AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation54861210.1155/2012/548612548612Research ArticleCocompact Open Sets and ContinuityAl GhourSamerSamarahSaltiSiegmundStefanDepartment of Mathematics and StatisticsJordan University of Science and TechnologyIrbid 22110Jordanjust.edu.jo20122612012201210102011051220112012Copyright © 2012 Samer Al Ghour and Salti Samarah.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Compact subsets of a topological space are used to define coc-open sets as new generalized open sets, and then coc-open sets are used to define (coc)-open sets as another type of generalized open sets. Several results and examples related to them are obtained; particularly a decomposition of open sets is given. Also, coc-open sets and (coc)-open sets are used to introduce coc-continuity and (coc)-continuity, respectively. As a main result, a decomposition theorem of continuity is obtained.

1. Introduction

Throughout this paper by a space we mean a topological space. Let (X,τ) be a space, and let A be a subset of X. A point xX is called a condensation point of A if, for each Uτ with xU, the set UA is uncountable. In 1982, Hdeib defined ω-closed sets and ω-open sets as follows: A is called ω-closed  if it contains all its condensation points. The complement of an ω-closed set is called ω-open. In 1989, Hdeib  introduced ω-continuity as a generalization of continuity as follows: A function f:(X,τ)(Y,σ) is called ω-continuous if the inverse image of each open set is ω-open. The authors in  proved that the family of all ω-open sets in a space (X,τ) forms a topology on X finer than τ and that the collection {U-C:U  τ and C is a countable subset of X} forms a base for that topology. Recently, in , the authors introduced a class of generalized open sets which is stronger ω-open as follows: a subset A of a space (X,τ) is called N-open if for each xA, there exists Uτ such that xU and U-A is finite. Throughout this paper, the family of all N-open sets in a space (X,τ) will be denoted by τn. The authors in  proved that τn is a topology on X that is finer than τ and they introduced several concepts related to N-open sets; in particular, they introduced N-continuity as a generalization of continuity and as a stronger form of ω-continuity as follows: a function f:(X,τ)(Y,σ) is called N-continuous if the inverse image of each open set is N-open. The authors in  continued the study of topological concepts via N-open sets. In the present work, we will generalize N-open sets by coc-open sets. Several results and concepts related to them will be introduced.

Throughout this paper, we use , , and to denote the set of real numbers, the set of rationals, and the set of natural numbers, respectively. For a subset A of a space (X,τ), The closure of A and the interior of A will be denoted by Clτ(A) and Intτ(A), respectively. Also, we write τ|A to denote the relative topology on A when A is nonempty. For a nonempty set X, τdisc,τind,τcof, and τcoc will denote, respectively, the discrete topology on X, the indiscrete topology on X, the cofinite topology on X, and the cocountable topology on X. For a subset A of we write (A,τu) to denote the subspace topology on A relative to the usual topology and we use τlr to denote the left ray topology on . For any two spaces (X,τ) and (Y,σ) we use τprod to denote the product topology on X×Y. The family of all compact subsets of a space (X,τ) will be denoted by C(X,τ).

2. Cocompact Open SetsDefinition 2.1.

A subset A of a space (X,τ) is called co-compact open set (notation: coc-open) if, for every xA, there exists an open set UX and a compact subset KC(X,τ) such that xU-KA. The complement of a coc-open subset is called coc-closed.

The family of all coc-open subsets of a space (X,τ) will be denoted by τk and the family {U-K:Uτ and KC(X,τ)} of coc-open sets will be denoted by k(τ).

Theorem 2.2.

Let (X,τ) be a space. Then the collection τk forms a topology on X.

Proof.

By the definition one has directly that τk. To see that Xτk, let xX, take U=X and K=. Then xU-KX.

Let U1,U2τk, and let xU1U2. For each i=1,2, we find an open set Vi and a compact subset Ki such that xVi-KiUi. Take V=V1V2 and K=K1K2. Then V is open, K is compact, and xV-KU1U2. It follows that U1U2 is coc-open.

Let {Uα:αΔ} be a collection of coc-open subsets of (X,τ) and xαΔUα. Then there exists α0Δ such that xUα. Since Uα is coc-open, then there exists an open set V and a compact subset K, such that xV-KUα. Therefore, we have xV-KUααΔUα. Hence, αΔUα is coc-open.

The following result follows directly from Definition 2.1.

Proposition 2.3.

Let (X,τ) be a space. Then the collection k(τ) forms a base for τk.

Corollary 2.4.

Let (X,τ) be a space. Then the collection τ{X-K:KC(X,τ)} forms a subbase for τk.

For a space (X,τ), the following example shows that the collection k(τ) is not a topology on X in general.

Example 2.5.

Let X=, and let τ=τlr. Consider the collection of elements of k(τ),Gn = (-,n+1/2)-{1,2,,n}, n. Then {Gn:n}=- which is not in k(τ).

Theorem 2.6.

Let (X,τ) be a space. Then ττnτk.

Proof.

Obvious.

Remark 2.7.

Each of the two inclusions in Theorem 2.6 is not equality in general; to see this, let X= and τ=τind. Then τn=τcof and τk=τdisc, and therefore ττn and τnτk.

In Remark 2.7, the space (X,τ) is an example on a compact space (X,τ) for which (X,τk) is not compact.

Definition 2.8 (see [<xref ref-type="bibr" rid="B6">6</xref>]).

A space (X,τ) is called CC if every compact set in X is closed.

It is well known that every T2 space is CC, but not conversely.

Theorem 2.9.

Let (X,τ) be a space. Then the following are equivalent:

(X,τ) is CC,

τ=k(τ),

τ=τn=τk.

Proof.

(a) (b) As is a compact subset of X, then, for every Uτ, U-=Uk(τ). Hence, we have τk(τ). Now let U-Kk(τ), where Uτ and K is a compact subset of X. As (X,τ) is CC, then K is closed and hence U-Kτ. Therefore, we have k(τ)τ.

(b) (c) By Theorem 2.6, it is sufficient to see that τkτ. Since by (b) τ=k(τ) and as k(τ) is a base for τk, then τkτ.

(c) (a) Let KC(X,τ). Then X-Kτk, and by (c), X-Kτ. Therefore, K is closed in X.

Corollary 2.10.

If (X,τ) is a T2-space, then τ=τn=τk.

Theorem 2.11.

For any space (X,τ), (X,τk) is CC.

Proof.

Let KC(X,τk). As ττk, then C(X,τk)C(X,τ) and hence KC(X,τ). Thus, we have X-Kτk, and hence K is closed in the space (X,τk).

Corollary 2.12.

For any space (X,τ), (τk)k=τk.

Proof.

Theorems 2.9 and 2.11.

Theorem 2.13.

If (X,τ) is a hereditarily compact space, then τk=τdisc.

Proof.

For every xX, X-{x} is compact and so {x}=X-(X-{x})k(τ)τk. Therefore, τk=τdisc.

Each of the following three examples shows that the converse of Theorem 2.13 is not true in general.

Example 2.14.

Let X= and τ=τlr. For every xX, take K=(-,x+1]-{x} and U=(-,x+1). Then KC(X,τ), U=(-,x+1)τ, and {x}=U-K. This shows that τk=τdisc. On the other hand, it is well known that (X,τ) is not hereditarily compact.

Example 2.15.

Let X= and τ={,}{Un:n}, where Un={1,2,,n}. Then the compact subsets of (X,τ) are the finite sets. For every n, Unτ, Un-1 is compact, and {n}=Un-Un-1. Therefore, τk=τdisc.

Example 2.16.

Let X= and τ be the topology on having the family {{2n-1,2n}:n} as a base. Then the compact subsets of (X,τ) are the finite sets. If x with x is odd, then {x}={x,x+1}-{x+1} and as {x,x+1}τ and {x+1} is compact, then {x}τk. Similarly, if x is even then {x}τk. Therefore, τk=τdisc.

The following question is natural: Is there a space (X,τ) for which τkτ and τkτdisc?

The following example shows that the answer of the above question is yes.

Example 2.17.

Let X= and τ={X}{UX:1U}. Then C(X,τ)={KX:1K}{KX:1K  and  K  is  finite}, hence τk=τ{UX:1U  and  X-U  is  finite}. Note that τkτ and τkτdisc.

Theorem 2.18.

Let (X,τ) be a space and A a nonempty subset of X. Then (τ|A)kτk|A.

Proof.

Let B(τ|A)k and xB. Then there exists Vτ|A and a compact subset KA such that  xV-KB. Since Vτ|A, then we can write V=UA, where U is open in X. Since U-Kτk, (U-K)Aτk|A. Hence, Bτk|A.

Question 1.

Let (X,τ) be a space and A a nonempty subset of X. Is it true that (τ|A)k=τk|A?

The following result is a partial answer for Question 1.

Theorem 2.19.

Let (X,τ) be a space and A  be a nonempty closed set in (X,τ). Then (τ|A)k=τk|A.

Proof.

By Theorem 2.18, (τ|A)kτk|A. Conversely, let Bτk|A and xB. Choose Hτk such that B=HA. As Hτk, there exists Uτ and KC(X,τ) such that xU-KH. Thus, we have x(UA)-(KA)B, UAτ|A, and KAC(A,τ|A). It follows that B(τ|A)k.

Definition 2.20.

Let (X,τ) be a space, and let AX. The coc-closure of A in (X,τ) is denoted by coc-Clτ(A) and defined as follows: coc-Clτ(A)={B:B  is  coc-closed  in  (X,τ)  and  AB}.

Remark 2.21.

Let (X,τ) be a space, and let AX. Then coc-Clτ(A)=Clτk(A) and coc-Clτ(A)Clτ(A).

Definition 2.22.

A space (X,τ) is called antilocally compact if any compact subset of X has empty interior.

For any infinite set X, (X,τcoc) is an anti-locally compact space. Also, (,τu) is an example of an anti-locally compact space.

Theorem 2.23.

Let (X,τ) be an anti-locally compact space. If Aτ then coc-Clτ(A)=Clτ(A).

Proof.

According to Remark 2.21, only we need to show that Clτ(A)Clτk(A). Suppose to the contrary that there is xClτ(A)-Clτk(A). As xClτk(A), there exists GA coc-closed such that xG and GA=. Take Uτ and KC(X,τ) such that xU-KG. Thus we have UAK. Since xClτ(A), it follows that UA and hence Int(K). This contradicts the assumption that (X,τ) is anti-locally compact.

In Theorem 2.23 the assumption “anti-locally compact” on the space cannot be dropped. As an example let X= and τ={,X,{0}}, then {0}τ, coc-Clτ({0})={0} while Clτ({0})=.

Theorem 2.24.

If f:(X,τ)(Y,σ) is injective, open, and continuous, then f:(X,τk)(Y,σk) is open.

Proof.

Let G=U-K where Uτ and KC(X,τ) be a basic element for τk. As f is injective, f(G)=f(U)-f(K). Also, as f:(X,τ)(Y,σ) is open, f(U)σ. And as f:(X,τ)(Y,σ) is continuous, f(K)C(Y,σ). This ends the proof.

Remark 2.25.

In Theorem 2.24, the continuity condition cannot be dropped. Take f:(,τind)(,τu), where f(x)=tan-1x. Then f is injective and open. On the other hand, as (,τind) is hereditarily compact we have (τind)k=τdisc, and as (,τu) is T2 we have (τu)k=τu. Thus, f:(,(τind)k)(,(τu)k) is not open.

3. <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M356"><mml:mrow><mml:msup><mml:mrow><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mtext>coc</mml:mtext></mml:mrow><mml:mo stretchy="false">)</mml:mo></mml:mrow></mml:mrow><mml:mrow><mml:mi>*</mml:mi></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula>-Open SetsDefinition 3.1.

A subset A of a space (X,τ) is called (coc)*-open if Intτk(A)=Intτ(A).

The family of all (coc)*-open subsets of a space (X,τ) will be denoted by k*(τ).

Theorem 3.2.

Let (X,τ) be a space. Then τk*(τ).

Proof.

Let Aτ. Then A=Intτ(A)Intτk(A)A. Thus Intτ(A)=Intτk(A) and hence Ak*(τ).

Theorem 3.3.

If (X,τ) is a CC space, then every subset of X is (coc)*-open.

Proof.

Let AX. Since (X,τ) is CC, then τ=τk, and so Intτk(A)=Intτ(A). Therefore, A is (coc)*-open.

Corollary 3.4.

If (X,τ) is a T2 space, then every subset of X is (coc)*-open.

According to Corollary 3.4, the inclusion in Theorem 3.2 is not equality in any T2 space that is not discrete for example, in (,τu) for the set A=(0,1){2}, we have Int(τu)k(A)=Intτu(A)=(0,1) and thus Ak*(τu)-τu.

Theorem 3.5.

If (X,τ) is a hereditarily compact space, then τ=k*(τ).

Proof.

By Theorem 3.2, we need only to show that k*(τ)τ. Let Ak*(τ). Then Intτk(A)=Intτ(A). Since (X,τ) is hereditarily compact, then, by Theorem 2.13, τk=τdisc and thus Intτ(A)=Intτk(A)=A. Therefore, Aτ.

The following result is a new decomposition of open sets in a space.

Theorem 3.6.

Let (X,τ) be a space. Then τ=τkk*(τ).

Proof.

By Theorems 2.6 and 3.2, it follows that ττkk*(τ). Conversely, let Aτkk*(τ). As Aτk, then Intτk(A)=A. Also, since Ak*(τ), then Intτk(A)=Intτ(A). It follows that Intτ(A)=A and hence Aτ.

Theorem 3.7.

For a space (X,τ), one has the following:

,Xk*(τ),

if A,Bk*(τ), then ABk*(τ).

Proof.

(a) The proof follows directly from Theorem 3.2.

(b) Let A,Bk*(τ). Then Intτk(A)=Intτ(A) and Intτk(B)=Intτ(B). Thus we haveIntτ(AB)=Intτ(A)Intτ(B)=Intτk(A)Intτk(B)=Intτk(AB). It follows that ABk*(τ).

The following example shows that arbitrary union of k*-open sets need not to be k*-open in general.

Example 3.8.

Consider the space defined in Example 2.17. For every natural number n3, put An=-{n,n+1,n+2,}. Then, for each n3, Intτk(An)=Intτ(An)=An-{1}, and thus An is k*-open. On the other hand, Intτk(n3An)=n3An while Int  τ(n3An)=(n3An)-{1}.

4. coc-Continuous FunctionsDefinition 4.1.

A function f:(X,τ)(Y,σ) is called coc-continuous at a point xX, if for every open set V containing f(x) there is a coc-open set U containing x such that f(U)V. If f is coc-continuous at each point of X, then f  is said to be coc-continuous.

The following theorem follows directly from the definition.

Theorem 4.2.

A function f:(X,τ)(Y,σ) is coc-continuous if and only if f:(X,τk)(Y,σ) is continuous.

Theorem 4.3.

Every N-continuous function is coc-continuous.

Proof.

Straightforward.

The identity function I:(,τind)(,τdisc) is a coc-continuous function that is not N-continuous.

The proof of the following result follows directly from Theorem 2.9.

Theorem 4.4.

Let f:(X,τ)(Y,σ) be a function for which (X,τ) is CC, then the following are equivalent.

f is continuous.

f is N-continuous.

f is coc-continuous.

The following example shows that the composition of two N-continuous functions need not to be even coc-continuous.

Example 4.5.

Let X=, Y={0,1,2}, Z={a,b},τ be as in Example 2.17, σ={,Y,{0},{0,1}}, and μ={,Z,{a}}. Define the function f:(X,τ)(Y,σ) by f(x)=2 if x{0,1} and f(x)=1 otherwise, and define the function g:(Y,σ)(Z,μ) by g(0)=g(2)=a and g(1)=b. Then f and g are N-continuous functions, but gf is not coc-continuous since (gf)-1({a})={0,1}τk.

Theorem 4.6.

(a) If f:(X,τ)(Y,σ) is N-continuous and if g:(Y,σ)(Z,μ) is continuous, then gf:(X,τ)(Z,μ) is N-continuous.

(b) If f:(X,τ)(Y,σ) is coc-continuous and if g:(Y,σ)(Z,μ) is continuous, then gf:(X,τ)(Z,μ) is coc-continuous.

Proof.

(a) It follows by noting that a function f:(X,τ)(Y,σ) is N-continuous if and only if f:(X,τn)(Y,σ).

(b) The proof follows directly from Theorem 4.2.

Theorem 4.7.

If f:(X,τ)(Y,σ) is coc-continuous and A  is a nonempty closed set in (X,τ), then the restriction of f to A,f|A:(A,τ|A)(Y,σ) is a coc-continuous function.

Proof.

Let V  be any open set in Y. Since f  is coc-continuous, then f-1(V) is coc-open in X and by Theorem 2.19, (f|A)-1(V)=f-1(V)A is coc-open in A. Therefore f|A is coc-continuous.

Theorem 4.8.

If f:(X,τ)(Y,σ) is coc-continuous and X=AB, where A and B are coc-closed subsets in (X,τ) and f|A:(A,τ|A)(Y,σ), f|B:(B,τ|B)(Y,σ) are coc-continuous functions, then f is coc-continuous.

Proof.

By Theorem 4.2 it is sufficient to show that f:(X,τk)(Y,σ) is continuous. Let C be a closed subset of (Y,σ). Then f-1(C) = f-1(C)X = f-1(C)(AB) = (f-1(C)A)(f-1(C)B). Since f|A:(A,τ|A)(Y,σ) is coc-continuous, then f-1(C)A=(f|A)-1(C) is coc-closed in (A,τ|A), and as A is coc-closed in (X,τ), it follows that f-1(C)A is coc-closed in (X,τ); similarly one can conclude that f-1(C)B is coc-closed in (X,τ). It follows that f-1(C) is closed in (X,τk) and hence f:(X,τk)(Y,σ) is continuous.

The following result follows directly from Theorem 4.2.

Theorem 4.9.

Let f:(X,τ)(Y,σ) and g:(X,τ)(Z,μ) be two functions. Then the function h:(X,τ)(Y×Z,τprod) defined by h(x)=(f(x),g(x)) is coc-continuous if and only if f and g are coc-continuous.

Corollary 4.10.

A function w:(X,τ)(Y,σ) is coc-continuous if and only if the graph function h:(X,τ)(X×Y,τprod), given by h(x)=(x,w(x)) for every xX, is coc-continuous.

Theorem 4.11.

Let f:(X,τ)(Y,σ) be a function. If there is a coc-open subset A of (X,τ) containing xX such that the restriction of f to A,f|A:(A,τ|A)(Y,σ) is coc-continuous at x, then f is coc-continuous at x.

Proof.

Let Vσ with f(x)V. Since f|A is coc-continuous at x, there is U(τ|A)kτk such that xU and (f|A)(U)=f(U)V.

Corollary 4.12.

Let f:(X,τ)(Y,σ) be a function, and let {Aα:αΔ}τk be a cover of X such that, for each αΔ,f|Aα:(Aα,τ|Aα)(Y,σ) is coc-continuous, then f is coc-continuous.

Proof.

Let xX. We show that f:(X,τ)(Y,σ) is coc-continuous at x. Since {Aα:αΔ} is a cover of X, then there exists αΔ such that xAα. Therefore, by Theorem 4.11, it follows that f is coc-continuous at x.

Definition 4.13.

A function f:(X,τ)(Y,σ) is called (coc)*-continuous if the inverse image of each open set is (coc)*-open.

Theorem 4.14.

Every continuous function is (coc)*-continuous.

Proof.

The proof follows directly from Theorem 3.2.

Theorem 4.15.

If (X,τ) is CC, then every function f:(X,τ)(Y,σ) is (coc)*-continuous.

Proof.

The proof follows directly from Theorem 3.3.

Corollary 4.16.

If (X,τ) is T2, then every function f:(X,τ)(Y,σ) is (coc)*-continuous.

By Corollary 4.16, it follows that the function f:(,τu)(,τu) where f(x)=0 for x is rational and f(x)=1 for x is irrational is (coc)*-continuous. On the other hand, it is well known that this function is discontinuous every where. Also, by Theorem 4.4, f is not coc-continuous.

Theorem 4.17.

Let f:(X,τ)(Y,σ) be a function with (X,τ) being a hereditarily compact space. Then f is continuous if and only if f is (coc)*-continuous.

Proof.

The proof follows directly from Theorem 3.5.

By Theorem 4.17, it follows that the identity function I:(,τind)(,τdisc) is not (coc)*-continuous. Therefore, this is an example of a coc-continuous function that is not (coc)*-continuous.

We end this section by the following decomposition of continuity via coc-continuity and (coc)*-continuity.

Theorem 4.18.

A function f:(X,τ)(Y,σ) is continuous if and only if it is coc-continuous and (coc)*-continuous.

Proof.

The proof follows directly from Theorem 3.6.

Acknowledgments

The authors are very grateful to the referees for their valuable comments and suggestions. Also, the authors would like to thank Jordan University of Science and Technology for the financial assistant of this paper.

HdeibH. Z.ω-closed mappingsRevista Colombiana de Matemáticas1982161-26578677814ZBL0574.54008HdeibH. Z.ω-continuous functionsDirasat Journal1989162136153Al-ZoubiK.Al-NashefB.The topology of ω-open subsetsAl-Manarah Journal200392169179Al-OmariA.SalmiM.New characterization of compact spacesProceedings of the 5th Asian Mathematical Conference2009Kuala Lumpur, Malaysia5360ZBL1207.54033Al-OmariA.NoiriT.NooraniM. S.Characterizations of strongly compact spacesInternational Journal of Mathematics and Mathematical Sciences200920099573038ZBL1179.54037LevineN.When are compact and closed equivalent?Mathematical Notes1965724144