ON GENERALIZED ω-CLOSED SETS KHALID Y. AL-ZOUBI Received 31 October 2004 and in revised form 3 July 2005

KHALID Y. AL-ZOUBI

Received 31 October 2004 and in revised form 3 July 2005

The class ofω-closed subsets of a space (X,τ) was defined to introduceω-closed func- tions. The aim of this paper is to introduce and study the class ofgω-closed sets. This class of sets is finer thang-closed sets andω-closed sets. We study the fundamental prop- erties of this class of sets. In the space (X,τω), the concepts closed set,g-closed set, and gω-closed set coincide. Further, we introduce and studygω-continuous andgω-irresolute functions.

1. Introduction

Throughout this work, a space will always mean a topological space on which no separa- tion axiom is assumed unless explicitly stated. Let (X,τ) be a space and letAbe a subset of X. A pointx∈Xis called a condensation point ofAif for eachU∈τwithx∈U, the set U∩Ais uncountable.Ais calledω-closed [10] if it contains all its condensation points.

The complement of anω-closed set is calledω-open. It is well known that a subsetWof a space (X,τ) isω-open if and only if for eachx∈W, there existsU∈τsuch thatx∈U andU−Wis countable. The family of allω-open subsets of a space (X,τ), denoted by τω, forms a topology onXfiner thanτ.

In 1970, Levine [13] introduced the notion of generalized closed sets. He defined a subsetAof a space (X,τ) to be generalized and closed (briefly g-closed) if clτ(A)⊆U wheneverU∈τandA⊆U.

Generalized semiclosed [4] (resp.,α-generalized closed [14],θ-generalized closed [8], generalized semi-preclosed [6], δ-generalized closed [7]) 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) operator.

InSection 2of the present work, we follow a similar line to introduce generalizedω- closed sets by utilizing theω-closure operator. We studyg-closed sets andgω-closed sets in the spaces (X,τ) and (X,τω). In particular, we show that a subsetAof a space (X,τ) is closed in (X,τω) if and only if it isg-closed in (X,τω) if and only if it isgω-closed in (X,τω).

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:13 (2005) 2011–2021 DOI:10.1155/IJMMS.2005.2011

In Section 3, we introducegω-continuity andgω-irresoluteness by using gω-closed sets and study some of their fundamental properties.

Now we begin to recall some known notions, definitions, and results which will be used in the work.

Let (X,τ) be a space and letAbe a subset ofX. The closure ofA, the interior ofA, and the relative topology onAwill be denoted by clτ(A), intτ(A), andτA, respectively. The ω-interior (ω-closure) of a subsetAof a space (X,τ) is the interior (closure) ofAin the space (X,τω), and is denoted by intτω(A)(clτω(A)).

Definition 1.1. A space (X,τ) is called

(a) locally countable [3] if each pointx∈Xhas a countable open neighborhood;

(b) anti-locally countable [1] if each nonempty open set is uncountable;

(c)T1/2-space [13] if everyg-closed set is closed (equivalently if every singleton is open or closed, see [8]).

Definition 1.2. A function f : (X,τ)→(Y,σ) is called

(a)g-continuous [5] if f⁻¹(V) isg-closed in (X,τ) for every closed setV of (Y,σ);

(b)g-irresolute [5] iff⁻¹(V) isg-closed in (X,τ) for everyg-closed setVof (Y,σ);

(c)ω-continuous [11] if f⁻¹(V) isω-open in (X,τ) for every open setVof (Y,σ);

(d)ω-irresolute [2] if f⁻¹(V) isω-open in (X,τ) for everyω-open setVof (Y,σ);

(e)α-continuous [15] if f⁻¹(V) isα-set in (X,τ) for every open setV of (Y,σ).

Lemma1.3 [3]. LetAbe a subset of a space(X,τ). Then, (a) (τω)ω=τω;

(b) (τA)ω=(τω)A. 2. Generalizedω-closed sets

Definition 2.1. A subsetAof a space (X,τ) is called generalizedω-closed (briefly,gω- closed) if clτω(A)⊆UwheneverU∈τandA⊆U.

We denote the family of all generalizedω-closed (generalized closed) subsets of a space (X,τ) byGωC(X,τ)(GC(X,τ)).

It is clear that if (X,τ) is a countable space, thenGωC(X,τ)=ᏼ(X), whereᏼ(X) is the power set ofX.

Proposition2.2. Everyg-closed set isgω-closed.

The proof follows immediately from the definitions and the fact thatτωis finer thanτ for any space (X,τ). However, the converse is not true in general as the following example shows.

Example 2.3. LetX= {a,b,c}with the topologyτ= {φ,X,{a},{a,b}}and letA= {a}. ThenA∈GωC(X,τ). ButA /∈GC(X,τ) sinceA⊆A∈τand clτ(A)=XA.

Lemma2.4. Let(A,τA)be an anti-locally countable subspace of a space(X,τ). Thenclτ(A)= clτω(A).

Proof. We need to prove that clτ(A)⊆clτω(A). Suppose that there exists x∈clτ(A)− clτω(A). Then,x /∈clτω(A), and so there existsWx∈τωsuch thatx∈WxandWx∩A= ∅.

Now chooseVx∈τsuch thatx∈VxandVx−Wx=Cxis countable. Then∅ =Vx∩A⊆ A∩(Wx∪Cx)=(A∩Wx)∪(A∩Cx)=A∩Cx⊆Vx∩A. ThusVx∩A=A∩Cx∈τA

(i.e.,Vx∩Ais a nonempty countable open set in (A,τA)), which is a contradiction and

the result follows.

Corollary2.5. Let(A,τA)be an anti-locally countable subspace of a space(X,τ). Then A∈GC(X,τ)if and only ifA∈GωC(X,τ).

Theorem2.6. Let(X,τ)be any space andA⊆X. Then the following are equivalent.

(a)Aisω-closed in(X,τ)(equivalentlyAis closed in(X,τω)).

(b)A∈GC(X,τω).

(c)A∈GωC(X,τω).

Proof. (a)⇒(b). It follows from the fact that every closed set isg-closed.

(b)⇒(c). It is obvious by usingProposition 2.2.

(c)⇒(a). We show that clτω(A)⊆A. Suppose thatx0∈/ A. ThenU=X− {x0}is anω- open set containingA. SinceA∈GωC(X,τω), cl(τω)ω(A)=clτω(A)⊆U(Lemma 1.3(a)), and thusx0∈/ clτω(A). Therefore, clτω(A)=A, that is,Aisω-closed in (X,τ).

In the same way, it can be shown that a subsetAof a space (X,τ) is closed if and only if clτ(A)⊆UwheneverU∈τωandA⊆U.

Proposition2.7. IfA∈GC(X,τω), thenA∈GωC(X,τ)but not conversely.

The proof is obvious.

Example 2.8. LetX=Rbe the set of all real numbers with the topologyτ= {φ,X,{1}}

and putA=R−Q. ThenAis anω-open subset of (X,τ) such that clτω(A)=R− {1}A (i.e.,A /∈GC(X,τω)). However,A∈GωC(X,τ) since the only open set in (X,τ) contain- ingAisX.

InExample 2.8,A∈GC(X,τ)−GC(X,τω). In the following, we give an example of a space (X,τ) and a subsetAofXsuch thatA∈GC(X,τω)−GC(X,τ). In other words, for a space (X,τ), the collectionsGC(X,τ) andGC(X,τω) are independent from each other.

Example 2.9. Consider X =R with the usual topology τu. Put A=(0, 1)∩Q. Then cl(τu)ω(A)=A (A is countable), and so A∈GC(R, (τu)ω). On the other hand, A /∈ GC(R,τu) sinceU=(0, 1) is open in (R,τu) such thatA⊆Uand clτu(A)=[0, 1]U.

Note that in Example 2.9, (R,τu) is anti-locally countable and A=(0, 1)Q∈ GωC(R,τu)−GC(R,τu). Thus the condition that (A,τA) is anti-locally countable in Corollary 2.5cannot be replaced by the condition that (X,τ) is anti-locally countable.

Proposition 2.10. Let A be agω-closed subset of a space(X,τ)andB⊆X. Then the following hold.

(a) clτω(A)−Acontains no nonempty closed set.

(b)IfA⊆B⊆clτω(A), thenB∈GωC(X,τ).

Proof. (a) Suppose by contrary that clτω(A)−Acontains a nonempty closed setC. Then A⊆X−CandX−Cis open in (X,τ). Thus, clτω(A)⊆X−Cor equivalently,C⊆X− clτω(A). Therefore,C⊆(X−clτω(A))∩(clτω(A)−A)= ∅.

(b) Let U ∈τ and B ⊆U. Then A⊆B⊆U. Since A∈GωC(X,τ), clτω(B)⊆ clτω(clτω(A))=clτω(A)⊆U, and the result follows.

Theorem2.11. If(X,τ)is a T1/2-space, then everygω-closed set in(X,τ)isg-closed in (X,τω).

Proof. Let A be a gω-closed subset of (X,τ). By Theorem 2.6, we show that A isω- closed in (X,τ). Suppose, to the contrary, that there exists x∈clτω(A)−A. Then, by Proposition 2.10(a),{x}is not closed. Since (X,τ) is aT1/2-space,{x}is open in (X,τ), and thus it isω-open. Therefore,{x} ∩A= ∅, a contradiction.

In the spaceXfromExample 2.3, everygω-closed set isω-closed while (X,τ) is not a T1/2-space. Thus, the converse ofTheorem 2.11is not true in general.

Theorem2.12. Let(X,τ)be an anti-locally countable space. Then(X,τ)is aT1-space if and only if everygω-closed set isω-closed.

Proof. We need to show the sufficiency part only. Letx∈Xand suppose that{x}is not closed. ThenA=X− {x}is not open, and thusAisgω-closed (the only open set con- tainingAisX). Therefore, by assumption,Aisω-closed, and thus{x}isω-open. So there existsU∈τsuch thatx∈UandU− {x}is countable. It follows thatUis a nonempty

countable open subset of (X,τ), a contradiction.

Proposition2.13. IfᏭ= {Aα:α∈I}is a locally finite collection ofgω-closed sets of a space(X,τ), thenA=

α∈IAαisgω-closed (in particular, a finite union ofgω-closed sets is gω-closed).

Proof. LetU be an open subset of (X,τ) such thatA⊆U. SinceAα∈GωC(X,τ) and Aα⊆U for eachα∈I, clτω(Aα)⊆U. Asτωis a topology onXfiner thanτ,Ꮽis locally finite in (X,τω). Therefore, clτω(A)=clτω(α∈IAα)=

α∈Iclτω(Aα)⊆U. Thus,Aisgω-

closed in (X,τ).

The following two examples show that a countable union ofgω-closed sets and a finite intersection ofgω-closed sets need not begω-closed.

Example 2.14. (a) ConsiderX=Rwith the usual topologyτu. For eachn∈N, putAn= [1/n, 1] andA=_∞

n∈NAn. ThenAis a countable union ofgω-closed sets butA is not gω-closed sinceU=(0, 2)∈τu,A⊆Uand clτω(A)=[0, 1]U.

(b) LetXbe an uncountable set and letAbe a subset ofXsuch thatAandX−Aare uncountable. Letτ= {∅,A,X}. Choosex0,x1∈/ Aandx0=x1. ThenA0=A∪ {x0}and A1=A∪ {x1}are twogω-closed subsets of (X,τ). ButA0A1=Ais notgω-closed since A⊆A∈τand clτω(A)=A.

Proposition2.15. IfA∈GωC(X,τ)andBis closed in(X,τ), thenA∩B∈GωC(X,τ).

Proof. LetU be an open set in (X,τ) such thatA∩B⊆U. PutW=X−B. ThenA⊆ U∪W∈τ. SinceA∈GωC(X,τ), clτω(A)⊆U∪W. Now, clτω(A∩B)⊆clτω(A)clτω(B)

⊆clτω(A)∩clτ(B)=clτω(A)∩B⊆(U∪W)∩B⊆U.

In [11], Hdeib shows that ifAis anω-open subset of a space (X,τ) andBis anω- open subset of a space (Y,σ), thenA×Bneed not beω-open in (X×Y,τ×σ), that is,

τω×σω(τ×σ)ω. To prove that the other inclusion always holds, we need the following lemma.

Lemma2.16. (a)IfAis anω-open subset of a space(X,τ), thenA−Cisω-open for every countable subsetCofX.

(b)The open image of anω-open set isω-open.

Proof. Part (a) is clear. To prove part (b), let f : (X,τ)→(Y,σ) be an open function and letW be anω-open subset of (X,τ). Let y∈f(W). There existsx∈W such that y= f(x). ChooseU∈τsuch thatx∈UandU−W=Cis countable. Since f is open,f(U) is open in (Y,σ) such that y= f(x)∈ f(U) and f(U)−f(W)⊆ f(U−W)= f(C) is

countable. Therefore, f(W) isω-open in (Y,σ).

Theorem2.17. Let(X,τ)and(Y,σ)be two topological spaces. Then(τ×σ)ω⊆τω×σω. Proof. LetW∈(τ×σ)ωand (x,y)∈W. There existU∈τandV∈σsuch that (x,y)∈ U×V andU×V−W=Cis countable. PutW1=(U∩pX(W))−(pX(C)− {x}) and W2=(V∩pY(W))−(pY(C)− {y}), where pX: (X×Y,τ×σ)→(X,τ) and pY: (X× Y,τ×σ)→(Y,σ) are the natural projections. ThenW1∈τω,W2∈σω (Lemma 2.16)

and (x,y)∈W1×W2⊆W. ThusW∈τω×σω.

Definition 2.18. A subsetAof a space (X,τ) is called generalizedω-open (briefly,gω- open) if its complementX−Aisgω-closed in (X,τ).

It is clear that a subset A of a space (X,τ) is gω-open if and only if F⊆intτω(A), wheneverF⊆AandFis closed in (X,τ).

Theorem2.19. IfA×Bis agω-open subset of(X×Y,τ×σ), thenAisgω-open in(X,τ) andBisgω-open in(Y,σ).

Proof. LetFAbe a closed subset of (X,τ) and letFBbe a closed subset of (Y,σ) such that FA⊆AandFB⊆B. ThenFA×FB is closed in (X×Y,τ×σ) such thatFA×FB⊆A×B.

By assumption,A×Bisgω-open in (X×Y,τ×σ), and soFA×FB⊆int(τ×σ)ω(A×B)⊆ intτω(A)×intσω(B) by usingTheorem 2.17. Therefore,FA⊆intτω(A) andFB⊆intσω(A), and the result follows.

The converse of the above theorem need not be true in general.

Example 2.20. LetX=Y =Rwith the usual topologyτu. LetA=R−QandB=(0, 3).

ThenAandBareω-open subsets of (R,τu), whileA×Bis notgω-open in (R×R,τu× τu), since int(τu×τu)ωA×B= ∅and{√

2} ×[1, 2] is a closed set in (R×R,τu×τu) con- tained inA×B.

Theorem2.21. Let(Y,τY)be a subspace of a space(X,τ)andA⊆Y. Then the following hold.

(a)IfA∈GωC(X,τ), thenA∈GωC(Y,τY).

(b)IfA∈GωC(Y,τY)andYisω-closed in(X,τ), thenA∈GωC(X,τ).

Proof. (a) LetVbe an open set of (Y,τY) such thatA⊆V. By usingLemma 1.3(b), there exists an open setU∈τsuch thatV=Y∩U. SinceA∈GωC(X,τ) andA⊆U, clτω(A)⊆ U. Now, cl(τY)ω(A)=cl(τω)Y(A)=clτω(A)∩Y⊆Y∩U=V. Therefore,A∈GωC(Y,τY).

(b) LetA⊆U, whereU∈τ. ThenA⊆Y∩U∈τY. SinceA∈GωC(Y,τY), cl(τY)ω(A)= cl(τω)Y(A)=clτω(A)∩Y⊆Y∩U. Finally, clτω(A)=clτω(A∩Y)⊆clτω(A)∩clτω(Y)=(Y isω-closed) clτω(A)∩Y⊆Y∩U⊆U. ThusA∈GωC(X,τ).

If we chooseA=Y inExample 2.14(b), thenA∈GωC(Y,τY)−GωC(X,τ). There- fore, the condition thatY isω-closed inTheorem 2.21(b) cannot be dropped.

3.gω-continuous functions

Definition 3.1. A function f : (X,τ)→(Y,σ) is called

(a)gω-continuous if f⁻¹(C)∈GωC(X,τ) for every closed subsetCof (Y,σ);

(b)gω-irresolute if f⁻¹(A)∈GωC(X,τ) for everyA∈GωC(Y,σ).

It follows from the definitions that a function f : (X,τ)→(Y,σ) isgω-continuous (gω-irresolute) if and only if f⁻¹(V) isgω-open in (X,τ) for every open (gω-open) subset Vof (Y,σ).

Proposition 3.2. Every g-continuous function and ω-continuous function is gω- continuous.

The proof follows from the definitions and Propositions2.2and2.7.

Example 3.3. (a) LetXbe an uncountable set and letAbe a proper uncountable sub- set ofX. Let f : (X,τindis)→(X,τdis) be the identity function. Then f isgω-continuous (GC(X,τindis)=ᏼ(X)). However, f is notω-continuous sinceAis closed in (X,τdis) and A= f⁻¹(A) is notω-closed in (X,τindis).

(b) Let (X,τ) be as inExample 2.3. Then, the identity function f : (X,τ)→(X,τdis) is gω-continuous but notg-continuous.

Let f : (X,τ)→(Y,σ) be a function. Then a function fω^ω: (X,τω)→(Y,σω) (resp., fω: (X,τω)→(Y,σ), f^ω: (X,τ)→(Y,σω)) associated with f is defined as follows: f_ω^ω(x)=

f(x) (resp., fω(x)= f(x), f^ω(x)= f(x)) for eachx∈X.

Theorem3.4. Let f : (X,τ)→(Y,σ)be a function. Then the following are equivalent.

(a) f_ω^ωis continuous.

(b) f_ω^ωisg-continuous.

(c) f_ω^ωisω-continuous.

(d) f_ω^ωisgω-continuous.

(e) f_ω^ωisgω-irresolute.

(f) f_ω^ωisω-irresolute.

(g) f_ω^ωisg-irresolute.

The proof follows fromTheorem 2.6.

The following result follows immediately from the definitions, Theorem 2.6, and Propositions2.2and2.7.

Theorem3.5. Let f : (X,τ)→(Y,σ)be a function. Then the following hold.

(a) fωisgω-continuous if and only if it isg-continuous.

(b)If fωisgω-irresolute, then f isg-irresolute.

(c)If fωisg-continuous, then f isgω-continuous.

(d)If f^ωisgω-continuous, then f isgω-continuous.

InExample 3.3(a), f is bothgω-continuous andg-irresolute. However, fωis neither g-continuous norgω-irresolute ((τindis)ω=τcocis the cocountable topology). Therefore, the converses of parts (b) and (c) ofTheorem 3.5are not true in general.Example 3.8 shows that also the converse of part (d) is not true.

Proposition3.6. Everygω-irresolute function isgω-continuous but not conversely.

The proof follows immediately from the definitions. For the converse, seeExample 3.8.

Theorem 3.7. If f : (X,τ)→(Y,σ)is closed and f^ω is gω-continuous, then f is gω- irresolute.

Proof. Assume thatAis agω-open subset of (Y,σ) and thatF⊆f⁻¹(A), whereFis closed in (X,τ). Then, f(F) is closed in (Y,σ) such that f(F)⊆A. SinceAisgω-open in (Y,σ), f(F)⊆intσω(A), and thusF⊆f⁻¹(intσω(A)). Since f^ω isgω-continuous and intσω(A) is open in (Y,σω), f⁻¹(intσω(A)) isgω-open in (X,τ). Therefore,F⊆intτω(f⁻¹(intσω(A)))⊆ intτω(f⁻¹(A)). This means that f⁻¹(A) isgω-open in (X,τ), and thus f isgω-irresolute.

The following example shows that the condition thatf^ωisgω-continuous inTheorem 3.7cannot be weakened to f beinggω-continuous.

Example 3.8. Let (X,τ) andA⊂Xbe as inExample 2.14(b). LetY=Rwith the topology σ= {U⊆R: 1∈U}

{∅}. Define f : (X,τ)→(Y,σ) as follows:

f(x)=





0, x∈X−A,

1, x∈A. (3.1)

Then f is closed, open, andgω-continuous. To show that f isgω-continuous, letU∈σ and letF be any closed set in (X,τ) such thatF⊆ f⁻¹(U). Then f⁻¹(U) must beX, and hence f⁻¹(U) isgω-open in (X,τ). But neither f isgω-irresolute nor f^ω isgω- continuous since{0}isω-open, hencegω-open in (Y,σ), while f⁻¹({0})=X−Ais not gω-open in (X,τ) (X−Ais closed but notω-open in (X,τ)).

Proposition3.9. If f : (X,τ)→(Y,σ)isgω-continuous, then for eachx∈X and each open setVin(Y,σ)with f(x)∈V, there exists agω-open setUin(X,τ)such thatx∈U and f(U)⊆V.

Proof. Letx∈X and letV be any open set in (Y,σ) containing f(x). PutU=f⁻¹(V).

Then, by assumption,Uis agω-open set in (X,τ) such thatx∈U and f(U)⊆V, and

the result follows.

The converse of the above proposition is not true in general as the following example shows.

Example 3.10. Let (X,τ) andA⊂Xbe as inExample 2.14(b) and letY = {0, 1}with the topologyσ= {∅,{0},Y}. Define f : (X,τ)→(Y,σ) as follows:

f(x)=





0, x∈X−A,

1, x∈A. (3.2)

Then f is notgω-continuous sinceX−A= f⁻¹({0}) is closed but notω-open in (X,τ).

On the other hand, f satisfies the property stated inProposition 3.9because{x}isgω- open in (X,τ) for eachx∈X.

Recall that a function f : (X,τ)→(Y,σ) is calledθ-continuous [9] (resp., almost con- tinuous [16], weakly continuous [12]) if for eachx∈Xand each open setVin (Y,σ) con- taining f(x), there exists an open setUin (X,τ) such thatx∈Uand f(clτ(U))⊆clσ(V) (resp., f(U)⊆intσ(clσ(V)), f(U)⊆clσ(V)).

The following result was obtained in [15].

Theorem3.11. Let f : (X,τ)→(Y,σ)be a function from a space(X,τ)into a regular space (Y,σ). Then the following are equivalent.

(a) f is continuous.

(b) f isθ-continuous.

(c) f is almost continuous.

(d) f isα-continuous.

(e) f is weakly continuous.

Theorem3.12. Let f : (X,τ)→(Y,σ)be a function from an anti-locally countable space (X,τ)onto a regular space(Y,σ). Then the following are equivalent.

(a) f is continuous.

(b) f isω-continuous.

(c)For eachx∈Xand each open setV in(Y,σ)with f(x)∈V, there exists anω-open setUin(X,τ)such thatx∈Uandf(U)⊆intσ(clσ(V)).

(d)For eachx∈Xand each open setV in(Y,σ)with f(x)∈V, there exists anω-open setUin(X,τ)such thatx∈Uandf(U)⊆intσω(clσ(V)).

(e)For eachx∈Xand each open setV in(Y,σ)with f(x)∈V, there exists anω-open setUin(X,τ)such thatx∈Uandf(U)⊆clσ(V).

Proof. In general, the implications (a)⇒(b)⇒(c)⇒(d)⇒(e) follow from the definitions and the fact that the topologyσωis finer thanσ.

(e)⇒(a). We show thatf is continuous at eachx∈X. Letx∈Xand letVbe any open set in (Y,σ) such that f(x)∈V. By regularity of (Y,σ), choose two open setsWandH in (Y,σ) such that f(x)∈H⊆clσ(H)⊆W⊆clσ(W)⊆V. By assumption, there exists anω-open set U in (X,τ) such thatx∈U and f(U)⊆clσ(H). Now, choose an open setG in (X,τ) such thatx∈G andG−U is countable. We claim that f(G)⊆clσ(W).

If not, chooset∈ f(G)−clσ(W). Therefore,t= f(g) for some g∈G. Now, t∈Y− clσ(W) which is an open set in (Y,σ), and so there existU1∈τωand an open setG1∈ τ such thatg ∈U1∩G1, f(U1)⊆clσ(Y−clσ(W)), andG1−U1 is countable. Finally, since f(U)f(U1)⊆clσ(H)∩clσ(Y−clσ(W))⊆W∩clσ(Y−clσ(W))= ∅,U∩U1=

∅, and sog∈G∩G1⊆(G−U)∪(G1−U1), that is, G∩G1 is a nonempty countable

open set in (X,τ), which contradicts the fact thatXis anti-locally countable. Thusf(G)⊆

clσ(W)⊆V, and hence f is continuous atx.

The following two examples show that the conditions thatXis anti-locally countable andYis regular inTheorem 3.12are essential.

Example 3.13. (a) Let (Y,σ) be as inExample 3.10. Then the function f : (R,τu)→(Y,σ) defined by

f(x)=





0, x∈R−Q,

1, x∈Q (3.3)

isω-continuous but not continuous. Here, (R,τu) is anti-locally countable and (Y,σ) is not regular.

(b) Let (Y,σ) be as in (a),Z= {0, 1}with the discrete topologyτdisand let f : (Y,σ)→ (Z,τdis) be the identity function. Clearly, (Y,σ) is not anti-locally countable, (Z,τdis) is regular, and f isω-continuous but not continuous.

Corollary3.14. Let f : (X,τ)→(Y,σ)be a function from an anti-locally countableT1/2- space (X,τ) onto a regular space (Y,σ). Then f is continuous if and only if it is gω- continuous.

The proof follows from Theorems2.11and3.12.

Example 3.3(a) shows that the assumption that (X,τ) is aT1/2-space in the above corollary cannot be dropped.

Theorem3.15. Let f : (X,τ)→(Y,σ)be agω-continuous function and letAbe a closed subset of(X,τ). Then, the restriction f |A: (A,τA)→(Y,σ)isgω-continuous.

Proof. LetFbe a closed subset of (Y,σ). Then (f |A)⁻¹(F)= f⁻¹(F)∩A. Since f isgω- continuous, f⁻¹(F)∈GωC(X,τ) and so, byProposition 2.15, f⁻¹(F)∩A∈GωC(X,τ).

Therefore, byTheorem 2.21(a), (f |A)⁻¹(F)∈GωC(A,τA) and the result follows.

Theorem3.16. Let(X,τ)be a topological space such thatX=A∪B, whereA,Bare both ω-closed in(X,τ). Let f : (X,τ)→(Y,σ)be given such that the restricions f |Aand f |Bare bothgω-continuous. Then f isgω-continuous.

Proof. LetFbe a closed subset of (Y,σ). Then, f⁻¹(F)=(f |A)⁻¹(F)∪(f |B)⁻¹(F). Since (f |A)⁻¹(F)∈GωC(A,τA) andAisω-closed in (X,τ), byTheorem 2.21(b), (f |A)⁻¹(F)∈ GωC(X,τ). Similarly, (f|B)⁻¹(F)∈GωC(X,τ). ByProposition 2.13,f⁻¹(F)∈GωC(X,τ).

Thus f isgω-continuous.

Theorem3.17. Let(X,τ)and(Y,σ)be topological spaces, where(Y,σ)is locally countable.

Then the projectionpX: (X×Y,τ×σ)→(X,τ)isgω-irresolute.

Proof. LetAbe agω-open subset of (X,τ) and letFbe a closed subset of (X×Y,τ×σ) such thatF⊆p_X⁻¹(A)=A×Y. For each (x,y)∈F, the closed set clτ{x}is contained inA.

By assumption, clτ{x} ⊆intτω(A). Therefore, (x,y)∈clτ{x} × {y} ⊆intτω(A)×Y. Now,

we show that intτω(A)×Y ⊆int(τ×σ)ω(A×Y). Let (s,t)∈intτω(A)×Y. ChooseU∈τ, W∈τω, and a countable open subsetVof (Y,σ) such that (s,t)∈(U∩W)×V,s∈W⊆ A, andU−Wis countable. SinceU×V−W×Y=(U−W)×Vis countable,W×Y∈ (τ×σ)ωand (s,t)∈W×Y⊆A×Y. Therefore, (s,t)∈int(τ_×σ)ω(A×Y), and hence the result follows. It follows that (x,y)∈int(τ×σ)ω(A×Y) for each (x,y)∈F, which means thatF⊆int(τ×σ)ω(A×Y). Therefore,p⁻_X¹(A)=A×Y isgω-open in (X×Y,τ×σ), and

hencepXisgω-irresolute.

To show that the condition (Y,σ) being locally countable inTheorem 3.17is essential, we consider the following example.

Example 3.18. Consider the projectionp: (R×R,τu×τu)→(R,τu) and letA=R−Q.

ThenA isω-open (and hence gω-open) in (R,τu) while p⁻¹(A)=(R−Q)×Ris not gω-open in (R×R,τu×τu) (seeExample 2.20). Thuspis notgω-irresolute.

The proof of the following theorem is left to the reader.

Theorem3.19. Let f : (X,τ)→(Y,σ)andg: (Y,σ)→(Z,γ)be two functions. Then the following hold.

(a)g◦f isgω-continuous ifgis continuous and f isgω-continuous.

(b)g◦f isgω-irresolute if f andgaregω-irresolute.

(c)g◦f is gω-continuous ifgisgω-continuous and f isgω-irresolute.

(d)Let(Y,σ)be aT1/2-space. Then,g◦f isgω-continuous ifgisgω-continuous andf^ω isgω-continuous.

The following example shows that the composition of twogω-continuous functions need not begω-continuous.

Example 3.20. Let (Y,σ) and f be as inExample 3.13, let (Y,γ) be the set{0, 1}with the topologyγ= {∅,{1},Y}and letg: (Y,σ)→(Y,γ) be the identity function. Then f and gare bothgω-continuous butg◦f is notgω-continuous.

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Khalid Y. Al-Zoubi: Department of Mathematics, Faculty of Science, Yarmouk University, Irbid- Jordan

E-mail address:[email protected]

Special Issue on Space Dynamics

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