#### Abstract

A topological space is said to be strongly compact if every preopen cover of admits a finite subcover. In this paper, we introduce a new class of sets called *??*-preopen sets which is weaker than both open sets and *??*-open sets. Where a subset
is said to be *??*-preopen if for each
there exists a preopen set
containing such that is a finite
set. We investigate some properties of the sets. Moreover, we obtain new characterizations and preserving theorems of strongly compact spaces.

#### 1. Introduction

It is well known that the effects of the investigation of properties of closed bounded intervals of real numbers, spaces of continuous functions, and solutions to differential equations are the possible motivations for the formation of the notion of compactness. Compactness is now one of the most important, useful, and fundamental notions of not only general topology but also other advanced branches of mathematics. Many researchers have pithily studied the fundamental properties of compactness and now the results can be found in any undergraduate textbook on analysis and general topology. The productivity and fruitfulness of the notion of compactness motivated mathematicians to generalize this notion. In 1982, Atia et al. [1] introduced a strong version of compactness defined in terms of preopen subsets of a topological space which they called strongly compact. A topological space is said to be strongly compact if every preopen cover of admits a finite subcover. Since then, many mathematicians have obtained several results concerning its properties. The notion of strongly compact relative to a topological space was introduced by Mashhour et al. [2] in 1984. They established several characterizations of these spaces. In 1987, Ganster [3] obtained an interesting result that answered the question: what type of a space do we get when we take the one-point-compactification of a discrete space? He showed that this space is strongly compact. He proved that a topological space is strongly compact if and only if it is compact and every infinite subset of has nonempty interior. In 1988, Jankovic et al. [4] showed that a topological space is strongly compact if and only if it is compact and the family of dense sets in is finite. Quite recently Jafari and Noiri [5, 6], by introducing the class of firmly precontinuous functions, found some new characterizations of strongly compact spaces. They also obtained properties of strongly compact spaces by using nets, filterbases, precomplete accumulation points. The notion of preopen sets plays an important role in the study of strongly compact spaces. In this paper, first we introduce and study the notion of -preopen sets as a generalization of preopen sets. Then, by using -preopen sets, we obtain new characterizations and further preservation theorems of strongly compact spaces. We improve some of the results established by Mashhour et al. [2]. Throughout this paper, and stand for topological spaces on which no separation axiom is assumed unless otherwise stated. For a subset of , the closure of and the interior of will be denoted by and , respectively. A subset of a topological space is said to be preopen [7] if . A subset is said to be -open [8] if for each there exists an open set containing such that is a finite set. The complement of an -open subset is said to be -closed.

#### 2. -Preopen Sets

In this section, we introduce and study the notion of -preopen sets.

*Definition 2.1. *A subset is said to be -*preopen* if for each there exists a preopen set containing such that is a finite set. The complement of an -preopen subset is said to be -preclosed.

The family of all -preopen (resp., preopen, preclosed) subsets of a space is denoted by (resp., , ).

Lemma 2.2. *For a subset of a topological space , both -openness and preopenness imply -preopenness.*

For a subset of a topological space, the following implications hold:

(2.1)

*Example 2.3. *Let be the closed interval with a topology ,. Then is an -open set which is not preopen.

*Example 2.4. *Let be the set of all real numbers with the usual topology. Then the set of all rational numbers is a preopen set which is not -open.

Lemma 2.5. *Let be a topological space. Then the union of any family of -preopen sets is -preopen.*

*Proof. *Let be a family of -preopen subsets of and . Then for some . This implies that there exists a preopen subset of containing such that is finite. Since , then is finite. Thus .

Recall that a space is called submaximal if every dense subset of is open.

Lemma 2.6 (see [9]). *For a topological space , the followings are equivalent. *(1)* is submaximal. *(2)*Every preopen set is open. *

*Definition 2.7 (see [10]). *A subset of a space is said to be -open if .

Lemma 2.8 (see [11]). *Let be a topological space. Then the intersection of an -open set and a preopen set is preopen.*

Theorem 2.9. *Let be a submaximal topological space. Then is a topological space.*

*Proof. *() We have .

() Let and . Then there exist preopen sets containing such that and are finite. And . Thus is finite, and since is submaximal by Lemma 2.6, the intersection of two preopen sets is preopen. Hence .

() Let be any family of -preopen sets of . Then, by Lemma 2.5, is -preopen.

The converse of above theorem is not true.

*Example 2.10. *Let with . Then is a topological space and is not a submaximal topological space.

Lemma 2.11. *Let be a topological space. Then the intersection of an -open set and an -preopen set is -preopen.*

*Proof. *Let be -open and -preopen. Then for every , there exists a preopen set containing such that is finite, and also by Lemma 2.8, is preopen. Now for each , there exists a preopen set containing and
Then is finite. Therefore is an -preopen set.

The following lemma is well known and will be stated without the proof.

Lemma 2.12. *A topological space is a -space if and only if every finite set is closed.*

Proposition 2.13. *If a topological space is a -space, then every nonempty -preopen set contains a nonempty preopen set.*

*Proof. *Let be a nonempty -preopen set and , then there exists a preopen set containing such that is finite. Let . Then and by Lemmas 2.8 and 2.12, is preopen.

The following example shows that if is not a -space, then there exists a nonempty -preopen set which does not contain a nonempty preopen set.

*Example 2.14. *Let with . Then is an -preopen set which does not contain any a nonempty preopen set.

Lemma 2.15 (see [12]). *Let and be subsets of a topological space . *(1)*If and , then . *(2)*If and , then . *

Lemma 2.16. *Let and be subsets of a topological space . *(1)*If and , then . *(2)*If and , then . *

*Proof. *() Let . Since is -preopen in , there exists a preopen set of containing such that is finite. Since is -open, by Lemma 2.8 we have . Since , by Lemma 2.15, , , and is finite. This shows that .

() If , for each there exists containing such that is finite. Since , by Lemma 2.15, and is finite and hence .

Lemma 2.17. *A subset of a space is -preopen if and only if for every , there exist a preopen subset containing and a finite subset such that .*

*Proof. *Let be -preopen and , then there exists a preopen subset containing such that is finite. Let ). Then . Conversely, let . Then there exist a preopen subset containing and a finite subset such that . Thus and is a finite set.

Theorem 2.18. *Let be a space and . If is -preclosed, then for some preclosed subset and a finite subset .*

*Proof. *If is -preclosed, then is -preopen, and hence for every , there exists a preopen set containing and a finite set such that . Thus . Let . Then is a preclosed set such that .

#### 3. Strongly Compact Spaces

*Definition 3.1. *() A topological space is said to be *strongly compact* [1] if every cover of by preopen sets admits a finite subcover.

() A subset of a space is said to be strongly compact relative to [2] if every cover of by preopen sets of admits a finite subcover.

Theorem 3.2. *If is a space such that every preopen subset of is strongly compact relative to , then every subset of is strongly compact relative to .*

*Proof. *Let be an arbitrary subset of and let be a cover of by preopen sets of . Then the family is a preopen cover of the preopen set . Hence by hypothesis there is a finite subfamily which covers where is a finite subset of the naturals . This subfamily is also a cover of the set .

Theorem 3.3. *A subset of a space is strongly compact relative to if and only if for any cover of by -preopen sets of , there exists a finite subset of such that .*

*Proof. *Let be a cover of and . For each , there exists such that . Since is -preopen, there exists a preopen set such that and is finite. The family is a preopen cover of . Since is strongly compact relative to , there exists a finite subset, says, such that , where . Now, we have
For each , is a finite set and there exists a finite subset of such that . Therefore, we have . Hence is strongly compact relative to .

Since every preopen set is -preopen, the proof of the converse is obvious.

Corollary 3.4. *For any space , the following properties are equivalent: *(1)* is strongly compact;*(2)*every -preopen cover of admits a finite subcover. *

Theorem 3.5. *For any space , the following properties are equivalent: *(1)* is strongly compact; *(2)*every proper -preclosed set is strongly compact with respect to . *

*Proof. *()() Let be a proper -preclosed subset of . Let be a cover of by preopen sets of . Now for each , there is a preopen set such that is finite. Then is a preopen cover of . Since is strongly compact, there exist a finite subset of and a finite number of points, says, in such that ; hence . Since is finite for each , there exists a finite subset of such that . Therefore, we obtain . This shows that is strongly compact relative to .

()() Let be any preopen cover of . We choose and fix one . Then is a preopen cover of a -preclosed set . There exists a finite subset of such that . Therefore, . This shows that is strongly compact.

Corollary 3.6 (see [2]). *If a space is strongly compact and is preclosed, then is strongly compact relative to .*

Theorem 3.7. *Let be a submaximal topological space. Then is strongly compact if and only if is compact.*

*Proof. *Let be an open cover of . For each , there exists such that . Since is -preopen, there exists a preopen set of such that and is finite. The family is a preopen cover of . Since is strongly compact, there exists a finite subset, says, such that , where . Now, we have
For each , is a finite set and there exists a finite subset of such that . Therefore, we have . Hence is compact.

Conversely, let be a preopen cover of . Then . Since is compact, there exists a finite subcover of for . Hence is strongly compact.

#### 4. Preservation Theorems

*Definition 4.1. *A function is said to be -*precontinuous* if is -preopen in for each open set in .

Theorem 4.2. *A function is -precontinuous if and only if for each point and each open set in with , there is an -preopen set in such that and .*

*Proof. **Sufficiency*. Let be open in and . Then and thus there exists a such that and . Then and . Then by Lemma 2.5, is -preopen.*Necessity*. Let and let be an open set of containing . Then since is -precontinuous. Let . Then and .

Proposition 4.3. *If is -precontinuous and is an -open set in , then the restriction is -precontinuous.*

*Proof. *Since is -precontinuous, for any open set in , is -preopen in . Hence by Lemma 2.11, is -preopen in . Therefore, by Lemma 2.16, is -preopen in . This implies that is -precontinuous.

Proposition 4.4. *Let be a function and let be an -open cover of . If the restriction is -precontinuous for each , then is -precontinuous.*

*Proof. *Suppose that is an arbitrary open set in . Then for each , we have because is -precontinuous. Hence by Lemma 2.16, for each . By Lemma 2.5, we obtain . This implies that is -precontinuous.

Theorem 4.5. *Let be an -precontinuous function from a space onto a space . If is strongly compact, then is compact.*

*Proof. *Let be an open cover of . Then is an -preopen cover of . Since is strongly compact, by Corollary 3.4, there exists a finite subset of such that ; hence . Therefore is compact.

*Definition 4.6 (see [12]). *A function is said to be * precontinuous* if is preopen in for each open set in .

It is clear that every precontinuous function is -precontinuous but not conversely.

*Example 4.7. *Let , , and ,. Let be the identity function. Then is an -precontinuous function which is not precontinuous; because there exists such that .

Corollary 4.8 (see [2]). *Let be a precontinuous function from a space onto a space . If is strongly compact, then is compact.*

*Definition 4.9 (see [13]). *A function is said to be -*preopen* if the image of each preopen set of is preopen in .

Proposition 4.10. *If is -preopen, then the image of an -preopen set of is -preopen in .*

*Proof. *Let be -preopen and an -preopen subset of . For any , there exists such that . Since is -preopen, there exists a preopen set such that and is finite. Since is -preopen, is preopen in such that and is finite. Therefore, is -preopen in .

*Definition 4.11 (see [14]). *A function is said to be *preirresolute* if is preopen in for each preopen set in .

Proposition 4.12. *If is a preirresolute injection and is -preopen in , then is -preopen in .*

*Proof. *Assume that is an -preopen subset of . Let . Then and there exists a preopen set containing such that is finite. Since is preirresolute, is a preopen set containing . Thus and it is finite. It follows that is -preopen in .

*Definition 4.13. *A function is said to be *-preclosed* if is -preclosed in for each preclosed set of .

Theorem 4.14. *If is an -preclosed surjection such that is strongly compact relative to for each , and is strongly compact, then is strongly compact.*

*Proof. *Let be any preopen cover of . For each , is strongly compact relative to and there exists a finite subset of such that . Now we put and . Then, since is -preclosed, is an -preopen set in containing such that . Since is an -preopen cover of , by Corollary 3.4 there exists a finite subset such that . Therefore, . This shows that is strongly compact.

*Definition 4.15 (see [2]). *A function is said to be -*preclosed* if is preclosed in for each preclosed set of .

It is clear that every -preclosed function is -preclosed but not conversely.

*Example 4.16. *Let with topology . Let be the function defined by setting , , , and . Then is an -preclosed function which is not -preclosed; because there exists such that .

Corollary 4.17 (see [2]). *If is an -preclosed surjection such that is strongly compact relative to for each , and is strongly compact, then is strongly compact.*

*Definition 4.18. *A function is said to be -*continuous* if for each and each preopen set of containing , there exists an -preopen set of containing such that .

It is clear that every preirresolute function is -continuous but the converse is not true.

*Example 4.19. *Let and . Then the function , defined as , , and , is -continuous but it is not preirresolute.

Theorem 4.20. *Let be a -continuous surjection from onto . If is strongly compact, then is strongly compact.*

*Proof. *Let be a preopen cover of . For each , there exists such that . Since is -continuous, there exists an -preopen set of containing such that . So is an -preopen cover of the strongly compact space , and by Corollary 3.4 there exists a finite subset such that . Therefore . This shows that is strongly compact.

Corollary 4.21 (see [2]). *Let be a preirresolute surjection from onto . If is strongly compact, then is strongly compact.*

#### Acknowledgments

This work is financially supported by the Ministry of Higher Education, Malaysia under FRGS Grant no: UKM-ST-06-FRGS0008-2008. The authors also would like to thank the referees for useful comments and suggestions.