-closed.
Theorem 3.3. Let (X,τ) be an antilocally countable space. Then (X,τ) is aT1-space if every rgω-closed set isω-closed.
Proof. Letx∈X, and suppose that {x}is not closed. ThenA=X− {x}is not open, and thusAisrgω-closed (the only regular open set containingA isX). Therefore, by assumption,Aisω-closed, and thus{x}isω-open. So there existsU∈τsuch thatx∈U andU− {x}is countable. It follows thatUis a nonempty countable open subset ofx∈X,
a contradiction.
Definition 3.4. A map f :X→Y is said to be
(i) approximately closed [16] (a-closed) provided thatf(F)⊆Int(A) wheneverFis a closed subset ofX,Ais ag-open subset ofY, and f(F)⊆A;
(ii) approximately continuous [16] (a-continuous) provided that Cl(A)⊆ f−1(V) wheneverVis an open subset ofY,Ais ag-closed subset ofX, andA⊆f−1(V).
Definition 3.5. A mapf :X→Yis said to be approximatelyω-closed (simply,a-ω-closed) provided thatf(F)⊆Intω(A) wheneverFis a regular closed subset ofX,Ais anrgω-open ofY, and f(F)⊆A.
Definition 3.6. A map f :X→Y is said to be approximatelyω-continuous (simply,a-ω- continuous) provided that Clω(A)⊆f−1(V) wheneverVis a regular open subset ofY,A is anrgω-closed subset ofX, andA⊆f−1(V).
The notions ofa-closed (resp.;a-continuous) anda-ω-closed (resp.;a-ω-continuous) are independent.
Example 3.7. LetX= {a,b,c,d}with the topologyτ= {φ,X,{a},{b},{a,b},{a,b,c}}. Let f : (X,τ)→(X,τ) be a function defined byf(a)=a, f(b)=d, f(c)=b, f(d)=c. Then f isa-ω-closed, sinceXis finite and thusτωis a discrete topology, and f is nota-closed function. Because the set A= {b,c} isg-open and F= {c,d}is closed, f(F)⊆A, but
f(F)Int(A).
Example 3.8. LetX=Rwith the topologyτ= {φ,X,R−Q}. Let f : (X,τ)→(X,τ) be a function defined by f(x)=0, for allx∈X. Thenf isa-closed, since for any closed setF ofX, the onlyg-open set containingf(F) isX. Andf is nota-ω-closed function. Because the setA=Qisrgω-open andF=Ris regular closed, f(F)⊆A, but f(F)Intω(A)=φ.
Theorem 3.9. A spaceXisrgω-T1/2-space if and only if every spaceY and every function f :X→Yarea-ω-continuous.
Proof. LetV be a regular open subset of Y and A is anrgω-closed subset of X such thatA⊆ f−1(V), sinceX isrgω-T1/2-space thenAisω-closed thusA=Clω(A), hence Clω(A)⊆ f−1(V) and f isa-ω-continuous. LetAbe a nonemptyrgω-closed subset ofX and letYbe the setXwith the topology{Y,A,φ}. Let f :X→Ybe the identity mapping.
By assumption f isa-ω-continuous. SinceAisrgω-closed subset inX and open inY such thatA⊆f−1(A), it follows that Clω(A)⊆f−1(A)=A. HenceAisω-closed inXand
thereforeXisrgω-T1/2-space.
Lemma 3.10. If the regular open and regular closed sets ofXcoincide, then all subsets ofX arergω-closed (and hence all arergω-open).
Proof. LetAbe any subset ofXsuch thatA⊆U andUis regular open, then Clω(A)⊆
Clω(U)⊆Cl(U)=U. ThereforeAisrgω-closed.
Theorem 3.11. If the regular open and regular closed sets ofY coincide, then a function f :X→Yisa-ω-closed if and only if f(F) isω-open for every regular closed subsetFofX.
Proof. Assume f isa-ω-closed byLemma 3.10all subsets of Y arergω-closed. So for any regular closed subsetFofX, f(F) isrgω-closed inY. Since f isa-ω-closed, f(F)⊆ Intω(f(F)), therefore f(F)=Intω(f(F)) thus f(F) is ω-open. Conversely if f(F)⊆A whereFis regular closed andAisrgω-open, then f(F)=Intω(f(F))⊆Intω(A) hence f
isa-ω-closed.
The proof of the following result for a-ω-continuous function is analogous and is omitted.
Theorem 3.12. If the regular open and regular closed sets ofX coincide, then a function f :X→Y isa-ω-continuous if and only if f−1(V) isω-closed for every regular open subset VofY.
4.rgω-continuity
In this section, we will introduce some new classes of maps and study some of their char- acterizations. In [11,3] a map f :X→Y is calledω-irresolute (resp.,R-map [17]) if the inverse image of everyω-closed (resp., regular closed) subset ofY isω-closed (resp., reg- ular closed) inX. In [3], a mapf :X→Y is calledgω-closed if the image of every closed subset ofXisgω-closed inY. Relatively new definitions are given next.
Definition 4.1. A mapf :X→Y is calledrgω-closed (resp., ro-preserving, pre-ω-closed) if f(V) isrgω-closed (resp., regular open,ω-closed) inY for every closed (resp., regular open,ω-closed) subsetVofX.
Example 4.2. LetX= {a,b,c,d}with the topologyτ= {φ,X,{a},{b},{a,b},{a,b,c}}. Let f : (X,τ)→(X,τ) be a function defined byf(a)=a, f(b)=b, f(c)=d, f(d)=c. Then f is ro-preserving, since the family of all regular open sets of Xis{φ,X,{a},{b}}. But if we definedg: (X,τ)→(X,τ) asg(a)=c,g(b)=d,g(c)=a,g(d)=b, theng is not ro-preserving function.
Definition 4.3. A map f :X→Y is calledrgω-continuous (resp.,rgω-irresolute) if the inverse image of everyω-closed (resp.,rgω-closed) subsetVofY isrgω-closed subset of X.
From the definition stated above we obtain the following diagram of implications:
continuous
ω-continuous gω-continuous rgω-continuous
ω-irresolute gω-irresolute rgω-irresolute
(4.1)
Theorem 4.4. Let f :X→Y be a surjective,rgω-irresolute, and pre-ω-closed map ifXis rgω-T1/2-space, thenY is also anrgω-T1/2-space.
Proof. LetAbergω-closed subset ofY. Since f is anrgω-irresolute map, then f−1(A) is anrgω-closed subset ofX. SinceXisrgω-T1/2-space, then f−1(A) is anω-closed subset of X. Since f is a pre-ω-closed map, then f(f−1(A))=Ais anω-closed subset ofY.
ThereforeY is alsorgω-T1/2-space.
Since everygω-closed set isrgω-closed, everygω-closed map isrgω-closed. Next we give new characterization ofgω-closed maps.
Theorem 4.5. A map f :X→Yisgω-closed if and only if for eachA⊆Yand each open set Ucontaining f−1(A), there exists agω-open subsetVofY such thatA⊆V and f−1(V)⊆ U.
Proof. LetF be agω-closed map,A⊆Y, and letU be an open set containing f−1(A).
Then V =Y − f(X−U) is gω-open subset of Y containing A and f−1(V)⊆U.
Conversely let F be closed subset of X and let H be an open subset of Y such that f(F)⊆H. Then f−1(Y− f(F))⊆X−F and X−F is open by hypothesis, there ex- ists a gω-open subset V of Y such thatY− f(F)⊆V and f−1(V)⊆X−F. There- fore,F⊆X−f−1(V) and hence f(F)⊆Y−V. SinceY−H⊆Y−f(F), f−1(Y−H)⊆ f−1(Y− f(F))⊆ f−1(V)⊆X−F, by taking complement, we get F⊆X−f−1(V)⊆ X− f−1(Y−f(F))⊆X−f−1(Y−H). Therefore f(F)⊆Y−V ⊆H. Since Y−V is gω-closed set and Clω(f(F))⊆Clω(Y−V)⊆H, hence f(F) isgω-closed. Thus f is a
gω-closed map.
Since everyω-closed set isrgω-closed, we have the following.
Theorem 4.6. Everyrgω-irresolute map isrgω-continuous map.
Definition 4.7. A subsetA⊆Xis said to beω-c-closed provided that there is a proper subset Bfor which A=Clω(B). A map f :X→Y is said to begω-c-closed if f(A) is gω-closed inYfor everyω-c-closed subsetA⊆X.
Since closed sets are obviouslyω-c-closed,gω-closed maps aregω-c-closed. In a sim- ilar manner, we say a map f :X→Y isrgω-c-closed if f(A) isrgω-closed inY for every ω-c-closed subsetA⊆X.
Theorem 4.8. Let f :X→Y be anR-map andrgω-c-closed. Then f(A) isrgω-closed in Y for everyrgω-closed subsetAofX.
Proof. Let A be an rgω-closed subset of X and let U be a regular open subset of Y such that f(A)⊆U. Since f is an R-map, f−1(U) is a regular open subset ofX and A⊆ f−1(U). AsAis anrgω-closed subset, Clω(A)⊆ f−1(U). Hence f(Clω(A))⊆(U).
Because Clω(A) isω-c-closed andFisrgω-c-closed map,f(Clω(A)) isrgω-closed. There- fore, Clω(f(A))⊆Clω(f(Clω(A)))⊆ f(Clω(A))⊆U. Hence f(A) is anrgω-closed subset
ofY.
Theorem 4.9. Let f :X→Y be ro-preserving andω-irresolute function, ifBisrgω-closed inY, thenf−1(B) isrgω-closed inX.
Proof. LetGbe a regular open subset ofX such that f−1(B)⊆G. ThenB⊆ f(G) and f(G) is regular open. SinceBisrgω-closed, then Clω(A)⊆ f(G) and f−1(Clω(B))⊆G.
Since f isω-irresolute then f−1(Clω(B)) isω-closed and Clω(f−1(Clω(B)))= f−1(Clω
(B)), therefore Clω(f−1((B)))⊆Clω(f−1(Clω(B)))⊆Gthus f−1(B) isrgω-closed inX.
Theorem 4.10. Letf :X→Ybea-ω-closed maps andω-irresolute maps, ifAisrgω-closed inY, thenf−1(A) isrgω-closed inX.
Proof. Assume thatAis anrgω-closed inY and f−1(A)⊆U, whereUis a regular open subset ofX. Taking complements we obtainX−U⊆X− f−1(A)⊆ f−1(Y−A) and f(X−U)⊆Y−A. Since f is a-ω-closed, f(X−U)⊆Intω(Y−A)=Y−Clω(A). It follows thatX−U⊆X− f−1(Clω(A)) and f−1(Clω(A))⊆U, since f is ω-irresolute, f−1(Clω(A)) is ω-closed thus we have f−1(A)⊆ f−1(Clω(A))⊆U and Clω(f−1(A))⊆ Clω(f−1(Clω(A)))= f−1(Clω(A))⊆U. Therefore Clω(f−1(A))⊆Uand f−1(A) isrgω-
closed inX.
Theorem 4.11. If f :X→Y isR-map andrgω-closed andAisg-closed subset ofX, then f(A) isrgω-closed.
Proof. Letf(A)⊆U, whereUis regular open subset ofXthenf−1(U) is regular open set containingA. SinceAisg-closed, we have then Cl(A)⊆ f−1(U) and f(Cl(A))⊆U. Since f isrgω-closed,f(Cl(A)) isrgω-closed. Therefore Clω(f(Cl(A)))⊆Uwhich implies that Clω(f(A))⊆U, hence f(A) isrgω-closed.
The proof ofTheorem 4.8can be easily modified to obtain the following result.
Theorem 4.12. Let f :X→Y bea-ω-map andrgω-c-closed. Then f(A) is rgω-closed subset ofYfor everyrgω-closed subsetAofX.
Theorem 4.13. Let f :X→Y beR-map and pre-ω-closed. Then f(A) isrgω-closed inY for everyrgω-closed subsetAofX.
Proof. LetAbe anyrgω-closed subset ofXand letU be any regular open subset ofY such that f(A)⊆U. Since f isR-map, f−1(U) is regular open andA⊆f−1(U). AsAis rgω-closed, Clω(A)⊆f−1(U). Hencef(Clω(A))⊆U. Therefore Clω(f(A))⊆Clω(f(Clω
(A)))=f(Clω(A))⊆U. Hence f(A) isrgω-closed inY. Definition 4.14. A map f :X→Y is said to beω-contra-R-map if for every regular open subsetVofY, f−1(V) isω-closed.
Example 4.15. LetX=Rwith the usual topologyτand letY= {a,b,c,d}, with the topol- ogyσ= {φ,Y,{a},{b},{a,b},{a,b,c}}. Then the function f : (X,τ)→(Y,σ) defined by
f(x)=
⎧⎨
⎩
a, ifx∈Q,
c, ifx /∈Q, (4.2)
isω-contra-R-map, sinceQisω-closed. But the function f(x) defined by
f(x)=
⎧⎨
⎩
a, ifx∈Q,
b, ifx /∈Q, (4.3)
is notω-contra-R-map, since the family of all regular open set in (Y,σ) is{φ,Y,{a},{b}}
and f−1({b}) is notω-closed.
Theorem 4.16. Let f :X→Y beω-contra-R-map andrgω-c-closed. Then f(A) isrgω- closed inYfor every subsetAofX.
Proof. LetAbe any subset ofXand letUbe any regular open subset ofYsuch thatf(A)⊆ U. ThenA⊆ f−1(U). Since f isω-contra-R-map, f−1(U) isω-closed and so Clω(A)⊆ Clω(f(U))= f(U). Hence f(Clω(A))⊆U. As Clω(A) isω-c-closed subset ofX and f isrgω-c-closed map, f(Clω(A)) isrgω-closed. Therefore Clω(f(A))⊆Clω(f(Cl(A)))⊆
f(Clω(A))⊆U. Thus f(A) isrgω-closed inY. Theorem 4.17. If map f :X→Y isrgω-continuous (resp.,rgω-irresolute) andXisrgω- T1/2, then f isω-continuous (resp.,rgω-irresolute).
Proof. LetAbe any closed (resp.,ω-closed) subset ofY. Since f is anrgω-continuous (resp.,rgω-irresolute) map, f−1(A) is anrgω-closed subset ofX. As (X,τ) isrgω-T1/2
space, f−1(A) is anω-closed subset ofX. Therefore, f is anω-continuous (resp.,rgω-
irresolute).
Theorem 4.18. Let f :X→Ybe a bijective, ro-preserving, andrgω-continuous map. Then f isrgω-irresolute map.
Proof. LetVbe anyrgω-closed subset ofXand letUbe any regular open subset ofYsuch that f−1(V)⊆U. ClearlyV⊆f(U). Sincef is a ro-preserving map, f(U) is regular open and, by assumption,V isrgω-closed set. Hence Clω(V)⊆ f(U) and f−1(Clω(V))⊆U.
Since f isrgω-continuous and Clω(V) isω-closed inY, thenf−1(Clω(V)) is argω-closed subset ofUand so Clω(f−1(Clω(V)))⊆U. Since Clω(f−1(V))⊆Clω(f−1(Clω(V)))⊆U, Clω(f−1(V))⊆U. Thereforef−1(V) is anrgω-closed subset. Hencef isf rgω-irresolute
map.
Theorem 4.19. A map f :X→Y is f rgω-closed if and only if for each subsetBofY and for each open setUcontaining f−1(B), there is anrgω-open setVofYsuch thatB⊆Vand
f−1(V)⊆U.
Proof. Suppose f isrgω-closed, letBbe a subset ofY, andUis an open set ofX such that f−1(B)⊆U. Then f(X−U) isrgω-closed inY. LetV=Y−f(X−U), thenV is rgω-open set and f−1(V)= f−1(Y−f(X−U))=X−(X−U)⊆U thereforeV is an rgω-open set containingBsuch that f−1(V)⊆U. Conversely suppose thatFis a closed set ofXthen f−1(Y−f(F))⊆X−F, andX−Fis open. By hypothesis, there is anrgω- open setV ofYsuch thatY−f(F)⊆V and f−1(V)⊆X−FthereforeF⊆X−f−1(V).
HenceY−V⊆ f(F)⊆ f(X− f−1(V))⊆Y−V implies that f(F)=Y−V, thus f is
rgω-closed.
Acknowledgment
The authors would like to thank the referees for useful comments and suggestions.
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Ahmad Al-Omari: School of Mathematical Sciences, Faculty of Science and Technology, National University of Malaysia (UKM), Selangor 43600, Malaysia
Email address:[email protected]
Mohd Salmi Md Noorani: School of Mathematical Sciences, Faculty of Science and Technology, National University of Malaysia (UKM), Selangor 43600, Malaysia
Email address:[email protected]
