# Weak and Strong Forms ofω-Continuous Functions

Authors: Ahmad Al-Omari; Takashi Noiri; Mohd. Salmi Md. Noorani
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We introduce weak and strong forms ofω-continuous
functions, namely, θ-ω-continuous and strongly θ-ω-continuous functions, and investigate their fundamental properties.


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## Body

## 1. Introduction

In 1943, Fomin [1] introduced the notion of θ-continuity. In 1968, the notions of θ-open subsets, θ-closed subsets, and θ-closure were introduced by Velic̆ko [2]. In 1989, Hdeib [3] introduced the notion of ω-continuity. The main purpose of the present paper is to introduce and investigate fundamental properties of weak and strong forms of ω-continuous functions. Throughout this paper, (X,τ) and (Y,σ) stand for topological spaces  (called simply spaces) with no separation axioms assumed unless otherwise stated. For a subset A of X, the closure of A and the interior of A will be denoted by Cl(A) and Int(A), respectively. Let (X,τ) be a space and A 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. However, A is said to be ω-closed [4] if it contains all its condensation points. The complement of an ω-closed set is said to be ω-open. It is well known that a subset W of a space (X,τ) is ω-open if and only if for each x∈W, there exists U∈τ such that x∈U and U-W is countable. The family of all ω-open subsets of a space (X,τ), denoted by τω or ωO(X), forms a topology on X finer than τ. The family of all ω-open sets of Xcontaining x∈X is denoted by ωO(X,x). The ω-closure and the ω-interior, that can be defined in the same way as Cl(A) and Int(A), respectively, will be denoted by ωCl(A) and ωInt(A). Several characterizations of ω-closed subsets were provided in [5–8].A pointx of X is called a θ-cluster points of A if Cl(U)∩A≠ϕ for every open set U of X containing x. The set of all θ-cluster points of A is called the θ-closure of A and is denoted by Clθ(A). A subset A is said to be θ-closed [2] if A=Clθ(A). The complement of a θ-closed set is said to be θ-open. A point x of X is called an ω-θ-cluster point of A if ωCl(U)∩A≠ϕ for every ω-open set U of X containing x. The set of all ω-θ-cluster points of A is called the ω-θ-closure of A and is denoted by ωClθ(A). A subset A is said to be ω-θ-closed if A=ωClθ(A). The complement of a ω-θ-closed set is said to be ω-θ-open. The ω-θ-interior of A is defined by the union of all ω-θ-open sets contained in A and is denoted by ωIntθ(A).


## 2.θ-ω-Continuous Functions

We begin by recalling the following definition. Next, we introduce a relatively new notion.Definition 2.1.
A functionf:X→Y is said to be ω-continuous (see [3]) (resp., almost weakly ω-continuous (see [9])) if for each x∈X and each open set V of Y containing f(x), there exists U∈ωO(X,x) such that f(U)⊆V (resp., f(U)⊆Cl(V)).Definition 2.2.
A functionf:X→Y is said to be θ-ω-continuous if for each x∈X and each open set V of Y containing f(x), there exists U∈ωO(X,x) such that f(ωCl(U))⊆Cl(V).Next, several characterizations ofθ-ω-continuous functions are obtained.Theorem 2.3.
For a functionf:X→Y, the following properties are equivalent:  (1)
f is θ-ω-continuous;(2)
ωClθ(f-1(B))⊆f-1(Clθ(B)) for every subset B of Y;(3)
f(ωClθ(A))⊆Clθ(f(A)) for every subset A of X.Proof.
(1)⇒(2) Let B be any subset of Y. Suppose that x∉f-1(Clθ(B)). Then f(x)∉Clθ(B) and there exists an open set V containing f(x) such that Cl(V)∩B=ϕ. Since f is θ-ω-continuous, there exists U∈ωO(X,x) such that f(ωCl(U))⊆Cl(V). Therefore, we have f(ωCl(U))∩B=ϕ and ωCl(U)∩f-1(B)=ϕ. This shows that x∉ωClθ(f-1(B)). Thus, we obtain ωClθ(f-1(B))⊆f-1(Clθ(B)).
(2)⇒(1) Let x∈X and V be an open set of Y containing f(x). Then we have Cl(V)∩(Y-Cl(V))=ϕ and f(x)∉Clθ(Y-Cl(V)). Hence, x∉f-1(Clθ(Y-Cl(V))) and x∉ωClθ(f-1(Y-Cl(V))). There exists U∈ωO(X,x) such that ωCl(U)∩f-1(Y-Cl(V))=ϕ and hence f(ωCl(U))⊆Cl(V). Therefore, f is θ-ω-continuous.
(2)⇒(3) Let A be any subset of X. Then we have ωClθ(A)⊆ωClθ(f-1(f(A)))⊆f-1(Clθ(f(A))) and hence f(ωClθ(A))⊆Clθ(f(A)).
(3)⇒(2) Let B be a subset of Y. We have f(ωClθ(f-1(B)))⊆Clθ(f(f-1(B)))⊆Clθ(B) and hence ωClθ(f-1(B))⊆f-1(Clθ(B)).Proposition 2.4.
A subsetU of a space X is ω-θ-open in X if and only if for each x∈U, there exists an ω-open set V with x∈V such that ωCl(V)⊆U.Proof.
Suppose thatU is ω-θ-open in X. Then X-U is ω-θ-closed. Let x∈U. Then x∉ωClθ(X-U)=X-U, and so there exists an ω-open set V with x∈V such that ωCl(V)∩(X-U)=ϕ. Thus ωCl(V)⊆U. Conversely, assume that U is not ω-θ-open. Then X-U is not ω-θ-closed, and so there exists x∈ωClθ(X-U) such that x∉X-U. Since x∈U, by hypothesis, there exists an ω-open set V with x∈V such that ωCl(V)⊆U. Thus ωCl(V)∩(X-U)=ϕ. This is a contradiction since x∈ωClθ(X-U).Theorem 2.5.
For a functionf:X→Y, the following properties are equivalent: (1)
f is θ-ω-continuous;(2)
f-1(V)⊆ωIntθ(f-1(Cl(V))) for every open set V of Y;(3)
ωClθ(f-1(V))⊆f-1(Cl(V)) for every open set V of Y.Proof.
(1)⇒(2) Suppose that V is any open set of Y and x∈f-1(V). Then f(x)∈V and there exists U∈ωO(X,x) such that f(ωCl(U))⊆Cl(V). Therefore, x∈U⊆ωCl(U)⊆f-1(Cl(V)). This shows that x∈ωIntθ(f-1(Cl(V))). Therefore, we obtain f-1(V)⊆ωIntθ(f-1(Cl(V))).
(2)⇒(3) Suppose that V is any open set of Y and x∉f-1(Cl(V)). Then f(x)∉Cl(V) and there exists an open set W containing f(x) such that W∩V=ϕ; hence Cl(W)∩V=ϕ. Therefore, we have f-1(Cl(W))∩f-1(V)=ϕ. Since x∈f-1(W), by (2) x∈ωIntθ(f-1(Cl(W))), there exists U∈ωO(X,x) such that ωCl(U)⊆f-1(Cl(W)). Thus we have ωCl(U)∩f-1(V)=ϕ and hence x∉ωClθ(f-1(V)). This shows that ωClθ(f-1(V))⊆f-1(Cl(V)).
(3)⇒(1) Suppose that x∈X and V are any open set of Y containing f(x). Then V∩(Y-Cl(V))=ϕ and f(x)∉Cl(Y-Cl(V)). Therefore x∉f-1(Cl(Y-Cl(V))) and by (3) x∉ωClθ(f-1(Y-Cl(V))). There exists U∈ωO(X,x) such that ωCl(U)∩f-1(Y-Cl(V))=ϕ. Therefore, we obtain f(ωCl(U))⊆Cl(V). This shows that f is θ-ω-continuous.A subsetA of X is said to be regular open (resp., regular closed) (see [10]) if A=Int(Cl(A)) (resp., A=Cl(Int(A))). Also, the family of all regular open (resp., regular closed) sets of X is denoted by RO(X) (resp., RC(X)).Theorem 2.6.
For a functionf:X→Y, the following properties are equivalent:  (1)
f is θ-ω-continuous;(2)
ωClθ[f-1(Int(Clθ(B)))]⊆f-1(Clθ(B)) for every subset B of Y;(3)
ωClθ[f-1(Int(Cl(V)))]⊆f-1(Cl(V)) for every open set V of Y;(4)
ωClθ[f-1(Int(K))]⊆f-1(K) for every closed set K of Y;(5)
ωClθ[f-1(Int(R))]⊆f-1(R) for every regular closed set R of Y.Proof.
(1)⇒(2) This follows immediately from Theorem 2.5(3) with V=Int(Clθ(B)).
(2)⇒(3) This is obvious since Clθ(V)=Cl(V) for every open set V of Y.
(3)⇒(4) For any closed set K of Y, Int(K)=Int(Cl(Int(K))) and by (3)
(2.1)ωClθ(f-1(Int(K)))=ωClθ(f-1(Int(Cl(Int(K)))))⊂f-1(Cl(Int(K)))⊂f-1(K).
(4)⇒(5) This is obvious.
(5)⇒(1) Let V be any open set of Y. Since Cl(V) is regular closed, by (5) ωClθ(f-1(V)))⊂ωClθ(f-1(Int(Cl(V)))))⊂f-1(Cl(V)). It follows from Theorem 2.5 that f is θ-ω-continuous.Definition 2.7.
A subsetA of a space X is said to be semi-open (see [11]) (resp., preopen (see [12]), β-open (see [13])) if A⊆Cl(Int(A)) (resp., A⊆Int(Cl(A)), A⊆Cl(Int(Cl(A)))).Theorem 2.8.
For a functionf:X→Y, the following properties are equivalent:  (1)
f is θ-ω-continuous;(2)
ωClθ[f-1(Int(Cl(G)))]⊆f-1(Cl(G)) for every β-open set G of Y;(3)
ωClθ[f-1(Int(Cl(G)))]⊆f-1(Cl(G)) for every semi-open set G of Y.Proof.
(1)⇒(2) This is obvious by Theorem 2.6(5) since Cl(G) is regular closed for every β-open set set G.
(2)⇒(3) This is obvious since every semi-open set is β-open.
(3)⇒(1) This follows immediately from Theorem 2.5(3) and since every open set is semi-open.Theorem 2.9.
For a functionf:X→Y, the following properties are equivalent: (1)
f is θ-ω-continuous;(2)
ωClθ[f-1(Int(Cl(G)))]⊆f-1(Cl(G)) for every preopen set G of Y;(3)
ωClθ[f-1(G)]⊆f-1(Cl(G)) for every preopen set G of Y;(4)
f-1(G)⊂ωIntθ(f-1(Cl(G))) for every preopen set G of Y.Proof.
(1)⇒(2) The proof follows from Theorem 2.8 (2) since every preopen set is β-open.
(2)⇒(3) This is obvious by the definition of a preopen set.
(3)⇒(4) Let G be any preopen set of Y. Then, by (3) we have
(2.2)X-ωIntθ(f-1(Cl(G)))=ωClθ(X-f-1(Cl(G)))=ωClθ(f-1(Y-Cl(G)))⊂f-1(Cl(Y-Cl(G)))=f-1(Y-Int(Cl(G)))⊂f-1(Y-G)=X-f-1(G).
Therefore, we obtain f-1(G)⊂ωIntθ(f-1(Cl(G))).
(4)⇒(1) This is obvious by Theorem 2.5 since every open set is preopen.Definition 2.10.
A functionf:X→Y is said to be almost ω-continuous if for each x∈X and each regular open set V of Y containing f(x), there exists U∈ωO(X,x) such that f(U)⊆V.Lemma 2.11.
For a functionf:X→Y, the following assertions are equivalent:      (1)
f is almost ω-continuous;(2)
for eachx∈X and each open set V of Y containing f(x), there exists U∈ωO(X,x) such that f(U)⊆Int(Cl(V));(3)
f-1(F)∈ωC(X) for every F∈RC(Y);(4)
f-1(V)∈ωO(X) for every V∈RO(Y).Proposition 2.12.
For a functionf:X→Y, the following properties hold: (1)
iff is almost ω-continuous, then it is θ-ω-continuous;(2)
iff is θ-ω-continuous, then it is almost weakly ω-continuous.Proof.
(1) Suppose thatx∈X and V is any open set of Y containing f(x). Since f is almost ω-continuous, f-1(Int(Cl(V))) is ω-open and f-1(Cl(V)) is ω-closed in X by Lemma 2.11. Now, set U=f-1(Int(Cl(V))). Then we have U∈ωO(X,x) and ωCl(U)⊆f-1(Cl(V)). Therefore, we obtain f(ωCl(U))⊆Cl(V). This shows that f is θ-ω-continuous.
(2) The proof follows immediately from the definition.Example 2.13.
LetX be an uncountable set and let A,B, and C be subsets of X such that each of them is uncountable and the family {A,B,C} is a partition of X. We define the topology τ={ϕ,X,{A},{B},{A,B}}. Then, the function f:(X,τ)→(X,τ) defined by f(A)=A,f(B)=C, and f(C)=A is θ-ω-continuous (and almost weakly ω-continuous) but is not almost ω-continuous since for xc∈C⊆X, A is regular open and f(xc)∈A but there is not open set Uxc containing xc such that f(Uxc)⊆A.Question 2.
Is the converse of Proposition2.12(2) true?It is clear that, for a subsetA of a space X, x∈ωCl(A) if and only if for any ω-open set U containing x, U∩A≠ϕ.Lemma 2.14.
For anω-open set U in a space X, ωCl(U)=ωClθ(U).Proof.
By definition,ωCl(U)⊆ωClθ(U). Let x∈ωClθ(U). Then for any ω-open set V containing x, ωCl(V)∩U≠ϕ. Let z∈ωCl(V)∩U. Then U∩V≠ϕ and x∈ωCl(U). Thus ωClθ(U)⊆ωCl(U).Definition 2.15.
A topological spaceX is said to be ω-regular (resp., ω*-regular) if for each ω-closed (resp., closed) set F and each point x∈X-F, there exist disjoint ω-open sets U and V such that x∈U and F⊆V.Lemma 2.16.
A topological spaceX is ω-regular (resp., ω*-regular) if and only if for each U∈ωO(X) (resp., U∈O(X)) and each point x∈U, there exists V∈ωO(X,x) such that x∈V⊆ωCl(V)⊆U.Proposition 2.17.
LetX be an ω-regular space. Then f:X→Y is θ-ω-continuous if and only if it is almost weakly ω-continuous.Proof.
Suppose thatf is almost weakly ω-continuous. Let x∈X and V be any open set of Y containing f(x). Then, there exists U∈ωO(X,x) such that f(U)⊆Cl(V). Since X is ω-regular, by Lemma 2.16 there exists W∈ωO(X,x) such that x∈W⊆ωCl(W)⊆U. Therefore, we obtain f(ωCl(W))⊆Cl(V). This shows that f is θ-ω-continuous.Theorem 2.18.
Letf:X→Y be a function and g:X→X×Y the graph function of f defined by g(x)=(x,f(x)) for each x∈X. Then g is θ-ω-continuous if and only if f is θ-ω-continuous.Proof
Necessity.
Suppose thatg is θ-ω-continuous. Let x∈X and V be an open set of Y containing f(x). Then X×V is an open set of X×Y containing g(x). Since g is θ-ω-continuous, there exists U∈ωO(X,x) such that g(ωCl(U))⊆Cl(X×V)=X×Cl(V). Therefore, we obtain f(ωCl(U))⊆Cl(V). This shows that f is θ-ω-continuous.Sufficiency.
Letx∈X and W be any open set of X×Y containing g(x). There exist open sets U1⊆X and V⊆Y such that g(x)=(x,f(x))∈U1×V⊆W. Since f is θ-ω-continuous, there exists U2∈ωO(X,x) such that f(ωCl(U2))⊆Cl(V). Let U=U1∩U2, then U∈ωO(X,x). Therefore, we obtain g(ωCl(U))⊆Cl(U1)×f(ωCl(U2))⊆Cl(U1)×Cl(V)⊆Cl(W). This shows that g is θ-ω-continuous.


## 3. Stronglyθ-ω-Continuous Functions

We introduce the following relatively new definition.Definition 3.1 (see [14]).
A functionf:X→Y is said to be strongly θ-continuous if for each x∈X and each open set V of Y containing f(x), there exists an open neighborhood U of x such that f(Cl(U))⊆V.Definition 3.2.
A functionf:X→Y is said to be strongly θ-ω-continuous if for each x∈X and each open set V of Y containing f(x), there exists U∈ωO(X,x) such that f(ωCl(U))⊆V.Clearly, the following holds and none of its implications is reversible:Remark 3.3.
Strongθ-ω-continuity is stronger than ω-continuity and is weaker than strong θ-continuity. Strong θ-ω-continuity and continuity are independent of each other as the following examples show.Example 3.4.
LetX={a,b,c}, τ={ϕ,X,{a,b}}, and σ={ϕ,X,{c}}. Define a function f:(X,τ)→(X,σ) as follows: f(a)=a,f(b)=f(c)=c. Then f is strongly θ-ω-continuous but it is not continuous.Example 3.5.
LetX be an uncountable set and let A,B, and C be subsets of X such that each of them is uncountable and the family {A,B,C} is a partition of X. We defined the topology τ={ϕ,X,{A},{B},{A,B}} and σ={ϕ,X,{A},{A,B}}. Then, the identity function f:(X,τ)→(X,σ) is continuous (and ω-continuous) but is not strongly θ-ω-continuous.Next, several characterizations of stronglyθ-ω-continuous functions are obtained.Theorem 3.6.
For a functionf:X→Y, the following properties are equivalent:     (1)
f is strongly θ-ω-continuous;(2)
f-1(V) is ω-θ-open in X for every open set V of Y;(3)
f-1(F) is ω-θ-closed in X for every closed set F of Y;(4)
f(ωClθ(A))⊆Cl(f(A)) for every subset A of X;(5)
ωClθ(f-1(B))⊆f-1(Cl(B)) for every subset B of Y.Proof.
(1)⇒(2) Let V be any open set of Y. Suppose that x∈f-1(V). Since f is strongly θ-ω-continuous, there exists U∈ωO(X,x) such that f(ωCl(U))⊆V. Therefore, we have x∈U⊆ωCl(U)⊆f-1(V). This shows that f-1(V) is ω-θ-open in X.
(2)⇒(3) This is obvious.
(3)⇒(4) Let A be any subset of X. Since Cl(f(A)) is closed in Y, by (3) f-1(Cl(f(A))) is ω-θ-closed, and we have ωClθ(A)⊆ωClθ(f-1(f(A)))⊆ωClθ(f-1(Cl(f(A))))=f-1(Cl(f(A))). Therefore, we obtain f(ωClθ(A))⊆Cl(f(A)).
(4)⇒(5) Let B be any subset of Y. By (4), we obtain f(ωClθ(f-1(B)))⊆Cl(f(f-1(B)))⊆Cl(B) and hence ωClθ(f-1(B))⊆f-1(Cl(B)).
(5)⇒(1) Let x∈X and V be any open neighborhood of f(x). Since Y-V is closed in Y, we have ωClθ(f-1(Y-V))⊆f-1(Cl(Y-V))=f-1(Y-V). Therefore, f-1(Y-V) is ω-θ-closed in X and f-1(V) is an ω-θ-open set containing x. There exists U∈ωO(X,x) such that ωCl(U)⊆f-1(V) and hence f(ωCl(U))⊆V. This shows that f is strongly θ-ω-continuous.Theorem 3.7.
LetY be a regular space. Then, for a function f:X→Y, the following properties are equivalent:      (1)
f is almost weakly ω-continuous;(2)
f is ω-continuous;(3)
f strongly θ-ω-continuous.Proof.
(1)⇒(2) Let x∈X and V be an open set of Y containing f(x). Since Y is regular, there exists an open set W such that f(x)∈W⊆Cl(W)⊆V. Since f is almost weakly ω-continuous, there exists U∈ωO(X,x) such that f(U)⊆Cl(W)⊆V. Therefore f is ω-continuous.
(2)⇒(3) Let x∈X and V be an open set of Y containing f(x). Since Y is regular, there exists an open set W such that f(x)∈W⊆Cl(W)⊆V. Since f is ω-continuous, f-1(W) is ω-open and f-1(Cl(W)) is ω-closed. Set U=f-1(W), then since x∈f-1(W)⊆f-1(Cl(W)), U∈ωO(X,x) and ωCl(U)⊆f-1(Cl(W)). Consequently, we have f(ωCl(U))⊆Cl(W)⊆V.
(3)⇒(1) The proof follows immediately from the definition.Corollary 3.8.
LetY be a regular space. Then, for a function f:X→Y, the following properties are equivalent:      (1)
f is strongly θ-ω-continuous;(2)
f is ω-continuous;(3)
f is almost ω-continuous;(4)
f is θ-ω-continuous;(5)
f is almost weakly ω-continuous.Theorem 3.9.
A spaceX is ω*-regular if and only if, for any space Y, any continuous function f:X→Y is strongly θ-ω-continuous.Proof
Sufficiency.
Letf:X→X be the identity function. Then f is continuous and strongly θ-ω-continuous by our hypothesis. For any open set U of X and any points x of U, we have f(x)=x∈U and there exists G∈ωO(X,x) such that f(ωCl(G))⊆U. Therefore, we have x∈G⊆ωCl(G)⊆U. It follows from Lemma 2.16, that is, X is ω*-regular.Necessity.
Suppose thatf:X→Y is continuous and X is ω*-regular. For any x∈X and any open neighborhood V of f(x), f-1(V) is an open set of X containing x. Since X is ω*-regular, there exists U∈ωO(X) such that x∈U⊆ωCl(U)⊆f-1(V) by Lemma 2.16. Therefore, we have f(ωCl(U))⊆V. This shows that f is strongly θ-ω-continuous.Theorem 3.10.
Letf:X→Y be a function and g:X→X×Y the graph function of f defined by g(x)=(x,f(x)) for each x∈X. If g is strongly θ-ω-continuous, then f is strongly θ-ω-continuous and X is ω*-regular.Proof.
Suppose thatg is strongly θ-ω-continuous. First, we show that f is strongly θ-ω-continuous. Let x∈X and V be an open set of Y containing f(x). Then X×V is an open set of X×Y containing g(x). Since g is strongly θ-ω-continuous, there exists U∈ωO(X,x) such that g(ωCl(U))⊆X×V. Therefore, we obtain f(ωCl(U))⊆V. Next, we show that X is ω*-regular. Let U be any open set of X and x∈U. Since g(x)∈U×Y and U×Y is open in X×Y, there exists G∈ωO(X,x) such that g(ωCl(G))⊆U×Y. Therefore, we obtain x∈G⊆ωCl(G)⊆U and hence X is ω*-regular.Proposition 3.11.
LetX be an ω-regular space. Then f:X→Y is strongly θ-ω-continuous if and only if f is ω-continuous.Proof.
Suppose thatf is ω-continuous. Let x∈X and V be any open set of Y containing f(x). By the ω-continuity of f, we have f-1(V)∈ωO(X,x) and hence there exists U∈ωO(X,x) such that ωCl(U)⊆f-1(V). Therefore, we obtain f(ωCl(U))⊆V. This shows that f is strongly θ-ω-continuous.Theorem 3.12.
Letf:X→Y be a function and g:X→X×Y the graph function of f defined by g(x)=(x,f(x)) for each x∈X. If f is strongly θ-ω-continuous and X is ω-regular, then g is strongly θ-ω-continuous.Proof.
Letx∈X and W be any open set of X×Y containing g(x). There exist open sets U1⊆X and V⊆Y such that g(x)=(x,f(x))∈U1×V⊆W. Since f is strongly θ-ω-continuous, there exists U2∈ωO(X,x) such that f(ωCl(U2))⊆V. Since X is ω-regular and U1∩U2∈ωO(X,x), there exists U∈ωO(X,x) such that x∈U⊆ωCl(U)⊆U1∩U2 (by Lemma 2.16). Therefore, we obtain g(ωCl(U))⊆U1×f(ωCl(U2))⊆U1×V⊆W. This shows that g is strongly θ-ω-continuous.Theorem 3.13.
Suppose that the product of twoω-open sets of X is ω-open. If f:X→Y is strongly θ-ω-continuous injection and Y is Hausdorff, then E={(x,y):f(x)=f(y)} is ω-θ-closed in X×X.Proof.
Suppose that(x,y)∉E. Then f(x)≠f(y). Since Y is Hausdorff, there exist open sets V and U containing f(x) and f(y), respectively, such that U∩V=ϕ. Since f is strongly θ-ω-continuous, there exist G∈ωO(X,x) and H∈ωO(X,y) such that f(ωCl(G))⊆V and f(ωCl(H))⊆U. Set D=G×H. It follows that (x,y)∈D∈ωO(X×Y) and ωCl(G×H)∩E⊆[ωCl(G)×ωCl(H)]∩E=ϕ. By Proposition 2.4, E is ω-θ-closed in X×X.Definition 3.14 (see [9]).
A spaceX is said to be ω-T2-space (resp., ω-Urysohn) if for each pair of distinct points x and y in X, there exist U∈ωO(X,x) and V∈ωO(X,y) such that U∩V=ϕ (resp., ωCl(U)∩ωCl(V)=ϕ).Theorem 3.15.
Iff:X→Y is strongly θ-ω-continuous injection and Y is T0-space (resp., Hausdorff), then X is ω-T2-space (resp., ω-Urysohn).Proof.
(1) Suppose thatY is T0-space. Let x and y be any distinct points of X. Since f is injective, f(x)≠f(y) and there exists either an open neighborhood V of f(x) not containing f(y) or an open neighborhood W of f(y) not containing f(x). If the first case holds, then there exists U∈ωO(X,x) such that f(ωCl(U))⊆V. Therefore, we obtain f(y)∉f(ωCl(U)) and hence X-ωCl(U)∈ωO(X,y). If the second case holds, then we obtain a similar result. Therefore, X is ω-T2.
(2) Suppose thatY is Hausdorff. Let x and y be any distinct points of X. Then f(x)≠f(y). Since Y is Hausdorff, there exist open sets V and U containing f(x) and f(y), respectively, such that U∩V=ϕ. Since f is strongly θ-ω-continuous, there exist G∈ωO(X,x) and H∈ωO(X,y) such that f(ωCl(G))⊆V and f(ωCl(H))⊆U. It follows that f(ωCl(G))∩f(ωCl(H))=ϕ, hence ωCl(G)∩ωCl(H)=ϕ. This shows that X is ω-Urysohn.A subsetK of a space X is said to be ω-closed relative to X if for every cover {Vα:α∈Λ} of K by ω-open sets of X, there exists a finite subset Λ0 of Λ such that K⊆∪{ωCl(Vα):α∈Λ0}.Theorem 3.16.
Letf:X→Y be strongly θ-ω-continuous and Kω-closed relative to X, then f(K) is a compact set of Y.Proof.
Suppose thatf:X→Y is a strongly θ-ω-continuous function and K is ω-closed relative to X. Let {Vα:α∈Λ} be an open cover of f(K). For each point x∈K, there exists α(x)∈Λ such that f(x)∈Vα(x). Since f is strongly θ-ω-continuous, there exists Ux∈ωO(X,x) such that f(ωCl(Ux))⊆Vα(x). The family {Ux:x∈K} is a cover of K by ω-open sets of X and hence there exists a finite subset K* of K such that K⊆⋃x∈K*ωCl(Ux). Therefore, we obtain f(K)⊆⋃x∈K*Vα(x). This shows that f(K) is compact.Recall that a subsetA of a space X is quasi H-closed relative to X if for every cover {Vα:α∈Λ} of A by open sets of X, there exist a finite subset Λ0 of Λ such that A⊆∪{Cl(Vα):α∈Λ0}. A space X is said to be quasi H-closed (see [15]) if X is quasi H-closed relative to X.Theorem 3.17.
Letf:X→Y be θ-ω-continuous and Kω-closed relative to X, then f(K) is quasi H-closed relative to Y.Proof.
Suppose thatf:X→Y is a θ-ω-continuous function and K is ω-closed relative to X. Let {Vα:α∈Λ} be an open cover of f(K). For each point x∈K, there exists α(x)∈Λ such that f(x)∈Vα(x). Since f is θ-ω-continuous, there exists Ux∈ωO(X,x) such that f(ωCl(Ux))⊆Cl(Vα(x)). The family {Ux:x∈K} is a cover of K by ω-open sets of X and hence there exists a finite subset K* of K such that K⊆⋃x∈K*ωCl(Ux). Therefore, we obtain f(K)⊆⋃x∈K*Cl(Vα(x)). This shows that f(K) is quasi H-closed relative to Y.Definition 3.18 (see [9]).
A functionf:X→Y is said to be pre-ω-open if f(U)∈ωO(Y) for every U∈ωO(X).Proposition 3.19.
Letf:X→Y and g:Y→Z be functions and let g∘f:X→Z be strongly θ-ω-continuous. If f:X→Y is pre-ω-open and bijective, then g is strongly θ-ω-continuous.Proof.
Lety∈Y and W be any open set of Z containing g(y). Since f is bijective, y=f(x) for some x∈X. Since (g∘f) is strongly θ-ω-continuous, there exists U∈ωO(X,x) such that (g∘f)(ωCl(U))⊆W. Since f is pre-ω-open and bijective, the image f(A) of an ω-closed set A of X is ω-closed in Y. Therefore, we have ωCl(f(U))⊆f(ωCl(U)) and hence g(ωCl(f(U)))⊆(g∘f)(ωCl(U))⊆W. Since f(U)∈ωO(Y,y), g is strongly θ-ω-continuous.Definition 3.20 (see [16]).
A functionf:X→Y is said to be ω-irresolute if f-1(V)∈ωO(X) for each V∈ωO(Y).Lemma 3.21.
Iff:X→Y is ω-irresolute and V is ω-θ-open in Y, then f-1(V) is ω-θ-open in X.Proof.
LetV be an ω-θ-open set of Y and x∈f-1(V). There exists W∈ωO(Y) such that f(x)∈W⊆ωCl(W)⊆V. Since f is ω-irresolute, we have f-1(W)∈ωO(X) and f-1(ωCl(W))∈ωC(X). Therefore, we obtain x∈f-1(W)⊆ωCl(f-1(W))⊆f-1(ωCl(W))⊆f-1(V). This shows that f-1(V) is ω-θ-open in X.Theorem 3.22.
Letf:X→Y and g:Y→Z be functions. Then, the following properties hold.      (1)
Iff is strongly θ-ω-continuous and g is continuous, then the composition g∘f is strongly θ-ω-continuous.(2)
Iff is ω-irresolute and g is strongly θ-ω-continuous, then the composition g∘f is strongly θ-ω-continuous.Proof.
(1) This is obvious from Theorem3.6.
(2) This follows immediately from Theorem3.6 and Lemma 3.21.Theorem 3.23 (see [3]).
For any spaceX, the following are equivalent:      (1)
X is Lindelöf;(2)
everyω-open cover of X has a countable subcover.Definition 3.24 (see [17]).
A spaceX is said to be nearly Lindelöf if every regular open cover of X has a countably subcover.Proposition 3.25.
Letf:X→Y be an almost ω-continuous surjection. If X is Lindelöf, then Y is nearly Lindelöf.Proof.
Let{Vα:α∈Λ} be a regular open cover of Y. Since f is almost ω-continuous, {f-1(Vα):α∈Λ} is an ω-open cover of X. Since X is Lindelöf, by Theorem 3.23 there exists a countable subcover {f-1(Vαn):n∈ℕ} of X. Hence {Vαn:n∈ℕ} is a countable subcover of Y.Definition 3.26 (see [18]).
A topological spaceX is said to be almost Lindelöf if for every open cover {Uα:α∈Λ} of X there exists a countable subset {αn:n∈ℕ}⊆Λ such that X=⋃n∈ℕCl(Uαn).Theorem 3.27.
Letf:X→Y be an almost weakly ω-continuous surjection. If X is Lindelöf, then Y is almost Lindelöf.Proof.
Let{Vα:α∈Λ} be an open cover of Y. Let x∈X and Vα(x) be an open set in Y such that f(x)∈Vα(x). Since f is almost weakly ω-continuous, there exists an ω-open set Uα(x) of X containing x such that f(Uα(x))⊆Cl(Vα(x)). Now {Uα(x):x∈X} is an ω-open cover of the Lindelöf space X. So by Theorem 3.23, there exists a countable subset {Uα(xn):n∈ℕ} such that X=⋃n∈ℕ(Uα(xn)). Thus Y=f(⋃n∈ℕ(Uα(xn)))⊆⋃n∈ℕf(Uα(xn))⊆⋃n∈ℕCl(Vα(xn)). This shows that Y is almost Lindelöf.We notice that a subspaceA of a space X is Lindelöf if and only if for every cover {Vα:α∈Λ} of A by open set of X, there exists a countable subset Λ0 of Λ such that {Vα:α∈Λ0} covers A.Definition 3.28 (see [4]).
A functionf:X→Y is said to be ω-closed if the image of every closed subset of X is ω-closed in Y.Theorem 3.29.
Iff:X→Y is an ω-closed surjection such that f-1(y) is a Lindelöf subspace for each y∈Y and Y is Lindelöf, then X is Lindelöf.Proof.
Let{Uα:α∈Λ} be an open cover of X. Since f-1(y) is a Lindelöf subspace for each y∈Y, there exists a countable subset Λ(y) of Λ such that f-1(y)⊆∪{Uα:α∈Λ(y)}. Let U(y)=∪{Uα:α∈Λ(y)} and V(y)=Y-f(X-U(y)). Since f is ω-closed, V(y) is an ω-open set containing y such that f-1(V(y))⊆U(y). Then {V(y):y∈Y} is an ω-open cover of the Lindelöf space Y. By Theorem 3.23, there exist countable points of Y, says, y1,y2,…,yn,… such that Y=⋃n∈NV(yn). Therefore, we have X=f-1(⋃n∈NV(yn))=⋃n∈Nf-1(V(yn))⊆⋃n∈N(U(yn))=⋃n∈N(∪{Uα:α∈Λ(yn)})=∪{Uα:α∈Λ(yn),n∈N}. This shows that X is Lindelöf.Theorem 3.30 (see [3]).
Letf be an ω-continuous function from a space X onto a space Y. If X is Lindelöf, then Y is Lindelöf.Corollary 3.31.
Letf:X→Y be an ω-closed and ω-continuous surjection such that f-1(y) is a Lindelöf subspace for each y∈Y. Then X is Lindelöf if and only if Y is Lindelöf.Proof.
LetX be Lindelöf. It follows from Theorem 3.30 that Y is Lindelöf. The converse is an immediate consequence of Theorem 3.29.



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*Source: 174042-2009-12-09.xml*