# Cocompact Open Sets and Continuity

Authors: Samer Al Ghour; Salti Samarah
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

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.


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## Body

## 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 x∈X is called a condensation point of A if, for each U∈τ with x∈U, the set U∩A is uncountable. In 1982, Hdeib defined ω-closed sets and ω-open sets as follows: A is called ω-closed [1] if it contains all its condensation points. The complement of an ω-closed set is called ω-open. In 1989, Hdeib [2] 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 [3] 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 [4], 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 x∈A, there exists U∈τ such that x∈U 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 [4] 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 [5] 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 useℝ, ℚ, and ℕ 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 ℝ 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 ℝ. 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 Sets

Definition 2.1.
A subsetA of a space (X,τ) is called co-compact open set (notation: coc-open) if, for every x∈A, there exists an open set U⊆X and a compact subset K∈C(X,τ) such that x∈U-K⊆A. The complement of a coc-open subset is called coc-closed.The family of allcoc-open subsets of a space (X,τ) will be denoted by τk and the family {U-K:U∈τ and K∈C(X,τ)} of coc-open sets will be denoted by ℬk(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 X∈τk, let x∈X, take U=X and K=⌀. Then x∈U-K⊆X.
LetU1,U2∈τk, and let x∈U1∩U2. For each i=1,2, we find an open set Vi and a compact subset Ki such that x∈Vi-Ki⊆Ui. Take V=V1∩V2 and K=K1∪K2. Then V is open, K is compact, and x∈V-K⊆U1∩U2. It follows that U1∩U2 is coc-open.
Let{Uα:α∈Δ} be a collection of coc-open subsets of (X,τ) and x∈⋃α∈ΔUα. Then there exists α0∈Δ such that x∈Uα∘. Since Uα∘ is coc-open, then there exists an open set V and a compact subset K, such that x∈V-K⊆Uα∘. Therefore, we have x∈V-K⊆Uα∘⊆⋃α∈ΔUα. Hence, ⋃α∈ΔUα is coc-open.The following result follows directly from Definition2.1.Proposition 2.3.
Let(X,τ) be a space. Then the collection ℬk(τ) forms a base for τk.Corollary 2.4.
Let(X,τ) be a space. Then the collection τ∪{X-K:K∈C(X,τ)} forms a subbase for τk.For a space(X,τ), the following example shows that the collection ℬk(τ) is not a topology on X in general.Example 2.5.
LetX=ℝ, and let τ=τlr. Consider the collection of elements of ℬk(τ),Gn = (-∞,n+1/2)-{1,2,…,n}, n∈ℕ. Then ∪{Gn:n∈ℕ}=ℝ-ℕ which is not in ℬk(τ).Theorem 2.6.
Let(X,τ) be a space. Then τ⊆τn⊆τk.Proof.
Obvious.Remark 2.7.
Each of the two inclusions in Theorem2.6 is not equality in general; to see this, let X=ℕ and τ=τind. Then τn=τcof and τk=τdisc, and therefore τ≠τn and τn≠τk.In Remark2.7, the space (X,τ) is an example on a compact space (X,τ) for which (X,τk) is not compact.Definition 2.8 (see [6]).
A space(X,τ) is called CC if every compact set in X is closed.It is well known that everyT2 space is CC, but not conversely.Theorem 2.9.
Let(X,τ) be a space. Then the following are equivalent:(a)
(X,τ) is CC,(b)
τ=ℬk(τ),(c)
τ=τn=τk.Proof.
(a)⇒ (b) As ⌀ is a compact subset of X, then, for every U∈τ, U-⌀=U∈ℬk(τ). Hence, we have τ⊆ℬk(τ). Now let U-K∈ℬk(τ), where U∈τ and K is a compact subset of X. As (X,τ) is CC, then K is closed and hence U-K∈τ. Therefore, we have ℬk(τ)⊆τ.
(b)⇒ (c) By Theorem 2.6, it is sufficient to see that τk⊆τ. Since by (b) τ=ℬk(τ) and as ℬk(τ) is a base for τk, then τk⊆τ.
(c)⇒ (a) Let K∈C(X,τ). Then X-K∈τk, and by (c), X-K∈τ. 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.
LetK∈C(X,τk). As τ⊆τk, then C(X,τk)⊆C(X,τ) and hence K∈C(X,τ). Thus, we have X-K∈τk, and hence K is closed in the space (X,τk).Corollary 2.12.
For any space(X,τ), (τk)k=τk.Proof.
Theorems2.9 and 2.11.Theorem 2.13.
If(X,τ) is a hereditarily compact space, then τk=τdisc.Proof.
For everyx∈X, X-{x} is compact and so {x}=X-(X-{x})∈ℬk(τ)⊆τk. Therefore, τk=τdisc.Each of the following three examples shows that the converse of Theorem2.13 is not true in general.Example 2.14.
LetX=ℝ and τ=τlr. For every x∈X, take K=(-∞,x+1]-{x} and U=(-∞,x+1). Then K∈C(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.
LetX=ℕ and τ={⌀,ℕ}∪{Un:n∈ℕ}, where Un={1,2,…,n}. Then the compact subsets of (X,τ) are the finite sets. For every n∈ℕ, Un∈τ, Un-1 is compact, and {n}=Un-Un-1. Therefore, τk=τdisc.Example 2.16.
LetX=ℕ and τ be the topology on ℕ having the family {{2n-1,2n}:n∈ℕ} as a base. Then the compact subsets of (X,τ) are the finite sets. If x∈ℕ 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 τk≠τ and τk≠τdisc?The following example shows that the answer of the above question is yes.Example 2.17.
LetX=ℝ and τ={X}∪{U⊆X:1∉U}. Then C(X,τ)={K⊆X:1∈K}∪{K⊆X:1∉KandKisfinite}, hence τk=τ∪{U⊆X:1∈UandX-Uisfinite}. Note that τk≠τ and τk≠τdisc.Theorem 2.18.
Let(X,τ) be a space and A a nonempty subset of X. Then (τ|A)k⊆τk|A.Proof.
LetB∈(τ|A)k and x∈B. Then there exists V∈τ|A and a compact subset K⊆A such thatx∈V-K⊆B. Since V∈τ|A, then we can write V=U∩A, where U is open in X. Since U-K∈τk, (U-K)∩A∈τk|A. Hence, B∈τk|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 Question1.Theorem 2.19.
Let(X,τ) be a space and Abe a nonempty closed set in (X,τ). Then (τ|A)k=τk|A.Proof.
By Theorem2.18, (τ|A)k⊆τk|A. Conversely, let B∈τk|A and x∈B. Choose H∈τk such that B=H∩A. As H∈τk, there exists U∈τ and K∈C(X,τ) such that x∈U-K⊆H. Thus, we have x∈(U∩A)-(K∩A)⊆B, U∩A∈τ|A, and K∩A∈C(A,τ|A). It follows that B∈(τ|A)k.Definition 2.20.
Let(X,τ) be a space, and let A⊆X. The coc-closure of A in (X,τ) is denoted by coc-Clτ(A) and defined as follows:
(2.1)coc-Clτ(A)=∩{B:Biscoc-closedin(X,τ)andA⊆B}.Remark 2.21.
Let(X,τ) be a space, and let A⊆X. 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 setX, (X,τcoc) is an anti-locally compact space. Also, (ℚ,τu) is an example of an anti-locally compact space.Theorem 2.23.
Let(X,τ) be an anti-locally compact space. If A∈τ then coc-Clτ(A)=Clτ(A).Proof.
According to Remark2.21, only we need to show that Clτ(A)⊆Clτk(A). Suppose to the contrary that there is x∉Clτ(A)-Clτk(A). As x∉Clτk(A), there exists G⊇A coc-closed such that x∈G and G∩A=⌀. Take U∈τ and K∈C(X,τ) such that x∈U-K⊆G. Thus we have U∩A⊆K. Since x∈Clτ(A), it follows that U∩A≠⌀ and hence Int(K)≠⌀. This contradicts the assumption that (X,τ) is anti-locally compact.In Theorem2.23 the assumption “anti-locally compact” on the space cannot be dropped. As an example let X=ℝ and τ={⌀,X,{0}}, then {0}∈τ, coc-Clτ({0})={0} while Clτ({0})=ℝ.Theorem 2.24.
Iff:(X,τ)→(Y,σ) is injective, open, and continuous, then f:(X,τk)→(Y,σk) is open.Proof.
LetG=U-K where U∈τ and K∈C(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 Theorem2.24, the continuity condition cannot be dropped. Take f:(ℝ,τind)→(ℝ,τu), where f(x)=tan-1x. Then f is injective and open. On the other hand, as (ℝ,τind) is hereditarily compact we have (τind)k=τdisc, and as (ℝ,τu) is T2 we have (τu)k=τu. Thus, f:(ℝ,(τind)k)→(ℝ,(τu)k) is not open.


## 3.(coc)*-Open Sets

Definition 3.1.
A subsetA 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 ℬk*(τ).Theorem 3.2.
Let(X,τ) be a space. Then τ⊆ℬk*(τ).Proof.
LetA∈τ. Then A=Intτ(A)⊆Intτk(A)⊆A. Thus Intτ(A)=Intτk(A) and hence A∈ℬk*(τ).Theorem 3.3.
If(X,τ) is a CC space, then every subset of X is (coc)*-open.Proof.
LetA⊆X. 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 Corollary3.4, the inclusion in Theorem 3.2 is not equality in any T2 space that is not discrete for example, in (ℝ,τu) for the set A=(0,1)∪{2}, we have Int(τu)k(A)=Intτu(A)=(0,1) and thus A∈ℬk*(τu)-τu.Theorem 3.5.
If(X,τ) is a hereditarily compact space, then τ=ℬk*(τ).Proof.
By Theorem3.2, we need only to show that ℬk*(τ)⊆τ. Let A∈ℬk*(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, A∈τ.The following result is a new decomposition of open sets in a space.Theorem 3.6.
Let(X,τ) be a space. Then τ=τk∩ℬk*(τ).Proof.
By Theorems2.6 and 3.2, it follows that τ⊆τk∩ℬk*(τ). Conversely, let A∈τk∩ℬk*(τ). As A∈τk, then Intτk(A)=A. Also, since A∈ℬk*(τ), then Intτk(A)=Intτ(A). It follows that Intτ(A)=A and hence A∈τ.Theorem 3.7.
For a space(X,τ), one has the following:(a)
⌀,X∈ℬk*(τ),(b)
ifA,B∈ℬk*(τ), then A∩B∈ℬk*(τ).Proof.
(a) The proof follows directly from Theorem3.2.
(b) LetA,B∈ℬk*(τ). Then Intτk(A)=Intτ(A) and Intτk(B)=Intτ(B). Thus we have(3.1)Intτ(A∩B)=Intτ(A)∩Intτ(B)=Intτk(A)∩Intτk(B)=Intτk(A∩B).
It follows that A∩B∈ℬk*(τ).The following example shows that arbitrary union ofk*-open sets need not to be k*-open in general.Example 3.8.
Consider the space defined in Example2.17. For every natural number n≥3, put An=ℝ-{n,n+1,n+2,…}. Then, for each n≥3, Intτk(An)=Intτ(An)=An-{1}, and thus An is k*-open. On the other hand, Intτk(⋃n≥3An)=⋃n≥3An while Intτ(⋃n≥3An)=(⋃n≥3An)-{1}.


## 4. coc-Continuous Functions

Definition 4.1.
A functionf:(X,τ)→(Y,σ) is called coc-continuous at a point x∈X, 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 fis said to be coc-continuous.The following theorem follows directly from the definition.Theorem 4.2.
A functionf:(X,τ)→(Y,σ) is coc-continuous if and only if f:(X,τk)→(Y,σ) is continuous.Theorem 4.3.
EveryN-continuous function is coc-continuous.Proof.
Straightforward.The identity functionI:(ℝ,τind)→(ℝ,τdisc) is a coc-continuous function that is not N-continuous.The proof of the following result follows directly from Theorem2.9.Theorem 4.4.
Letf:(X,τ)→(Y,σ) be a function for which (X,τ) is CC, then the following are equivalent.(a)
f is continuous.(b)
f is N-continuous.(c)
f is coc-continuous.The following example shows that the composition of twoN-continuous functions need not to be even coc-continuous.Example 4.5.
LetX=ℝ, 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 g∘f is not coc-continuous since (g∘f)-1({a})={0,1}∉τk.Theorem 4.6.
(a) Iff:(X,τ)→(Y,σ) is N-continuous and if g:(Y,σ)→(Z,μ) is continuous, then g∘f:(X,τ)→(Z,μ) is N-continuous.
(b) Iff:(X,τ)→(Y,σ) is coc-continuous and if g:(Y,σ)→(Z,μ) is continuous, then g∘f:(X,τ)→(Z,μ) is coc-continuous.Proof.
(a) It follows by noting that a functionf:(X,τ)→(Y,σ) is N-continuous if and only if f:(X,τn)→(Y,σ).
(b) The proof follows directly from Theorem4.2.Theorem 4.7.
Iff:(X,τ)→(Y,σ) is coc-continuous and Ais 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.
LetVbe any open set in Y. Since fis 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.
Iff:(X,τ)→(Y,σ) is coc-continuous and X=A∪B, 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 Theorem4.2 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)∩(A∪B) = (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 Theorem4.2.Theorem 4.9.
Letf:(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 functionw:(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 x∈X, is coc-continuous.Theorem 4.11.
Letf:(X,τ)→(Y,σ) be a function. If there is a coc-open subset A of (X,τ) containing x∈X 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.
LetV∈σ with f(x)∈V. Since f|A is coc-continuous at x, there is U∈(τ|A)k⊆τk such that x∈U and (f|A)(U)=f(U)⊆V.Corollary 4.12.
Letf:(X,τ)→(Y,σ) be a function, and let {Aα:α∈Δ}⊆τk be a cover of X such that, for each α∈Δ,f|Aα:(Aα,τ|Aα)→(Y,σ) is coc-continuous, then f is coc-continuous.Proof.
Letx∈X. We show that f:(X,τ)→(Y,σ) is coc-continuous at x. Since {Aα:α∈Δ} is a cover of X, then there exists α∘∈Δ such that x∈Aα∘. Therefore, by Theorem 4.11, it follows that f is coc-continuous at x.Definition 4.13.
A functionf:(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 Theorem3.2.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 Theorem3.3.Corollary 4.16.
If(X,τ) is T2, then every function f:(X,τ)→(Y,σ) is (coc)*-continuous.By Corollary4.16, it follows that the function f:(ℝ,τu)→(ℝ,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.
Letf:(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 Theorem3.5.By Theorem4.17, it follows that the identity function I:(ℝ,τind)→(ℝ,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 viacoc-continuity and (coc)*-continuity.Theorem 4.18.
A functionf:(X,τ)→(Y,σ) is continuous if and only if it is coc-continuous and (coc)*-continuous.Proof.
The proof follows directly from Theorem3.6.



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