() CUBO A Mathematical Journal Vol.17, No¯ 01, (99–106). March 2015 Semi Open sets in bispaces Amar Kumar Banerjee Department of Mathematics, Burdwan University, Burdwan-713104, W.B., India. akbanerjee1971@gmail.com Pratap Kumar Saha Behala College, Kolkata,W.B., India. pratapsaha2@gmail.com ABSTRACT The notions of semi open sets in a topological space were introduced by N.Levine in 1963. Here we study the same using the idea of τ1(τ2) semi open sets with respect to τ2(τ1), pairwise semi open sets in a more general structure of a bispace and investigate how far several results as valid in a bitopological space are affected in bispaces. RESUMEN Las nociones de conjuntos semiabiertos en un espacio topológico se introdujeron por N. Levine en 1963. Aqúı estudiamos lo mismo usando la idea de conjuntos semiabiertos τ1(τ2) respecto de conjuntos abiertos semiabiertos dos a dos τ2(τ1), en una estructura más general de biespacio e investigamos cómo varios resultados válidos en un espacio bitopológico cambian en biespacios. Keywords and Phrases: bispaces, semi open sets, τ1 semi open sets with respect to τ2. 2010 AMS Mathematics Subject Classification: 54A05, 54E55, 54E99 100 Amar Kumar Banerjee & Pratap Kumar Saha CUBO 17, 1 (2015) 1 Introduction The notion of a σ space or simply a space was introduced by A.D.Alexandroff [1] in 1940 generalising the idea of a topological space where only countable unions of open sets were taken to be open. In 2001 the idea of space was used by Lahiri and Das [8] to generalise the notion of a bitopological space to a bispace. N.Levine [9] introduced the concept of semi open sets in a topological space in 1963 and this idea was generalised by S.Bose [3] in the setting of a bitopological space (X, P, Q) using the idea of P(Q) semi open sets with respect to Q(P) etc. Later the same was studied in a space by Lahiri and Das [7] and they critically took the matter of generalisation in this setting. Here we have studied the concept of τ1(τ2) semi open sets with respect to τ2(τ1) and some other properties in the setting of a bispace and have shown with typical examples how far several results as valid in[3] are affected in bispaces. Also we have given a necessary and sufficient condition for a bispace to be a bitopological space in terms of τ1(τ2) semi open sets with respect to τ2(τ1). 2 Preliminaries Definition 2.1. [1] A set X is called an Alexandroff space or simply a space if in it is chosen a system of subsets F satisfying the following axioms: (1) The intersection of a countable number of sets from F is a set in F. (2) The union of a finite number of sets from F is a set in F. (3) The void set φ is a set in F. (4) The whole set X is a set in F. . Sets of F are called closed sets. Their complementary sets are called open sets. It is clear that instead of closed set in the definition of the space, one may put open sets with subject to the condition of countable summability, finite intersectibility and the condition that X and φ should be open. The collection of all such open sets will sometimes be denoted by τ and the space by (X, τ). Note that, in general τ is not a topology as can be easily seen by taking X = R, the set of real numbers and τ as the collection of all Fσ sets in R. Definition 2.2. [1] To every set M of (X, τ) we correlate its closure M = the intersection of all closed sets containing M. Sometimes the closure of a set M will be denoted by τclM or simply clM when there is no confusion about τ. Generally the closure of a set in a space is not a closed set. From the axioms, it easily follows that 1) M ∪ N = M ∪ N; 2) M ⊂ M ; 3) M = M ; 4) φ = φ. CUBO 17, 1 (2015) Semi open sets in bispaces 101 Definition 2.3. [7] The interior of a set M in (X, τ) is defined as the union of all open sets contained in M and is denoted by τintM or intM when there is no confusion. Definition 2.4. [6] A non empty set X on which are defined two arbitrary topologies P, Q is called a bitopological space and denoted by (X, P, Q). Definition 2.5. [8] Let X be a nonempty set. If τ1 and τ2 be two collections of subsets of X such that (X, τ1) and (X, τ2) are two spaces, then X is called a bispace and is denoted by (X, τ1, τ2). 3 Pairwise Semi Open sets Definition 3.1. (cf. Definition 1[3] ): Let (X, τ1, τ2) be a bispace. We say that a subset A of X is τ1 semi open with respect to τ2 (in short τ1 s.o.w.r.to τ2 ) if and only if there exists a τ1 open set O such that O ⊂ A ⊂ τ2clO. Similarly A ⊂ X, is τ2 semi open with respect to τ1 (in short τ2 s.o.w.r.to τ1) if and only if there exists a τ2 open set O such that O ⊂ A ⊂ τ1clO. We say that A is pairwise semi open if and only if it is both τ1 s.o.w.r.to τ2 and τ2 s.o.w.r.to τ1. Note that a τ1(τ2) open set is τ1(τ2) s.o.w.r.to τ2(τ1). Throughout our discussion, (X, τ1, τ2) or simply X stands for a bispace, R stands for the set of real numbers, Q for the set of rational numbers and N stands for the set of natural numbers and sets are always subsets of X unless otherwise stated. Theorem 3.2. Let (X, τ1, τ2) be a bispace. Let A ⊂ X, and A is τ1 s.o.w.r.to τ2 then τ2clA = τ2cl(τ1intA). Proof. Let A is τ1 s.o.w.r.to τ2 then there exists a τ1 open set O such that O ⊂ A ⊂ τ2clO. Also O ⊂ τ1intA. Therefore, A ⊂ τ2clO ⊂ τ2cl(τ1intA) and hence τ2clA ⊂ τ2cl(τ2cl(τ1intA)) = τ2cl(τ1intA). Also τ2cl(τ1intA) ⊂ τ2clA. Therefore, τ2clA = τ2cl(τ1intA). Corollary 3.3. If A is τ1 s.o.w.r.to τ2 and A 6= φ then τ1intA 6= φ. Corollary 3.4. Let A is τ1 s.o.w.r.to τ2 and A ⊂ B then A ⊂ τ2cl(τ1intB). If (X, τ1, τ2) is a bitopological space then converse part of the theorem 3.2 also holds which is seen in [3] . But in a bispace, this may not be true as shown below: Example 3.5. Let X = [0, 2] and {Gi} be the collection of all countable subsets of irrational numbers in [0, 1]. Let τ1 be the collection of all sets of the form Gi ∪ { √ 2} together with X and φ, and τ2 be the collection of all sets Gi together with X and φ. Then (X, τ1, τ2) is a bispace. Now consider a subset A = [0, 1] ∪ { √ 2} then τ2clA = X and τ1intA is set of all irrational numbers in [0, 1] together with √ 2 and hence τ2cl(τ1intA) = X. Therefore, τ2clA = τ2cl(τ1intA). But for any τ1 102 Amar Kumar Banerjee & Pratap Kumar Saha CUBO 17, 1 (2015) open set G(6= X, φ), τ2clG = G ∪ Q1 ∪ [1, 2], where Q1 is the set of all rational numbers in [0, 1]. Clearly τ2clG does not contain A. Therefore, there does not exist any τ1 open set G satisfying G ⊂ A ⊂ τ2clG. So A is not τ1 s.o.w.r.to τ2. However we observe in the following theorem that the converse part of theorem 3.2 holds under an additional condition. Theorem 3.6. In a bispace (X, τ1, τ2), let τ2clA = τ2cl(τ1intA). Then A is τ1 s.o.w.r.to τ2 for any subset A of X if the condition C1 is satisfied. C1 : Arbitrary union of τ1 open sets is τ1 s.o.w.r.to τ2. Proof. Let O = τ1intA. Then by the condition C1, O is τ1 s.o.w.r.to τ2. So there exists a τ1 open set G such that G ⊂ O ⊂ τ2clG. Now since τ2clO = τ2cl(τ1intA) = τ2clA and O ⊂ τ2clG, it follows that τ2clO ⊂ τ2cl(τ2clG) = τ2clG and hence τ2clA ⊂ τ2clG. Therefore, G ⊂ O ⊂ A ⊂ τ2clA ⊂ τ2clG and so A is τ1 s.o.w.r.to τ2. Remark 3.7. We see in the Example 3.8 below that there is a bispace which is not a bitopological space where the condition C1 holds good. Example 3.8. Let X = [0, 2], τ1 be the collection of all sets Gi together with X and φ and τ2 be the collection of all sets Fi together with X and φ where {Gi} and {Fi} are the collection of all countable subsets of irrational numbers in [0, 1] and [1, 2] respectively. Then (X, τ1, τ2) is a bispace but not a bitopological space. Now consider all τ1 open sets {Gi}. Then ∪Gi is the set of all irrational numbers in [0, 1] which is not τ1 open. But since, for any τ1 open set Gi, τ2clGi = [0, 1] ∪ Q2 where Q2 is set of all rational numbers in [1, 2]. It follows that Gi ⊂ ∪Gi ⊂ τ2clGi. This implies that ∪Gi is τ1 s.o.w.r.to τ2 although it is not τ1 open. Theorem 3.9. Countable union of τ1 s.o.sets w.r.to τ2 is τ1 s.o.w.r.to τ2. Proof. Let {An : n ∈ N} be a countable collection of τ1 s.o.sets w.r.to τ2. Then for each n ∈ N there exists a τ1 open set On such that On ⊂ An ⊂ τ2clOn. This implies that ∪{On : n ∈ N} ⊂ ∪{An : n ∈ N} ⊂ ∪{τ2clOn : n ∈ N} ⊂ τ2cl(∪{On : n ∈ N}) i.e, O ⊂ ∪{An : n ∈ N} ⊂ τ2clO, where O = ∪{On : n ∈ N}, a τ1 open set. Hence ∪{An : n ∈ N} is τ1 s.o.w.r.to τ2. Remark 3.10. In [3] it was proved that in a bitopological space arbitrary union of τ1 s.o.sets w.r.to τ2 is τ1 s.o.w.r.to τ2. But this may not be true in a bispace as shown in the Example 3.11 below. Example 3.11. Consider X = R. Let τ1 open sets are X, φ and all sets Gi where {Gi} is the collection of all countable subsets of irrational numbers in R and τ2 open sets are X, φ and all Fσ sets in R. Clearly τ2 closed sets are the Gδ sets. So for any subset G in R we have τ2clG = G. Therefore, for any set A which is τ1 s.o.w.r.to τ2 in X, there exists a τ1 open set Gi such that Gi ⊂ A ⊂ τ2clGi = Gi. This implies that A = Gi, i.e., A is τ1 open set. So τ1 open sets are the only τ1 s.o.sets w.r.to τ2. Since the union of all τ1 open sets Gi (Gi 6= X) is precisely the set CUBO 17, 1 (2015) Semi open sets in bispaces 103 of all irrational numbers in R which is not τ1 open, it follows that arbitrary union of τ1 s.o.sets w.r.to τ2 may not be τ1 s.o.w.r.to τ2. However the additional condition C1 ensures the result in theorem 3.9 for arbitrary union. Theorem 3.12. Arbitrary union of τ1 s.o.sets w.r.to τ2 is τ1 s.o.w.r.to τ2 if and only if the condition C1 is satisfied. Proof. Assume first that the arbitrary union of τ1 s.o.sets w.r.to τ2 is τ1 s.o.w.r.to τ2. Since every τ1 open set is τ1 s.o.w.r.to τ2, arbitrary union of τ1 open sets is τ1 s.o.w.r.to τ2, i.e., the condition C1 holds. Next assume that the condition C1 holds. Let {Ai} be an arbitrary collection of τ1 s.o.sets w.r.to τ2 and A = ∪Ai. For each i, there exists a τ1 open set Gi such that Gi ⊂ Ai ⊂ τ2clGi. Therefore, ∪Gi ⊂ ∪Ai = A ⊂ ∪τ2clGi ⊂ τ2cl(∪Gi). Since ∪Gi, by assumption, is τ1 s.o.w.r.to τ2, there exists a τ1 open set G such that G ⊂ ∪Gi ⊂ τ2clG. Therefore, G ⊂ ∪Gi ⊂ A ⊂ ∪τ2clGi ⊂ τ2cl(∪Gi) ⊂ τ2clτ2clG = τ2clG. This proves that A is τ1 s.o.w.r.to τ2. Theorem 3.13. Let A be τ1 s.o.w.r.to τ2 in a bispace (X, τ1, τ2) and let A ⊂ B ⊂ τ2clA. Then B is τ1 s.o.w.r.to τ2. Proof. Since A is τ1 s.o.w.r.to τ2, there exists a τ1 open set O such that O ⊂ A ⊂ τ2clO. Therefore, O ⊂ A ⊂ B ⊂ τ2clA ⊂ τ2cl(τ2clO) = τ2clO and hence B is τ1 s.o.w.r.to τ2. Theorem 3.14. Let (X, τ1, τ2) be a bispace and let A ⊂ Y ⊂ X. If A is pairwise s.o. in X, it is pairwise s.o. in Y. Proof. Let A be τ1 s.o.w.r.to τ2. Then there exists a τ1 open set O such that O ⊂ A ⊂ τ2clO. Let OY = Y ∩ O which is τ1 open in Y. So OY = Y ∩ O ⊂ Y ∩ A ⊂ Y ∩ τ2clO = τ2clOY in Y. Interchanging the role of τ1 and τ2 we get the result. We denote the class of all τ1 s.o.w.r.to τ2 by τ1S.O.(X)τ2. Theorem 3.15. Let B = {Bα} be a collection of subsets of X such that (i) τ1 ⊂ B and (ii) B ∈ B and B ⊂ D ⊂ τ2clB imply D ∈ B, then τ1S.O.(X)τ2 ⊂ B. Proof. Let A ∈ τ1S.O.(X)τ2 , then there exists a τ1 open set O such that O ⊂ A ⊂ τ2clO. Therefore, O ∈ B and O ⊂ A ⊂ τ2clO imply that A ∈ B and hence the result follows. We denote the set {τ1intA : A ∈ τ1S.O.(X)τ2} by τ1int(τ1S.O.(X)τ2 ). Interchanging the role of τ1 and τ2 we may denote other such classes at our will. From the construction, it is obvious that τ1 ⊂ τ1int(τ1S.O.(X)τ2). But in general, τ1 may not be equal to τ1int(τ1S.O.(X)τ2) as shown in the following example. 104 Amar Kumar Banerjee & Pratap Kumar Saha CUBO 17, 1 (2015) Example 3.16. Let (X, τ1, τ2) be the bispace as in Example 3.8. Now for any τ1 open set Gi, τ2clGi = [0, 1] ∪ Q2, where Q2 is set of all rational numbers in [1, 2]. Let A = [0, 1]. Then for any τ1 open set Gi, Gi ⊂ A ⊂ [0, 1] ∪ Q2 = τ2clGi. This implies that A is τ1 s.o.w.r.to τ2, i.e., A ∈ τ1S.O.(X)τ2. But τ1intA is the set of all irrational numbers in [0,1] which is not τ1 open. However equality τ1 = τ1int(τ1S.O.(X)τ2) holds if an additional condition holds. Theorem 3.17. In a bispace (X, τ1, τ2), τ1 = τ1int(τ1S.O.(X)τ2) if and only if the condition C2 is satisfied. C2 : For any A ⊂ X which is τ1 s.o.w.r.to τ2, there exists a maximal τ1 open set O such that O ⊂ A ⊂ τ2clO. Proof. First assume that τ1 = τ1int(τ1S.O.(X)τ2), and let A be any subset of X which is τ1 s.o.w.r.to τ2. Then τ1intA ∈ τ1. Also by theorem 3.2, A ⊂ τ2cl(τ1intA). Again if G is any τ1 open set satisfying G ⊂ A ⊂ τ2clG, then G ⊂ τ1intA. Hence τ1intA is the maximal τ1 open set contained in A such that τ1intA ⊂ A ⊂ τ2cl(τ1intA). Taking O = τ1intA, we get O ⊂ A ⊂ τ2clO. Conversely let A ∈ τ1S.O.(X)τ2. By the condition, there exists a maximal τ1 open set O such that O ⊂ A ⊂ τ2clO......(1). If possible let O 6= τ1intA. Then there exists a τ1 open set G ⊂ A such that G is not contained in O . Since O ∪ G is τ1 open set and O ∪ G ⊂ A ⊂ τ2clO ⊂ τ2cl(O ∪ G), this contradicts that O is maximal satisfying the condition (1). Hence τ1intA = O and so τ1intA is a τ1 open set, i.e., τ1intA ∈ τ1. Therefore τ1int(τ1S.O.(X)τ2 ) ⊂ τ1 and consequently τ1 = τ1int(τ1S.O.(X)τ2). Remark 3.18. We see that there is a bispace which is not bitopological space where the condition C2 holds. For, consider the bispace (X, τ1, τ2) as in Example 3.11 where the τ1 open sets are the only τ1 s.o.sets w.r.to τ2. So for any set A which τ1 s.o.w.r.to τ2 in X, there exists a maximal τ1 open set O(= A) such that O ⊂ A ⊂ τ2clO. We now give a necessary and sufficient condition in terms of semi open sets for a bispace to be a bitopological space. Theorem 3.19. A bispace (X, τ1, τ2) is a bitopological space if and only if following condition holds: (i) arbitrary union of τ1(τ2) s.o.sets w.r.to τ2(τ1) is τ1(τ2) s.o.w.r.to τ2(τ1) (ii) τ1 = τ1int(τ1S.O.(X)τ2 ) and τ2 = τ2int(τ2S.O.(X)τ1). Proof. If (X, τ1, τ2) is a bitopological space then (i) holds[3] . For (ii), let O ∈ τ1. Then O ∈ τ1S.O.(X)τ2 and since O = τ1intO, O ∈ τ1int(τ1S.O.(X)τ2). Therefore, τ1 ⊂ τ1int(τ1S.O.(X)τ2 ). On the other hand, let O ∈ τ1int(τ1S.O.(X)τ2). Then O = τ1intA for some A ∈ τ1S.O.(X)τ2 and hence O ∈ τ1. Therefore, τ1int(τ1S.O.(X)τ2 ) ⊂ τ1. Therefore, τ1 = τ1int(τ1S.O.(X)τ2). Similarly we can prove that τ2 = τ2int(τ2S.O.(X)τ1). Conversely, it suffices to show that an arbitrary union of τ1(τ2) open sets is τ1(τ2) open in CUBO 17, 1 (2015) Semi open sets in bispaces 105 (X, τ1, τ2). Let {Gi} be an arbitrary collection of τ1 open sets and G = ∪Gi. Each Gi being τ1 open is τ1 s.o.w.r.to τ2. So by (i) G is τ1 s.o.w.r.to τ2. Then by (ii) τ1intG is τ1 open. So τ1intG = G and similarly arbitrary union of τ2 open sets is τ2 open set and this proves the theorem. Definition 3.20. (cf. [7] ): Let (X, τ1, τ2) be a bispace. Two non empty subsets A and B are said to be (i) pairwise weakly separated if there exist a τ1 open set U and a τ2 open set V such that A ⊂ U, B ⊂ V, A ∩ V = φ, B ∩ U = φ. (ii) pairwise strongly separated if there exists a τ1 open set U and a τ2 open set V such that A ⊂ U, B ⊂ V, U ∩ V = φ. Definition 3.21. (cf.[7] ): A subset A in a bispace (X, τ1, τ2) is said to be pairwise connected if it can not be expressed as the unions of two pairwise weakly separated sets. Remark 3.22. In [3] it is proved that in a bitopological space (X, τ1, τ2) if A = O ∪ B where (i) O 6= φ is τ1 open (ii) A is pairwise connected and (iii) B ′ τ 1 , the derived set of B w.r.to τ1 is empty, then A is τ1 s.o.w.r.to τ2. But this is not true in a bispace as shown in the following example. Example 3.23. Let X = ([0, 1] − Q) ∪ { √ 2} and {Gi} be the collection of all countable subsets of [0, 1] − Q where Q is the set of all rational numbers. Let τ1 be the collection of all sets Gi ∪ { √ 2} together with X and φ, and τ2 be the collection of all sets Gi together with X and φ where Gi ∈ {Gi}. Then (X, τ1, τ2) is a bispace. Let A = ([0, 1 2 ] − Q) ∪ { √ 2}. Then A is pairwise connected, because if A = A1 ∪ A2, then at least one of A1 and A2 say A1 is uncountable and X is the only τ1 open set containing A1. Let G be a nonempty countable subset of [0, 1 2 ] − Q and O = G ∪ { √ 2}. Then A = O ∪ (A − O) where O 6= φ, is τ1 open. Also (A − O) ′ τ 1 , the set of all τ1 limit points of A − O is empty. Indeed if α ∈ [0, 1] − Q, then {α, √ 2} is a τ1 open set containing α satisfying {α, √ 2} ∩ ((A − O) − {α}) = φ. Again if α = √ 2 then for any p ∈ G, {p, α} is a τ1 open set containing α satisfying {p, α} ∩ ((A − O) − {α}) = φ. Thus no point of X can be a τ1 limit point of A − O. So all the conditions stated in the Remark 3.22 above are satisfied, but A is not τ1 s.o.w.r.to τ2, because if Gi ∪ { √ 2} is any τ1 open set contained in A, then τ2cl(Gi ∪ { √ 2}) = Gi ∪ { √ 2}. However the following theorem is true. Theorem 3.24. Let (X, τ1, τ2) be a bispace. If A = O ∪ B where (i) O(6= φ) is τ1 open (ii) A is pairwise connected (iii) there exists a τ2 closed set F1 ⊃ O such that B ∩ F1 ⊂ G ⊂ τ2clO for some τ1 open set G, then A is τ1 s.o.w.r.to τ2. Proof. We show that B ⊂ τ2clO. If B 6⊂ τ2clO then there exists a τ2 closed set F ⊃ O such that B 6⊂ F. Let B1 = B ∩ F, B2 = B − B1. Then B2 ⊂ X − F and B2 6= φ. Further, let B∗ 1 = B1 ∩ F1 and B∗∗1 = B1 ∩ (X − F1). Then A = O ∪ B = O ∪ B1 ∪ B2 = (O ∪ B ∗ 1 ) ∪ (B∗∗ 1 ∪ B2). Now B∗∗ 1 ∪ B2 ⊂ (X − F1) ∪ (X − F) = X − (F ∩ F1) = G2 (say), which is τ2 open. Since B∗1 = 106 Amar Kumar Banerjee & Pratap Kumar Saha CUBO 17, 1 (2015) B1 ∩F1 = B∩F∩F1 ⊂ B∩F1 ⊂ G, we have O∪B∗1 ⊂ O∪G = G1 (say), which is τ1 open set. 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