Weak and Strong Forms of 𝜔-Continuous Functions

Abstract

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

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 𝜏_𝜔 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 point 𝑥 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 Cl_𝜃(𝐴). A subset 𝐴 is said to be 𝜃-closed [2] if 𝐴=Cl_𝜃(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 𝜔Cl_𝜃(𝐴). A subset 𝐴 is said to be 𝜔-𝜃-closed if 𝐴=𝜔Cl_𝜃(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 𝜔Int_𝜃(𝐴).

2. 𝜃-𝜔-Continuous Functions

We begin by recalling the following definition. Next, we introduce a relatively new notion.

Definition 2.1. A function 𝑓∶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 function 𝑓∶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 of 𝜃-𝜔-continuous functions are obtained.

Theorem 2.3. For a function 𝑓∶𝑋→𝑌, the following properties are equivalent: (1)𝑓 is 𝜃-𝜔-continuous; (2)𝜔Cl_𝜃(𝑓⁻¹(𝐵))⊆𝑓⁻¹(Cl_𝜃(𝐵)) for every subset 𝐵 of 𝑌; (3)𝑓(𝜔Cl_𝜃(𝐴))⊆Cl_𝜃(𝑓(𝐴)) for every subset 𝐴 of 𝑋.

Proof. (1)⇒(2) Let 𝐵 be any subset of 𝑌. Suppose that 𝑥∉𝑓⁻¹(Cl_𝜃(𝐵)). Then 𝑓(𝑥)∉Cl_𝜃(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∉𝜔Cl_𝜃(𝑓⁻¹(𝐵)). Thus, we obtain 𝜔Cl_𝜃(𝑓⁻¹(𝐵))⊆𝑓⁻¹(Cl_𝜃(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)∉Cl_𝜃(𝑌−Cl(𝑉)). Hence, 𝑥∉𝑓⁻¹(Cl_𝜃(𝑌−Cl(𝑉))) and 𝑥∉𝜔Cl_𝜃(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 𝜔Cl_𝜃(𝐴)⊆𝜔Cl_𝜃(𝑓⁻¹(𝑓(𝐴)))⊆𝑓⁻¹(Cl_𝜃(𝑓(𝐴))) and hence 𝑓(𝜔Cl_𝜃(𝐴))⊆Cl_𝜃(𝑓(𝐴)).
(3)⇒(2) Let 𝐵 be a subset of 𝑌. We have 𝑓(𝜔Cl_𝜃(𝑓⁻¹(𝐵)))⊆Cl_𝜃(𝑓(𝑓⁻¹(𝐵)))⊆Cl_𝜃(𝐵) and hence 𝜔Cl_𝜃(𝑓⁻¹(𝐵))⊆𝑓⁻¹(Cl_𝜃(𝐵)).

Proposition 2.4. A subset 𝑈 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 that 𝑈 is 𝜔-𝜃-open in 𝑋. Then 𝑋−𝑈 is 𝜔-𝜃-closed. Let 𝑥∈𝑈. Then 𝑥∉𝜔Cl_𝜃(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∈𝜔Cl_𝜃(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∈𝜔Cl_𝜃(𝑋−𝑈).

Theorem 2.5. For a function 𝑓∶𝑋→𝑌, the following properties are equivalent: (1)𝑓 is 𝜃-𝜔-continuous; (2)𝑓⁻¹(𝑉)⊆𝜔Int_𝜃(𝑓⁻¹(Cl(𝑉))) for every open set 𝑉 of 𝑌; (3)𝜔Cl_𝜃(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∈𝜔Int_𝜃(𝑓⁻¹(Cl(𝑉))). Therefore, we obtain 𝑓⁻¹(𝑉)⊆𝜔Int_𝜃(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∈𝜔Int_𝜃(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∉𝜔Cl_𝜃(𝑓⁻¹(𝑉)). This shows that 𝜔Cl_𝜃(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∉𝜔Cl_𝜃(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 subset 𝐴 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 function 𝑓∶𝑋→𝑌, the following properties are equivalent: (1)𝑓 is 𝜃-𝜔-continuous; (2)𝜔Cl_𝜃[𝑓⁻¹(Int(Cl_𝜃(𝐵)))]⊆𝑓⁻¹(Cl_𝜃(𝐵)) for every subset 𝐵 of 𝑌;(3)𝜔Cl_𝜃[𝑓⁻¹(Int(Cl(𝑉)))]⊆𝑓⁻¹(Cl(𝑉)) for every open set 𝑉 of 𝑌; (4)𝜔Cl_𝜃[𝑓⁻¹(Int(𝐾))]⊆𝑓⁻¹(𝐾) for every closed set 𝐾 of 𝑌;(5)𝜔Cl_𝜃[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(Cl_𝜃(𝐵)).
(2)⇒(3) This is obvious since Cl_𝜃(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) 𝜔Cl_𝜃(𝑓⁻¹(Int(𝐾)))=𝜔Cl_𝜃(𝑓⁻¹(Int(Cl(Int(𝐾)))))⊂𝑓⁻¹(Cl(Int(𝐾)))⊂𝑓⁻¹(𝐾). (2.1)
(4)⇒(5) This is obvious.
(5)⇒(1) Let 𝑉 be any open set of 𝑌. Since Cl(𝑉) is regular closed, by (5) 𝜔Cl_𝜃(𝑓⁻¹(𝑉)))⊂𝜔Cl_𝜃(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 subset 𝐴 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 function 𝑓∶𝑋→𝑌, the following properties are equivalent: (1)𝑓 is 𝜃-𝜔-continuous;(2)𝜔Cl_𝜃[𝑓⁻¹(Int(Cl(𝐺)))]⊆𝑓⁻¹(Cl(𝐺)) for every 𝛽-open set 𝐺 of 𝑌;(3)𝜔Cl_𝜃[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 function 𝑓∶𝑋→𝑌, the following properties are equivalent: (1)𝑓 is 𝜃-𝜔-continuous;(2)𝜔Cl_𝜃[𝑓⁻¹(Int(Cl(𝐺)))]⊆𝑓⁻¹(Cl(𝐺)) for every preopen set 𝐺 of 𝑌;(3)𝜔Cl_𝜃[𝑓⁻¹(𝐺)]⊆𝑓⁻¹(Cl(𝐺)) for every preopen set 𝐺 of 𝑌; (4)𝑓⁻¹(𝐺)⊂𝜔Int_𝜃(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 𝑋−𝜔Int_𝜃(𝑓⁻¹(Cl(𝐺)))=𝜔Cl_𝜃(𝑋−𝑓⁻¹(Cl(𝐺)))=𝜔Cl_𝜃(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). (2.2) Therefore, we obtain 𝑓⁻¹(𝐺)⊂𝜔Int_𝜃(𝑓⁻¹(Cl(𝐺))).
(4)⇒(1) This is obvious by Theorem 2.5 since every open set is preopen.

Definition 2.10. A function 𝑓∶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 function 𝑓∶𝑋→𝑌, the following assertions are equivalent: (1)𝑓 is almost 𝜔-continuous; (2) for each 𝑥∈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 function 𝑓∶𝑋→𝑌, the following properties hold: (1) if 𝑓 is almost 𝜔-continuous, then it is 𝜃-𝜔-continuous; (2) if 𝑓 is 𝜃-𝜔-continuous, then it is almost weakly 𝜔-continuous.

Proof. (1) Suppose that 𝑥∈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. Let 𝑋 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 𝑥_𝑐∈𝐶⊆𝑋, 𝐴 is regular open and 𝑓(𝑥_𝑐)∈𝐴 but there is not open set 𝑈_𝑥𝑐 containing 𝑥_𝑐 such that 𝑓(𝑈_𝑥𝑐)⊆𝐴.

Question 2. Is the converse of Proposition 2.12(2) true?

It is clear that, for a subset 𝐴 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 an 𝜔-open set 𝑈 in a space 𝑋, 𝜔Cl(𝑈)=𝜔Cl_𝜃(𝑈).

Proof. By definition, 𝜔Cl(𝑈)⊆𝜔Cl_𝜃(𝑈). Let 𝑥∈𝜔Cl_𝜃(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 𝜔Cl_𝜃(𝑈)⊆𝜔Cl(𝑈).

Definition 2.15. A topological space 𝑋 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 space 𝑋 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. Let 𝑋 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 that 𝑓 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. Let 𝑓∶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 that 𝑔 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. Let 𝑥∈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. Strongly 𝜃-𝜔-Continuous Functions

We introduce the following relatively new definition.

Definition 3.1 (see [14]). A function 𝑓∶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 function 𝑓∶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. Strong 𝜃-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. Let 𝑋={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. Let 𝑋 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 strongly 𝜃-𝜔-continuous functions are obtained.

Theorem 3.6. For a function 𝑓∶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(𝜔Cl_𝜃(𝐴))⊆Cl(𝑓(𝐴)) for every subset 𝐴 of 𝑋; (5)𝜔Cl_𝜃(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 𝜔Cl_𝜃(𝐴)⊆𝜔Cl_𝜃(𝑓⁻¹(𝑓(𝐴)))⊆𝜔Cl_𝜃(𝑓⁻¹(Cl(𝑓(𝐴))))=𝑓⁻¹(Cl(𝑓(𝐴))). Therefore, we obtain 𝑓(𝜔Cl_𝜃(𝐴))⊆Cl(𝑓(𝐴)).
(4)⇒(5) Let 𝐵 be any subset of 𝑌. By (4), we obtain 𝑓(𝜔Cl_𝜃(𝑓⁻¹(𝐵)))⊆Cl(𝑓(𝑓⁻¹(𝐵)))⊆Cl(𝐵) and hence 𝜔Cl_𝜃(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 𝜔Cl_𝜃(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. Let 𝑌 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. Let 𝑌 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 space 𝑋 is 𝜔^∗-regular if and only if, for any space 𝑌, any continuous function 𝑓∶𝑋→𝑌 is strongly 𝜃-𝜔-continuous.

Proof
Sufficiency. Let 𝑓∶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 that 𝑓∶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. Let 𝑓∶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 that 𝑔 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. Let 𝑋 be an 𝜔-regular space. Then 𝑓∶𝑋→𝑌 is strongly 𝜃-𝜔-continuous if and only if 𝑓 is 𝜔-continuous.

Proof. Suppose that 𝑓 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. Let 𝑓∶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. Let 𝑥∈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 two 𝜔-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 space 𝑋 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. If 𝑓∶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 that 𝑌 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 that 𝑌 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 subset 𝐾 of a space 𝑋 is said to be 𝜔-closed relative to 𝑋 if for every cover {𝑉_𝛼∶𝛼∈Λ} of 𝐾 by 𝜔-open sets of 𝑋, there exists a finite subset Λ₀ of Λ such that 𝐾⊆∪{𝜔Cl(𝑉_𝛼)∶𝛼∈Λ₀}.

Theorem 3.16. Let 𝑓∶𝑋→𝑌 be strongly 𝜃-𝜔-continuous and 𝐾𝜔-closed relative to 𝑋, then 𝑓(𝐾) is a compact set of 𝑌.

Proof. Suppose that 𝑓∶𝑋→𝑌 is a strongly 𝜃-𝜔-continuous function and 𝐾 is 𝜔-closed relative to 𝑋. Let {𝑉_𝛼∶𝛼∈Λ} be an open cover of 𝑓(𝐾). For each point 𝑥∈𝐾, there exists 𝛼(𝑥)∈Λ such that 𝑓(𝑥)∈𝑉_𝛼(𝑥). Since 𝑓 is strongly 𝜃-𝜔-continuous, there exists 𝑈_𝑥∈𝜔𝑂(𝑋,𝑥) such that 𝑓(𝜔Cl(𝑈_𝑥))⊆𝑉_𝛼(𝑥). The family {𝑈_𝑥∶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(𝑈_𝑥). Therefore, we obtain 𝑓(𝐾)⊆⋃_{𝑥∈𝐾∗}𝑉_𝛼(𝑥). This shows that 𝑓(𝐾) is compact.

Recall that a subset 𝐴 of a space 𝑋 is quasi 𝐻-closed relative to 𝑋 if for every cover {𝑉_𝛼∶𝛼∈Λ} of 𝐴 by open sets of 𝑋, there exist a finite subset Λ₀ of Λ such that 𝐴⊆∪{Cl(𝑉_𝛼)∶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. Let 𝑓∶𝑋→𝑌 be 𝜃-𝜔-continuous and 𝐾𝜔-closed relative to 𝑋, then 𝑓(𝐾) is quasi 𝐻-closed relative to 𝑌.

Proof. Suppose that 𝑓∶𝑋→𝑌 is a 𝜃-𝜔-continuous function and 𝐾 is 𝜔-closed relative to 𝑋. Let {𝑉_𝛼∶𝛼∈Λ} be an open cover of 𝑓(𝐾). For each point 𝑥∈𝐾, there exists 𝛼(𝑥)∈Λ such that 𝑓(𝑥)∈𝑉_𝛼(𝑥). Since 𝑓 is 𝜃-𝜔-continuous, there exists 𝑈_𝑥∈𝜔𝑂(𝑋,𝑥) such that 𝑓(𝜔Cl(𝑈_𝑥))⊆Cl(𝑉_𝛼(𝑥)). The family {𝑈_𝑥∶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(𝑈_𝑥). Therefore, we obtain 𝑓(𝐾)⊆⋃_{𝑥∈𝐾∗}Cl(𝑉_𝛼(𝑥)). This shows that 𝑓(𝐾) is quasi 𝐻-closed relative to 𝑌.

Definition 3.18 (see [9]). A function 𝑓∶𝑋→𝑌 is said to be pre-𝜔-open if 𝑓(𝑈)∈𝜔𝑂(𝑌) for every 𝑈∈𝜔𝑂(𝑋).

Proposition 3.19. Let 𝑓∶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. Let 𝑦∈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 function 𝑓∶𝑋→𝑌 is said to be 𝜔-irresolute if 𝑓⁻¹(𝑉)∈𝜔𝑂(𝑋) for each 𝑉∈𝜔𝑂(𝑌).

Lemma 3.21. If 𝑓∶𝑋→𝑌 is 𝜔-irresolute and 𝑉 is 𝜔-𝜃-open in 𝑌, then 𝑓⁻¹(𝑉) is 𝜔-𝜃-open in 𝑋.

Proof. Let 𝑉 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. Let 𝑓∶𝑋→𝑌 and 𝑔∶𝑌→𝑍 be functions. Then, the following properties hold. (1)If 𝑓 is strongly 𝜃-𝜔-continuous and 𝑔 is continuous, then the composition 𝑔∘𝑓 is strongly 𝜃-𝜔-continuous. (2)If 𝑓 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 Theorem 3.6.
(2) This follows immediately from Theorem 3.6 and Lemma 3.21.

Theorem 3.23 (see [3]). For any space 𝑋, the following are equivalent: (1)𝑋 is Lindelöf; (2) every 𝜔-open cover of 𝑋 has a countable subcover.

Definition 3.24 (see [17]). A space 𝑋 is said to be nearly Lindelöf if every regular open cover of 𝑋 has a countably subcover.

Proposition 3.25. Let 𝑓∶𝑋→𝑌 be an almost 𝜔-continuous surjection. If 𝑋 is Lindelöf, then 𝑌 is nearly Lindelöf.

Proof. Let {𝑉_𝛼∶𝛼∈Λ} be a regular open cover of Y. Since 𝑓 is almost 𝜔-continuous, {𝑓⁻¹(𝑉_𝛼)∶𝛼∈Λ} is an 𝜔-open cover of 𝑋. Since 𝑋 is Lindelöf, by Theorem 3.23 there exists a countable subcover {𝑓⁻¹(𝑉_𝛼𝑛)∶𝑛∈ℕ} of 𝑋. Hence {𝑉_𝛼𝑛∶𝑛∈ℕ} is a countable subcover of 𝑌.

Definition 3.26 (see [18]). A topological space 𝑋 is said to be almost Lindelöf if for every open cover {𝑈_𝛼∶𝛼∈Λ} of 𝑋 there exists a countable subset {𝛼_𝑛∶𝑛∈ℕ}⊆Λ such that 𝑋=⋃_𝑛∈ℕCl(𝑈_𝛼𝑛).

Theorem 3.27. Let 𝑓∶𝑋→𝑌 be an almost weakly 𝜔-continuous surjection. If 𝑋 is Lindelöf, then 𝑌 is almost Lindelöf.

Proof. Let {𝑉_𝛼∶𝛼∈Λ} be an open cover of 𝑌. Let 𝑥∈𝑋 and 𝑉_𝛼(𝑥) be an open set in 𝑌 such that 𝑓(𝑥)∈𝑉_𝛼(𝑥). Since 𝑓 is almost weakly 𝜔-continuous, there exists an 𝜔-open set 𝑈_𝛼(𝑥) of 𝑋 containing 𝑥 such that 𝑓(𝑈_𝛼(𝑥))⊆Cl(𝑉_𝛼(𝑥)). Now {𝑈_𝛼(𝑥)∶𝑥∈𝑋} is an 𝜔-open cover of the Lindelöf space 𝑋. So by Theorem 3.23, there exists a countable subset {𝑈_{𝛼(𝑥𝑛)}∶𝑛∈ℕ} such that 𝑋=⋃_𝑛∈ℕ(𝑈_{𝛼(𝑥𝑛)}). Thus 𝑌=𝑓(⋃_𝑛∈ℕ(𝑈_{𝛼(𝑥𝑛)}))⊆⋃_𝑛∈ℕ𝑓(𝑈_{𝛼(𝑥𝑛)})⊆⋃_𝑛∈ℕCl(𝑉_{𝛼(𝑥𝑛)}). This shows that 𝑌 is almost Lindelöf.

We notice that a subspace 𝐴 of a space 𝑋 is Lindelöf if and only if for every cover {𝑉_𝛼∶𝛼∈Λ} of 𝐴 by open set of 𝑋, there exists a countable subset Λ₀ of Λ such that {𝑉_𝛼∶𝛼∈Λ₀} covers 𝐴.

Definition 3.28 (see [4]). A function 𝑓∶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. If 𝑓∶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 {𝑈_𝛼∶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)⊆∪{𝑈_𝛼∶𝛼∈Λ(𝑦)}. Let 𝑈(𝑦)=∪{𝑈_𝛼∶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,…,𝑦_𝑛,… such that 𝑌=⋃_𝑛∈𝐍𝑉(𝑦_𝑛). Therefore, we have 𝑋=𝑓⁻¹(⋃_𝑛∈𝐍𝑉(𝑦_𝑛))=⋃_𝑛∈𝐍𝑓⁻¹(𝑉(𝑦_𝑛))⊆⋃_𝑛∈𝐍(𝑈(𝑦_𝑛))=⋃_𝑛∈𝐍(∪{𝑈_𝛼∶𝛼∈Λ(𝑦_𝑛)})=∪{𝑈_𝛼∶𝛼∈Λ(𝑦_𝑛),𝑛∈𝐍}. This shows that 𝑋 is Lindelöf.

Theorem 3.30 (see [3]). Let 𝑓 be an 𝜔-continuous function from a space 𝑋 onto a space 𝑌. If 𝑋 is Lindelöf, then 𝑌 is Lindelöf.

Corollary 3.31. Let 𝑓∶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. Let 𝑋 be Lindelöf. It follows from Theorem 3.30 that 𝑌 is Lindelöf. The converse is an immediate consequence of Theorem 3.29.

Acknowledgments

This work is financially supported by the Ministry of Higher Education, Malaysia, under FRGS grant no. UKM-ST-06-FRGS0008-2008. The authors would like to thank the referees for useful comments and suggestions. Theorems 2.6, 2.8, and 2.9 are established by suggestions of one of the referees.