**Advances in Pure Mathematics** Vol.3 No.1(2013), Article ID:27001,7 pages DOI:10.4236/apm.2013.31007

Bc-Open Sets in Topological Spaces

Department of Mathematics, Faculty of Science, University of Zakho, Zakho, Iraq

Email: hariwan_math@yahoo.com

Received August 1, 2012; revised September 24, 2012; accepted October 3, 2012

**Keywords:** Closed; b-Open; Bc-Open

ABSTRACT

In this paper, we introduce a new class of b-open sets called Bc-open, this class of sets lies strictly between the classes of θ-semi open and b-open sets. We also study its fundamental properties and compare it with some other types of sets and we investigate further topological properties of sets and we introduce and investigate new class of space named Bccompact.

1. Introduction

In 1937, regular open sets were introduced and used to define the semi-regularization space of a topological space. Throughout this paper, and stand for topological spaces with no separation axioms assumed unless otherwise stated. For a subset A of X, the closure of A and the interior of A will be denoted by and respectively. Stone [1] defined a subset A of a space X to be a regular open if . Norman Levine [2] defined a subset A of a space X to be a semi-open if, or equivalently, a set A of a space X will be termed semiopen if and only if there exists an open set such that. Mashhour et al. [3] defined a subset A of a space X to be a preopen if. Njastad [4] defined a subset A of a space X to be an -open if. The complement of a semi-open (resp., regular open) set is said to be semi-closed [5] (resp., regular closed). The intersection of all semi-closed sets of X containing A is called the semi-closure [6] of A. The union of semi-open sets of X contained in A is called the semi-interior of A. Joseph and Kwack [7] introduced the concept of -semi open sets using semi-open sets to improve the notion of -closed spaces. Also Joseph and Kwack [7] introduced that a subset A of a space X is called -semi-open if for each, there exists a semi-open set such that. It is well-known that, a space X is called if to each pair of distinct points x, y of X, there exists a pair of open sets, one containing x but not y and the other containing y but not x, as well as is if and only if for any point, the singleton set is closed. A space X is regular if for each and each open set G containing x, there exists an open set H such that

. Ahmed [8] defined a topological space to be s^{**}-normal if and only if for every semi-closed set F and every semi-open set G containing F, there exists an open set H such that . In 1968, Velicko [9], defined the concepts of -open and -open as, a subset A of a space X is called -open (resp., -open) if for each, there exists an open set _{ }such that (resp.,). Di Maio and Noiri [10] introduced that a subset A of a space X is called semi--open if for each, there exists a semi-open set G such that. The family of all open (resp., semi-open, -open, preopen, θ-semi-open, semi-θ-open, θ-open, δ-open, regular open, semi-closed and regular closed) subsets of a topological space are denoted by (resp., , , , , , , , , and).

Definition 1.1. [11] A subset A of a space X is called b-open if. The family of all b-open subsets of a topological space is denoted by or (Briefly.).

In 1999, J. Dontchev and T. Noiri [12] have shown the following lemma:

Lemma 1.2. For a subset A of a space, the following conditions are equivalent:

1)

2)

3)

4)

Theorem 1.3. [13] If is s^{**}-normal, then

.

We recall that a topological space X is said to be extremally disconnected [14] if is open for every open set G of X.

Definition 1.4. [15] A space X is called locally indiscrete if every open subset of X is closed.

Theorem 1.5. [13] A space X is extremally disconnected if and only if.

Theorem 1.6. [15] A space is extremally disconnected if and only if.

2. Bc-Open Sets

In this section, we introduce a new class of b-open sets called Bc-open sets in topological spaces.

Definition 2.1. A subset A of a space X is called Bcopen if for each, there exists a closed set F such that. The family of all Bc-open subsets of a topological space is denoted by or (Briefly.).

Proposition 2.2. A subset A of a space X is Bc-open if and only if A is b-open and it is a union of closed sets. That is where A is b-open set and is closed sets for each.

Proof. Obvious.

It is clear from the definition that every Bc-open subset of a space X is b-open, but the converse is not true in general as shown by the following example.

Example 2.3. Consider with the topology. Then the family of closed sets are:. We can find easily the following families:

and

.

Then but

The next example notices that a Bc-open set need not be a closed set.

Example 2.4. Consider the space R with usual topology, if such that, then is Bc-open set, but it is not closed.

The following result shows that the arbitrary union of Bc-open sets in a topological space is Bc-open.

Proposition 2.5. Let be a collection of Bc-open sets in a topological space X. Then

is Bc-open.

Proof. Let be a Bc-open set for each, then is -open and hence is b-open. Let, there exist such that. Since is b-open for each, there exists a closed set such that

so Therefore, is Bc-open set.

The following example shows that the intersection of two Bc-open sets need not be Bc-open set.

Example 2.6. Consider the space as in example 2.3, There and, but

From the above example we notice that the family of all Bc-open subsets of a space X is a supratopology and need not be a topology in general.

The following result shows that the family of all Bcopen sets will be a topology on X.

Proposition 2.7. If the family of all b-open sets of a space X is a topology on X, then the family of Bc-open is also a topology on X.

Proof. Clearly and by Proposition 2.5 the union of any family of Bc-open sets is Bc-open. To complete the proof it is enough to show that the finite intersection of Bc-open sets is Bc-open set. Let A and B be two Bc-open sets then A and B are -open sets. Since is a topology on X, so is b-open. Let, then and, so there exists F and E such that and this implies that. Since any intersection of closed sets is closed, is closed set. Thus is Bc-open set. This completes the proof.

Proposition 2.8. The set A is Bc-open in the space if and only if for each, there exists a Bcopen set B such that.

Proof. Assume that A is Bc-open set in the, then for each, put is Bc-open set containing x such that.

Conversely, suppose that for each, there exists a Bc-open set B such that, thus where for each x, therefore A is Bc-open set.

In the following proposition, the family of b-open sets is identical to the family of Bc-open sets.

Proposition 2.9. If a space X is -space, then the families

Proof. Let A be any subset of a space X and , if, then , then for each. Since a space X is, then every singleton is closed set and hence. Therefore. Hence, but

generally, therefore

.

Proposition 2.10. Every -semi open set of a space X is Bc-open set.

Proof. Let A be a -semi open set in X, then for each, there exists a semi-open set G such that

, so for each implies that which is semi-open set and is a union of closed sets, by Proposition 2.2, A is Bc-open set.

The following example shows that the converse of the above Proposition may not be true in general.

Example 2.11. Since a space X with cofinite topology is T1, and then the family of b-open and Bc-open sets are identical. Hence any open set G is Bc-open but not - semi open.

The proof of the following corollaries is clear from their definitions.

Corollary 2.12. Every -open set is Bc-open.

Corollary 2.13. Every regular-closed is Bc-open set.

Proposition 2.14. If a topological space is locally indiscrete, then.

Proof. Let A be any subset of a space X and , if, then. If, then. Since X is locally indiscrete, then is closed and hence, this implies that for each,. Therefore, A is Bc-open set. Hence.

Remark 2.15. Since every open set is semi-open, it follows that if a topological space is or locally indiscrete, then

Proposition 2.16. Let be a topological space, if X is regular, then

Proof. Let A be any subset of a space X, and A is open, if, then. If, since X is regular, so for each, there exists an open set G such that. Thus we have . Since and hence, therefore.

Proposition 2.17. Let be an extremally disconnected space. If, then.

Proof. Let. If, then . If. Since a space X is extremally disconnected, then by Theorem 1.5,. Hence. But in general. Therefore,.

Corollary 2.18. Letbe an extremally disconnected space. If, then.

Proof. The proof is directly from Proposition 2.28 and the fact that

Proposition 2.19. Letbe an s^{**}-normal space. If, then.

Proof. Let. If, then . If, since a space X is s^{**}-normal, then by Theorem 1.3,. Hence . But in general. Therefore,.

Proposition 2.20. For any subset A of a space and. The following conditions are equivalent:

1) A is regular closed.

2) A is closed and Bc-open.

3) A is closed and b-open.

4) A is α-closed and b-open.

5) A is pre-closed and b-open.

Proof. Follows from Lemma 1.2.

Definition 2.21. A subset B of a space X is called Bcclosed if is Bc-open. The family of all Bc-closed subsets of a topological space is denoted by or (Briefly,).

Proposition 2.22. A subset B of a space X is Bc-closed if and only if B is a b-closed set and it is an intersection of open sets.

Proof. Clear.

Proposition 2.23. Let be a collection of Bc-closed sets in a topological space X. Then is Bc-closed.

Proof. Follows from Proposition 2.5.

The union of two Bc-closed sets need not be Bc-closed as is shown by the following counterexample.

Example 2.24. In Example 2.3, the family of Bcclosed subset of X is:. Here and, but

.

All of the following results are true by using complement.

Proposition 2.25. If a space X is, then

Proposition 2.26. For any subset B of a space X. If, then.

Corollary 2.27. Each -closed set is Bc-closed.

Corollary 2.28. Each regular open set is Bc-closed.

Proposition 2.29. If a topological space is locally indiscrete, then.

Proposition 2.30. Let be a topological space, if X is regular or locally indiscrete, then the family of closed sets is a subset of the family of Bc-closed sets.

Proposition 2.31. Let be any extremally disconnected space. If, then.

Corollary 2.32. Let be an extremally disconnected space. If, then.

Proposition 2.33. Let be a s^{**}-normal space. If, then.

Proposition 2.34. For any subset B of a space and. The following conditions are equivalent:

1) B is regular open.

2) B is open and Bc-closed.

3) B is open and b-closed.

4) B is α-open and b-closed.

5) B is preopen and b-closed.

Diagram 1 shows the relations among, , , , , , , and.

Diagram 1.

3. Some Properties of Bc-Open Sets

In this section, we define and study topological properties of Bc-neighborhood, Bc-interior, Bc-closure and Bcderived of a set using the concept of Bc-open sets.

Definition 3.1. Let be a topological space and, then a subset N of X is said to be Bc-neighborhood of, if there exists a -open set U in X such that.

Proposition 3.2. In a topological space, a subset A of X is Bc-open if and only if it is a Bcneighbourhood of each of its points.

Proof. Let be a Bc-open set, since for every and A is Bc-open. this shows A is a Bc-neighborhood of each of its points.

Conversely, suppose that A is a Bc-neighborhood of each of its points. Then for each, there exists such that. Then

. Since each Bx is Bc-open. It follows that A is Bc-open set.

Proposition 3.3. For any two subsets A, B of a topological space and, if A is a Bc-neighborhood of a point, then B is also Bc-neighborhood of the same point.

Proof. let A be a Bc-neighborhood of, and, then by Definition 2.1, there exists a Bc-open set U such that, this implies that B is also a Bc-neighborhood of x.

Remark 3.4. Every Bc-neighborhood of a point is bneighborhood, it follows from every Bc-open set is bopen.

Definition 3.5. Let A be a subset of a topological space, a point is said to be Bc-interior point of, if there exist a Bc-open set such that. The set of all Bc-interior points of A is called Bc-interior of A and is denoted by.

Some properties of the Bc-interior of a set are investigated in the following theorem.

Theorem 3.6. For subsets A, B of a space X, the following statements hold.

1) is the union of all Bc-open sets which are contained in A.

2) is -open set in X.

3) is -open if and only if.

4).

5) and.

6).

7) If, then.

8) If, then.

9)

10).

Proof. 7) Let and, then by Definition 3.5, there exists a Bc-open set such that implies that. thus .

The other parts of the theorem can be proved easily.

Proposition 3.7. For a subset A of a topological space, then.

Proof. This follows immediately since all -open set is b-open.

Definition 3.8. Let A be a subset of a space X. A point is said to be Bc-limit point of A if for each Bcopen set U containing. The set of all Bc-limit points of A is called a Bc-derived set of A and is denoted by.

Proposition 3.9. Let A be a subset of X, if for each closed set F of X containing x such that , then a point is Bc-limit point of A.

Proof. Let U be any Bc-open set containing x, then for each, there exists a closed set F such that. By hypothesis, we have . Hence. Therefore, a point is Bc-limit point of A.

Some properties of Bc-derived set are stated in the following theorem.

Theorem 3.10. Let A and B be subsets of a space X. Then we have the following properties:

1).

2) If, then.

3) If, then.

4).

5).

6).

7).

Proof. We only prove 6), 7), and the other parts can be proved obviously.

6) If and is a Bc-open set containing x, then. Let . Then, since and. Let. Then, for and. Hence,. Therefore,

7) Let. If, the result is obvious. So, let, then, for Bcopen set containing. Thus, or. Now, it follows similarly from 1) that . Hence,. Therefore, in any case,.

Corollary 3.11. For a subset A of a space X, then.

Proof. It is sufficient to recall that every Bc-open set is b-open.

Definition 3.12. For any subset A in the space X, the Bc-closure of A, denoted by, is defined by the intersection of all Bc-closed sets containing A.

Proposition 3.13. A subset A of a topological space X is Bc-closed if and only if it contains the set of its Bclimit points.

Proof. Assume that A is Bc-closed and if possible that x is a Bc-limit point of A which belongs to, then is Bc-open set containing the Bc-limit point of A, therefore, which is a contradiction.

Conversely, assume that A contains the set of its Bclimit points. For each, there exists a Bc-open set U containing x such that, that is by Proposition 2.8, is Bc-open set and hence A is Bc-closed set.

Proposition 3.14. Let A be a subset of a space X, then.

Proof. Since, then.

On the other hand. To show that , since is the smallest Bc-closed set containing A, so it is enough to prove that is Bc-closed. Let. This implies thatand. Since, there exists a Bc-open set of which contains no point of A other than x but. So Gx contains no point of A, which implies. Again, Gx is a Bc-open set of each of its points. But as Gx does not contain any point of A, nopoint of Gx can be a Bc-limit point of A. Therefore, nopoint of Gx can belong to. This implies that. Hence, it follows that

Therefore, is Bc-closed. Hence

. Thus

.

Corollary 3.15. Let A be a set in a space X. A point is in the Bc-closure of A if and only if for every Bc-open set U containing x.

Proof. Let. Then, where F is Bc-closed with. So and is a Bc-open set containing x and hence

Conversely, suppose that there exists a Bc-open set containing x with. Then and is a Bc-closed. Hence.

Proposition 3.16. Let A be any subset of a space X. If for every closed set F of X containing x, then the point x is in the Bc-closure of A.

Proof. Suppose that U be any Bc-open set containing x, then by Definition 2.1, there exists a closed set F such that. So by hypothesis implies for every Bc-open set U containing x. Therefore.

Here we introduce some properties of Bc-closure of the sets.

Theorem 3.17. For subsets A, B of a space X, the following statements are true.

1) The Bc-closure of A is the intersection of all Bcclosed sets containing A.

2)

3)_{ }-closed set in X 4) is Bc-closed set if and only if

5)

6) and.

7) If, then

8) If

9)

10).

Proof. Obvious.

Proposition 3.18. For any subset A of a topological space X. The following statements are true.

1)

2)

3)

4)

Proof. We only prove 1), the other parts can be proved similarly. For any point, implies that, then for each containing, then. Thus.

Conversely, by reverse the above steps, we can prove this part.

Remark 3.19. If is a subset of a topological space X. Then

Proof. Obvious.

4. Bc-Compactness

In this section, we introduce and investigate new class of space named Bc-compact.

Definition 4.1. A filter base in a topological space Bc-converges to a point if for every Bcopen set V containing x, there exists an such that.

Definition 4.2. A filter base in a topological space Bc-accumulates to a point if, for every -open set V containing and every.

Proposition 4.3. Let be a filter base in a topological space. If Bc-converges to a point, then rc-converges to a point x.

Proof. Suppose that Bc-converges to a point. Let V be any regular closed set containing x, then. Since Bc-converges to a point, there exists an such that. This shows that rc-converges to a point x.

In general the converse of the above proposition is not necessarily true, as the following example shows.

Example 4.4. Consider the space. Let . Then rc-converges to 0, but does not Bc-converges to 0, because the set is Bc-open containing 0, there exist no such that.

Corollary 4.5. Let be a filter base in a topological space. If Bc-accumulates to a point, then rc-accumulates to a point x.

Proof. Similar to Proposition 4.3.

Proposition 4.6. Let be a filter base in a topological space and E is any closed set containing. If there exists an _{ }such that, then Bcconverges to a point.

Proof. Let be any Bc-open set containing, then for each, there exists a closed set E such that. By hypothesis, there exists an such that which implies that. Hence Bc-converges to a point.

Proposition 4.7. Let be a filter base in a topological space and E is any closed set containing, such that for each, then is Bcaccumulation to a point.

Proof. The proof is similar to Proposition 4.6.

Definition 4.8. We say that a topological space is Bc-compact if for every Bc-open cover of X, there exists a finite subset of such that.

Theorem 4.9. If every closed cover of a space X has a finite subcover, then X is Bc-compact.

Proof. Let be any Bc-open cover of X, and, then for each, there exists a closed set such that. So the family is a cover of X by closed set, then by hypothesis, this family has a finite subcover such that

.

Therefore,. Hence X is Bccompact.

Proposition 4.10. If a topological space is bcompact, then it is Bc-compact.

Proof. Let be any Bc-open cover of X. Then is b-open cover of X. Since X is bcompact, there exists a finite subset of such that. Hence X is Bc-compact.

Proposition 4.11. Every Bc-compact -space is bcompact.

Proof. Suppose that X is and Bc-compact space. Let be any b-open cover of X. Then for every, there exists such that. Since X is, by Since X is Bc-compact, so there exists a finite subset of in X such that . Hence X is b-compact.

The next corollary is an immediate consequence of Proposition 4.10 and 4.11.

Corollary 4.12. Let X be a -space. Then X is Bccompact if and only if X is b-compact.

Proposition 4.13. Let a topological space be locally indiscrete. If X is Bc-compact then X is s-compact.

Proof. Follows from Proposition 2.14.

Proposition 4.14. If a topological space is Bccompact, then it is rc-compact.

Proof. Let be any regular closed cover of X. Then is a Bc-open cover of X. Since X is -compact, there exists a finite subset of such that. Hence X is rc-compact.

Proposition 4.15. Let a topological space be regular. If X is Bc-compact, then it is compact.

Proof. Let be any open cover of X. By Proposition 2.16, forms a Bc-open cover of X. Since X is Bc-compact, there exists a finite subset of such that. Hence X is compact.

Proposition 4.16. Let X be an almost regular space. If X is Bc-compact, then it is nearly compact.

Proof. Let be any regular open cover of X. Since X is almost regular space, then, for each and regular open there exists an open set Gx such that But is regulaclosed for each, this implies that the family is Bc-open cover of X, since X is Bc - compact, then there exists a subfamily

such that

. Thus X is nearly compact.

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