EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 2, 2017, 323-334 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global bTµ- compactness and bTµ - connectedness in supra topological spaces K.Krishnaveni1,∗, M.Vigneshwaran 2 1 Research Scholar, Department of Mathematics, Coimbatore, Tamilnadu, India. 2 Department of Mathematics, Coimbatore, Tamilnadu, India Abstract. In this paper we newly originate the notion of bTµ - compact space and inspected its several effects and characterizations. Also we newly originate and study the concept of bTµ - Lindelof spaces and Connected spaces. 2010 Mathematics Subject Classifications: 54D05, 54D20, 54D30 Key Words and Phrases: bTµ -open sets; bTµ -compact spaces; bTµ - Lindelof spaces and bTµ - connected spaces. 1. Introduction The supra topological spaces was introduced by Mashhour.et.al [6] in 1983. They studied S - continuous maps and S*- continuous maps. The supra b- open set and supra b-continuity was brought out by Sayed.et.al [8] in 2010. Recently Krishnaveni and Vig- neshwaran [4] came out with supra bT -closed sets and defined their properties. In 2013, Jamal M.Mustafa.et.al[3] came out with the concect of supra b- connected and supra b- Lindelof spaces. Now we bring up with the new concepts of supra bT -compact, supra bT - Lindelof, Countably supra bT - compact and supra bT -connected spaces and reviewed several properties for these concepts. 2. Preliminaries Definition 1 (6,8). A subfamily of µ of X is said to be a supra topology on X, if (i) X, φεµ (ii) if Aiεµ for all iε J then ∪Aiεµ. The pair (X,µ) is called supra topological space. The elements of µ are called supra open sets in (X,µ) and complement of a supra open set is called a supra closed set. ∗Corresponding author. Email addresses: krishnavenikaliswami@gmail.com (K.Krishna), vignesh.mat@gmail.com (M.Vignesh) http://www.ejpam.com 323 c© 2017 EJPAM All rights reserved. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 324 Definition 2 (6). (i) The supra closure of a set A is denoted by clµ(A) and is defined as clµ(A) = ∩{B : B is a supra closed set and A ⊆ B}. (ii) The supra interior of a set A is denoted by intµ(A) and defined as intµ(A) = ∪{B :B is a supra open set and A ⊇ B}. Definition 3 (8). Let (X,τ) be a topological spaces and µ be a supra topolgy on X. We call µ a supra topology associated with τ if τ ⊂ µ. Definition 4 (8). Let (X,µ) be a supra topological space. A set A is called a supra b-open set if A ⊆ clµ(intµ(A)) ∪ intµ(clµ(A)). The complement of a supra b-open set is called a supra b-closed set. Definition 5 (4). A subset A of a supra topological space(X,µ) is called bTµ-closed set if bclµ(A) ⊂ U whenever A⊂ U and U is Tµ- open in (X,µ). Definition 6 (4). Let (X,τ) and (Y,σ) be two topological spaces and µ be an associated supra topology with τ . A function f:(X,τ) →(Y, σ) is called bTµ - continuous if f−1(V) is bTµ - closed in (X,τ) for every supra closed set V of (Y,σ). Definition 7 (4). Let (X,τ) and (Y,σ) be two topological spaces and µ be an associated supra topology with τ . A function f:(X,τ)→(Y, σ) is called bTµ - irresolute if f−1(V) is bTµ - closed in (X,τ) for every bTµ - closed set V of (Y,σ). Definition 8 (4). A supra topological space (X,µ) is called bTT µ c - space,if every bTµ- closed set is supra closed set. Definition 9 (5). Let (X,τ) and (Y,σ) be two topological spaces and µ be an associated supra topology with τ . A function f:(X,τ) →(Y, σ) is called strongly bTµ - continuous if the inverse image of every bTµ-closed in Y is supra closed in X. Definition 10 (5). Let (X,τ) and (Y,σ) be two topological spaces and µ be an associated supra topology with τ . A function f:(X,τ) →(Y,σ) is called perfectly bTµ - continuous if the inverse image of every bTµ-closed in Y is both supra closed and supra open in X. 3. supra bT - Compactness Definition 11. A collection {Ai : i ∈ I} of bTµ - open sets in a supra topological space (X,µ) is called a bTµ-open cover of a subset B of X if B⊂ ⋃ {Ai : i ∈ I} holds. Definition 12. A supra topological space (X,µ) is bTµ - compact if every bTµ - open cover of X has a finite subcover. Definition 13. A subset B of a supra topological space (X,µ) is said to be bTµ - compact relative to (X,µ) if, for every collection {Ai : i ∈ I} of bTµ - open subsets of X such that B⊂ ⋃ {Ai : i ∈ I} there exist a finite subset Io of I such that B ⊆ ⋃ {Ai : i ∈ Io}. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 325 Definition 14. A subset B of a supra topological space (X,µ) is said to be bTµ - compact if B is bTµ - compact as a subspace of X. Theorem 1. Every bTµ - compact space is supra compact. Proof. Let {Ai : i ∈ I} be a supra open cover of (X,µ). By [4] {Ai : i ∈ I} is a bTµ - open cover of (X,µ). Since (X,µ) is bTµ - compact, bTµ - open cover {Ai : i ∈ I} of (X,µ) has a finite subcover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,µ) is a supra compact space. Theorem 2. Every bTµ - closed subset of a bTµ - compact space is bTµ - compact relative to X. Proof. Let A be a bTµ - closed subset of a supra topological space (X,µ). Then Ac is bTµ - open in (X,µ). Let S = {Ai : i ∈ I} be an bTµ - open cover of A by bTµ - open subset in (X,µ). Let S* = S∪Ac is a bTµ - open cover of (X,µ). That is X = ( ⋃ i∈I Ai) ⋃ Ac. By hypothesis (X,µ) is a bTµ- compact and hence S* is reducible to a finite sub cover of (X,µ) say X = Ai1 ∪ Ai2∪, · · · ,∪Ain ∪ Ac, Aik ∈ S∗. But A and Ac are disjoint. Hence A ⊂ Ai1 ∪Ai2∪, · · · ,∪Ain ∈ S. Thus a bTµ -open cover S of A contains a finite subcover. Hence A is bTµ- compact relative to (X,µ). Theorem 3. A bTµ - continuous image of a bTµ - compact space is supra compact. Proof. Let f: X → Y be a bTµ- continuous map from a bTµ - compact X onto a supra topological space Y. Let {Ai : i ∈ I} be a supra open cover of Y. Then f−1 {Ai : i ∈ I} is a bTµ - open cover of X, as f is bTµ - continuous. Since X is bTµ - compact, the bTµ - open cover of X, f−1 {Ai : i ∈ I} has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = ⋃n i=1 f −1(Ai), which implies f(X) = n⋃ i=1 (Ai), then Y = n⋃ i=1 (Ai). That is {A1, A2, · · · , An} is a finite sub cover of {Ai : i ∈ I} for Y. Hence Y is supra compact. Theorem 4. If a map f : (X,τ)→ (Y, σ) is bTµ - irresolute and a subset S of X is bTµ - compact relative to (X,τ), then the image f(S) is bTµ - compact relative to (Y,σ). Proof. Let {Ai : i ∈ I} be a collection of bTµ - open cover of (Y,σ ), such that f(S)⊆⋃ i∈I Ai. Then S ⊆ n⋃ i=1 f−1(Ai), where { f−1(Ai) : i ∈ I) } is bTµ- open set in (X,τ). Since S is bTµ -compact relative to (X,τ), there exist finite subcollection {A1, A2, · · · , An} such that S ⊆ n⋃ i=1 f−1(Ai). That is f(S) ⊆ n⋃ i=1 Ai. Hence f(S) is bTµ - compact relative to (Y,σ). Theorem 5. If a map f : (X,τ)→ (Y, σ) is strongly bTµ - continuous map from a supra compact space (X,τ) onto a supra topological space (Y,σ), then (Y,σ) is bTµ - compact. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 326 Proof. Let {Ai : i ∈ I} be a bTµ - open cover of (Y,σ). Since f is strongly bTµ - continuous, { f−1(Ai : i ∈ I) } is an supra open cover of (X,τ). Again, since (X,τ) is supra compact, the supra open cover { f−1(Ai) : i ∈ I) } of (X,τ) has a finite sub cover say{ f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 f−1(Ai), which implies f(X) = n⋃ i=1 (Ai), so that Y = n⋃ i=1 (Ai). That is A1, A2, · · · , An is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is bTµ -compact. Theorem 6. If a map f : (X,τ)→ (Y, σ) is perfectly bTµ- continuous map from a compact space (X,τ) onto a supra topological space (Y,σ), then (Y,σ) is bTµ - compact. Proof. Let {Ai : i ∈ I} be a bTµ - open cover of (Y,σ). Since f is perfectly bTµ- continuous, { f−1(Ai) : i ∈ I) } is a supra open cover of (X,τ). Again, since (X,τ) is supra compact, the supra open cover { f−1(Ai) : i ∈ I } of (X,τ) has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 f−1(Ai), which implies f(X) = n⋃ i=1 Ai, so that Y = n⋃ i=1 Ai. That is A1, A2, · · · , An is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is bTµ -compact. Theorem 7. If a map f : (X,τ) → (Y, σ) be bTµ - irresolute map from bTµ - compact space (X,τ) onto supra topological space (Y,σ) then (Y,σ) bTµ - compact. Proof. If a map f : (X,τ)→ (Y, σ) is bTµ - irresolute map from a bTµ- compact space (X,τ) onto a supra topological space (Y,σ). Let {Ai : i ∈ I} be a bTµ - open cover of (Y,σ). Then { f−1(Ai) : i ∈ I } is an bTµ - open cover of (X,τ), since f is bTµ - irresolute. As (X,τ) is bTµ - compact, the bTµ - open cover { f−1(Ai) : i ∈ I } of (X,τ) has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 f−1(Ai), which implies f(X) = n⋃ i=1 Ai, so that Y = n⋃ i=1 Ai. That is A1, A2, · · · , An is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is bTµ -compact. Theorem 8. If (X,τ) is compact and bTT µ c space, then (X,τ) is bTµ - compact. Proof. Let (X,τ) is bTµ - compact space. Let {Ai : i ∈ I} be a bTµ - open cover of (X, ). Since by bTT µ c -space,{Ai : i ∈ I} is a supra open cover of (X,τ). Since (X,τ) is compact, supra open cover {Ai : i ∈ I} of (X,τ) has a finite sub cover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,τ) is a bTµ - compact space. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 327 Theorem 9. A supra topological space (X,τ) is bTµ - compact if and only if every family of bTµ-closed sets of (X,τ) having finite intersection property has a non empty intersection. Proof. Suppose (X,τ) is bTµ- compact, Let {Ai : i ∈ I} be a family of bTµ- closed sets with finite intersection property. Suppose ⋂ i∈I Ai = φ, then X - ⋂ i∈I Ai= X. This implies⋃ i∈I (X−Ai) = X. Thus the cover {X −Ai : i ∈ I} is a bTµ- open cover of (X,τ). Then, the bTµ-open cover {X −Ai : i ∈ I} has a finite sub cover say X-{X −Ai : i = 1, 2, · · · , n} . This implies X = ⋃ i∈I (X − Ai) which implies X = X - n⋂ i=1 Ai, which implies X-X = X -[ X − n⋂ i=1 Ai ] which implies φ = n⋂ i=1 Ai. This disproves the assumption. Hence n⋂ i=1 Ai 6= φ Conversely suppose (X,τ ) is not bTµ- compact. Then there exit an bTµ- open cover of (X,τ) say {Gi : i ∈ I} having no finite sub cover. This implies for any finite sub family Gi : i = 1, 2, · · · , n of {Gi : i ∈ I}, we have n⋃ i=1 Gi 6= X, which implies X - n⋃ i=1 Gi 6= X - X , therefore ⋂ i∈I (X − Gi) 6= φ. Then the family {X −Gi : i ∈ I} of bTµ - closed sets has a finite intersection property. Also by assumption ⋂ i∈I (X − Gi) 6= φ which implies X - n⋃ i=1 Gi 6= φ, so that n⋃ i=1 Gi 6= X. This implies {Gi : i ∈ I} is not a cover of (X,τ). This disproves the fact that {Gi : i ∈ I} is a cover for (X,τ). Therefore a bTµ - open cover {Gi : i ∈ I} of (X,τ) has a finite sub cover {Gi : i = 1, 2, · · · , n}. Hence (X,τ) is bTµ - compact. Theorem 10. Let A be a bTµ - compact set relative to a supra topological space X and B be a bTµ -closed subset of X. Then A∩ B is bTµ - compact relative to X. Proof. Let A is bTµ - compact relative to X. Suppose that {Ai : i ∈ I} is a cover of A∩ B by bTµ - open sets in X. Then {Ai : i ∈ I} ∪ {Bc} is a cover of A by bTµ -open sets in X, but A is bTµ - compact relative to X, so there exist i1, i2, · · · , in such that A ⊆ ⋃ {Aij : j = 1, 2, · · · , n} ∪ Bc . Then A∩B ⊆ ⋃{⋃ Aij ∩B, j = 1, 2, · · · , n } ⊆⋃ {Aij : j = 1, 2, · · · , n}. Hence A∩ B is bTµ - compact relative to X. Theorem 11. If a function f : (X,τ) → (Y, σ) is bTµ - irresolute and a subset of X is bTµ - compact relative to X, then f(B) is bTµ - compact relative to Y. Proof. Let {Ai : i ∈ I} be a cover of f(B) by bTµ -open subsets of Y. Then { f−1(Ai) : i ∈ I } is a cover of B by bTµ -open subsets of X. Since B is bTµ -compact relative to X,{ f−1(Ai) : i ∈ I } has a finite subcover say { f−1(A1), f −1(A2), · · · , f−1(An) } for B. Now K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 328 {A1, A2, · · · , An} is a finite subcover of {Ai : i ∈ I} for f(B). So f(B) is bTµ -compact relative to Y. 4. Countably supra bT - Compactness in supra topological spaces In this section,we concentrate on the concept of countably bTµ- Compactness and their properties. Definition 15. A supra topological space (X,τ) is said to be countably bTµ - compact if every countable bTµ - open cover of X has a finite subcover. Theorem 12. If (X,τ) is a countably bTµ - compact space, then (X,τ) is countably supra compact. Proof. Let (X,τ) is countably bTµ - compact space. Let {Ai : i ∈ I} be a countable supra open cover of (X,τ). By [4], {Ai : i ∈ I} is a countable bTµ - open cover of (X,τ). Since (X,τ) is countably bTµ - compact, countable bTµ - open cover {Ai : i ∈ I} of (X,τ) has a finite subcover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,τ) is a countably supra compact space. Theorem 13. If (X,τ) is countably supra compact and bTT µ c -space, then (X,τ) is count- ably bTµ - compact. Proof. Let (X,τ) is countably bTµ - compact space. Let {Ai : i ∈ I} be a countable bTµ - open cover of (X,τ). Since by bTT µ c - space {Ai : i ∈ I} is a countable open cover of (X,τ ). Since (X,τ) is countably supra compact, countable supra open cover {Ai : i ∈ I} of (X,τ) has a finite sub cover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,τ) is a countably bTµ - compact space. Theorem 14. Every bTµ - compact space is countably bTµ -compact. Proof. Let (X,τ) is bTµ-compact space. Let {Ai : i ∈ I} be a countable bTµ- pen cover of (X,τ). Since (X,τ) is bTµ- compact, bTµ- pen cover{Ai : i ∈ I} of (X,τ) has a finite subcover say {Ai : i = 1, 2, · · · , n} for (X,τ). Hence (X,τ) is a countably bTµ-compact space. Theorem 15. Let f : (X,τ) → (Y, σ) be a bTµ - continuous injective mapping. If X is countably bTµ - compact space then (Y,σ) is countably supra compact. Proof. Let f : (X,τ) → (Y, σ) be a bTµ - continuous map from a countably bTµ - compact (X,τ) onto a supra topological space (Y,σ). Let {Ai : i ∈ I} be a countable supra open cover of Y. Then { f−1(Ai) : i ∈ I } is a countable bTµ - open cover of X, as f is bTµ - continuous. Since X is countably bTµ- compact, the countable bTµ-open cover{ f−1(Ai) : i ∈ I } of X has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 329 = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 Ai, then Y = n⋃ i=1 Ai. That is {A1, A2, · · · , An} is a finite sub cover of {Ai : i ∈ I} for Y. Hence Y is countably supra compact. Theorem 16. If a map f : (X,τ) → (Y, σ) is perfectly bTµ - continuous map from a countably supra compact space (X,τ) onto a supra topological space (Y,σ), then (Y,σ) is countably bTµ - compact. Proof. Let {Ai : i ∈ I} be a countable bTµ - open cover of (Y,σ). Since f is perfectly bTµ - continuous, { f−1(Ai) : i ∈ I } is a countable supra open cover of (X,τ). Again, since (X,τ ) is countably supra compact, the countable supra open cover { f−1(Ai) : i ∈ I } of (X,τ) has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 {(Ai)}, so that Y = n⋃ i=1 {(Ai)}. That is {A1, A2, · · · , An} is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is countably bTµ -compact. Theorem 17. If a map f : (X,τ) → (Y, σ) is strongly bTµ- continuous map from a countably supra compact space (X,τ) onto a supra topological space (Y,σ), then (Y,σ) is countably bTµ - compact. Proof. Let {Ai : i ∈ I} be a countable bTµ - open cover of (Y,σ ). Since f is strongly bTµ - continuous, { f−1(Ai) : i = 1, 2, · · · , n } is an countable supra open cover of (X,τ). Again, since (X,τ) is countably supra compact, the countable supra open cover { f−1(Ai) : i ∈ I } of (X,τ) has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 Ai, so that Y= n⋃ i=1 Ai. That is {A1, A2, · · · , An} is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is countably bTµ -compact. Theorem 18. The image of a countably bTµ - compact space under a bTµ- irresolute map is countably bTµ- compact. Proof. If a map f : (X,τ)→ (Y, σ) is bTµ - irresolute map from a countably bTµ - com- pact space (X,τ) onto a supra topological space (Y,σ ). Let {Ai : i ∈ I} be a countable bTµ - open cover of (Y,σ ).Then { f−1(Ai) : i = 1, 2, · · · , n } is an countable bTµ - open cover of (X,τ), since f is bTµ - irresolute. As (X,τ) is countably bTµ - compact, the countable bTµ- open cover { f−1(Ai) : i ∈ I } of (X,τ) has a finite sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 {(Ai)}, so that Y = n⋃ i=1 {(Ai)}. That is {A1, A2, · · · , An} is a finite sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is countably bTµ -compact. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 330 5. supra bT-Lindelof space In this section, we concentrate on the concept of bTµ- Lindelof space and their prop- erties. Definition 16. A supra topological space (X,τ) is said to be bTµ - Lindelof space if every bTµ - open cover of X has a countable subcover. Theorem 19. Every bTµ - Lindelof space is supra Lindelof space. Proof. Let {Ai : i ∈ I} be a supra open cover of (X,τ). By [4], {Ai : i ∈ I} is a bTµ - open cover of (X,τ). Since (X,τ) is bTµ - Lindelof space, bTµ- open cover {Ai : i ∈ I} of (X,τ) has a countable subcover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,τ) is a supra Lindelof space. Theorem 20. If (X,τ) is supra Lindelof space and bTT µ c -space, then (X,τ ) is bTµ - Lindelof space. Proof. Let {Ai : i ∈ I} be a bTµ - open cover of (X,τ). Since by bTT µ c -space,{Ai : i ∈ I} is a supra open cover of (X,τ). Since (X,τ) is compact, supra open cover {Ai : i ∈ I} of (X,τ) has a countable sub cover say {Ai : i = 1, 2, · · · , n} for X. Hence (X,τ) is a bTµ - Lindelof space. Theorem 21. Every bTµ - compact space is bTµ - Lindelof space. Proof. Let {Ai : i ∈ I} be a bTµ - open cover of (X,τ). Then {Ai : i ∈ I} has a finite subcover say {Ai : i = 1, 2, · · · , n}. Since (X,τ) is bTµ - compact space. Since every finite subcover is always countable subcover and therefore {Ai : i = 1, 2, · · · , n} is countable subcover of {Ai : i ∈ I}. Hence (X,τ) is bTµ - Lindelof space. Theorem 22. A bTµ - continuous image of a bTµ - Lindelof space is supra Lindelof space. Proof. Let f : (X,τ)→ (Y,σ)be a bTµ - continuous map from a bTµ - Lindelof space X onto a supra topological space Y. Let {Ai : i ∈ I} be a supra open cover of Y. Then{ f−1(Ai) : i ∈ I } is a bTµ - open cover of X, as f is bTµ - continuous. Since X is bTµ - Lindelof space, the bTµ - open cover { f−1(Ai) : i ∈ I } of X has a countable sub cover say{ f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 Ai, then Y = n⋃ i=1 Ai. That is {A1, A2, · · · , An} is a countable sub cover of {Ai : i ∈ I} for Y. Hence Y is supra Lindelof space. Theorem 23. The image of a bTµ - Lindelof space under a bTµ - irresolute map is bTµ - Lindelof space. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 331 Proof. If a map f : (X,τ)→ (Y, σ) is bTµ - irresolute map from a bTµ - Lindelof space (X,τ) onto a supra topological space (Y,σ). Let {Ai : i ∈ I} be a bTµ - open cover of (Y,σ).Then { f−1(Ai) : i ∈ I } is an bTµ - open cover of (X,τ). Since f is bTµ - irresolute. As (X,τ) is bTµ - Lindelof space, the bTµ - open cover { f−1(Ai) : i ∈ I } of (X,τ) has a countable sub cover say { f−1(Ai) : i = 1, 2, · · · , n } . Therefore X = n⋃ i=1 { f−1(Ai) } , which implies f(X) = n⋃ i=1 Ai, so that Y= n⋃ i=1 Ai. That is {A1, A2, · · · , An} is a countable sub cover of {Ai : i ∈ I} for (Y,σ). Hence (Y,σ) is bTµ - Lindelof space. Theorem 24. If (X,τ) is bTµ - Lindelof space and countably bTµ - compact space then (X,τ) is bTµ - compact space. Proof. Suppose (X,τ) is bTµ - Lindelof space and countably bTµ - compact space. Let {Ai : i ∈ I} be a bTµ - open cover of (X,τ). Since (X,τ) is bTµ - Lindelof space, {Ai : i ∈ I} has a countable subcover say {Ain : i ∈ I, n ∈ N}, therefore {Ain : i ∈ I, n ∈ N} is a count- able subcover of (X,τ) and {Ain : i ∈ I, n ∈ N} is subfamily of {Ai : i ∈ I} and so {Ain : i ∈ I, n ∈ N} is a countable bTµ - open cover of (X,τ). Again, since (X,τ) is countably bTµ - com- pact, {Ain : i ∈ I, n ∈ N} has a finite subcover and {Aik : i ∈ I, k = 1, 2n}. Therefore {Aik : i ∈ I, k = 1, 2, · · · , n} is a finite subcover of {Ai : i ∈ I} for (X,τ). Hence (X,τ) is bTµ - compact space. Theorem 25. If a function f : (X,τ) → (Y, σ) is bTµ - irresolute and a subset of X is bTµ - Lindelof relative to X, then f(B) is bTµ - Lindelof relative to Y. Proof. Let {Ai : i ∈ I} be a cover of f(B) by bTµ -open subsets of Y. Then { f−1(Ai) : i ∈ I } is a cover of B by bTµ -open subsets of X. Since B is bTµ -Lindelof relative to X,{ f−1(Ai) : i ∈ I } has a countable subcover say { f−1(A1), f −1(A2), · · · , f−1(An) } for B. Now {A1, A2, · · · , An} is a countable subcover of {Ai : i ∈ I} for f(B). So f(B) is bTµ -Lindelof relative to Y. 6. supra bT-Connectedness in Supra Topological space Definition 17. A supra topological space (X,µ) is said to be bTµ - Connected if X cannot be written as a disjoint union of two non empty bTµ -open sets. A subsets of (X,µ) is bTµ -connected if it is bTµ -connected as a subspace. Theorem 26. Every bTµ -connected space is supra connected. Proof. Let A and B are supra open sets in X. Since every supra open sets is bTµ -open set. Therefore A and B are bTµ -open and X is bTµ - connected space. Therefore X 6= A ∪B. Therefore X is supra connected. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 332 Example 1. Let X={a, b, c} and τ = {X,φ, {a}}. Then it is bTµ -connected. Remark 1. The converse of the above theorem need not be true in general, which follows from the following example. Example 2. Let X={a, b, c} and τ = {X,φ, {a} , {b} , {a, b}}.Clearly (X,τ) is supra con- nected . The bTµ - open sets of X are {X,φ, {b, c} , {a, c} , {a, b} , {b} , {a}}. Therefore (X,τ) is not a bTµ -connected space, since X = {b, c} ∪ {a} where {b, c} and {a} are non empty bTµ -open sets. Theorem 27. For a supra topological space (X,τ) the following are equivalent (i) (X,τ) is bTµ -connected. (ii) The only subset of (X,τ) which are both bTµ - open and bTµ -closed are the empty set X and φ. (iii) Each bTµ -continuous map of (X,τ) into a discrete space (Y,σ) with atleast two points is a constant map. Proof. (1)⇒(2) Let G be a bTµ-open and bTµ- closed subset of (X,τ).Then X-G is also both bTµ-open and bTµ -closed. Then X = G∪(X-G) a disjoint union of two non empty bTµ -open sets which contradicts the fact that (X,τ) is bTµ-connected. Hence G = φ (or) X. (2)⇒(1)Suppose that X = A∪ B where A and B are disjoint non empty bTµ-open subsets of (X,τ). Since A = X-B, then A is both bTµ -open and bTµ- closed. By assumption A =φ or X, which is a contradiction. Hence (X,τ) is bTµ-connected. (2)⇒(3) Let f : (X,τ)→ (Y, σ) be a bTµ-continuous map, where (Y,σ) is discrete space with atleast two points. Then f−1(y) is bTµ -closed and bTµ -open for each y∈ Y. That is (X,τ) is covered by bTµ-closed and bTµ-open covering { f−1 {y} : y ∈ Y } .By assumption, { f−1 {y} } = φ or X for each y∈ Y. If f−1 {y} = φ for each y ∈ Y, then f fails to be a map. Therefore their exist atleast one point say f−1 {y1} 6= φ, y1 ∈ Y such that f−1({y1}) = X. This shows that f is a constant map. (3)⇒(2) Let G be both bTµ -open and bTµ -closed in (X,τ). Suppose G 6= φ. Let f : (X,τ)→ (Y, σ) be a bTµ -continuous map defined by f(G) = {a} and f(X-G) = {b} where a 6= b and a,b∈ Y. By assumption , f is constant so G = X. Theorem 28. Let f : (X,τ)→ (Y, σ) be a bTµ -continuous surjection and (X,τ) is bTµ - connected, then (Y,σ) is supra connected . Proof. Suppose (Y,σ) is not supra connected. Let Y = A∪ B, where A and B are disjoint non empty supra open subsets in (Y,σ). Since f is bTµ -continuous, X = f−1(A) ⋃ f−1(B), where f−1(A) and f−1(B) are disjoint non empty bTµ -open subsets in (X,τ). This disproves the fact that (X,τ) is bTµ -connected. Hence (Y,σ) is supra connected. K.Krishna, M.Vignesh / Eur. J. Pure Appl. Math, 10 (2) (2017), 323-334 333 Theorem 29. If f : (X,τ)→ (Y, σ) is a bTµ-irresolute surjection and X is bTµ-connected, then Y is bTµ-connected. Proof. Suppose that Y is bTµ-connected. Let Y = A∪B, where A and B are non empty bTµ-open set in Y. Since f is bTµ -irresolute and onto, X = f−1(A)∪f−1(B), where f−1(A) and f−1(B) are disjoint non empty bTµ-open sets in (X,τ). This contradicts the fact that (X,τ) is bTµ- connected. Hence (Y,σ) is bTµ -connected. Theorem 30. Suppose that X is a bTT µ c -space and X is supra connected then bTµ - connected. Proof. Suppose that X is supra connected. Then X cannot be expressed as disjoint union of two non empty proper subset of X. Suppose X is not bTµ -connected space. Let A and B be any two bTµ-open subsets of X such that X = A∪ B, where A∩B = φ and A⊂X, B⊂ X. Since X is bTT µ c -space and A, B are bTµ -open. A,B are open subsets of X, which contradicts that X is supra connected. Therefore X is bTµ -connected. Theorem 31. If the bTµ -open sets C and D form a separation of X and if Y is bTµ- connected subspace of X, then Y lies entirely within C or D. Proof. Since C and D are both bTµ -open in X. The set C∩Y and D∩Y are bTµ -open in Y, these two sets are disjoint and their union is Y. If they were both non empty, they would constitute a separation of Y. Therefore, one of them is empty. Hence Y must lie entirely in C or D. Theorem 32. Let A be a bTµ-connected subspace of X. If A ⊂ B ⊂ bTµcl(A),then B is also bTµ-connected. Proof. Let A be bTµ -connected.Let A ⊂ B ⊂ bTµcl(A). Suppose that B = C ∪D is a separation of B by bTµ -open sets.Thus by previous theorem above A must lie entirely in C or D.Suppose that A⊂ C,then bTµcl(A) ⊆ bTµcl(C).Since bTµcl(C) and D are disjoint, B cannot intersect D.This disproves the fact that D is non empty subset of B.So D=φwhich implies B is bTµ-connected. 7. 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