On Semicompact Sets and Associated Properties

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 𝑆𝐶𝑆 spaces is given, where a space 𝑋 is 𝑆𝐶𝑆 if every subset of 𝑋 which is semicompact in 𝑋 is semiclosed; we study several properties of such spaces, it is mainly shown that a semi-𝑇₂ semicompact space is 𝑆𝐶𝑆 if and only if it is extremally disconnected. It is also shown that in an 𝑜𝑠-regular space 𝑋 if every point has an 𝑆𝐶𝑆 neighborhood, then 𝑋 is 𝑆𝐶𝑆.

1. Introduction and Preliminaries

A subset 𝐴 of a space 𝑋 is called semi-open [1] if 𝐴⊂Int𝐴, or equivalently, if there exists an open subset 𝑈 of 𝑋 such that 𝑈⊂𝐴⊂𝑈; 𝐴 is called semiclosed if 𝑋⧵𝐴 is semi-open. The semiclosure scl(𝐴) of a subset 𝐴 of a space 𝑋 is the intersection of all semiclosed subsets of 𝑋 that contain 𝐴 or equivalently the smallest semiclosed subset of 𝑋 that contains 𝐴. Clearly, 𝐴 is semiclosed if and only if scl(𝐴)=𝐴; it is also clear that if 𝐴 is a subset of a space 𝑋 and 𝑥∈𝑋, then 𝑥∈scl(𝐴) if and only if 𝑆∩𝐴≠𝜙 for each semi-open subset 𝑆 of 𝑋 containing 𝑥. A subset 𝐴 of a space 𝑋 is called preopen [2] (resp., 𝛼-open [3]) if 𝐴⊂Int𝐴 (resp., 𝐴⊂IntInt𝐴). Njastad [3] pointed out that the family of all 𝛼-open subsets of a space (𝑋,𝜏), denoted by 𝜏^𝛼, is a topology on 𝑋 finer than 𝜏. We will denote the families of semi-open (resp., preopen, 𝛼-open) subsets of a space 𝑋 by 𝑆𝑂(𝑋) (resp., 𝑃𝑂(𝑋), 𝛼𝑂(𝑋)). If (𝑋,𝜏) is a topological space, we will denote the space (𝑋,𝜏^𝛼) by 𝑋^𝛼. Janković [4] pointed out that 𝑃𝑂(𝑋)=𝑃𝑂(𝑋^𝛼), 𝑆𝑂(𝑋)=𝑆𝑂(𝑋^𝛼) and 𝛼𝑂(𝑋)=𝛼𝑂(𝑋^𝛼). Reilly and Vamanamurthy observed in [5] that 𝜏^𝛼=𝑆𝑂(𝑋)∩𝑃𝑂(𝑋). It is known that the intersection of a semi-open (resp., preopen) set with an 𝛼-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 𝑋 is called semicompact [6] (resp., semi-Lindelöf [7]) if any semi-open cover of X has a finite (resp., countable) subcover. A subset 𝐴 of a space 𝑋 will be called semicompact (resp., semi-Lindelöf) if it is semicompact (resp., semi-Lindelöf) as a subspace.

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

A space 𝑋 is called semi-𝑇₂ [9] if for each distinct points 𝑥 and 𝑦 of 𝑋, there exist two disjoint semi-open subsets 𝑈 and 𝑉 of 𝑋 containing 𝑥 and 𝑦, respectively.

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

Throughout this paper, a space 𝑋 stands for a topological space, and if 𝑋 is a space and 𝐴⊂𝑋, then 𝐴 and Int𝐴 stand respectively for the closure of 𝐴 in 𝑋 and the interior of 𝐴 in 𝑋. 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 𝐴⊂𝐵⊂𝑋, where 𝑋 is a space. Then(i)If 𝐴 is semi-open in 𝑋, then 𝐴 is semi-open in 𝐵;(ii)[12] If 𝐴 is semi-open in 𝐵 and 𝐵 is semi-open in 𝑋, then 𝐴 is semi-open in 𝑋.

Proposition 1.2. Let 𝐴⊂𝐵⊂𝑋, where 𝑋 is a space and 𝐵 is preopen in 𝑋. Then 𝐴 is semi-open (resp., semiclosed) in 𝐵 if and only if 𝐴=𝑆∩𝐵, where 𝑆 is semi-open (resp., semi-closed) in 𝑋.

2. Semicompact Sets

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

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

By semicompact in 𝑋, we will mean semicompact relative to 𝑋.

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

Remark 2.3. It is easy to see from the fact that 𝑆𝑂(𝑋)=𝑆𝑂(𝑋^𝛼), that a subset 𝐴 of a space 𝑋 is semicompact (resp., semi-Lindelöf) in 𝑋 if and only if it is semicompact (resp., semi-Lindelöf) in 𝑋^𝛼.

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

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

Proposition 2.5. Let 𝐵 be a preopen subset of a space 𝑋 and 𝐴⊂𝐵. If 𝐴 is semicompact (resp., semi-Lindelöf) in 𝑋, then 𝐴 is semicompact (resp., semi-Lindelöf) in 𝐵.

Proof. We will show the case when 𝐴 is semicompact in 𝑋, the other case is similar. Suppose that 𝒜={𝐴_𝛼∶𝛼∈Λ} is a cover of 𝐴 by semi-open sets in 𝐵. By Proposition 1.2, 𝐴_𝛼=𝑆_𝛼∩𝐵 for each 𝛼∈Λ, where 𝑆_𝛼 is semi-open in 𝑋 for each 𝛼∈Λ. Thus 𝒮={𝑆_𝛼∶𝛼∈Λ} is a cover of 𝐴 by semi-open sets in 𝑋, but 𝐴 is semicompact in 𝑋, so there exist 𝛼₁,𝛼₂,…,𝛼_𝑛∈Λ such that 𝐴⊂⋃^𝑖=𝑛_𝑖=1𝑆_𝛼𝑖, and thus 𝐴⊂⋃^𝑖=𝑛_𝑖=1(𝑆_𝛼𝑖∩𝐵)=⋃^𝑖=𝑛_𝑖=1𝐴_𝛼𝑖. Hence, 𝐴 is semicompact in 𝐵.

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

Proposition 2.7. Let 𝐵 be a preopen subset of a space 𝑋 and 𝐴⊂𝐵. Then 𝐴 is semicompact (resp., semi-Lindelöf) in 𝑋 if and only if 𝐴 is semicompact (resp., semi-Lindelöf) in 𝐵.

Proof. Necessity. It follows from Proposition 2.5.
Sufficiency. We will show the case when 𝐴 is semicompact in 𝐵, the other case is similar. Suppose that 𝒮={𝑆_𝛼∶𝛼∈Λ} is a cover of 𝐴 by semi-open sets in 𝑋. Then 𝒜={𝑆_𝛼∩𝐵∶𝛼∈Λ} is a cover of 𝐴. Since 𝑆_𝛼 is semi-open in 𝑋 for each 𝛼∈Λ and 𝐵 is preopen in 𝑋, it follows from Proposition 1.2 that 𝑆_𝛼∩𝐵 is semi-open in 𝐵 for each 𝛼∈Λ, but 𝐴 is semicompact in 𝐵, so there exist 𝛼₁,𝛼₂,…𝛼_𝑛∈Λ such that 𝐴⊂⋃^𝑖=𝑛_𝑖=1(𝑆_𝛼𝑖∩𝐵)⊂⋃^𝑖=𝑛_𝑖=1𝑆_𝛼𝑖. Hence, 𝐴 is semicompact in 𝑋.

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

Proposition 2.9. Let 𝐴 be a semicompact (resp., semi-Lindelöf) set in a space 𝑋 and 𝐵 be a semi-closed subset of 𝑋. Then 𝐴∩𝐵 is semicompact (resp., semi-Lindelöf) in 𝑋. In particular, a semi-closed subset 𝐴 of a semicompact (resp., semi-Lindelöf) space 𝑋 is semicompact (resp., semi-Lindelöf) in 𝑋.

Proof. We will show the case when 𝐴 is semicompact in 𝑋, the other case is similar. Suppose that 𝒮={𝑆_𝛼∶𝛼∈Λ} is a cover of 𝐴∩𝐵 by semi-open sets in 𝑋. Then 𝒜={𝑆_𝛼∶𝛼∈Λ}∪{𝑋⧵𝐵} is a cover of 𝐴 by semi-open sets in 𝑋, but 𝐴 is semicompact in 𝑋, so there exist 𝛼₁,𝛼₂,…,𝛼_𝑛∈Λ such that 𝐴⊂(⋃^𝑖=𝑛_𝑖=1𝑆_𝛼𝑖)∪(𝑋⧵𝐵). Thus 𝐴∩𝐵⊂⋃^𝑖=𝑛_𝑖=1(𝑆_𝛼𝑖∩𝐵)⊂⋃^𝑖=𝑛_𝑖=1𝑆_𝛼𝑖. Hence, 𝐴∩𝐵 is strongly compact in 𝑋.

Proposition 2.10. Let 𝑓∶𝑋→𝑌 be an irresolute function. Then(i)[13] If 𝐴 is semicompact in 𝑋, then 𝑓(𝐴) is semicompact in 𝑌;(ii)If 𝐴 is semi-Lindelöf in 𝑋, then 𝑓(𝐴) is semi-Lindelöf in 𝑌.

Proof. (ii) The proof is similar to that of (i). We will, however, show it for the convenience of the reader. Suppose that 𝒮={𝑆_𝛼∶𝛼∈Λ} is a cover of 𝑓(𝐴) by semi-open sets in 𝑌. Then 𝒜={𝑓⁻¹(𝑆_𝛼)∶𝛼∈Λ} is a cover of 𝐴, but 𝑓 is irresolute, so 𝑓⁻¹(𝑆_𝛼) is semi-open in 𝑋 for each 𝛼∈Λ. Since 𝐴 is semi-Lindelöf in 𝑋, there exist 𝛼₁,𝛼₂,𝛼₃,…∈Λ such that 𝐴⊂⋃^𝑖=∞_𝑖=1𝑓⁻¹(𝑆_𝛼𝑖). Thus 𝑓(𝐴)⊂⋃^𝑖=∞_𝑖=1𝑓(𝑓⁻¹(𝑆_𝛼𝑖))⊂⋃^𝑖=∞_𝑖=1𝑆_𝛼𝑖. Hence, 𝑓(𝐴) is semi-Lindelöf in 𝑋.

Proposition 2.11. Let 𝑓∶𝑋→𝑌 be a pre-semi-closed surjection. If for each 𝑦∈𝑌, 𝑓⁻¹(𝑦) is semicompact (resp., semi-Lindelöf) in 𝑋, then 𝑓⁻¹(𝐴) is semicompact (resp., semi-Lindelöf) in 𝑋 whenever 𝐴 is semicompact (resp., semi-Lindelöf) in 𝑌.

Proof. We will show the case when 𝐴 is semicompact in 𝑋, the other case is similar. Suppose that 𝒮={𝑆_𝛼∶𝛼∈Λ} is a cover of 𝑓⁻¹(𝐴) by semi-open sets in 𝑋. Then it follows by assumption that for each 𝑦∈𝐴, there exists a finite subcollection 𝒮^𝑦 of 𝒮 such that 𝑓⁻¹(𝑦)⊂⋃𝒮^𝑦. Let 𝑉_𝑦=∪𝒮^𝑦. Then 𝑉_𝑦 is semi-open in 𝑋 as any union of semi-open sets is semi-open. Let 𝐻_𝑦=𝑌⧵𝑓(𝑋⧵𝑉_𝑦). Then 𝐻_𝑦 is semi-open in 𝑌 as 𝑓 is pre-semi-closed, also 𝑦∈𝐻_𝑦 for each 𝑦∈𝐴 as 𝑓⁻¹(𝑦)⊂𝑉_𝑦. Thus, ℋ={𝐻_𝑦∶𝑦∈𝐴} is a cover of 𝐴 by semi-open sets in 𝑌, but 𝐴 is semicompact in 𝑌, so there exist 𝑦₁,𝑦₂,…,𝑦_𝑛∈𝐴 such that 𝐴⊂⋃^𝑖=𝑛_𝑖=1𝐻_𝑦𝑖. Thus, 𝑓⁻¹(𝐴)⊂⋃^𝑖=𝑛_𝑖=1𝑓⁻¹(𝐻_𝑦𝑖)⊂⋃^𝑖=𝑛_𝑖=1𝑉_𝑦𝑖. Since S𝑦_𝑖 is a finite subcollection of 𝒮 for each 𝑖∈{1,2,…,𝑛}, it follows that ⋃^𝑖=𝑛_𝑖=1𝒮^𝑦𝑖 is a finite subcollection of 𝒮. Hence, 𝑓⁻¹(𝐴) is semicompact in 𝑋.

3. 𝑆𝐶𝑆 Spaces

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

Remark 3.2. It follows from Remark 2.3, that a space 𝑋 is 𝑆𝐶𝑆 if and only if 𝑋^𝛼 is 𝑆𝐶𝑆.

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

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

Theorem 3.4. Let 𝑋 be a semi-𝑇₂ extremally disconnected space. Then 𝑋 is 𝑆𝐶𝑆.

Proof. Let 𝐹 be a subset of 𝑋 which is semicompact in 𝑋 and let 𝑥∉𝐹. Then for each 𝑦∈𝐹 there exist two disjoint semi-open sets 𝑈 and 𝑉 containing 𝑥 and 𝑦 respectively (as 𝑋 is semi-𝑇₂). Since 𝐹 is semicompact in 𝑋, there exist 𝑦₁,𝑦₂,…,𝑦_𝑛∈𝐹 such that 𝐹⊂⋃^𝑛_𝑖=1𝑉_𝑦𝑖. Let 𝑈=⋂^𝑛_𝑖=1𝑈_𝑦𝑖. Then 𝑈 is a semi-open subset of 𝑋 that contains 𝑥 and disjoint from 𝐹 (as 𝑋 is extremally disconnected using Proposition 3.3). Thus, 𝑥∉scl(𝐹). Hence, 𝐹 is semi-closed in 𝑋.

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

Proof. Let 𝐹=𝐴∪𝐵, where 𝐴 and 𝐵 are semi-closed in 𝑋. It follows by assumption that 𝐴 and 𝐵 are semicompact in 𝑋 and thus by Proposition 2.4, 𝐹 is semicompact in 𝑋, but 𝑋 is 𝑆𝐶𝑆, so 𝐹 is semi-closed in 𝑋. Hence by Proposition 3.3, 𝑋 is extremally disconnected. The last part follows by Proposition 2.9.

Corollary 3.6. For a semi-𝑇₂ semicompact space, the followings are equivalent:(i)𝑋 is 𝑆𝐶𝑆.(ii)𝑋 is extremally disconnected.

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

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

Theorem 3.8. Let 𝑓 be a pre-semi-closed function from a space 𝑋 onto a space 𝑌 such that for each 𝑦∈𝑌, 𝑓⁻¹(𝑦) is semicompact in 𝑋. If 𝑋 is 𝑆𝐶𝑆, then so is 𝑌.

Proof. Let 𝐹 be a semicompact set in 𝑌. Then by Proposition 2.11, 𝑓⁻¹(𝐹) is semicompact in 𝑋, but 𝑋 is 𝑆𝐶𝑆, so 𝑓⁻¹(𝐹) is semi-closed in 𝑋, but 𝑓 is a pre-semi-closed surjection, so 𝐹=𝑓(𝑓⁻¹(𝐹)) is semi-closed. Hence, 𝑌 is 𝑆𝐶𝑆.

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

Proof. Let 𝐹 be a semicompact set in 𝑋. Then it follows from Proposition 2.10(i) that 𝑓(𝐹) is semicompact in 𝑌, but 𝑌 is 𝑆𝐶𝑆, so 𝑓(𝐹) is semi-closed in 𝑌. Since 𝑓 is one-to-one and irresolute, 𝐹=𝑓⁻¹(𝑓(𝐹)) is semi-closed in 𝑋. Hence, 𝑋 is 𝑆𝐶𝑆.

Lemma 3.10. A subset 𝐴 of ⊕𝑋_𝛼 is semi-open if and only if 𝐴∩𝑋_𝛼 is semi-open in 𝑋_𝛼 for each 𝛼. Thus a subset 𝐴 of ⊕𝑋_𝛼 is semi-closed if and only if 𝐴∩𝑋_𝛼 is semi-closed in 𝑋_𝛼 for each 𝛼.

Proof. Since 𝑋_𝛼 is open in ⊕𝑋_𝛼, it follows that if 𝐴 is semi-open in ⊕𝑋_𝛼, then 𝐴∩𝑋_𝛼 is semi-open in ⊕𝑋_𝛼 and thus semi-open in 𝑋_𝛼 for each 𝛼. Now suppose that 𝐴∩𝑋_𝛼 is semi-open in 𝑋_𝛼 for each 𝛼. Then 𝐴∩𝑋_𝛼 is semi-open in ⊕𝑋_𝛼 for each 𝛼 because 𝑋_𝛼 is open and thus semi-open in ⊕𝑋_𝛼. Thus, 𝐴=∪(𝐴∩𝑋_𝛼) is semi-open in ⊕𝑋_𝛼 as the arbitrary union of semi-open sets is semi-open.

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

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

Corollary 3.12. ⊕𝑋_𝛼 is 𝑆𝐶𝑆 if and only if 𝑋_𝛼 is 𝑆𝐶𝑆 for each 𝛼.

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

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

Definition 3.13. A space 𝑋 is called 𝑜𝑠-regular if whenever 𝑈 is an open subset of 𝑋 and 𝑥∈𝑈, there exists an open subset 𝐾 of 𝑋 and a semi-closed subset 𝑆 of 𝑋 such that 𝑥∈𝐾⊂𝑆⊂𝑈.

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

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

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

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

Proof. Let 𝐹 be a semi-closed subset of 𝑋. By assumption, 𝐹 is semicompact in 𝑋. Since 𝑓 is irresolute, it follows by Proposition 2.10 that 𝑓(𝐹) is semicompact in 𝑌. Since 𝑌 is 𝑆𝐶𝑆, it follows that 𝑓(𝐹) is semi-closed in 𝑌. 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 𝑓∶𝑋→𝑌 be irresolute and 𝐴 be an 𝛼-open subspace of 𝑋. Then the restriction function 𝑓|_𝐴∶𝐴→𝑌 is irresolute.

Theorem 3.18. Let 𝑋 be an 𝑆𝐶𝑆 space and 𝑌 be any space. If 𝑓∶𝑋→𝑌 is a function whose graph 𝐺_𝑓 is an 𝛼-open subspace of 𝑋×𝑌 in which every semi-closed subset is semicompact in 𝐺_𝑓, then 𝑓 is irresolute. In particular, any function having an 𝑆𝐶𝑆 domain and an 𝛼-open, semicompact graph is irresolute.

Proof. Let 𝑃_𝑋∶𝑋×𝑌→𝑋 and 𝑃_𝑌∶𝑋×𝑌→𝑌 be the projection functions. Since 𝐺_𝑓 is an 𝛼-open subspace of 𝑋×𝑌, it follows from Lemma 3.17 that 𝑃_𝑋|_𝐺𝑓 is irresolute. Thus it follows from Theorem 3.16 that 𝑃_𝑋|_𝐺𝑓 is pre-semi-closed, that is, (𝑃_𝑋|_𝐺𝑓)⁻¹ is irresolute. Also, 𝑃_𝑌 is irresolute. Thus, 𝑓=𝑃_𝑌∘(𝑃_𝑋|_𝐺𝑓)⁻¹ is irresolute. The last part follows from Proposition 2.9.

4. 𝑆𝐿𝑆 Spaces

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

Definition 4.1. A space 𝑋 is said to be 𝑆𝐿𝑆 if any subset of 𝑋 which is semi-Lindelöf in 𝑋 is semi-closed.

Remark 4.2. It follows from Remark 2.3, that a space 𝑋 is 𝑆𝐿𝑆 if and only if 𝑋^𝛼 is 𝑆𝐿𝑆.

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

Theorem 4.3. Let 𝑋 be a semi-𝑇₂𝜔-extremally disconnected. Then 𝑋 is 𝑆𝐿𝑆.

Theorem 4.4. If 𝑋 is an 𝑆𝐿𝑆 space such that every semi-closed subset 𝐴 of 𝑋 is semi-Lindelöf in 𝑋, then 𝑋 is 𝜔-extremally disconnected. In particular, an 𝑆𝐿𝑆 semi-Lindelöf space is 𝜔-extremally disconnected.

Corollary 4.5. For a semi-𝑇₂ semi-Lindelöf space, the followings are equivalent:(i)𝑋 is 𝑆𝐿𝑆.(ii)𝑋 is 𝜔-extremally disconnected.

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

Theorem 4.7. Let 𝑓 be a pre-semi-closed function from a space 𝑋 onto a space 𝑌 such that for each 𝑦∈𝑌, 𝑓⁻¹(𝑦) is semi-Lindelöf in 𝑋. If 𝑋 is 𝑆𝐿𝑆, then so is 𝑌.

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

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

Corollary 4.10. ⊕𝑋_𝛼 is 𝑆𝐿𝑆 if and only if 𝑋_𝛼 is 𝑆𝐿𝑆 for each 𝛼.

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

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

Theorem 4.13. Let 𝑋 be a space in which every semi-closed subset is semi-Lindelöf in 𝑋, 𝑌 be an 𝑆𝐿𝑆 space. Then any irresolute function 𝑓 from 𝑋 into 𝑌 is pre-semi-closed. In particular, any irresolute function from a semi-Lindelöf space 𝑋 into an 𝑆𝐿𝑆 space 𝑌 is pre-semi-closed.

Theorem 4.14. Let 𝑋 be an 𝑆𝐿𝑆 space and 𝑌 be any space. If 𝑓∶𝑋→𝑌 is a function whose graph 𝐺_𝑓 is an 𝛼-open subspace of 𝑋×𝑌 in which every semi-closed subset is semi-Lindelöf in 𝐺_𝑓, then 𝑓 is irresolute. In particular, any function having an 𝑆𝐿𝑆 domain and an 𝛼-open, semi-Lindelöf graph is irresolute.