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

Throughout this paper, we use , , and to denote the set of real numbers, the set of rationals, and the set of natural numbers, respectively. For a subset of a space , The closure of and the interior of will be denoted by and , respectively. Also, we write to denote the relative topology on when is nonempty. For a nonempty set , , and will denote, respectively, the discrete topology on , the indiscrete topology on , the cofinite topology on , and the cocountable topology on . For a subset of we write to denote the subspace topology on relative to the usual topology and we use to denote the left ray topology on . For any two spaces and we use to denote the product topology on . The family of all compact subsets of a space will be denoted by .

#### 2. Cocompact Open Sets

Definition 2.1. A subset of a space is called co-compact open set (notation: -open) if, for every , there exists an open set and a compact subset such that . The complement of a -open subset is called -closed.

The family of all -open subsets of a space will be denoted by and the family and of -open sets will be denoted by .

Theorem 2.2. Let be a space. Then the collection forms a topology on .

Proof. By the definition one has directly that . To see that , let , take and . Then .
Let , and let . For each , we find an open set and a compact subset such that . Take and . Then is open, is compact, and . It follows that is -open.
Let be a collection of -open subsets of and . Then there exists such that . Since is -open, then there exists an open set and a compact subset , such that . Therefore, we have . Hence, is -open.

The following result follows directly from Definition 2.1.

Proposition 2.3. Let be a space. Then the collection forms a base for .

Corollary 2.4. Let be a space. Then the collection forms a subbase for .

For a space , the following example shows that the collection is not a topology on in general.

Example 2.5. Let , and let . Consider the collection of elements of = , . Then which is not in .

Theorem 2.6. Let be a space. Then .

Proof. Obvious.

Remark 2.7. Each of the two inclusions in Theorem 2.6 is not equality in general; to see this, let and . Then and , and therefore and .

In Remark 2.7, the space is an example on a compact space for which is not compact.

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

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

Theorem 2.9. Let be a space. Then the following are equivalent:(a) is CC,(b),(c).

Proof. (a) (b) As is a compact subset of , then, for every , . Hence, we have . Now let , where and is a compact subset of . As is CC, then is closed and hence . Therefore, we have .
(b) (c) By Theorem 2.6, it is sufficient to see that . Since by (b) and as is a base for , then .
(c) (a) Let . Then , and by (c), . Therefore, is closed in .

Corollary 2.10. If is a T2-space, then .

Theorem 2.11. For any space , is CC.

Proof. Let . As , then and hence . Thus, we have , and hence is closed in the space .

Corollary 2.12. For any space , .

Proof. Theorems 2.9 and 2.11.

Theorem 2.13. If is a hereditarily compact space, then .

Proof. For every , is compact and so . Therefore, .

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

Example 2.14. Let and . For every , take and . Then , , and . This shows that . On the other hand, it is well known that is not hereditarily compact.

Example 2.15. Let and , where . Then the compact subsets of are the finite sets. For every , , is compact, and . Therefore, .

Example 2.16. Let and be the topology on having the family as a base. Then the compact subsets of are the finite sets. If with is odd, then and as and is compact, then . Similarly, if is even then . Therefore, .

The following question is natural: Is there a space for which and ?

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

Example 2.17. Let and . Then , hence . Note that and .

Theorem 2.18. Let be a space and a nonempty subset of . Then .

Proof. Let and . Then there exists and a compact subset such that. Since , then we can write , where is open in . Since , . Hence, .

Question 1. Let be a space and a nonempty subset of . Is it true that ?

The following result is a partial answer for Question 1.

Theorem 2.19. Let be a space and be a nonempty closed set in . Then .

Proof. By Theorem 2.18, . Conversely, let and . Choose such that . As , there exists and such that . Thus, we have , , and . It follows that .

Definition 2.20. Let be a space, and let . The -closure of in is denoted by - and defined as follows:

Remark 2.21. Let be a space, and let . Then and .

Definition 2.22. A space is called antilocally compact if any compact subset of has empty interior.

For any infinite set , is an anti-locally compact space. Also, is an example of an anti-locally compact space.

Theorem 2.23. Let be an anti-locally compact space. If then -.

Proof. According to Remark 2.21, only we need to show that . Suppose to the contrary that there is . As , there exists coc-closed such that and . Take and such that . Thus we have . Since , it follows that and hence Int. This contradicts the assumption that is anti-locally compact.

In Theorem 2.23 the assumption “anti-locally compact” on the space cannot be dropped. As an example let and , then , - while .

Theorem 2.24. If is injective, open, and continuous, then is open.

Proof. Let where and be a basic element for . As is injective, . Also, as is open, . And as is continuous, . This ends the proof.

Remark 2.25. In Theorem 2.24, the continuity condition cannot be dropped. Take , where . Then is injective and open. On the other hand, as is hereditarily compact we have , and as is we have . Thus, is not open.

#### 3. -Open Sets

Definition 3.1. A subset of a space is called -open if .

The family of all -open subsets of a space will be denoted by .

Theorem 3.2. Let be a space. Then .

Proof. Let . Then . Thus and hence .

Theorem 3.3. If is a CC space, then every subset of is -open.

Proof. Let . Since is CC, then , and so . Therefore, is -open.

Corollary 3.4. If is a space, then every subset of is -open.

According to Corollary 3.4, the inclusion in Theorem 3.2 is not equality in any space that is not discrete for example, in for the set , we have and thus .

Theorem 3.5. If is a hereditarily compact space, then .

Proof. By Theorem 3.2, we need only to show that . Let . Then . Since is hereditarily compact, then, by Theorem 2.13, and thus . Therefore, .

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

Theorem 3.6. Let be a space. Then .

Proof. By Theorems 2.6 and 3.2, it follows that . Conversely, let . As , then . Also, since , then . It follows that and hence .

Theorem 3.7. For a space , one has the following:(a),(b)if , then .

Proof. (a) The proof follows directly from Theorem 3.2.
(b) Let . Then and . Thus we have It follows that .

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

Example 3.8. Consider the space defined in Example 2.17. For every natural number , put . Then, for each , , and thus is -open. On the other hand, while Int.

#### 4. coc-Continuous Functions

Definition 4.1. A function is called -continuous at a point , if for every open set containing there is a -open set containing such that . If is -continuous at each point of , then is said to be -continuous.

The following theorem follows directly from the definition.

Theorem 4.2. A function is -continuous if and only if is continuous.

Theorem 4.3. Every -continuous function is -continuous.

Proof. Straightforward.

The identity function is a -continuous function that is not -continuous.

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

Theorem 4.4. Let be a function for which is CC, then the following are equivalent.(a) is continuous.(b) is -continuous.(c) is -continuous.

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

Example 4.5. Let , , be as in Example 2.17, , and . Define the function by if and otherwise, and define the function by and . Then and are -continuous functions, but is not -continuous since .

Theorem 4.6. (a) If is -continuous and if is continuous, then is -continuous.
(b) If is -continuous and if is continuous, then is -continuous.

Proof. (a) It follows by noting that a function is -continuous if and only if .
(b) The proof follows directly from Theorem 4.2.

Theorem 4.7. If is -continuous and is a nonempty closed set in , then the restriction of to , is a -continuous function.

Proof. Let be any open set in . Since is -continuous, then is -open in and by Theorem 2.19, is -open in . Therefore is -continuous.

Theorem 4.8. If is -continuous and , where and are -closed subsets in and , are -continuous functions, then is -continuous.

Proof. By Theorem 4.2 it is sufficient to show that is continuous. Let be a closed subset of . Then = = = . Since is -continuous, then is -closed in , and as is -closed in , it follows that is -closed in ; similarly one can conclude that is -closed in . It follows that is closed in and hence is continuous.

The following result follows directly from Theorem 4.2.

Theorem 4.9. Let and be two functions. Then the function defined by is -continuous if and only if and are -continuous.

Corollary 4.10. A function is -continuous if and only if the graph function , given by for every , is -continuous.

Theorem 4.11. Let be a function. If there is a -open subset of containing such that the restriction of to is -continuous at , then is -continuous at .

Proof. Let with . Since is -continuous at , there is such that and .

Corollary 4.12. Let be a function, and let be a cover of such that, for each is -continuous, then is -continuous.

Proof. Let . We show that is -continuous at . Since is a cover of , then there exists such that . Therefore, by Theorem 4.11, it follows that is -continuous at .

Definition 4.13. A function is called -continuous if the inverse image of each open set is -open.

Theorem 4.14. Every continuous function is -continuous.

Proof. The proof follows directly from Theorem 3.2.

Theorem 4.15. If is CC, then every function is -continuous.

Proof. The proof follows directly from Theorem 3.3.

Corollary 4.16. If is , then every function is -continuous.

By Corollary 4.16, it follows that the function where for is rational and for is irrational is -continuous. On the other hand, it is well known that this function is discontinuous every where. Also, by Theorem 4.4, is not -continuous.

Theorem 4.17. Let be a function with being a hereditarily compact space. Then is continuous if and only if is -continuous.

Proof. The proof follows directly from Theorem 3.5.

By Theorem 4.17, it follows that the identity function is not -continuous. Therefore, this is an example of a -continuous function that is not -continuous.

We end this section by the following decomposition of continuity via -continuity and -continuity.

Theorem 4.18. A function is continuous if and only if it is -continuous and -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.