#### Abstract

We continue the study of semicompact sets in a topological space. Several properties, mapping properties of semicompact sets are studied. A special interest to spaces is given, where a space is if every subset of which is semicompact in is semiclosed; we study several properties of such spaces, it is mainly shown that a semi- semicompact space is if and only if it is extremally disconnected. It is also shown that in an -regular space if every point has an neighborhood, then is .

#### 1. Introduction and Preliminaries

A subset of a space is called semi-open [1] if , or equivalently, if there exists an open subset of such that ; is called semiclosed if is semi-open. The semiclosure of a subset of a space is the intersection of all semiclosed subsets of that contain or equivalently the smallest semiclosed subset of that contains . Clearly, is semiclosed if and only if ; it is also clear that if is a subset of a space and , then if and only if for each semi-open subset of containing . A subset of a space is called preopen [2] (resp., -open [3]) if (resp., ). Njastad [3] pointed out that the family of all -open subsets of a space , denoted by , is a topology on finer than . We will denote the families of semi-open (resp., preopen, -open) subsets of a space by (resp., , ). If is a topological space, we will denote the space by . JankoviΔ [4] pointed out that , and . Reilly and Vamanamurthy observed in [5] that . It is known that the intersection of a semi-open (resp., preopen) set with an -open set is semi-open (resp., preopen) and that the arbitrary union of semi-open (resp., preopen) sets is semi-open (resp., preopen).

A space is called semicompact [6] (resp., semi-LindelΓΆf [7]) if any semi-open cover of X has a finite (resp., countable) subcover. A subset of a space will be called semicompact (resp., semi-LindelΓΆf) if it is semicompact (resp., semi-LindelΓΆf) as a subspace.

A function from a space into a space is called semi-continuous [1] if the inverse image of each open subset of is semi-open in , irresolute [8] if the inverse image of each semi-open subset of is semi-open in and is called pre-semi-open (resp., pre-semiclosed [8]) if it maps semi-open (resp., semiclosed) subsets of onto semi-open (resp., semiclosed) subsets of .

A space is called semi- [9] if for each distinct points and of , there exist two disjoint semi-open subsets and of containing and respectively.

A space is called extremally disconnected [10] if the closure of each open subset of is open or equivalently if every regular closed subset of is preopen.

Throughout this paper, a space stands for a topological space, and if is a space and , then and stand respectively for the closure of in and the interior of in . For the concepts not defined here, we refer the reader to [11].

In concluding this section, we recall the following facts for their importance in the material of our paper.

Proposition 1.1. *Let , where is a space. Then*(i)*If is semi-open in , then is semi-open in ;*(ii)*[12] If is semi-open in and is semi-open in , then is semi-open in .*

Proposition 1.2. *Let , where is a space and is preopen in . Then is semi-open (resp., semiclosed) in if and only if , where is semi-open (resp., semi-closed) in .*

#### 2. Semicompact Sets

This section is mainly devoted to continue the study of semicompact sets. We also introduce and study semi-LindelΓΆf sets.

*Definition 2.1 . (see [13]). *A subset of a space is called semicompact relative to if any semi-open cover of in has a finite subcover of .

By semicompact in , we will mean semicompact relative to .

*Definition 2.2. *A subset of a space is called semi-LindelΓΆf in if any semi-open cover of in has a countable subcover of .

*Remark 2.3. *It is easy to see from the fact that , that a subset of a space is semicompact (resp., semi-LindelΓΆf) in if and only if it is semicompact (resp., semi-LindelΓΆf) in .

The proof of the following proposition is straightforward, and thus omitted.

Proposition 2.4. *The finite (resp., countable) union of semicompact (resp., semi-LindelΓΆf) sets in a space is semicompact (resp., semi-LindelΓΆf) in .*

Proposition 2.5. *Let be a preopen subset of a space and . If is semicompact (resp., semi-LindelΓΆf) in , then is semicompact (resp., semi-LindelΓΆf) in .*

*Proof. *We will show the case when is semicompact in , the other case is similar. Suppose that is a cover of by semi-open sets in . By Proposition 1.2, for each , where is semi-open in for each . Thus is a cover of by semi-open sets in , but is semicompact in , so there exist such that , and thus . Hence, is semicompact in .

Corollary 2.6. *Let be subset of a space . If is semicompact (resp., semi-LindelΓΆf) in , then is semicompact (resp., semi-LindelΓΆf).*

Proposition 2.7. *Let be a preopen subset of a space and . Then is semicompact (resp., semi-LindelΓΆf) in if and only if is semicompact (resp., semi-LindelΓΆf) in .*

*Proof. **Necessity*. It follows from Proposition 2.5.*Sufficiency*. We will show the case when is semicompact in , the other case is similar. Suppose that is a cover of by semi-open sets in . Then is a cover of . Since is semi-open in for each and is preopen in , it follows from Proposition 1.2 that is semi-open in for each , but is semicompact in , so there exist such that . Hence, is semicompact in .

Corollary 2.8. *A preopen subset of a space is semicompact (resp., semi-LindelΓΆf) if and only if is semicompact (resp., semi-LindelΓΆf) in .*

Proposition 2.9. *Let be a semicompact (resp., semi-LindelΓΆf) set in a space and be a semi-closed subset of . Then is semicompact (resp., semi-LindelΓΆf) in . In particular, a semi-closed subset of a semicompact (resp., semi-LindelΓΆf) space is semicompact (resp., semi-LindelΓΆf) in .*

*Proof. *We will show the case when is semicompact in , the other case is similar. Suppose that is a cover of by semi-open sets in . Then is a cover of by semi-open sets in , but is semicompact in , so there exist such that . Thus . Hence, is strongly compact in .

Proposition 2.10. *Let be an irresolute function. Then*(i)*[13] If is semicompact in , then is semicompact in ;*(ii)*If is semi-LindelΓΆf in , then is semi-LindelΓΆf in .*

*Proof. *(ii) The proof is similar to that of (i). We will, however, show it for the convenience of the reader. Suppose that is a cover of by semi-open sets in . Then is a cover of , but is irresolute, so is semi-open in for each . Since is semi-LindelΓΆf in , there exist such that . Thus . Hence, is semi-LindelΓΆf in .

Proposition 2.11. *Let be a pre-semi-closed surjection. If for each , is semicompact (resp., semi-LindelΓΆf) in , then is semicompact (resp., semi-LindelΓΆf) in whenever is semicompact (resp., semi-LindelΓΆf) in .*

*Proof. *We will show the case when is semicompact in , the other case is similar. Suppose that is a cover of by semi-open sets in . Then it follows by assumption that for each , there exists a finite subcollection of such that . Let . Then is semi-open in as any union of semi-open sets is semi-open. Let . Then is semi-open in as is pre-semi-closed, also for each as . Thus, is a cover of by semi-open sets in , but is semicompact in , so there exist such that . Thus, . Since S is a finite subcollection of for each , it follows that is a finite subcollection of . Hence, is semicompact in .

#### 3. Spaces

*Definition 3.1. *A space is said to be if any subset of which is semicompact in is semi-closed.

*Remark 3.2. *It follows from Remark 2.3, that a space is if and only if is .

We recall the following result from [3], it will be helpful to show the next two theorems.

Proposition 3.3. *A space is extremally disconnected if and only if the intersection of any two semi-open subsets of is semi-open.*

Theorem 3.4. *Let be a semi- extremally disconnected space. Then is .*

*Proof. *Let be a subset of which is semicompact in and let . Then for each there exist two disjoint semi-open sets and containing and respectively (as is semi-). Since is semicompact in , there exist such that . Let . Then is a semi-open subset of that contains and disjoint from (as is extremally disconnected using Proposition 3.3). Thus, . Hence, is semi-closed in .

Theorem 3.5. *If is an space such that every semi-closed subset of is semicompact in , then is extremally disconnected. In particular, an semicompact space is extremally disconnected.*

*Proof. *Let where and are semi-closed in . It follows by assumption that and are semicompact in and thus by Proposition 2.4, is semicompact in , but is , so is semi-closed in . Hence by Proposition 3.3, is extremally disconnected. The last part follows by Proposition 2.9.

Corollary 3.6. *For a semi- semicompact space, the followings are equivalent:*(i)* is .*(ii)* is extremally disconnected.*

Observing that a singleton of a space is semi-open if and only if it is open, the following proposition seems clear.

Proposition 3.7. *If every subset of a space is semicompact in , then is if and only if is a finite discrete space.*

Theorem 3.8. *Let be a pre-semi-closed function from a space onto a space such that for each , is semicompact in . If is , then so is .*

*Proof. *Let be a semicompact set in . Then by Proposition 2.11, is semicompact in , but is , so is semi-closed in , but is a pre-semi-closed surjection, so is semi-closed. Hence, is .

Theorem 3.9. *Let f be an irresolute one-to-one function from a space into an space . Then is .*

*Proof. *Let be a semicompact set in . Then it follows from Proposition 2.10(i) that is semicompact in , but is , so is semi-closed in . Since is one-to-one and irresolute, is semi-closed in . Hence, is .

Lemma 3.10. *A subset of is semi-open if and only if is semi-open in for each . Thus a subset of is semi-closed if and only if is semi-closed in for each .*

*Proof. *Since is open in , it follows that if is semi-open in , then is semi-open in and thus semi-open in for each . Now suppose that is semi-open in for each . Then is semi-open in for each because is open and thus semi-open in . Thus, is semi-open in as the arbitrary union of semi-open sets is semi-open.

Corollary 3.11. *Being is hereditary with respect to preopen subsets.*

*Proof. *Let be a preopen subset of an space and let be semicompact in . Then by Proposition 2.7, is semicompact in , but is , so is semi-closed in . By Proposition 1.2, is semi-closed in . Hence, is .

Corollary 3.12. * is if and only if is for each .*

*Proof. **Necessity.* It follows from Corollary 3.11 since is open and thus preopen in .*Sufficiency.* Suppose that is an space for each and let be a subset of which is semicompact in . Since is closed and thus semi-closed in , it follows from Proposition 2.9 that is semicompact in , but is preopen in , so it follows from Proposition 2.7 that is semicompact in . Since is , is semi-closed in for each , thus by Lemma 3.10, is semi-closed in . Hence, is .

Recall that a space is called -regular [14] if whenever is an open subset of and , there exists a semi-open subset of and a semi-closed subset of such that . We now define a type of regularity which is stronger than -regularity and weaker than regularity.

*Definition 3.13. *A space is called -regular if whenever is an open subset of and , there exists an open subset of and a semi-closed subset of such that .

Theorem 3.14. *If is an -regular space in which every point has an neighborhood, then is .*

*Proof. *Let be a subset of which is semicompact in and let . Then by assumption there exists an neighborhood of . Since being is hereditary with respect to preopen sets (Corollary 3.11), it follows that has an open neighborhood . Now since is -regular, there exists an open subset of and a semi-closed subset of such that . Since is semicompact in and is a semi-closed subset of , it follows from Proposition 2.9 that is semicompact in , thus by Proposition 2.5, is semicompact in , but is , so is semi-closed in , that is, is semi-open in and thus semi-open in by Proposition 1.1(ii) as is open and thus semi-open in . Thus is a semi-open subset of that contains and disjoint from and therefore, . Hence, is semi-closed in , and therefore, is .

Corollary 3.15. *If is a regular space in which every point has an neighborhood, then is .*

Theorem 3.16. *Let be a space in which every semi-closed subset is semicompact in , be an space. Then any irresolute function from into is pre-semi-closed. In particular, any irresolute function from a semicompact space into an space is pre-semi-closed.*

*Proof. *Let be a semi-closed subset of . By assumption, is semicompact in . Since is irresolute, it follows by Proposition 2.10 that is semicompact in . Since is , it follows that is semi-closed in . The last part follows from Proposition 2.9.

The following lemma will be helpful to show the next result, the easy proof is omitted.

Lemma 3.17. *(i) The projection function is irresolute.**(ii) Let be irresolute and be an -open subspace of . Then the restriction function is irresolute.*

Theorem 3.18. *Let be an space and be any space. If is a function whose graph is an -open subspace of in which every semi-closed subset is semicompact in , then is irresolute. In particular, any function having an domain and an -open, semicompact graph is irresolute.*

*Proof. *Let and be the projection functions. Since is an -open subspace of , it follows from Lemma 3.17 that is irresolute. Thus it follows from Theorem 3.16 that is pre-semi-closed, that is, is irresolute. Also, is irresolute. Thus, is irresolute. The last part follows from Proposition 2.9.

#### 4. Spaces

The study of this section is analogous to that of the preceding section, similar proofs are omitted.

*Definition 4.1. *A space is said to be if any subset of which is semi-LindelΓΆf in is semi-closed.

*Remark 4.2. *It follows from Remark 2.3, that a space is if and only if is .

Following Proposition 3.3, we will call β a space -extremally disconnected if the countable intersection of semi-open subsets of is semi-open.

Theorem 4.3. *Let be a semi--extremally disconnected. Then is .*

Theorem 4.4. *If is an space such that every semi-closed subset of is semi-LindelΓΆf in , then is -extremally disconnected. In particular, an semi-LindelΓΆf space is -extremally disconnected.*

Corollary 4.5. *For a semi- semi-LindelΓΆf space, the followings are equivalent:*(i)* is .*(ii)* is -extremally disconnected.*

Proposition 4.6. *If every subset of a space is semi-LindelΓΆf in , then is if and only if is a countable discrete space.*

Theorem 4.7. *Let be a pre-semi-closed function from a space onto a space such that for each , is semi-LindelΓΆf in . If is , then so is .*

Theorem 4.8. *Let f be an irresolute one-to-one function from a space into an space . Then is .*

Proposition 4.9. *Being is hereditary with respect to preopen subsets.*

Corollary 4.10. * is if and only if is for each .*

Theorem 4.11. *If is an -regular space in which every point has an neighborhood, then is .*

Corollary 4.12. *If is a regular space in which every point has an neighborhood, then is .*

Theorem 4.13. *Let be a space in which every semi-closed subset is semi-LindelΓΆf in , be an space. Then any irresolute function from into is pre-semi-closed. In particular, any irresolute function from a semi-LindelΓΆf space into an space is pre-semi-closed.*

Theorem 4.14. *Let be an space and be any space. If is a function whose graph is an -open subspace of in which every semi-closed subset is semi-LindelΓΆf in , then is irresolute. In particular, any function having an domain and an -open, semi-LindelΓΆf graph is irresolute.*