Research Article | Open Access

Volume 2009 |Article ID 174042 | https://doi.org/10.1155/2009/174042

Ahmad Al-Omari, Takashi Noiri, Mohd. Salmi Md. Noorani, "Weak and Strong Forms of -Continuous Functions", International Journal of Mathematics and Mathematical Sciences, vol. 2009, Article ID 174042, 12 pages, 2009. https://doi.org/10.1155/2009/174042

Weak and Strong Forms of -Continuous Functions

Accepted04 Nov 2009
Published09 Dec 2009

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 [5â€“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 LindelĂ¶f; (2) every -open cover of has a countable subcover.

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

Proposition 3.25. Let be an almost -continuous surjection. If is LindelĂ¶f, then is nearly LindelĂ¶f.

Proof. Let be a regular open cover of Y. Since is almost -continuous, is an -open cover of . Since is LindelĂ¶f, 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 LindelĂ¶f 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 LindelĂ¶f, then is almost LindelĂ¶f.

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 LindelĂ¶f space . So by Theorem 3.23, there exists a countable subset such that . Thus . This shows that is almost LindelĂ¶f.

We notice that a subspace of a space is LindelĂ¶f 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 LindelĂ¶f subspace for each and is LindelĂ¶f, then is LindelĂ¶f.

Proof. Let be an open cover of . Since is a LindelĂ¶f 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 LindelĂ¶f space . By Theorem 3.23, there exist countable points of , says, such that . Therefore, we have . This shows that is LindelĂ¶f.

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

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

Proof. Let be LindelĂ¶f. It follows from Theorem 3.30 that is LindelĂ¶f. 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.

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