# On Semicompact Sets and Associated Properties

Authors: Mohammad S. Sarsak
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We continue the study of semicompact sets in a topological space. Several
properties, mapping properties of semicompact sets are studied.
A special interest toSCS 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.


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## Body

## 1. Introduction and Preliminaries

A subsetA 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 τα, 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,τα) by Xα. Janković [4] pointed out that PO(X)=PO(Xα), SO(X)=SO(Xα) and αO(X)=αO(Xα). Reilly and Vamanamurthy observed in [5] that τα=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 spaceX 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 functionf 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 spaceX 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 spaceX 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 spaceX 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.
LetA⊂B⊂X, where X is a space. Then(i)
IfA 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.
LetA⊂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 subsetA 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 inX, we will mean semicompact relative to X.Definition 2.2.
A subsetA 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 thatSO(X)=SO(Xα), 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 Xα.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 spaceX is semicompact (resp., semi-Lindelöf) in X.Proposition 2.5.
LetB 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 whenA is semicompact in X, the other case is similar. Suppose that 𝒜={Aα:α∈Λ} is a cover of A by semi-open sets in B. By Proposition 1.2, Aα=Sα∩B for each α∈Λ, where Sα is semi-open in X for each α∈Λ. Thus 𝒮={Sα:α∈Λ} is a cover of A by semi-open sets in X, but A is semicompact in X, so there exist α1,α2,…,αn∈Λ such that A⊂⋃i=1i=nSαi, and thus A⊂⋃i=1i=n(Sαi∩B)=⋃i=1i=nAαi. Hence, A is semicompact in B.Corollary 2.6.
LetA 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.
LetB 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={Sα:α∈Λ} is a cover of A by semi-open sets in X. Then 𝒜={Sα∩B:α∈Λ} is a cover of A. Since Sα is semi-open in X for each α∈Λ and B is preopen in X, it follows from Proposition 1.2 that Sα∩B is semi-open in B for each α∈Λ, but A is semicompact in B, so there exist α1,α2,…αn∈Λ such that A⊂⋃i=1i=n(Sαi∩B)⊂⋃i=1i=nSαi. Hence, A is semicompact in X.Corollary 2.8.
A preopen subsetA of a space X is semicompact (resp., semi-Lindelöf) if and only if A is semicompact (resp., semi-Lindelöf) in X.Proposition 2.9.
LetA 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 whenA is semicompact in X, the other case is similar. Suppose that 𝒮={Sα:α∈Λ} is a cover of A∩B by semi-open sets in X. Then 𝒜={Sα: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,…,αn∈Λ such that A⊂(⋃i=1i=nSαi)∪(X∖B). Thus A∩B⊂⋃i=1i=n(Sαi∩B)⊂⋃i=1i=nSαi. Hence, A∩B is strongly compact in X.Proposition 2.10.
Letf:X→Y be an irresolute function. Then(i)
[13] If A is semicompact in X, then f(A) is semicompact in Y;(ii)
IfA 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 that𝒮={Sα:α∈Λ} is a cover of f(A) by semi-open sets in Y. Then 𝒜={f-1(Sα):α∈Λ} is a cover of A, but f is irresolute, so f-1(Sα) 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=1i=∞f-1(Sαi). Thus f(A)⊂⋃i=1i=∞f(f-1(Sαi))⊂⋃i=1i=∞Sαi. Hence, f(A) is semi-Lindelöf in X.Proposition 2.11.
Letf: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 whenA is semicompact in X, the other case is similar. Suppose that 𝒮={Sα: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 𝒮y of 𝒮 such that f-1(y)⊂⋃𝒮y. Let Vy=∪𝒮y. Then Vy is semi-open in X as any union of semi-open sets is semi-open. Let Hy=Y∖f(X∖Vy). Then Hy is semi-open in Y as f is pre-semi-closed, also y∈Hy for each y∈A as f-1(y)⊂Vy. Thus, ℋ={Hy:y∈A} is a cover of A by semi-open sets in Y, but A is semicompact in Y, so there exist y1,y2,…,yn∈A such that A⊂⋃i=1i=nHyi. Thus, f-1(A)⊂⋃i=1i=nf-1(Hyi)⊂⋃i=1i=nVyi. Since Syi is a finite subcollection of 𝒮 for each i∈{1,2,…,n}, it follows that ⋃i=1i=n𝒮yi is a finite subcollection of 𝒮. Hence, f-1(A) is semicompact in X.


## 3.SCS Spaces

Definition 3.1.
A spaceX is said to be SCS if any subset of X which is semicompact in X is semi-closed.Remark 3.2.
It follows from Remark2.3, that a space X is SCS if and only if Xα is SCS.We recall the following result from [3],  it will be helpful to show the next two theorems.Proposition 3.3.
A spaceX is extremally disconnected if and only if the intersection of any two semi-open subsets of X is semi-open.Theorem 3.4.
LetX be a semi-T2 extremally disconnected space. Then X is SCS.Proof.
LetF 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,…,yn∈F such that F⊂⋃i=1nVyi. Let U=⋂i=1nUyi. 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.
IfX 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.
LetF=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 spaceX is semi-open if and only if it is open, the following proposition seems clear.Proposition 3.7.
If every subset of a spaceX is semicompact in X, then X is SCS if and only if X is a finite discrete space.Theorem 3.8.
Letf 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.
LetF 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 spaceX into an SCS space Y. Then X is SCS.Proof.
LetF 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 subsetA of ⊕Xα is semi-open if and only if A∩Xα is semi-open in Xα for each α. Thus a subset A of ⊕Xα is semi-closed if and only if A∩Xα is semi-closed in Xα for each α.Proof.
SinceXα is open in ⊕Xα, it follows that if A is semi-open in ⊕Xα, then A∩Xα is semi-open in ⊕Xα and thus semi-open in Xα for each α. Now suppose that A∩Xα is semi-open in Xα for each α. Then A∩Xα is semi-open in ⊕Xα for each α because Xα is open and thus semi-open in ⊕Xα. Thus, A=∪(A∩Xα) is semi-open in ⊕Xα as the arbitrary union of semi-open sets is semi-open.Corollary 3.11.
BeingSCS is hereditary with respect to preopen subsets.Proof.
LetA 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.
⊕Xα is SCS if and only if Xα is SCS for each α.Proof.
Necessity. It follows from Corollary 3.11 since Xα is open and thus preopen in ⊕Xα.
Sufficiency. Suppose that Xα is an SCS space for each α and let F be a subset of ⊕Xα which is semicompact in ⊕Xα. Since Xα is closed and thus semi-closed in ⊕Xα, it follows from Proposition 2.9 that F∩Xα is semicompact in ⊕Xα, but Xα is preopen in ⊕Xα, so it follows from Proposition 2.7 that F∩Xα is semicompact in Xα. Since Xα is SCS, F∩Xα is semi-closed in Xα for each α, thus by Lemma 3.10, F is semi-closed in ⊕Xα. Hence, ⊕Xα is SCS.Recall that a spaceX 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 spaceX 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.
IfX is an os-regular space in which every point has an SCS neighborhood, then X is SCS.Proof.
LetF 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.
IfX is a regular space in which every point has an SCS neighborhood, then X is SCS.Theorem 3.16.
LetX 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.
LetF 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) Letf: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.
LetX be an SCS space and Y be any space. If f:X→Y is a function whose graph Gf is an α-open subspace of X×Y in which every semi-closed subset is semicompact in Gf, then f is irresolute. In particular, any function having an SCS domain and an α-open, semicompact graph is irresolute.Proof.
LetPX:X×Y→X and PY:X×Y→Y be the projection functions. Since Gf is an α-open subspace of X×Y, it follows from Lemma 3.17 that PX|Gf is irresolute. Thus it follows from Theorem 3.16 that PX|Gf is pre-semi-closed, that is, (PX|Gf)-1 is irresolute. Also, PY is irresolute. Thus, f=PY∘(PX|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 spaceX 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 Remark2.3, that a space X is SLS if and only if Xα is SLS.Following Proposition3.3, we will call   a space Xω-extremally disconnected if the countable intersection of semi-open subsets of X is semi-open.Theorem 4.3.
LetX be a semi-T2ω-extremally disconnected. Then X is SLS.Theorem 4.4.
IfX 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 spaceX is semi-Lindelöf in X, then X is SLS if and only if X is a countable discrete space.Theorem 4.7.
Letf 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 spaceX into an SLS space Y. Then X is SLS.Proposition 4.9.
BeingSLS is hereditary with respect to preopen subsets.Corollary 4.10.
⊕Xα is SLS if and only if Xα is SLS for each α.Theorem 4.11.
IfX is an os-regular space in which every point has an SLS neighborhood, then X is SLS.Corollary 4.12.
IfX is a regular space in which every point has an SLS neighborhood, then X is SLS.Theorem 4.13.
LetX 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.
LetX be an SLS space and Y be any space. If f:X→Y is a function whose graph Gf is an α-open subspace of X×Y in which every semi-closed subset is semi-Lindelöf in Gf, then f is irresolute. In particular, any function having an SLS domain and an α-open, semi-Lindelöf graph is irresolute.



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