#### Abstract

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

#### 1. Introduction

In 1943, Fomin [1] introduced the notion of -continuity. In 1968, the notions of -open subsets, -closed subsets, and -closure were introduced by Veliko [2]. In 1989, Hdeib [3] introduced the notion of -continuity. The main purpose of the present paper is to introduce and investigate fundamental properties of weak and strong forms of -continuous functions. Throughout this paper, and stand for topological spaces (called simply spaces) with no separation axioms assumed unless otherwise stated. For a subset of , the closure of and the interior of will be denoted by and , respectively. Let be a space and a subset of . A point is called a condensation point of if for each with , the set is uncountable. However, is said to be -closed [4] if it contains all its condensation points. The complement of an -closed set is said to be -open. It is well known that a subset of a space is -open if and only if for each , there exists such that and is countable. The family of all -open subsets of a space , denoted by or , forms a topology on finer than . The family of all -open sets of containing is denoted by . The -closure and the -interior, that can be defined in the same way as and , respectively, will be denoted by and . Several characterizations of -closed subsets were provided in [5–8].

A point of is called a -cluster points of if for every open set of containing . The set of all -cluster points of is called the -closure of and is denoted by . A subset is said to be -closed [2] if . The complement of a -closed set is said to be -open. A point of is called an --cluster point of if for every -open set of containing . The set of all --cluster points of is called the --closure of and is denoted by . A subset is said to be --closed if . The complement of a --closed set is said to be --open. The --interior of is defined by the union of all --open sets contained in and is denoted by .

#### 2. --Continuous Functions

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

*Definition 2.1. *A function is said to be -continuous (see [3]) (resp., almost weakly -continuous (see [9])) if for each and each open set of containing , there exists such that (resp., .

*Definition 2.2. *A function is said to be --continuous if for each and each open set of containing , there exists such that .

Next, several characterizations of --continuous functions are obtained.

Theorem 2.3. *For a function , the following properties are equivalent: *(1)* is --continuous; *(2)* for every subset of ; *(3)* for every subset of .*

*Proof. *(1)(2) Let be any subset of . Suppose that . Then and there exists an open set containing such that . Since is --continuous, there exists such that . Therefore, we have and . This shows that . Thus, we obtain .

(2)(1) Let and be an open set of containing . Then we have and . Hence, and . There exists such that and hence . Therefore, is --continuous.

(2)(3) Let be any subset of . Then we have and hence .

(3)(2) Let be a subset of . We have and hence .

Proposition 2.4. *A subset of a space is --open in if and only if for each , there exists an -open set with such that .*

*Proof. *Suppose that is --open in . Then is --closed. Let . Then , and so there exists an -open set with such that . Thus . Conversely, assume that is not --open. Then is not --closed, and so there exists such that . Since , by hypothesis, there exists an -open set with such that . Thus . This is a contradiction since .

Theorem 2.5. *For a function , the following properties are equivalent: *(1)* is --continuous; *(2)* for every open set of ; *(3)* for every open set of .*

*Proof. *(1)(2) Suppose that is any open set of and . Then and there exists such that . Therefore, . This shows that . Therefore, we obtain .

(2)(3) Suppose that is any open set of and . Then and there exists an open set containing such that ; hence . Therefore, we have . Since , by (2) , there exists such that . Thus we have and hence . This shows that .

(3)(1) Suppose that and are any open set of containing . Then and . Therefore and by (3) . There exists such that . Therefore, we obtain . This shows that is --continuous.

A subset of is said to be regular open (resp., regular closed) (see [10]) if (resp., ). Also, the family of all regular open (resp., regular closed) sets of is denoted by (resp., ).

Theorem 2.6. *For a function , the following properties are equivalent: *(1)* is --continuous; *(2)* for every subset of ;*(3)* for every open set of ; *(4)* for every closed set of ;*(5)* for every regular closed set of .*

*Proof. *(1)(2) This follows immediately from Theorem 2.5(3) with .

(2)(3) This is obvious since for every open set V of Y.

(3)(4) For any closed set of , and by (3)

(4)(5) This is obvious.

(5)(1) Let be any open set of . Since is regular closed, by (5) . It follows from Theorem 2.5 that is --continuous.

*Definition 2.7. *A subset of a space is said to be semi-open (see [11]) (resp., preopen (see [12]), -open (see [13])) if (resp., , ).

Theorem 2.8. *For a function , the following properties are equivalent: *(1)* is --continuous;*(2)* for every -open set of ;*(3)* for every semi-open set of .*

*Proof. *(1)(2) This is obvious by Theorem 2.6(5) since is regular closed for every -open set set .

(2)(3) This is obvious since every semi-open set is -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 , the following properties are equivalent: *(1)* is --continuous;*(2)* for every preopen set of ;*(3)* for every preopen set of ; *(4)* for every preopen set of .*

*Proof. *(1)(2) The proof follows from Theorem 2.8 (2) since every preopen set is -open.

(2)(3) This is obvious by the definition of a preopen set.

(3)(4) Let be any preopen set of . Then, by (3) we have
Therefore, we obtain .

(4)(1) This is obvious by Theorem 2.5 since every open set is preopen.

*Definition 2.10. *A function is said to be almost -continuous if for each and each regular open set of containing , there exists such that .

Lemma 2.11. *For a function , the following assertions are equivalent: *(1)* is almost -continuous; *(2)* for each and each open set of containing , there exists such that ;*(3)* for every ; *(4)* for every .*

Proposition 2.12. *For a function , the following properties hold: *(1)* if is almost -continuous, then it is --continuous; *(2)* if is --continuous, then it is almost weakly -continuous.*

*Proof. *(1) Suppose that and is any open set of containing . Since is almost -continuous, is -open and is -closed in by Lemma 2.11. Now, set . Then we have and . Therefore, we obtain . This shows that is --continuous.

(2) The proof follows immediately from the definition.

*Example 2.13. *Let be an uncountable set and let and be subsets of such that each of them is uncountable and the family is a partition of . We define the topology . Then, the function defined by *, * and is --continuous (and almost weakly -continuous) but is not almost -continuous since for , is regular open and but there is not open set containing such that *. *

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

It is clear that, for a subset of a space , if and only if for any -open set containing , .

Lemma 2.14. *For an -open set in a space , .*

*Proof. *By definition, . Let . Then for any -open set containing , . Let . Then and . Thus .

*Definition 2.15. *A topological space is said to be -regular (resp., -regular) if for each -closed (resp., closed) set and each point , there exist disjoint -open sets and such that and .

Lemma 2.16. *A topological space is -regular (resp., -regular) if and only if for each (resp., ) and each point , there exists such that .*

Proposition 2.17. *Let be an -regular space. Then is --continuous if and only if it is almost weakly -continuous.*

*Proof. *Suppose that is almost weakly -continuous. Let and be any open set of containing . Then, there exists such that . Since is -regular, by Lemma 2.16 there exists such that . Therefore, we obtain . This shows that is --continuous.

Theorem 2.18. *Let be a function and the graph function of defined by for each . Then is --continuous if and only if is --continuous.*

*Proof**Necessity. *Suppose that is --continuous. Let and be an open set of containing . Then is an open set of containing . Since is --continuous, there exists such that . Therefore, we obtain . This shows that is --continuous.*Sufficiency. *Let and be any open set of containing . There exist open sets and such that . Since is --continuous, there exists such that . Let , then . Therefore, we obtain . This shows that is --continuous.

#### 3. Strongly --Continuous Functions

We introduce the following relatively new definition.

*Definition 3.1 (see [14]). *A function is said to be strongly -continuous if for each and each open set of containing , there exists an open neighborhood of such that .

*Definition 3.2. *A function is said to be strongly --continuous if for each and each open set of containing , there exists such that .

Clearly, the following holds and none of its implications is reversible:

*Remark 3.3. *Strong --continuity is stronger than -continuity and is weaker than strong -continuity. Strong --continuity and continuity are independent of each other as the following examples show.

*Example 3.4. *Let , , and . Define a function as follows: *, **.* Then is strongly --continuous but it is not continuous.

*Example 3.5. *Let be an uncountable set and let and be subsets of such that each of them is uncountable and the family is a partition of *.* We defined the topology and *.* Then, the identity function is continuous (and -continuous) but is not strongly --continuous.

Next, several characterizations of strongly --continuous functions are obtained.

Theorem 3.6. *For a function , the following properties are equivalent: *(1)* is strongly --continuous; *(2)* is --open in for every open set of ; *(3)* is --closed in for every closed set of ; *(4)* for every subset of ; *(5)* for every subset of .*

*Proof. *(1)(2) Let be any open set of . Suppose that . Since is strongly --continuous, there exists such that . Therefore, we have . This shows that is --open in .

(2)(3) This is obvious.

(3)(4) Let be any subset of . Since is closed in , by (3) is --closed, and we have . Therefore, we obtain .

(4)(5) Let be any subset of . By (4), we obtain and hence .

(5)(1) Let and be any open neighborhood of . Since is closed in , we have . Therefore, is --closed in and is an --open set containing . There exists such that and hence . This shows that is strongly --continuous.

Theorem 3.7. *Let be a regular space. Then, for a function , the following properties are equivalent: *(1)* is almost weakly -continuous; *(2)* is -continuous; *(3)* strongly --continuous.*

*Proof. *(1)(2) Let and be an open set of containing . Since is regular, there exists an open set such that . Since is almost weakly -continuous, there exists such that . Therefore is -continuous.

(2)(3) Let and be an open set of containing . Since is regular, there exists an open set such that . Since is -continuous, is -open and is -closed. Set , then since , and . Consequently, we have .

(3)(1) The proof follows immediately from the definition.

Corollary 3.8. *Let be a regular space. Then, for a function , the following properties are equivalent: *(1)* is strongly --continuous; *(2)* is -continuous; *(3)* is almost -continuous; *(4)* is --continuous; *(5)* is almost weakly -continuous.*

Theorem 3.9. *A space is -regular if and only if, for any space , any continuous function is strongly --continuous.*

*Proof**Sufficiency. *Let be the identity function. Then is continuous and strongly --continuous by our hypothesis. For any open set of and any points of , we have and there exists such that . Therefore, we have . It follows from Lemma 2.16, that is, is -regular.*Necessity. *Suppose that is continuous and is -regular. For any and any open neighborhood of , is an open set of containing . Since is -regular, there exists such that by Lemma 2.16. Therefore, we have . This shows that is strongly --continuous.

Theorem 3.10. *Let be a function and the graph function of defined by for each . If is strongly --continuous, then is strongly --continuous and is -regular.*

*Proof. *Suppose that is strongly --continuous. First, we show that is strongly --continuous. Let and be an open set of containing . Then is an open set of containing . Since is strongly --continuous, there exists such that . Therefore, we obtain . Next, we show that is -regular. Let be any open set of and . Since and is open in , there exists such that . Therefore, we obtain and hence is -regular.

Proposition 3.11. *Let be an -regular space. Then is strongly --continuous if and only if is -continuous.*

*Proof. *Suppose that is -continuous. Let and be any open set of containing . By the -continuity of , we have and hence there exists such that . Therefore, we obtain . This shows that is strongly --continuous.

Theorem 3.12. *Let be a function and the graph function of defined by for each . If is strongly --continuous and is -regular, then is strongly --continuous.*

*Proof. *Let and be any open set of containing . There exist open sets and such that . Since is strongly --continuous, there exists such that . Since is -regular and , there exists such that (by Lemma 2.16). Therefore, we obtain . This shows that is strongly --continuous.

Theorem 3.13. *Suppose that the product of two -open sets of is -open. If is strongly --continuous injection and is Hausdorff, then is --closed in .*

*Proof. *Suppose that . Then . Since is Hausdorff, there exist open sets and containing and , respectively, such that . Since is strongly --continuous, there exist and such that and . Set . It follows that and . By Proposition 2.4, is --closed in .

*Definition 3.14 (see [9]). *A space is said to be --space (resp., -*Urysohn*) if for each pair of distinct points and in , there exist and such that (resp., ).

Theorem 3.15. *If is strongly --continuous injection and is -space (resp., Hausdorff), then is --space (resp., -Urysohn).*

*Proof. *(1) Suppose that is -space. Let and be any distinct points of . Since is injective, and there exists either an open neighborhood of not containing or an open neighborhood of not containing . If the first case holds, then there exists such that . Therefore, we obtain and hence . If the second case holds, then we obtain a similar result. Therefore, is -.

(2) Suppose that is Hausdorff. Let and be any distinct points of . Then . Since is Hausdorff, there exist open sets and containing and , respectively, such that . Since is strongly --continuous, there exist and such that and . It follows that , hence . This shows that is -*Urysohn*.

A subset of a space is said to be -closed relative to if for every cover of by -open sets of , there exists a finite subset of such that .

Theorem 3.16. *Let be strongly --continuous and -closed relative to , then is a compact set of .*

*Proof. *Suppose that is a strongly --continuous function and is -closed relative to . Let be an open cover of . For each point , there exists such that . Since is strongly --continuous, there exists such that . The family is a cover of by -open sets of and hence there exists a finite subset of such that . Therefore, we obtain . This shows that is compact.

Recall that a subset of a space is quasi -closed relative to if for every cover of by open sets of , there exist a finite subset of such that . A space is said to be quasi -closed (see [15]) if is quasi -closed relative to .

Theorem 3.17. *Let be --continuous and -closed relative to , then is quasi -closed relative to .*

*Proof. *Suppose that is a --continuous function and is -closed relative to . Let be an open cover of . For each point , there exists such that . Since is --continuous, there exists such that . The family is a cover of by -open sets of and hence there exists a finite subset of such that . Therefore, we obtain . This shows that is quasi -closed relative to .

*Definition 3.18 (see [9]). *A function is said to be pre--open if for every .

Proposition 3.19. *Let and be functions and let be strongly --continuous. If is pre--open and bijective, then is strongly --continuous.*

*Proof. *Let and be any open set of containing . Since is bijective, for some . Since is strongly --continuous, there exists such that . Since is pre--open and bijective, the image of an -closed set of is -closed in . Therefore, we have and hence . Since , is strongly --continuous.

*Definition 3.20 (see [16]). *A function is said to be -irresolute if for each .

Lemma 3.21. *If is -irresolute and is --open in , then is --open in .*

*Proof. *Let be an --open set of and . There exists such that . Since is -irresolute, we have and . Therefore, we obtain . This shows that is --open in .

Theorem 3.22. *Let and be functions. Then, the following properties hold. *(1)*If is strongly --continuous and is continuous, then the composition is strongly --continuous. *(2)*If is -irresolute and is strongly --continuous, then the composition is strongly --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 , the following are equivalent: *(1)* is Lindelöf; *(2)* every -open cover of has a countable subcover.*

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

Proposition 3.25. *Let be an almost -continuous surjection. If is Lindelöf, then is nearly Lindelöf.*

*Proof. *Let be a regular open cover of Y. Since is almost -continuous, is an -open cover of . Since is Lindelöf, by Theorem 3.23 there exists a countable subcover of . Hence is a countable subcover of .

*Definition 3.26 (see [18]). *A topological space is said to be almost Lindelöf if for every open cover of there exists a countable subset such that .

Theorem 3.27. *Let be an almost weakly -continuous surjection. If is Lindelöf, then is almost Lindelöf.*

*Proof. *Let be an open cover of . Let and be an open set in such that . Since is almost weakly -continuous, there exists an -open set of containing such that . Now is an -open cover of the Lindelöf space . So by Theorem 3.23, there exists a countable subset such that . Thus . This shows that is almost Lindelöf.

We notice that a subspace of a space is Lindelöf if and only if for every cover of by open set of , there exists a countable subset of such that covers .

*Definition 3.28 (see [4]). *A function is said to be -closed if the image of every closed subset of is -closed in .

Theorem 3.29. *If is an -closed surjection such that is a Lindelöf subspace for each and is Lindelöf, then is Lindelöf.*

*Proof. *Let be an open cover of . Since is a Lindelöf subspace for each , there exists a countable subset of such that . Let and . Since is -closed, is an -open set containing such that . Then is an -open cover of the Lindelöf space . By Theorem 3.23, there exist countable points of , says, such that . Therefore, we have . This shows that is Lindelöf.

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

Corollary 3.31. *Let be an -closed and -continuous surjection such that is a Lindelöf subspace for each . Then is Lindelöf if and only if is Lindelöf.*

*Proof. *Let be Lindelöf. It follows from Theorem 3.30 that 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.