Cocompact Open Sets and Continuity

Abstract

Compact subsets of a topological space are used to define coc-open sets as new generalized open sets, and then coc-open sets are used to define (coc)^∗-open sets as another type of generalized open sets. Several results and examples related to them are obtained; particularly a decomposition of open sets is given. Also, coc-open sets and (coc)^∗-open sets are used to introduce coc-continuity and (coc)^∗-continuity, respectively. As a main result, a decomposition theorem of continuity is obtained.

1. Introduction

Throughout this paper by a space we mean a topological space. Let (𝑋,𝜏) be a space, and let 𝐴 be a subset of 𝑋. A point 𝑥∈𝑋 is called a condensation point of 𝐴 if, for each 𝑈∈𝜏 with 𝑥∈𝑈, the set 𝑈∩𝐴 is uncountable. In 1982, Hdeib defined 𝜔-closed sets and 𝜔-open sets as follows: 𝐴 is called 𝜔-closed [1] if it contains all its condensation points. The complement of an 𝜔-closed set is called 𝜔-open. In 1989, Hdeib [2] introduced 𝜔-continuity as a generalization of continuity as follows: A function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is called 𝜔-continuous if the inverse image of each open set is 𝜔-open. The authors in [3] proved that the family of all 𝜔-open sets in a space (𝑋,𝜏) forms a topology on 𝑋 finer than 𝜏 and that the collection {𝑈−𝐶∶𝑈∈𝜏 and 𝐶 is a countable subset of 𝑋} forms a base for that topology. Recently, in [4], the authors introduced a class of generalized open sets which is stronger 𝜔-open as follows: a subset 𝐴 of a space (𝑋,𝜏) is called 𝑁-open if for each 𝑥∈𝐴, there exists 𝑈∈𝜏 such that 𝑥∈𝑈 and 𝑈−𝐴 is finite. Throughout this paper, the family of all 𝑁-open sets in a space (𝑋,𝜏) will be denoted by 𝜏^𝑛. The authors in [4] proved that 𝜏^𝑛 is a topology on 𝑋 that is finer than 𝜏 and they introduced several concepts related to 𝑁-open sets; in particular, they introduced 𝑁-continuity as a generalization of continuity and as a stronger form of 𝜔-continuity as follows: a function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is called 𝑁-continuous if the inverse image of each open set is 𝑁-open. The authors in [5] continued the study of topological concepts via 𝑁-open sets. In the present work, we will generalize 𝑁-open sets by coc-open sets. Several results and concepts related to them will be introduced.

Throughout this paper, we use ℝ, ℚ, and ℕ to denote the set of real numbers, the set of rationals, and the set of natural numbers, respectively. For a subset 𝐴 of a space (𝑋,𝜏), The closure of 𝐴 and the interior of 𝐴 will be denoted by Cl_𝜏(𝐴) and Int_𝜏(𝐴), respectively. Also, we write 𝜏|_𝐴 to denote the relative topology on 𝐴 when 𝐴 is nonempty. For a nonempty set 𝑋, 𝜏_disc,𝜏_ind,𝜏_cof, and 𝜏_coc will denote, respectively, the discrete topology on 𝑋, the indiscrete topology on 𝑋, the cofinite topology on 𝑋, and the cocountable topology on 𝑋. For a subset 𝐴 of ℝ we write (𝐴,𝜏_𝑢) to denote the subspace topology on 𝐴 relative to the usual topology and we use 𝜏_lr to denote the left ray topology on ℝ. For any two spaces (𝑋,𝜏) and (𝑌,𝜎) we use 𝜏_prod to denote the product topology on 𝑋×𝑌. The family of all compact subsets of a space (𝑋,𝜏) will be denoted by 𝐶(𝑋,𝜏).

2. Cocompact Open Sets

Definition 2.1. A subset 𝐴 of a space (𝑋,𝜏) is called co-compact open set (notation: coc-open) if, for every 𝑥∈𝐴, there exists an open set 𝑈⊆𝑋 and a compact subset 𝐾∈𝐶(𝑋,𝜏) such that 𝑥∈𝑈−𝐾⊆𝐴. The complement of a coc-open subset is called coc-closed.

The family of all coc-open subsets of a space (𝑋,𝜏) will be denoted by 𝜏^𝑘 and the family {𝑈−𝐾∶𝑈∈𝜏 and 𝐾∈𝐶(𝑋,𝜏)} of coc-open sets will be denoted by ℬ^𝑘(𝜏).

Theorem 2.2. Let (𝑋,𝜏) be a space. Then the collection 𝜏^𝑘 forms a topology on 𝑋.

Proof. By the definition one has directly that Ø∈𝜏^𝑘. To see that 𝑋∈𝜏^𝑘, let 𝑥∈𝑋, take 𝑈=𝑋 and 𝐾=Ø. Then 𝑥∈𝑈−𝐾⊆𝑋.
Let 𝑈₁,𝑈₂∈𝜏^𝑘, and let 𝑥∈𝑈₁∩𝑈₂. For each 𝑖=1,2, we find an open set 𝑉_𝑖 and a compact subset 𝐾_𝑖 such that 𝑥∈𝑉_𝑖−𝐾_𝑖⊆𝑈_𝑖. Take 𝑉=𝑉₁∩𝑉₂ and 𝐾=𝐾₁∪𝐾₂. Then 𝑉 is open, 𝐾 is compact, and 𝑥∈𝑉−𝐾⊆𝑈₁∩𝑈₂. It follows that 𝑈₁∩𝑈₂ is coc-open.
Let {𝑈_𝛼∶𝛼∈Δ} be a collection of coc-open subsets of (𝑋,𝜏) and 𝑥∈⋃_𝛼∈Δ𝑈_𝛼. Then there exists 𝛼₀∈Δ such that 𝑥∈𝑈_𝛼∘. Since 𝑈_𝛼∘ is coc-open, then there exists an open set 𝑉 and a compact subset 𝐾, such that 𝑥∈𝑉−𝐾⊆𝑈_𝛼∘. Therefore, we have 𝑥∈𝑉−𝐾⊆𝑈_𝛼∘⊆⋃_𝛼∈Δ𝑈_𝛼. Hence, ⋃_𝛼∈Δ𝑈_𝛼 is coc-open.

The following result follows directly from Definition 2.1.

Proposition 2.3. Let (𝑋,𝜏) be a space. Then the collection ℬ^𝑘(𝜏) forms a base for 𝜏^𝑘.

Corollary 2.4. Let (𝑋,𝜏) be a space. Then the collection 𝜏∪{𝑋−𝐾∶𝐾∈𝐶(𝑋,𝜏)} forms a subbase for 𝜏^𝑘.

For a space (𝑋,𝜏), the following example shows that the collection ℬ^𝑘(𝜏) is not a topology on 𝑋 in general.

Example 2.5. Let 𝑋=ℝ, and let 𝜏=𝜏_lr. Consider the collection of elements of ℬ^𝑘(𝜏),𝐺_𝑛 = (−∞,𝑛+1/2)−{1,2,…,𝑛}, 𝑛∈ℕ. Then ∪{𝐺_𝑛∶𝑛∈ℕ}=ℝ−ℕ which is not in ℬ^𝑘(𝜏).

Theorem 2.6. Let (𝑋,𝜏) be a space. Then 𝜏⊆𝜏^𝑛⊆𝜏^𝑘.

Proof. Obvious.

Remark 2.7. Each of the two inclusions in Theorem 2.6 is not equality in general; to see this, let 𝑋=ℕ and 𝜏=𝜏_ind. Then 𝜏^𝑛=𝜏_cof and 𝜏^𝑘=𝜏_disc, and therefore 𝜏≠𝜏^𝑛 and 𝜏^𝑛≠𝜏^𝑘.

In Remark 2.7, the space (𝑋,𝜏) is an example on a compact space (𝑋,𝜏) for which (𝑋,𝜏^𝑘) is not compact.

Definition 2.8 (see [6]). A space (𝑋,𝜏) is called CC if every compact set in 𝑋 is closed.

It is well known that every 𝑇₂ space is CC, but not conversely.

Theorem 2.9. Let (𝑋,𝜏) be a space. Then the following are equivalent:(a)(𝑋,𝜏) is CC,(b)𝜏=ℬ^𝑘(𝜏),(c)𝜏=𝜏^𝑛=𝜏^𝑘.

Proof. (a) ⇒ (b) As Ø is a compact subset of 𝑋, then, for every 𝑈∈𝜏, 𝑈−Ø=𝑈∈ℬ^𝑘(𝜏). Hence, we have 𝜏⊆ℬ^𝑘(𝜏). Now let 𝑈−𝐾∈ℬ^𝑘(𝜏), where 𝑈∈𝜏 and 𝐾 is a compact subset of 𝑋. As (𝑋,𝜏) is CC, then 𝐾 is closed and hence 𝑈−𝐾∈𝜏. Therefore, we have ℬ^𝑘(𝜏)⊆𝜏.
(b) ⇒ (c) By Theorem 2.6, it is sufficient to see that 𝜏^𝑘⊆𝜏. Since by (b) 𝜏=ℬ^𝑘(𝜏) and as ℬ^𝑘(𝜏) is a base for 𝜏^𝑘, then 𝜏^𝑘⊆𝜏.
(c) ⇒ (a) Let 𝐾∈𝐶(𝑋,𝜏). Then 𝑋−𝐾∈𝜏^𝑘, and by (c), 𝑋−𝐾∈𝜏. Therefore, 𝐾 is closed in 𝑋.

Corollary 2.10. If (𝑋,𝜏) is a T₂-space, then 𝜏=𝜏^𝑛=𝜏^𝑘.

Theorem 2.11. For any space (𝑋,𝜏), (𝑋,𝜏^𝑘) is CC.

Proof. Let 𝐾∈𝐶(𝑋,𝜏^𝑘). As 𝜏⊆𝜏^𝑘, then 𝐶(𝑋,𝜏^𝑘)⊆𝐶(𝑋,𝜏) and hence 𝐾∈𝐶(𝑋,𝜏). Thus, we have 𝑋−𝐾∈𝜏^𝑘, and hence 𝐾 is closed in the space (𝑋,𝜏^𝑘).

Corollary 2.12. For any space (𝑋,𝜏), (𝜏^𝑘)^𝑘=𝜏^𝑘.

Proof. Theorems 2.9 and 2.11.

Theorem 2.13. If (𝑋,𝜏) is a hereditarily compact space, then 𝜏^𝑘=𝜏_disc.

Proof. For every 𝑥∈𝑋, 𝑋−{𝑥} is compact and so {𝑥}=𝑋−(𝑋−{𝑥})∈ℬ^𝑘(𝜏)⊆𝜏^𝑘. Therefore, 𝜏^𝑘=𝜏_disc.

Each of the following three examples shows that the converse of Theorem 2.13 is not true in general.

Example 2.14. Let 𝑋=ℝ and 𝜏=𝜏_lr. For every 𝑥∈𝑋, take 𝐾=(−∞,𝑥+1]−{𝑥} and 𝑈=(−∞,𝑥+1). Then 𝐾∈𝐶(𝑋,𝜏), 𝑈=(−∞,𝑥+1)∈𝜏, and {𝑥}=𝑈−𝐾. This shows that 𝜏^𝑘=𝜏_disc. On the other hand, it is well known that (𝑋,𝜏) is not hereditarily compact.

Example 2.15. Let 𝑋=ℕ and 𝜏={Ø,ℕ}∪{𝑈_𝑛∶𝑛∈ℕ}, where 𝑈_𝑛={1,2,…,𝑛}. Then the compact subsets of (𝑋,𝜏) are the finite sets. For every 𝑛∈ℕ, 𝑈_𝑛∈𝜏, 𝑈_𝑛−1 is compact, and {𝑛}=𝑈_𝑛−𝑈_𝑛−1. Therefore, 𝜏^𝑘=𝜏_disc.

Example 2.16. Let 𝑋=ℕ and 𝜏 be the topology on ℕ having the family {{2𝑛−1,2𝑛}∶𝑛∈ℕ} as a base. Then the compact subsets of (𝑋,𝜏) are the finite sets. If 𝑥∈ℕ with 𝑥 is odd, then {𝑥}={𝑥,𝑥+1}−{𝑥+1} and as {𝑥,𝑥+1}∈𝜏 and {𝑥+1} is compact, then {𝑥}∈𝜏^𝑘. Similarly, if 𝑥 is even then {𝑥}∈𝜏^𝑘. Therefore, 𝜏^𝑘=𝜏_disc.

The following question is natural: Is there a space (𝑋,𝜏) for which 𝜏^𝑘≠𝜏 and 𝜏^𝑘≠𝜏_disc?

The following example shows that the answer of the above question is yes.

Example 2.17. Let 𝑋=ℝ and 𝜏={𝑋}∪{𝑈⊆𝑋∶1∉𝑈}. Then 𝐶(𝑋,𝜏)={𝐾⊆𝑋∶1∈𝐾}∪{𝐾⊆𝑋∶1∉𝐾and𝐾isfinite}, hence 𝜏^𝑘=𝜏∪{𝑈⊆𝑋∶1∈𝑈and𝑋−𝑈isfinite}. Note that 𝜏^𝑘≠𝜏 and 𝜏^𝑘≠𝜏_disc.

Theorem 2.18. Let (𝑋,𝜏) be a space and 𝐴 a nonempty subset of 𝑋. Then (𝜏|_𝐴)^𝑘⊆𝜏^𝑘|_𝐴.

Proof. Let 𝐵∈(𝜏|_𝐴)^𝑘 and 𝑥∈𝐵. Then there exists 𝑉∈𝜏|_𝐴 and a compact subset 𝐾⊆𝐴 such that𝑥∈𝑉−𝐾⊆𝐵. Since 𝑉∈𝜏|_𝐴, then we can write 𝑉=𝑈∩𝐴, where 𝑈 is open in 𝑋. Since 𝑈−𝐾∈𝜏^𝑘, (𝑈−𝐾)∩𝐴∈𝜏^𝑘|_𝐴. Hence, 𝐵∈𝜏^𝑘|_𝐴.

Question 1. Let (𝑋,𝜏) be a space and 𝐴 a nonempty subset of 𝑋. Is it true that (𝜏|_𝐴)^𝑘=𝜏^𝑘|_𝐴?

The following result is a partial answer for Question 1.

Theorem 2.19. Let (𝑋,𝜏) be a space and 𝐴be a nonempty closed set in (𝑋,𝜏). Then (𝜏|_𝐴)^𝑘=𝜏^𝑘|_𝐴.

Proof. By Theorem 2.18, (𝜏|_𝐴)^𝑘⊆𝜏^𝑘|_𝐴. Conversely, let 𝐵∈𝜏^𝑘|_𝐴 and 𝑥∈𝐵. Choose 𝐻∈𝜏^𝑘 such that 𝐵=𝐻∩𝐴. As 𝐻∈𝜏^𝑘, there exists 𝑈∈𝜏 and 𝐾∈𝐶(𝑋,𝜏) such that 𝑥∈𝑈−𝐾⊆𝐻. Thus, we have 𝑥∈(𝑈∩𝐴)−(𝐾∩𝐴)⊆𝐵, 𝑈∩𝐴∈𝜏|_𝐴, and 𝐾∩𝐴∈𝐶(𝐴,𝜏|_𝐴). It follows that 𝐵∈(𝜏|_𝐴)^𝑘.

Definition 2.20. Let (𝑋,𝜏) be a space, and let 𝐴⊆𝑋. The coc-closure of 𝐴 in (𝑋,𝜏) is denoted by coc-Cl_𝜏(𝐴) and defined as follows: coc-Cl_𝜏(𝐴)=∩{𝐵∶𝐵iscoc-closedin(𝑋,𝜏)and𝐴⊆𝐵}. (2.1)

Remark 2.21. Let (𝑋,𝜏) be a space, and let 𝐴⊆𝑋. Then coc-Cl_𝜏(𝐴)=Cl_𝜏𝑘(𝐴) and coc-Cl_𝜏(𝐴)⊆Cl_𝜏(𝐴).

Definition 2.22. A space (𝑋,𝜏) is called antilocally compact if any compact subset of 𝑋 has empty interior.

For any infinite set 𝑋, (𝑋,𝜏_coc) is an anti-locally compact space. Also, (ℚ,𝜏_𝑢) is an example of an anti-locally compact space.

Theorem 2.23. Let (𝑋,𝜏) be an anti-locally compact space. If 𝐴∈𝜏 then coc-Cl_𝜏(𝐴)=Cl_𝜏(𝐴).

Proof. According to Remark 2.21, only we need to show that Cl_𝜏(𝐴)⊆Cl_𝜏𝑘(𝐴). Suppose to the contrary that there is 𝑥∉Cl_𝜏(𝐴)−Cl_𝜏𝑘(𝐴). As 𝑥∉Cl_𝜏𝑘(𝐴), there exists 𝐺⊇𝐴 coc-closed such that 𝑥∈𝐺 and 𝐺∩𝐴=Ø. Take 𝑈∈𝜏 and 𝐾∈𝐶(𝑋,𝜏) such that 𝑥∈𝑈−𝐾⊆𝐺. Thus we have 𝑈∩𝐴⊆𝐾. Since 𝑥∈Cl_𝜏(𝐴), it follows that 𝑈∩𝐴≠Ø and hence Int(𝐾)≠Ø. This contradicts the assumption that (𝑋,𝜏) is anti-locally compact.

In Theorem 2.23 the assumption “anti-locally compact” on the space cannot be dropped. As an example let 𝑋=ℝ and 𝜏={Ø,𝑋,{0}}, then {0}∈𝜏, coc-Cl_𝜏({0})={0} while Cl_𝜏({0})=ℝ.

Theorem 2.24. If 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is injective, open, and continuous, then 𝑓∶(𝑋,𝜏^𝑘)→(𝑌,𝜎^𝑘) is open.

Proof. Let 𝐺=𝑈−𝐾 where 𝑈∈𝜏 and 𝐾∈𝐶(𝑋,𝜏) be a basic element for 𝜏^𝑘. As 𝑓 is injective, 𝑓(𝐺)=𝑓(𝑈)−𝑓(𝐾). Also, as 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is open, 𝑓(𝑈)∈𝜎. And as 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is continuous, 𝑓(𝐾)∈𝐶(𝑌,𝜎). This ends the proof.

Remark 2.25. In Theorem 2.24, the continuity condition cannot be dropped. Take 𝑓∶(ℝ,𝜏_ind)→(ℝ,𝜏_𝑢), where 𝑓(𝑥)=tan⁻¹𝑥. Then 𝑓 is injective and open. On the other hand, as (ℝ,𝜏_ind) is hereditarily compact we have (𝜏_ind)^𝑘=𝜏_disc, and as (ℝ,𝜏_𝑢) is 𝑇₂ we have (𝜏_𝑢)^𝑘=𝜏_𝑢. Thus, 𝑓∶(ℝ,(𝜏_ind)^𝑘)→(ℝ,(𝜏_𝑢)^𝑘) is not open.

3. (coc)^∗-Open Sets

Definition 3.1. A subset 𝐴 of a space (𝑋,𝜏) is called (coc)^∗-open if Int_𝜏𝑘(𝐴)=Int_𝜏(𝐴).

The family of all (coc)^∗-open subsets of a space (𝑋,𝜏) will be denoted by ℬ^𝑘∗(𝜏).

Theorem 3.2. Let (𝑋,𝜏) be a space. Then 𝜏⊆ℬ^𝑘∗(𝜏).

Proof. Let 𝐴∈𝜏. Then 𝐴=Int_𝜏(𝐴)⊆Int_𝜏𝑘(𝐴)⊆𝐴. Thus Int_𝜏(𝐴)=Int_𝜏𝑘(𝐴) and hence 𝐴∈ℬ^𝑘∗(𝜏).

Theorem 3.3. If (𝑋,𝜏) is a CC space, then every subset of 𝑋 is (coc)^∗-open.

Proof. Let 𝐴⊆𝑋. Since (𝑋,𝜏) is CC, then 𝜏=𝜏^𝑘, and so Int_𝜏𝑘(𝐴)=Int_𝜏(𝐴). Therefore, 𝐴 is (coc)^∗-open.

Corollary 3.4. If (𝑋,𝜏) is a 𝑇₂ space, then every subset of 𝑋 is (coc)^∗-open.

According to Corollary 3.4, the inclusion in Theorem 3.2 is not equality in any 𝑇₂ space that is not discrete for example, in (ℝ,𝜏_𝑢) for the set 𝐴=(0,1)∪{2}, we have Int_{(𝜏𝑢)𝑘}(𝐴)=Int_𝜏𝑢(𝐴)=(0,1) and thus 𝐴∈ℬ^𝑘∗(𝜏_𝑢)−𝜏_𝑢.

Theorem 3.5. If (𝑋,𝜏) is a hereditarily compact space, then 𝜏=ℬ^𝑘∗(𝜏).

Proof. By Theorem 3.2, we need only to show that ℬ^𝑘∗(𝜏)⊆𝜏. Let 𝐴∈ℬ^𝑘∗(𝜏). Then Int_𝜏𝑘(𝐴)=Int_𝜏(𝐴). Since (𝑋,𝜏) is hereditarily compact, then, by Theorem 2.13, 𝜏^𝑘=𝜏_disc and thus Int_𝜏(𝐴)=Int_𝜏𝑘(𝐴)=𝐴. Therefore, 𝐴∈𝜏.

The following result is a new decomposition of open sets in a space.

Theorem 3.6. Let (𝑋,𝜏) be a space. Then 𝜏=𝜏^𝑘∩ℬ^𝑘∗(𝜏).

Proof. By Theorems 2.6 and 3.2, it follows that 𝜏⊆𝜏^𝑘∩ℬ^𝑘∗(𝜏). Conversely, let 𝐴∈𝜏^𝑘∩ℬ^𝑘∗(𝜏). As 𝐴∈𝜏^𝑘, then Int_𝜏𝑘(𝐴)=𝐴. Also, since 𝐴∈ℬ^𝑘∗(𝜏), then Int_𝜏𝑘(𝐴)=Int_𝜏(𝐴). It follows that Int_𝜏(𝐴)=𝐴 and hence 𝐴∈𝜏.

Theorem 3.7. For a space (𝑋,𝜏), one has the following:(a)Ø,𝑋∈ℬ^𝑘∗(𝜏),(b)if 𝐴,𝐵∈ℬ^𝑘∗(𝜏), then 𝐴∩𝐵∈ℬ^𝑘∗(𝜏).

Proof. (a) The proof follows directly from Theorem 3.2.
(b) Let 𝐴,𝐵∈ℬ^𝑘∗(𝜏). Then Int_𝜏𝑘(𝐴)=Int_𝜏(𝐴) and Int_𝜏𝑘(𝐵)=Int_𝜏(𝐵). Thus we haveInt_𝜏(𝐴∩𝐵)=Int_𝜏(𝐴)∩Int_𝜏(𝐵)=Int_𝜏𝑘(𝐴)∩Int_𝜏𝑘(𝐵)=Int_𝜏𝑘(𝐴∩𝐵). (3.1) It follows that 𝐴∩𝐵∈ℬ^𝑘∗(𝜏).

The following example shows that arbitrary union of 𝑘^∗-open sets need not to be 𝑘^∗-open in general.

Example 3.8. Consider the space defined in Example 2.17. For every natural number 𝑛≥3, put 𝐴_𝑛=ℝ−{𝑛,𝑛+1,𝑛+2,…}. Then, for each 𝑛≥3, Int_𝜏𝑘(𝐴_𝑛)=Int_𝜏(𝐴_𝑛)=𝐴_𝑛−{1}, and thus 𝐴_𝑛 is 𝑘^∗-open. On the other hand, Int_𝜏𝑘(⋃_𝑛≥3𝐴_𝑛)=⋃_𝑛≥3𝐴_𝑛 while Int_𝜏(⋃_𝑛≥3𝐴_𝑛)=(⋃_𝑛≥3𝐴_𝑛)−{1}.

4. coc-Continuous Functions

Definition 4.1. A function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is called coc-continuous at a point 𝑥∈𝑋, if for every open set 𝑉 containing 𝑓(𝑥) there is a coc-open set 𝑈 containing 𝑥 such that 𝑓(𝑈)⊆𝑉. If 𝑓 is coc-continuous at each point of 𝑋, then 𝑓is said to be coc-continuous.

The following theorem follows directly from the definition.

Theorem 4.2. A function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous if and only if 𝑓∶(𝑋,𝜏^𝑘)→(𝑌,𝜎) is continuous.

Theorem 4.3. Every 𝑁-continuous function is coc-continuous.

Proof. Straightforward.

The identity function 𝐼∶(ℝ,𝜏_ind)→(ℝ,𝜏_disc) is a coc-continuous function that is not 𝑁-continuous.

The proof of the following result follows directly from Theorem 2.9.

Theorem 4.4. Let 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) be a function for which (𝑋,𝜏) is CC, then the following are equivalent.(a)𝑓 is continuous.(b)𝑓 is 𝑁-continuous.(c)𝑓 is coc-continuous.

The following example shows that the composition of two 𝑁-continuous functions need not to be even coc-continuous.

Example 4.5. Let 𝑋=ℝ, 𝑌={0,1,2}, 𝑍={𝑎,𝑏},𝜏 be as in Example 2.17, 𝜎={Ø,𝑌,{0},{0,1}}, and 𝜇={Ø,𝑍,{𝑎}}. Define the function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) by 𝑓(𝑥)=2 if 𝑥∈{0,1} and 𝑓(𝑥)=1 otherwise, and define the function 𝑔∶(𝑌,𝜎)→(𝑍,𝜇) by 𝑔(0)=𝑔(2)=𝑎 and 𝑔(1)=𝑏. Then 𝑓 and 𝑔 are 𝑁-continuous functions, but 𝑔∘𝑓 is not coc-continuous since (𝑔∘𝑓)⁻¹({𝑎})={0,1}∉𝜏^𝑘.

Theorem 4.6. (a) If 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is 𝑁-continuous and if 𝑔∶(𝑌,𝜎)→(𝑍,𝜇) is continuous, then 𝑔∘𝑓∶(𝑋,𝜏)→(𝑍,𝜇) is 𝑁-continuous.
(b) If 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous and if 𝑔∶(𝑌,𝜎)→(𝑍,𝜇) is continuous, then 𝑔∘𝑓∶(𝑋,𝜏)→(𝑍,𝜇) is coc-continuous.

Proof. (a) It follows by noting that a function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is 𝑁-continuous if and only if 𝑓∶(𝑋,𝜏^𝑛)→(𝑌,𝜎).
(b) The proof follows directly from Theorem 4.2.

Theorem 4.7. If 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous and 𝐴is a nonempty closed set in (𝑋,𝜏), then the restriction of 𝑓 to 𝐴,𝑓|_𝐴∶(𝐴,𝜏|_𝐴)→(𝑌,𝜎) is a coc-continuous function.

Proof. Let 𝑉be any open set in 𝑌. Since 𝑓is coc-continuous, then 𝑓⁻¹(𝑉) is coc-open in 𝑋 and by Theorem 2.19, (𝑓|_𝐴)⁻¹(𝑉)=𝑓⁻¹(𝑉)∩𝐴 is coc-open in 𝐴. Therefore 𝑓|_𝐴 is coc-continuous.

Theorem 4.8. If 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous and 𝑋=𝐴∪𝐵, where 𝐴 and 𝐵 are coc-closed subsets in (𝑋,𝜏) and 𝑓|_𝐴∶(𝐴,𝜏|_𝐴)→(𝑌,𝜎), 𝑓|_𝐵∶(𝐵,𝜏|_𝐵)→(𝑌,𝜎) are coc-continuous functions, then 𝑓 is coc-continuous.

Proof. By Theorem 4.2 it is sufficient to show that 𝑓∶(𝑋,𝜏^𝑘)→(𝑌,𝜎) is continuous. Let 𝐶 be a closed subset of (𝑌,𝜎). Then 𝑓⁻¹(𝐶) = 𝑓⁻¹(𝐶)∩𝑋 = 𝑓⁻¹(𝐶)∩(𝐴∪𝐵) = (𝑓⁻¹(𝐶)∩𝐴)∪(𝑓⁻¹(𝐶)∩𝐵). Since 𝑓|_𝐴∶(𝐴,𝜏|_𝐴)→(𝑌,𝜎) is coc-continuous, then 𝑓⁻¹(𝐶)∩𝐴=(𝑓|_𝐴)⁻¹(𝐶) is coc-closed in (𝐴,𝜏|_𝐴), and as 𝐴 is coc-closed in (𝑋,𝜏), it follows that 𝑓⁻¹(𝐶)∩𝐴 is coc-closed in (𝑋,𝜏); similarly one can conclude that 𝑓⁻¹(𝐶)∩𝐵 is coc-closed in (𝑋,𝜏). It follows that 𝑓⁻¹(𝐶) is closed in (𝑋,𝜏^𝑘) and hence 𝑓∶(𝑋,𝜏^𝑘)→(𝑌,𝜎) is continuous.

The following result follows directly from Theorem 4.2.

Theorem 4.9. Let 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) and 𝑔∶(𝑋,𝜏)→(𝑍,𝜇) be two functions. Then the function ℎ∶(𝑋,𝜏)→(𝑌×𝑍,𝜏_prod) defined by ℎ(𝑥)=(𝑓(𝑥),𝑔(𝑥)) is coc-continuous if and only if 𝑓 and 𝑔 are coc-continuous.

Corollary 4.10. A function 𝑤∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous if and only if the graph function ℎ∶(𝑋,𝜏)→(𝑋×𝑌,𝜏_prod), given by ℎ(𝑥)=(𝑥,𝑤(𝑥)) for every 𝑥∈𝑋, is coc-continuous.

Theorem 4.11. Let 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) be a function. If there is a coc-open subset 𝐴 of (𝑋,𝜏) containing 𝑥∈𝑋 such that the restriction of 𝑓 to 𝐴,𝑓|_𝐴∶(𝐴,𝜏|_𝐴)→(𝑌,𝜎) is coc-continuous at 𝑥, then 𝑓 is coc-continuous at 𝑥.

Proof. Let 𝑉∈𝜎 with 𝑓(𝑥)∈𝑉. Since 𝑓|_𝐴 is coc-continuous at 𝑥, there is 𝑈∈(𝜏|_𝐴)^𝑘⊆𝜏^𝑘 such that 𝑥∈𝑈 and (𝑓|_𝐴)(𝑈)=𝑓(𝑈)⊆𝑉.

Corollary 4.12. Let 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) be a function, and let {𝐴_𝛼∶𝛼∈Δ}⊆𝜏^𝑘 be a cover of 𝑋 such that, for each 𝛼∈Δ,𝑓|_𝐴𝛼∶(𝐴_𝛼,𝜏|_𝐴𝛼)→(𝑌,𝜎) is coc-continuous, then 𝑓 is coc-continuous.

Proof. Let 𝑥∈𝑋. We show that 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is coc-continuous at 𝑥. Since {𝐴_𝛼∶𝛼∈Δ} is a cover of 𝑋, then there exists 𝛼_∘∈Δ such that 𝑥∈𝐴_𝛼∘. Therefore, by Theorem 4.11, it follows that 𝑓 is coc-continuous at 𝑥.

Definition 4.13. A function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is called (coc)^∗-continuous if the inverse image of each open set is (coc)^∗-open.

Theorem 4.14. Every continuous function is (coc)^∗-continuous.

Proof. The proof follows directly from Theorem 3.2.

Theorem 4.15. If (𝑋,𝜏) is CC, then every function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is (coc)^∗-continuous.

Proof. The proof follows directly from Theorem 3.3.

Corollary 4.16. If (𝑋,𝜏) is 𝑇₂, then every function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is (coc)^∗-continuous.

By Corollary 4.16, it follows that the function 𝑓∶(ℝ,𝜏_𝑢)→(ℝ,𝜏_𝑢) where 𝑓(𝑥)=0 for 𝑥 is rational and 𝑓(𝑥)=1 for 𝑥 is irrational is (coc)^∗-continuous. On the other hand, it is well known that this function is discontinuous every where. Also, by Theorem 4.4, 𝑓 is not coc-continuous.

Theorem 4.17. Let 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) be a function with (𝑋,𝜏) being a hereditarily compact space. Then 𝑓 is continuous if and only if 𝑓 is (coc)^∗-continuous.

Proof. The proof follows directly from Theorem 3.5.

By Theorem 4.17, it follows that the identity function 𝐼∶(ℝ,𝜏_ind)→(ℝ,𝜏_disc) is not (coc)^∗-continuous. Therefore, this is an example of a coc-continuous function that is not (coc)^∗-continuous.

We end this section by the following decomposition of continuity via coc-continuity and (coc)^∗-continuity.

Theorem 4.18. A function 𝑓∶(𝑋,𝜏)→(𝑌,𝜎) is continuous if and only if it is coc-continuous and (coc)^∗-continuous.

Proof. The proof follows directly from Theorem 3.6.

Acknowledgments

The authors are very grateful to the referees for their valuable comments and suggestions. Also, the authors would like to thank Jordan University of Science and Technology for the financial assistant of this paper.