Research Article | Open Access

Volume 2009 |Article ID 465387 | https://doi.org/10.1155/2009/465387

Mohammad S. Sarsak, "On Semicompact Sets and Associated Properties", International Journal of Mathematics and Mathematical Sciences, vol. 2009, Article ID 465387, 8 pages, 2009. https://doi.org/10.1155/2009/465387

# On Semicompact Sets and Associated Properties

Accepted24 Dec 2009
Published18 Feb 2010

#### 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.

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