# Characterizations of Strongly Compact Spaces

Authors: Ahmad Al-Omari; Takashi Noiri; Mohd. Salmi Md. Noorani
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

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


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## Body

## 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 X is said to be strongly compact if every preopen cover of X 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 X 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 X has nonempty interior. In 1988, Janković et al. [4] showed that a topological space (X,τ) is strongly compact if and only if it is compact and the family of dense sets in (X,τ) 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 𝒩-preopen sets as a generalization of preopen sets. Then, by using 𝒩-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, (X,τ) and (Y,σ) stand for topological spaces on which no separation axiom is assumed unless otherwise stated. For a subset A of X, the closure of A and the interior of A will be denoted by Cl(A) and Int(A), respectively. A subset A of a topological space X is said to be preopen [7] if A⊆Int(Cl(A)). A subset A is said to be 𝒩-open [8] if for each x∈A there exists an open set Ux containing x such that Ux-A is a finite set. The complement of an 𝒩-open subset is said to be 𝒩-closed.


## 2.N-Preopen Sets

In this section, we introduce and study the notion of𝒩-preopen sets.Definition 2.1.
A subsetA is said to be 𝒩-preopen if for each x∈A there exists a preopen set Ux containing x such that Ux-A is a finite set. The complement of an 𝒩-preopen subset is said to be 𝒩-preclosed.The family of all𝒩-preopen (resp., preopen, preclosed) subsets of a space (X,τ) is denoted by 𝒩PO(X) (resp., PO(X), PC(X)).Lemma 2.2.
For a subset of a topological space(X,τ), both 𝒩-openness and preopenness imply 𝒩-preopenness.For a subset of a topological space, the following implications hold:(2.1)Example 2.3.
LetX=[0,7] be the closed interval with a topology τ={ϕ,X,{1},{2},{1,2},{1,2,[3,5]}. Then {2,[3,5]} is an 𝒩-open set which is not preopen.Example 2.4.
Letℝ be the set of all real numbers with the usual topology. Then the set ℚ of all rational numbers is a preopen set which is not 𝒩-open.Lemma 2.5.
Let(X,τ) be a topological space. Then the union of any family of 𝒩-preopen sets is 𝒩-preopen.Proof.
Let{Ui:i∈I} be a family of 𝒩-preopen subsets of X and x∈⋃i∈IUi. Then x∈Uj for some j∈I. This implies that there exists a preopen subset V of X containing x such that V∖Uj is finite. Since V∖⋃i∈IUi⊆V∖Uj, then V∖⋃i∈IUi is finite. Thus ⋃i∈IUi∈𝒩PO(X).Recall that a space(X,τ) is called submaximal if every dense subset of X is open.Lemma 2.6 (see [9]).
For a topological space(X,τ), the followings are equivalent.  (1)
X is submaximal.(2)
Every preopen set is open.Definition 2.7 (see [10]).
A subsetA of a space X is said to be α-open if A⊆Int(Cl(Int(A))).Lemma 2.8 (see [11]).
Let(X,τ) be a topological space. Then the intersection of an α-open set and a preopen set is preopen.Theorem 2.9.
Let(X,τ) be a submaximal topological space. Then (X,𝒩PO(X)) is a topological space.Proof.
(1) We have ϕ,X∈𝒩PO(X).
(2) Let U,V∈𝒩PO(X) and x∈U∩V. Then there exist preopen sets G,H∈X containing x such that G∖U and H∖V are finite. And (G∩H)∖(U∩V)=(G∩H)∩((X∖U)∪(X∖V))⊆(G∩(X∖U))∪(H∩(X∖V)). Thus (G∩H)∖(U∩V) is finite, and since X is submaximal by Lemma 2.6, the intersection of two preopen sets is preopen. Hence U∩V∈𝒩PO(X).
(3) Let {Ui:i∈I} be any family of 𝒩-preopen sets of X. Then, by Lemma 2.5, ⋃i∈IUi is 𝒩-preopen.The converse of above theorem is not true.Example 2.10.
LetX={a,b,c} with τ={ϕ,X,{a},{a,b}}. Then (X,𝒩PO(X)) is a topological space and (X,τ) is not a submaximal topological space.Lemma 2.11.
Let(X,τ) be a topological space. Then the intersection of an α-open set and an 𝒩-preopen set is 𝒩-preopen.Proof.
LetU be α-open and A𝒩-preopen. Then for every x∈A, there exists a preopen set Vx⊆X containing x such that Vx-A is finite, and also by Lemma 2.8, U∩Vx is preopen. Now for each x∈U∩A, there exists a preopen set U∩Vx⊆X containing x and
(2.2)(U∩Vx)-(U∩A)=(U∩Vx)∩[(X-U)∪(X-A)]=[(U∩Vx)∩(X-U)]∪[(U∩Vx)∩(X-A)]=(U∩Vx)-A⊆Vx-A.
Then (U∩Vx)-(U∩A) is finite. Therefore U∩A is an 𝒩-preopen set.The following lemma is well known and will be stated without the proof.Lemma 2.12.
A topological space is aT1-space if and only if every finite set is closed.Proposition 2.13.
If a topological spaceX is a T1-space, then every nonempty 𝒩-preopen set contains a nonempty preopen set.Proof.
LetA be a nonempty 𝒩-preopen set and x∈A, then there exists a preopen set Ux containing x such that Ux-A is finite. Let C=Ux-A=Ux∩(X-A). Then x∈Ux-C⊆A and by Lemmas 2.8 and 2.12,  Ux-C=Ux∩(X-C) is preopen.The following example shows that ifX is not a T1-space, then there exists a nonempty 𝒩-preopen set which does not contain a nonempty preopen set.Example 2.14.
LetX={a,b,c} with τ={ϕ,X,{a},{b},{a,b}}. Then {c} is an 𝒩-preopen set which does not contain any a nonempty preopen set.Lemma 2.15 (see [12]).
LetA and X0 be subsets of a topological space X.  (1)
IfA∈PO(X) and X0∈SO(X), then A∩X0∈PO(X0).(2)
IfA∈PO(X0) and X0∈PO(X), then A∈PO(X).Lemma 2.16.
LetA and X0 be subsets of a topological space X.  (1)
IfA∈𝒩PO(X) and X0∈αO(X), then A∩X0∈𝒩PO(X0).(2)
IfA∈𝒩PO(X0) and X0∈PO(X), then A∈𝒩PO(X).Proof.
(1) Let x∈A∩X0. Since A is 𝒩-preopen in X, there exists a preopen set H of X containing x such that H-A is finite. Since X0 is α-open, by Lemma 2.8 we have H∩X0∈PO(X). Since X0∈αO(X)⊆SO(X), by Lemma 2.15, H∩X0∈PO(X0), x∈H∩X0, and (H∩X0)-(A∩X0) is finite. This shows that A∩X0∈𝒩PO(X0).
(2) If A∈𝒩PO(X0), for each x∈A there exists Vx∈PO(X0) containing x such that Vx-A is finite. Since X0∈PO(X), by Lemma 2.15, Vx∈PO(X) and Vx-A is finite and hence A∈𝒩PO(X).Lemma 2.17.
A subsetA of a space X is 𝒩-preopen if and only if for every x∈A, there exist a preopen subset Ux containing x and a finite subset C such that Ux-C⊆A.Proof.
LetA be 𝒩-preopen and x∈A, then there exists a preopen subset Ux containing x such that Ux-A is finite. Let C=Ux-A=Ux∩(X-A). Then Ux-C⊆A. Conversely, let x∈A. Then there exist a preopen subset Ux containing x and a finite subset C such that Ux-C⊆A. Thus Ux-A⊆C and Ux-A is a finite set.Theorem 2.18.
LetX be a space and F⊆X. If F is 𝒩-preclosed, then F⊆K∪C for some preclosed subset K and a finite subset C.Proof.
IfF is 𝒩-preclosed, then X-F is 𝒩-preopen, and hence for every x∈X-F, there exists a preopen set U containing x and a finite set C such that U-C⊆X-F. Thus F⊆X-(U-C)=X-(U∩(X-C))=(X-U)∪C. Let K=X-U. Then K is a preclosed set such that F⊆K∪C.


## 3. Strongly Compact Spaces

Definition 3.1.
(1) A topological space X is said to be strongly compact [1] if every cover of X by preopen sets admits a finite subcover.
(2) A subset A of a space X is said to be strongly compact relative to X [2] if every cover of A by preopen sets of X admits a finite subcover.Theorem 3.2.
IfX is a space such that every preopen subset of X is strongly compact relative to X, then every subset of X is strongly compact relative to X.Proof.
LetB be an arbitrary subset of X and let {Ui:i∈I} be a cover of B by preopen sets of X. Then the family {Ui:i∈I} is a preopen cover of the preopen set ∪{Ui:i∈I}. Hence by hypothesis there is a finite subfamily {Uij:j∈ℕ0} which covers ∪{Ui:i∈I} where ℕ0 is a finite subset of the naturals ℕ. This subfamily is also a cover of the set B.Theorem 3.3.
A subsetA of a space X is strongly compact relative to X if and only if for any cover {Vα:α∈Λ} of A by 𝒩-preopen sets of X, there exists a finite subset Λ0 of Λ such that A⊆∪{Vα:α∈Λ0}.Proof.
Let{Vα:α∈Λ} be a cover of A and Vα∈𝒩PO(X). For each x∈A, there exists α(x)∈Λ such that x∈Vα(x). Since Vα(x) is 𝒩-preopen, there exists a preopen set Uα(x) such that x∈Uα(x) and Uα(x)∖Vα(x) is finite. The family {Uα(x):x∈A} is a preopen cover of A. Since A is strongly compact relative to X, there exists a finite subset, says, x1,x2,…,xn such that A⊆∪{Uα(xi):i∈F}, where F={1,2,…,n}. Now, we have
(3.1)A⊆⋃i∈F((Uα(xi)∖Vα(xi))∪Vα(xi))=(⋃i∈F(Uα(xi)∖Vα(xi)))∪(⋃i∈FVα(xi)).
For each xi, Uα(xi)∖Vα(xi) is a finite set and there exists a finite subset Λ(xi) of Λ such that (Uα(xi)∖Vα(xi))∩A⊆∪{Vα:α∈Λ(xi)}. Therefore, we have A⊆(⋃i∈F(∪{Vα:α∈Λ(xi)}))∪(⋃i∈FVα(xi)). Hence A is strongly compact relative to X.
Since every preopen set is𝒩-preopen, the proof of the converse is obvious.Corollary 3.4.
For any spaceX, the following properties are equivalent: (1)
X is strongly compact;(2)
every𝒩-preopen cover of X admits a finite subcover.Theorem 3.5.
For any spaceX, the following properties are equivalent:  (1)
X is strongly compact;(2)
every proper𝒩-preclosed set is strongly compact with respect to X.Proof.
(1)⇒(2) Let A be a proper 𝒩-preclosed subset of X. Let {Uα:α∈Λ} be a cover of A by preopen sets of X. Now for each x∈X-A, there is a preopen set Vx such that Vx∩A is finite. Then {Uα:α∈Λ}∪{Vx:x∈X-A} is a preopen cover of X. Since X is strongly compact, there exist a finite subset Λ1 of Λ and a finite number of points, says, x1,x2,…,xn in X-A such that X=(∪{Uα:α∈Λ1})∪(∪{Vxi:1≤i≤n}); hence A⊂(∪{Uα:α∈Λ1})∪(∪{A∩Vxi:1≤i≤n}). Since A∩Vxi is finite for each i, there exists a finite subset Λ2 of Λ such that ∪{A∩Vxi:1≤i≤n}⊂∪{Uα:α∈Λ2}. Therefore, we obtain A⊂∪{Uα:α∈Λ1∪Λ2}. This shows that A is strongly compact relative to X.
(2)⇒(1) Let {Vα:α∈Λ} be any preopen cover of X. We choose and fix one α0∈Λ. Then ∪{Vα:α∈Λ-{α0}} is a preopen cover of a 𝒩-preclosed set X-Vα0. There exists a finite subset Λ0 of Λ-{α0} such that X-Vα0⊂∪{Vα:α∈Λ0}. Therefore, X=∪{Vα:α∈Λ0∪{α0}}. This shows that X is strongly compact.Corollary 3.6 (see [2]).
If a spaceX is strongly compact and A is preclosed, then A is strongly compact relative to X.Theorem 3.7.
Let(X,τ) be a submaximal topological space. Then (X,τ) is strongly compact if and only if (X,𝒩PO(X)) is compact.Proof.
Let{Vα:α∈Λ} be an open cover of (X,𝒩PO(X)). For each x∈X, there exists α(x)∈Λ such that x∈Vα(x). Since Vα(x) is 𝒩-preopen, there exists a preopen set Uα(x) of X such that x∈Uα(x) and Uα(x)∖Vα(x) is finite. The family {Uα(x):x∈X} is a preopen cover of (X,τ). Since (X,τ) is strongly compact, there exists a finite subset, says, x1,x2,…,xn such that X=∪{Uα(xi):i∈F}, where F={1,2,…,n}. Now, we have
(3.2)X=⋃i∈F((Uα(xi)∖Vα(xi))∪Vα(xi))=(⋃i∈F(Uα(xi)∖Vα(xi)))∪(⋃i∈FVα(xi)).
For each xi, Uα(xi)∖Vα(xi) is a finite set and there exists a finite subset Λ(xi) of Λ such that (Uα(xi)∖Vα(xi))⊆∪{Vα:α∈Λ(xi)}. Therefore, we have X=(⋃i∈F(∪(Vα:α∈Λ(xi))))∪(⋃i∈FVα(xi)). Hence 𝒩PO(X) is compact.
Conversely, let𝒰 be a preopen cover of (X,τ). Then 𝒰⊆𝒩PO(X). Since (X,𝒩PO(X)) is compact, there exists a finite subcover of 𝒰 for X. Hence (X,τ) is strongly compact.


## 4. Preservation Theorems

Definition 4.1.
A functionf:X→Y is said to be 𝒩-precontinuous if f-1(V) is 𝒩-preopen in X for each open set V in Y.Theorem 4.2.
A functionf:X→Y is 𝒩-precontinuous if and only if for each point x∈X and each open set V in Y with f(x)∈V, there is an 𝒩-preopen set U in X such that x∈U and f(U)⊆V.Proof.
Sufficiency. Let V be open in Y and x∈f-1(V). Then f(x)∈V and thus there exists a Ux∈𝒩PO(X) such that x∈Ux and f(Ux)⊆V. Then x∈Ux⊆f-1(V) and f-1(V)=∪{Ux:x∈f-1(V)}. Then by Lemma 2.5, f-1(V) is 𝒩-preopen.
Necessity. Let x∈X and let V be an open set of Y containing f(x). Then x∈f-1(V)∈𝒩PO(X) since f is 𝒩-precontinuous. Let U=f-1(V). Then x∈U and f(U)⊆V.Proposition 4.3.
Iff:X→Y is 𝒩-precontinuous and X0 is an α-open set in X, then the restriction f|X0:X0→Y is 𝒩-precontinuous.Proof.
Sincef is 𝒩-precontinuous, for any open set V in Y, f-1(V) is 𝒩-preopen in X. Hence by Lemma 2.11, f-1(V)∩X0 is 𝒩-preopen in X. Therefore, by Lemma 2.16, (f|X0)-1(V)=f-1(V)∩X0 is 𝒩-preopen in X0. This implies that f|X0 is 𝒩-precontinuous.Proposition 4.4.
Letf:X→Y be a function and let {Aα:α∈Λ} be an α-open cover of X. If the restriction f|Aα:Aα:→Y is 𝒩-precontinuous for each α∈Λ, then f is 𝒩-precontinuous.Proof.
Suppose thatV is an arbitrary open set in Y. Then for each α∈Λ, we have (f|Aα)-1(V)=f-1(V)∩Aα∈𝒩PO(Aα) because f|Aα is 𝒩-precontinuous. Hence by Lemma 2.16,  f-1(V)∩Aα∈𝒩PO(X) for each α∈Λ. By Lemma 2.5, we obtain ∪{f-1(V)∩Aα:α∈Λ}=f-1(V)∈𝒩PO(X). This implies that f is 𝒩-precontinuous.Theorem 4.5.
Letf be an 𝒩-precontinuous function from a space X onto a space Y. If X is strongly compact, then Y is compact.Proof.
Let{Vα:α∈Λ} be an open cover of Y. Then {f-1(Vα):α∈Λ} is an 𝒩-preopen cover of X. Since X is strongly compact, by Corollary 3.4, there exists a finite subset Λ0 of Λ such that X=∪{f-1(Vα):α∈Λ0}; hence Y=∪{Vα:α∈Λ0}. Therefore Y is compact.Definition 4.6 (see [12]).
A functionf:X→Y is said to be precontinuous if f-1(V) is preopen in X for each open set V in Y.It is clear that every precontinuous function is𝒩-precontinuous but not conversely.Example 4.7.
LetX={a,b,c,d}, τ={ϕ,X,{a},{b},{a,b},{a,b,c}}, and σ={ϕ,X,{a,d},{b,c}}. Let f:(X,τ)→(X,σ) be the identity function. Then f is an 𝒩-precontinuous function which is not precontinuous; because there exists {b,c}∈σ such that f-1({b,c})∉PO(X,τ).Corollary 4.8 (see [2]).
Letf be a precontinuous function from a space X onto a space Y. If X is strongly compact, then Y is compact.Definition 4.9 (see [13]).
A functionf:X→Y is said to be M-preopen if the image of each preopen set U of X is preopen in Y.Proposition 4.10.
Iff:X→Y is M-preopen, then the image of an 𝒩-preopen set of X is 𝒩-preopen in Y.Proof.
Letf:X→Y be M-preopen and W an 𝒩-preopen subset of X. For any y∈f(W), there exists x∈W such that f(x)=y. Since W is 𝒩-preopen, there exists a preopen set U such that x∈U and U-W=C is finite. Since f is M-preopen, f(U) is preopen in Y such that y=f(x)∈f(U) and f(U)-f(W)⊆f(U-W)=f(C) is finite. Therefore, f(W) is 𝒩-preopen in Y.Definition 4.11 (see [14]).
A functionf:X→Y is said to be preirresolute if f-1(V) is preopen in X for each preopen set V in Y.Proposition 4.12.
Iff:X→Y is a preirresolute injection and A is 𝒩-preopen in Y, then f-1(A) is 𝒩-preopen in X.Proof.
Assume thatA is an 𝒩-preopen subset of Y. Let x∈f-1(A). Then f(x)∈A and there exists a preopen set V containing f(x) such that V-A is finite. Since f is preirresolute, f-1(V) is a preopen set containing x. Thus f-1(V)-f-1(A)=f-1(V-A) and it is finite. It follows that f-1(A) is 𝒩-preopen in X.Definition 4.13.
A functionf:X→Y is said to be 𝒩-preclosed if f(A) is 𝒩-preclosed in Y for each preclosed set A of X.Theorem 4.14.
Iff:X→Y is an 𝒩-preclosed surjection such that f-1(y) is strongly compact relative to X for each y∈Y, and Y is strongly compact, then X is strongly compact.Proof.
Let{Uα:α∈Λ} be any preopen cover of X. For each y∈Y, f-1(y) is strongly compact relative to X and there exists a finite subset Λ(y) of Λ such that f-1(y)⊂∪{Uα:α∈Λ(y)}. Now we put U(y)=∪{Uα:α∈Λ(y)} and V(y)=Y-f(X-U(y)). Then, since f is 𝒩-preclosed, V(y) is an 𝒩-preopen set in Y containing y such that f-1(V(y))⊂U(y). Since {V(y):y∈Y} is an 𝒩-preopen cover of Y, by Corollary 3.4 there exists a finite subset {yk:1≤k≤n}⊆Y such that Y=⋃k=1nV(yk). Therefore, X=f-1(Y)=⋃k=1nf-1(V(yk))⊆⋃k=1nU(yk)=⋃k=1n{Uα:α∈Λ(yk)}. This shows that X is strongly compact.Definition 4.15 (see [2]).
A functionf:X→Y is said to be M-preclosed if f(A) is preclosed in Y for each preclosed set A of X.It is clear that everyM-preclosed function is 𝒩-preclosed but not conversely.Example 4.16.
LetX={a,b,c,d} with topology τ={ϕ,X,{a},{b},{a,b},{a,b,c}}. Let f:(X,τ)→(X,τ) be the function defined by setting f(a)=c, f(b)=d, f(c)=a, and f(d)=b. Then f is an 𝒩-preclosed function which is not M-preclosed; because there exists {c}∈PC(X) such that f({c})={a}∉PC(X).Corollary 4.17 (see [2]).
Iff:X→Y is an M-preclosed surjection such that f-1(y) is strongly compact relative to X for each y∈Y, and Y is strongly compact, then X is strongly compact.Definition 4.18.
A functionf:X→Y is said to be δ𝒩-continuous if for each x∈X and each preopen set V of Y containing f(x), there exists an 𝒩-preopen set U of X containing x such that f(U)⊆V.It is clear that every preirresolute function isδ𝒩-continuous but the converse is not true.Example 4.19.
LetX={a,b,c,d} and τ={ϕ,X,{a},{b},{a,b},{a,b,c}}. Then the function f:(X,τ)→(X,τ), defined as f(a)=c, f(b)=d, f(c)=a, and f(d)=b, is δN-continuous but it is not preirresolute.Theorem 4.20.
Letf:X→Y be a δ𝒩-continuous surjection from X onto Y. If X is strongly compact, then Y is strongly compact.Proof.
Let{Vα:α∈Λ} be a preopen cover of Y. For each x∈X, there exists α(x)∈Λ such that f(x)∈Vα(x). Since f is δ𝒩-continuous, there exists an 𝒩-preopen set Uα(x) of X containing x such that f(Uα(x))⊆Vα(x). So {Uα(x):x∈X} is an 𝒩-preopen cover of the strongly compact space X, and by Corollary 3.4 there exists a finite subset {xk:1≤k≤n}⊆X such that X=⋃k=1nUα(xk). Therefore Y=f(X)=f(⋃k=1nUα(xk))=⋃k=1nf(Uα(xk))⊆⋃k=1nVα(xk). This shows that Y is strongly compact.Corollary 4.21 (see [2]).
Letf:X→Y be a preirresolute surjection from X onto Y. If X is strongly compact, then Y is strongly compact.



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