Characterizations of Strongly Compact Spaces

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 ??-preopen sets which is weaker than both open sets and ??-open sets. Where a subset 𝐴 is said to be ??-preopen if for each 𝑥∈𝐴 there exists a preopen set 𝑈_𝑥 containing 𝑥 such that 𝑈_𝑥−𝐴 is a finite set. We investigate some properties of the sets. Moreover, we obtain new characterizations and preserving theorems of strongly compact spaces.

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, Jankovic 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 𝑈_𝑥 containing 𝑥 such that 𝑈_𝑥−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 of 𝒩-preopen sets.

Definition 2.1. A subset 𝐴 is said to be 𝒩-preopen if for each 𝑥∈𝐴 there exists a preopen set 𝑈_𝑥 containing 𝑥 such that 𝑈_𝑥−𝐴 is a finite set. The complement of an 𝒩-preopen subset is said to be 𝒩-preclosed.

The family of all 𝒩-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. Let 𝑋=[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. Let ℝ 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 {𝑈_𝑖∶𝑖∈𝐼} be a family of 𝒩-preopen subsets of 𝑋 and 𝑥∈⋃_𝑖∈𝐼𝑈_𝑖. Then 𝑥∈𝑈_𝑗 for some 𝑗∈𝐼. This implies that there exists a preopen subset 𝑉 of 𝑋 containing 𝑥 such that 𝑉⧵𝑈_𝑗 is finite. Since 𝑉⧵⋃_𝑖∈𝐼𝑈_𝑖⊆𝑉⧵𝑈_𝑗, then 𝑉⧵⋃_𝑖∈𝐼𝑈_𝑖 is finite. Thus ⋃_𝑖∈𝐼𝑈_𝑖∈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 subset 𝐴 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 {𝑈_𝑖∶𝑖∈𝐼} be any family of 𝒩-preopen sets of 𝑋. Then, by Lemma 2.5, ⋃_𝑖∈𝐼𝑈_𝑖 is 𝒩-preopen.

The converse of above theorem is not true.

Example 2.10. Let 𝑋={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. Let 𝑈 be 𝛼-open and 𝐴𝒩-preopen. Then for every 𝑥∈𝐴, there exists a preopen set 𝑉_𝑥⊆𝑋 containing 𝑥 such that 𝑉_𝑥−𝐴 is finite, and also by Lemma 2.8, 𝑈∩𝑉_𝑥 is preopen. Now for each 𝑥∈𝑈∩𝐴, there exists a preopen set 𝑈∩𝑉_𝑥⊆𝑋 containing 𝑥 and (𝑈∩𝑉_𝑥)−(𝑈∩𝐴)=(𝑈∩𝑉_𝑥)∩[(𝑋−𝑈)∪(𝑋−𝐴)]=[(𝑈∩𝑉_𝑥)∩(𝑋−𝑈)]∪[(𝑈∩𝑉_𝑥)∩(𝑋−𝐴)]=(𝑈∩𝑉_𝑥)−𝐴⊆𝑉_𝑥−𝐴. (2.2) Then (𝑈∩𝑉_𝑥)−(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 a 𝑇₁-space if and only if every finite set is closed.

Proposition 2.13. If a topological space 𝑋 is a 𝑇₁-space, then every nonempty 𝒩-preopen set contains a nonempty preopen set.

Proof. Let 𝐴 be a nonempty 𝒩-preopen set and 𝑥∈𝐴, then there exists a preopen set 𝑈_𝑥 containing 𝑥 such that 𝑈_𝑥−𝐴 is finite. Let 𝐶=𝑈_𝑥−𝐴=𝑈_𝑥∩(𝑋−𝐴). Then 𝑥∈𝑈_𝑥−𝐶⊆𝐴 and by Lemmas 2.8 and 2.12, 𝑈_𝑥−𝐶=𝑈_𝑥∩(𝑋−𝐶) is preopen.

The following example shows that if 𝑋 is not a 𝑇₁-space, then there exists a nonempty 𝒩-preopen set which does not contain a nonempty preopen set.

Example 2.14. Let 𝑋={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]). Let 𝐴 and 𝑋₀ be subsets of a topological space 𝑋. (1)If 𝐴∈PO(𝑋) and 𝑋₀∈SO(𝑋), then 𝐴∩𝑋₀∈PO(𝑋₀). (2)If 𝐴∈PO(𝑋₀) and 𝑋₀∈PO(𝑋), then 𝐴∈PO(𝑋).

Lemma 2.16. Let 𝐴 and 𝑋₀ be subsets of a topological space 𝑋. (1)If 𝐴∈𝒩PO(𝑋) and 𝑋₀∈𝛼𝑂(𝑋), then 𝐴∩𝑋₀∈𝒩PO(𝑋₀). (2)If 𝐴∈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 𝑉_𝑥∈PO(𝑋₀) containing 𝑥 such that 𝑉_𝑥−𝐴 is finite. Since 𝑋₀∈PO(𝑋), by Lemma 2.15, 𝑉_𝑥∈PO(𝑋) and 𝑉_𝑥−𝐴 is finite and hence 𝐴∈𝒩PO(𝑋).

Lemma 2.17. A subset 𝐴 of a space 𝑋 is 𝒩-preopen if and only if for every 𝑥∈𝐴, there exist a preopen subset 𝑈_𝑥 containing 𝑥 and a finite subset 𝐶 such that 𝑈_𝑥−𝐶⊆𝐴.

Proof. Let 𝐴 be 𝒩-preopen and 𝑥∈𝐴, then there exists a preopen subset 𝑈_𝑥 containing 𝑥 such that 𝑈_𝑥−𝐴 is finite. Let 𝐶=𝑈_𝑥−𝐴=𝑈_𝑥∩(𝑋−𝐴). Then 𝑈_𝑥−𝐶⊆𝐴. Conversely, let 𝑥∈𝐴. Then there exist a preopen subset 𝑈_𝑥 containing 𝑥 and a finite subset 𝐶 such that 𝑈_𝑥−𝐶⊆𝐴. Thus 𝑈_𝑥−𝐴⊆𝐶 and 𝑈_𝑥−𝐴 is a finite set.

Theorem 2.18. Let 𝑋 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. If 𝐹 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. If 𝑋 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. Let 𝐵 be an arbitrary subset of 𝑋 and let {𝑈_𝑖∶𝑖∈𝐼} be a cover of 𝐵 by preopen sets of 𝑋. Then the family {𝑈_𝑖∶𝑖∈𝐼} is a preopen cover of the preopen set ∪{𝑈_𝑖∶𝑖∈𝐼}. Hence by hypothesis there is a finite subfamily {𝑈_𝑖𝑗∶𝑗∈ℕ₀} which covers ∪{𝑈_𝑖∶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 subset 𝐴 of a space 𝑋 is strongly compact relative to 𝑋 if and only if for any cover {𝑉_𝛼∶𝛼∈Λ} of 𝐴 by 𝒩-preopen sets of 𝑋, there exists a finite subset Λ₀ of Λ such that 𝐴⊆∪{𝑉_𝛼∶𝛼∈Λ₀}.

Proof. Let {𝑉_𝛼∶𝛼∈Λ} be a cover of 𝐴 and 𝑉_𝛼∈𝒩PO(𝑋). For each 𝑥∈𝐴, there exists 𝛼(𝑥)∈Λ such that 𝑥∈𝑉_𝛼(𝑥). Since 𝑉_𝛼(𝑥) is 𝒩-preopen, there exists a preopen set 𝑈_𝛼(𝑥) such that 𝑥∈𝑈_𝛼(𝑥) and 𝑈_𝛼(𝑥)⧵𝑉_𝛼(𝑥) is finite. The family {𝑈_𝛼(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,…,𝑥_𝑛 such that 𝐴⊆∪{𝑈_{𝛼(𝑥𝑖)}∶𝑖∈𝐹}, where 𝐹={1,2,…,𝑛}. Now, we have 𝐴⊆_𝑖∈𝐹((𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)})∪𝑉_{𝛼(𝑥𝑖)})=(_𝑖∈𝐹(𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)}))∪(_𝑖∈𝐹𝑉_{𝛼(𝑥𝑖)}). (3.1) For each 𝑥_𝑖, 𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)} is a finite set and there exists a finite subset Λ(𝑥_𝑖) of Λ such that (𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)})∩𝐴⊆∪{𝑉_𝛼∶𝛼∈Λ(𝑥_𝑖)}. Therefore, we have 𝐴⊆(⋃_𝑖∈𝐹(∪{𝑉_𝛼∶𝛼∈Λ(𝑥_𝑖)}))∪(⋃_𝑖∈𝐹𝑉_{𝛼(𝑥𝑖)}). Hence 𝐴 is strongly compact relative to 𝑋.
Since every preopen set is 𝒩-preopen, the proof of the converse is obvious.

Corollary 3.4. For any space 𝑋, the following properties are equivalent: (1)𝑋 is strongly compact;(2)every 𝒩-preopen cover of 𝑋 admits a finite subcover.

Theorem 3.5. For any space 𝑋, the following properties are equivalent: (1)𝑋 is strongly compact; (2)every proper 𝒩-preclosed set is strongly compact with respect to 𝑋.

Proof. (1)⇒(2) Let 𝐴 be a proper 𝒩-preclosed subset of 𝑋. Let {𝑈_𝛼∶𝛼∈Λ} be a cover of 𝐴 by preopen sets of 𝑋. Now for each 𝑥∈𝑋−𝐴, there is a preopen set 𝑉_𝑥 such that 𝑉_𝑥∩𝐴 is finite. Then {𝑈_𝛼∶𝛼∈Λ}∪{𝑉_𝑥∶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,…,𝑥_𝑛 in 𝑋−𝐴 such that 𝑋=(∪{𝑈_𝛼∶𝛼∈Λ₁})∪(∪{𝑉_𝑥𝑖∶1≤𝑖≤𝑛}); hence 𝐴⊂(∪{𝑈_𝛼∶𝛼∈Λ₁})∪(∪{𝐴∩𝑉_𝑥𝑖∶1≤𝑖≤𝑛}). Since 𝐴∩𝑉_𝑥𝑖 is finite for each 𝑖, there exists a finite subset Λ₂ of Λ such that ∪{𝐴∩𝑉_𝑥𝑖∶1≤𝑖≤𝑛}⊂∪{𝑈_𝛼∶𝛼∈Λ₂}. Therefore, we obtain 𝐴⊂∪{𝑈_𝛼∶𝛼∈Λ₁∪Λ₂}. This shows that 𝐴 is strongly compact relative to 𝑋.
(2)⇒(1) Let {𝑉_𝛼∶𝛼∈Λ} be any preopen cover of 𝑋. We choose and fix one 𝛼₀∈Λ. Then ∪{𝑉_𝛼∶𝛼∈Λ−{𝛼₀}} is a preopen cover of a 𝒩-preclosed set 𝑋−𝑉_𝛼0. There exists a finite subset Λ₀ of Λ−{𝛼₀} such that 𝑋−𝑉_𝛼0⊂∪{𝑉_𝛼∶𝛼∈Λ₀}. Therefore, 𝑋=∪{𝑉_𝛼∶𝛼∈Λ₀∪{𝛼₀}}. This shows that 𝑋 is strongly compact.

Corollary 3.6 (see [2]). If a space 𝑋 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 {𝑉_𝛼∶𝛼∈Λ} be an open cover of (𝑋,𝒩PO(𝑋)). For each 𝑥∈𝑋, there exists 𝛼(𝑥)∈Λ such that 𝑥∈𝑉_𝛼(𝑥). Since 𝑉_𝛼(𝑥) is 𝒩-preopen, there exists a preopen set 𝑈_𝛼(𝑥) of 𝑋 such that 𝑥∈𝑈_𝛼(𝑥) and 𝑈_𝛼(𝑥)⧵𝑉_𝛼(𝑥) is finite. The family {𝑈_𝛼(𝑥)∶𝑥∈𝑋} is a preopen cover of (𝑋,𝜏). Since (𝑋,𝜏) is strongly compact, there exists a finite subset, says, 𝑥₁,𝑥₂,…,𝑥_𝑛 such that 𝑋=∪{𝑈_{𝛼(𝑥𝑖)}∶𝑖∈𝐹}, where 𝐹={1,2,…,𝑛}. Now, we have 𝑋=_𝑖∈𝐹((𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)})∪𝑉_{𝛼(𝑥𝑖)})=(_𝑖∈𝐹(𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)}))∪(_𝑖∈𝐹𝑉_{𝛼(𝑥𝑖)}). (3.2) For each 𝑥_𝑖, 𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)} is a finite set and there exists a finite subset Λ(𝑥_𝑖) of Λ such that (𝑈_{𝛼(𝑥𝑖)}⧵𝑉_{𝛼(𝑥𝑖)})⊆∪{𝑉_𝛼∶𝛼∈Λ(𝑥_𝑖)}. Therefore, we have 𝑋=(⋃_𝑖∈𝐹(∪(𝑉_𝛼∶𝛼∈Λ(𝑥_𝑖))))∪(⋃_𝑖∈𝐹𝑉_{𝛼(𝑥𝑖)}). Hence 𝒩PO(𝑋) is compact.
Conversely, let 𝒰 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 function 𝑓∶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 function 𝑓∶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 𝑈_𝑥∈𝒩PO(𝑋) such that 𝑥∈𝑈_𝑥 and 𝑓(𝑈_𝑥)⊆𝑉. Then 𝑥∈𝑈_𝑥⊆𝑓⁻¹(𝑉) and 𝑓⁻¹(𝑉)=∪{𝑈_𝑥∶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. If 𝑓∶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. Since 𝑓 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. Let 𝑓∶𝑋→𝑌 be a function and let {𝐴_𝛼∶𝛼∈Λ} be an 𝛼-open cover of 𝑋. If the restriction 𝑓_|𝐴𝛼∶𝐴_𝛼∶→𝑌 is 𝒩-precontinuous for each 𝛼∈Λ, then 𝑓 is 𝒩-precontinuous.

Proof. Suppose that 𝑉 is an arbitrary open set in 𝑌. Then for each 𝛼∈Λ, we have (𝑓_|𝐴𝛼)⁻¹(𝑉)=𝑓⁻¹(𝑉)∩𝐴_𝛼∈𝒩PO(𝐴_𝛼) because 𝑓_|𝐴𝛼 is 𝒩-precontinuous. Hence by Lemma 2.16, 𝑓⁻¹(𝑉)∩𝐴_𝛼∈𝒩PO(𝑋) for each 𝛼∈Λ. By Lemma 2.5, we obtain ∪{𝑓⁻¹(𝑉)∩𝐴_𝛼∶𝛼∈Λ}=𝑓⁻¹(𝑉)∈𝒩PO(𝑋). This implies that 𝑓 is 𝒩-precontinuous.

Theorem 4.5. Let 𝑓 be an 𝒩-precontinuous function from a space 𝑋 onto a space 𝑌. If 𝑋 is strongly compact, then 𝑌 is compact.

Proof. Let {𝑉_𝛼∶𝛼∈Λ} be an open cover of 𝑌. Then {𝑓⁻¹(𝑉_𝛼)∶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(𝑉_𝛼)∶𝛼∈Λ₀}; hence 𝑌=∪{𝑉_𝛼∶𝛼∈Λ₀}. Therefore 𝑌 is compact.

Definition 4.6 (see [12]). A function 𝑓∶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 is 𝒩-precontinuous but not conversely.

Example 4.7. Let 𝑋={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]). Let 𝑓 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 function 𝑓∶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. If 𝑓∶𝑋→𝑌 is 𝑀-preopen, then the image of an 𝒩-preopen set of 𝑋 is 𝒩-preopen in 𝑌.

Proof. Let 𝑓∶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 function 𝑓∶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. If 𝑓∶𝑋→𝑌 is a preirresolute injection and 𝐴 is 𝒩-preopen in 𝑌, then 𝑓⁻¹(𝐴) is 𝒩-preopen in 𝑋.

Proof. Assume that 𝐴 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 function 𝑓∶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. If 𝑓∶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 {𝑈_𝛼∶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)⊂∪{𝑈_𝛼∶𝛼∈Λ(𝑦)}. Now we put 𝑈(𝑦)=∪{𝑈_𝛼∶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 {𝑦_𝑘∶1≤𝑘≤𝑛}⊆𝑌 such that 𝑌=⋃^𝑛_𝑘=1𝑉(𝑦_𝑘). Therefore, 𝑋=𝑓⁻¹(𝑌)=⋃^𝑛_𝑘=1𝑓⁻¹(𝑉(𝑦_𝑘))⊆⋃^𝑛_𝑘=1𝑈(𝑦_𝑘)=⋃^𝑛_𝑘=1{𝑈_𝛼∶𝛼∈Λ(𝑦_𝑘)}. This shows that 𝑋 is strongly compact.

Definition 4.15 (see [2]). A function 𝑓∶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 every 𝑀-preclosed function is 𝒩-preclosed but not conversely.

Example 4.16. Let 𝑋={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]). If 𝑓∶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 function 𝑓∶𝑋→𝑌 is said to be 𝛿_𝒩-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 is 𝛿_𝒩-continuous but the converse is not true.

Example 4.19. Let 𝑋={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 𝛿_𝑁-continuous but it is not preirresolute.

Theorem 4.20. Let 𝑓∶𝑋→𝑌 be a 𝛿_𝒩-continuous surjection from 𝑋 onto 𝑌. If 𝑋 is strongly compact, then 𝑌 is strongly compact.

Proof. Let {𝑉_𝛼∶𝛼∈Λ} be a preopen cover of 𝑌. For each 𝑥∈𝑋, there exists 𝛼(𝑥)∈Λ such that 𝑓(𝑥)∈𝑉_𝛼(𝑥). Since 𝑓 is 𝛿_𝒩-continuous, there exists an 𝒩-preopen set 𝑈_𝛼(𝑥) of 𝑋 containing 𝑥 such that 𝑓(𝑈_𝛼(𝑥))⊆𝑉_𝛼(𝑥). So {𝑈_𝛼(𝑥)∶𝑥∈𝑋} is an 𝒩-preopen cover of the strongly compact space 𝑋, and by Corollary 3.4 there exists a finite subset {𝑥_𝑘∶1≤𝑘≤𝑛}⊆𝑋 such that 𝑋=⋃^𝑛_𝑘=1𝑈_{𝛼(𝑥𝑘)}. Therefore 𝑌=𝑓(𝑋)=𝑓(⋃^𝑛_𝑘=1𝑈_{𝛼(𝑥𝑘)})=⋃^𝑛_𝑘=1𝑓(𝑈_{𝛼(𝑥𝑘)})⊆⋃^𝑛_𝑘=1𝑉_{𝛼(𝑥𝑘)}. This shows that 𝑌 is strongly compact.

Corollary 4.21 (see [2]). Let 𝑓∶𝑋→𝑌 be a preirresolute surjection from 𝑋 onto 𝑌. If 𝑋 is strongly compact, then 𝑌 is strongly compact.

Acknowledgments

This work is financially supported by the Ministry of Higher Education, Malaysia under FRGS Grant no: UKM-ST-06-FRGS0008-2008. The authors also would like to thank the referees for useful comments and suggestions.