6_kilicman.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 5, No. 3, 2012, 357-364 ISSN 1307-5543 – www.ejpam.com On Characterizations of New Separation Axioms and Topological Properties S. Pious Missier1, M. J. Jeyanthi2, Adem Kılıçman3,∗ 1 Department of Mathematics, V. O. Chidambaram College, Thoothukudi-628 008 (T.N.), India 2 Aditanar College of Arts and Science, Tiruchendur(T.N.), India 3 Institute of Mathematical Research (INSPEM) and Department of Mathematics, University Putra Malaysia, 43400 UPM, Serdang, Selangor, Malaysia Abstract. In this paper, we introduce new separations axioms Λr−R0, Λr −R1 and Λr −Dk, and study their properties. 2010 Mathematics Subject Classifications: 54D10, 54D15 Key Words and Phrases: Λr − R0, Λr − R1 and Λr − Dk, k = 0,1,2 1. Introduction Caldas and Jafari [1] introduced the notions of Λδ − R0 and Λδ − R1 topological spaces. In this paper, we define Λr -open sets, that is, if (X ,τ) is a topological space and A ⊂ X . Then Λr -kernel of A is defined by Λr − ker(A) = ∩ � G/G ∈ ΛrO(X ,τ) and A⊂ G � . Then we introduce some Λr -separation axioms, we call these axioms as Λr − R0,Λr − R1 and study the properties of these axioms. We also define Λr -difference sets and utilize them to define the Λr − Dk, k = 0,1,2 axioms. Throughout the paper (X ,τ) (or simply X ) will always denote a topological space. Let (X ,τ) be a topological space and S ⊂ X . Then S is called regularly-open if S = Int(clS). The complement Sc(= X \ S) of a regularly-open set S is called the regularly-closed set. The family of all regularly-open sets(resp. regularly-closed sets) will be denoted by RO(X ,τ) (resp. RC(X ,τ)). A subset S of a topological space (X ,τ) is called Λr -set if S = Λr(S) where Λr(S) = ∩{G/G ∈ RO(X ,τ) and S ⊆ G}. ∗Corresponding author. Email addresses: spmissier�yahoo. om (S. Missier), jeyanthimani karaj�gmail. om (M. Jeyanthi), akili man�putra.upm.edu.my (A. Kılıçman) http://www.ejpam.com 357 c© 2012 EJPAM All rights reserved. S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 358 The collection of all Λr -sets is denoted by Λr(X ,τ). Throughout this paper, we let A be a subset of a space (X ,τ). Then A is called a Λr -closed set if A= T ∩ C where T is a Λr -set and C is a closed set. The complement of a Λr -closed set is called Λr -open. The collection of all Λr -open sets is denoted by ΛrO(X ,τ). The collection of all Λr -closed sets is denoted by Λr C(X ,τ). A point x ∈ X is called a Λr -cluster point of A if for every Λr -open set U containing x , A∩ U 6= ;. The set of all Λr -cluster points of A is called the Λr -closure of A and is denoted by Λr − cl(A). 2. Λr − R0 Spaces Definition 1. The topological space (X ,τ) is said to be Λr − R0 if for each Λr -open set G, x ∈ G⇒ Λr − cl({x}) ⊆ G. Theorem 1. For a topological space (X ,τ), the following statements are equivalent: (1) (X ,τ) is Λr − R0, (2) For any Λr -closed set F and a point x /∈ F, ∃U ∈ ΛrO(X ,τ) such that x /∈ U and F ⊆ U, (3) For any Λr -closed set F and a point x /∈ F, Λr − cl({x})∩ F = ;. Proof. (1)⇒ (2) Let F be a Λr -closed set and x /∈ F . Then F c is Λr -open and x ∈ F c . Since X is Λr −R0, Λr − cl({x}) ⊆ F c and hence F ⊆ X − (Λr − cl({x})). Thus X − (Λr − cl({x})) is a Λr -open set containing F and x /∈ X − (Λr − cl({x})). (2) ⇒ (3) Let F be a Λr -closed set and x /∈ F . Then ∃U ∈ ΛrO(X ,τ) such that x /∈ U and F ⊆ U . Claim: U ∩ Λr − cl({x}) = ;. For, if U ∩ Λr − cl({x}) 6= ;, then ∃ a point y in X such that y ∈ U and y ∈ Λr − cl({x}). That implies y is a Λr -cluster point of {x}. That implies for every Λr -open set G containing y, G ∩ {x} 6= ;. That is, x ∈ G. Here U is a Λr -open set containing y. Hence x ∈ U , which is a contradiction. Therefore U ∩ Λr − cl({x}) = ; and hence F ∩Λr − cl({x}) = ;. (3) ⇒ (1) Let G be a Λr -open set and x ∈ G. Then Gc is Λr -closed and x /∈ Gc. By (3), Λr − cl({x})∩ Gc = ; and hence Λr − cl({x}) ⊆ G. Therefore (X ,τ) is Λr − R0. Theorem 2. A space (X ,τ) is Λr − R0 iff for each pair x , y of distinct points in X , Λr − cl({x})∩Λr − cl({y}) = ; or {x , y} ⊆ Λr − cl({x})∩Λr − cl({y}). Proof. Let (X ,τ) be a Λr − R0 space. Let x , y ∈ X such that x 6= y. Then we have Case(i): Suppose Λr − cl({x})∩Λr − cl({y}) 6= ;. If {x , y} is not a subset of Λr − cl({x})∩Λr − cl({y}) and x /∈ Λr − cl({y}), then x ∈ X − (Λr − cl({y})) and S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 359 X − (Λr − cl({y})) is Λr -open. Since (X ,τ) is Λr − R0, Λr − cl({x}) ⊆ X − (Λr − cl({y})). Therefore Λr − cl({x})∩Λr − cl({y}) = ;. This is a contradiction. Hence {x , y} ⊆ Λr − cl({x})∩Λr − cl({y}). Case(ii): Suppose {x , y} is not a subset of Λr− cl({x})∩Λr− cl({y}) and let x /∈ Λr− cl({y}). Then x ∈ X − (Λr − cl({y})) and X − (Λr − cl({y})) is Λr -open. Since (X ,τ) is Λr − R0, Λr − cl({x}) ⊆ X − (Λr − cl({y})) and hence Λr − cl({x})∩Λr − cl({y}) = ;. Conversely, let U be a Λr -open set and x ∈ U . Suppose Λr − cl({x}) is not a subset of U . Then ∃ a point y ∈ Λr − cl({x}) such that y /∈ U . That implies y ∈ X − U and X − U is Λr -closed. Since Λr− cl({y}) is the smallest Λr -closed set containing y, Λr− cl({y}) ⊆ X −U and hence Λr − cl({y}) ∩ U = ;. Since x ∈ U , x /∈ Λr − cl({y}) and hence {x , y} is not a subset of Λr − cl({x}) ∩ Λr − cl({y}). Also y ∈ Λr − cl({x}) ∩ Λr − cl({y}). That implies Λr − cl({x})∩Λr − cl({y}) 6= ;. So by this contradiction, Λr − cl({x}) ⊆ U and hence (X ,τ) is Λr − R0. Theorem 3. For any points x and y in a topological space (X ,τ), the following are equivalent: (1) Λr − ker({x}) 6= Λr − ker({y}), (2) Λr − cl({x}) 6= Λr − cl({y}). Proof. (1)⇒ (2) Suppose Λr − ker({x}) 6= Λr − ker({y}). Then ∃ a point z in X such that z ∈ Λr − ker({x}) and z /∈ Λr − ker({y})⇒ Λr − cl({z}) ∩ {x} 6= ; and Λr − cl({z})∩{y} = ; ⇒ x ∈ Λr − cl({z}) and y /∈ Λr − cl({z})⇒ Λr − cl({x}) ⊆ Λr − cl({z}) and y /∈ Λr − cl({z})⇒ y /∈ Λr − cl({x})⇒ Λr − cl({x}) 6= Λr − cl({y}). (2)⇒ (1) Suppose Λr−cl({x}) 6= Λr−cl({y}). Then ∃ a point z in X such that z ∈ Λr−cl({x}) and z /∈ Λr−cl({y}). That implies ∃ a Λr -open set V containing z such that x v ∈ V and y /∈ V . That is, V is a Λr -open set containing x but not y. If y ∈ Λr−ker({x}), then x ∈ Λr−cl({y}). That implies for every Λr -open set G containing x , G ∩ {y} 6= ;. That is, y ∈ G. By this contradiction, y /∈ Λr − ker({x}) and hence Λr − ker({x}) 6= Λr − ker({y}). Theorem 4. For a topological space (X ,τ), the following are equivalent: (1) (X ,τ) is Λr − R0, (2) For any non-empty set A and G ∈ ΛrO(X ,τ) such that A∩ G 6= ;, ∃F ∈ Λr C(X ,τ) such that A∩ F 6= ; and F ⊆ G, (3) For any G ∈ ΛrO(X ,τ), G = ∪{F/F ∈ Λr C(X ,τ) and F ⊆ G}, (4) For any F ∈ Λr C(X ,τ), F = ∩{G/G ∈ ΛrO(X ,τ) and F ⊆ G}, (5) For any x ∈ X , Λr − cl({x}) ⊆ Λr − ker({x}), S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 360 (6) For any x , y ∈ X , y ∈ Λr − cl({x})⇔ x ∈ Λr − cl({y}). Proof. (1) ⇒ (2) Let A be any nonempty subset of X and G be a Λr -open set such that A∩ G 6= ;. Let x ∈ A∩ G. Since (X ,τ) is Λr − R0, x ∈ G ⇒ Λr − cl({x}) ⊆ G. Since x ∈ A, Λr−cl({x})∩A 6= ;. ThusΛr−cl({x}) is a Λr -closed set contained in G and A∩Λr−cl({x}) 6= ;. (2)⇒ (3) Let G ∈ ΛrO(X ,τ) and x ∈ G. Then by (2), ∃F ∈ Λr C(X ,τ) such that {x} ∩ F 6= ; and F ⊆ G. That implies x ∈ F where F ∈ Λr C(X ,τ) and F ⊆ G and hence x ∈ ∪{F/F ∈ Λr C(X ,τ) and F ⊆ G}. Therefore G ⊆ ∪{F/F ∈ Λr C(X ,τ) and F ⊆ G}. Also ∪{F/F ∈ Λr C(X ,τ) and F ⊆ G} ⊆ G. Hence G = ∪{F/F ∈ Λr C(X ,τ) and F ⊆ G}. (3)⇒ (4) Let F ∈ Λr C(X ,τ). Then F c ∈ ΛrO(X ,τ). By (3), F c = ∪{Gc/Gc ∈ Λr C(X ,τ) and Gc ⊆ F c}. That implies F = ∩{G/G ∈ ΛrO(X ,τ) and F ⊆ G}. (4) ⇒ (5) Let y /∈ Λr − ker({x}). Then x /∈ Λr − cl({y}). That implies ∃ a Λr -open set V containing x such that V ∩ {y} = ; and hence Λr − cl({y}) ∩ V = ;. By (4), Λr− cl({y}) = ∩{G/G ∈ ΛrO(X ,τ) and Λr− cl({y}) ⊆ G}. Since x ∈ V , x /∈ Λr− cl({y}) and hence ∃G ∈ ΛrO(X ,τ) such that Λr − cl({y}) ⊆ G and x /∈ G. Therefore Λr − cl({x})∩G = ; and hence y /∈ Λr − cl({x}). Therefore Λr − cl({x}) ⊆ Λr − ker({x}). (5)⇒ (6) If y ∈ Λr − cl({x}), then y ∈ Λr − ker({x}) by (5). That implies x ∈ Λr − cl({y}). Similarly, if x ∈ Λr − cl({y}), then by (5), x ∈ Λr − ker({y}) and hence y ∈ Λr − cl({x}). Thus x ∈ Λr − cl({y})⇔ y ∈ Λr − cl({x}). (6)⇒ (1) Let G ∈ ΛrO(X ,τ) and x ∈ G. If y /∈ G, then y ∈ X − G and hence Λr − cl({y}) ⊆ X −G since Λr − cl({y}) is the smallest Λr -closed set containing y. Therefore Λr − cl({y}) ∩ G = ; and hence x /∈ Λr − cl({y}). By (6), y /∈ Λr − cl({x}). Therefore Λr − cl({x}) ⊆ G and hence (X ,τ) is Λr − R0. Corollary 1. For a topological space (X ,τ), the following properties are equivalent: (1) (X ,τ) is Λr − R0, (2) For any x ∈ X , Λr − cl({x}) = Λr − ker({x}). Theorem 5. For a topological space (X ,τ), the following properties are equivalent: (1) (X ,τ) is Λr − R0 (2) If F is Λr -closed, then F = Λr − ker(F) (3) If F is Λr -closed and x ∈ F, then Λr − ker({x})⊆ F (4) If x ∈ X , then Λr − ker({x})⊆ Λr − cl({x}). S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 361 Proof. (1)⇒ (2) Let F be a Λr -closed set and x /∈ F . Then X − F is Λr -open and x ∈ X − F . By (1), Λr − cl({x}) ⊆ X − F and hence Λr − cl({x}) ∩ F = ;. That implies x /∈ Λr − ker(F). Therefore Λr − ker(F) ⊆ F . Also F ⊂ Λr − ker(F). Hence F = Λr − ker(F). (2)⇒ (3) Let F be a Λr -closed set and x ∈ F . ThenΛr − ker({x})⊆ Λr − ker(F) = F . (3)⇒ (4) Since x ∈ Λr − cl({x}) and Λr − cl({x}) is Λr -closed, by (3), Λr − ker({x})⊆ Λr − cl({x}). (4)⇒ (1) Since x ∈ Λr − cl({y})⇔ y ∈ Λr − cl({x}), (X ,τ) is Λr − R0. The following Examples 1 and 2 show that Λr − T0 and Λr − R0 are independent. Example 1. Let X = {a, b, c} and τ = {X ,;, {a}, {b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {b, c}}. Here (X ,τ) is Λr − R0 but it is not Λr − T0. Example 2. Let X = {a, b, c} and τ = {X ,;, {a}, {a, b}}. Then ΛrO(X ,τ) = {X ,;, {a}, {a, b}}. Here (X ,τ) is Λr − T0 but it is not Λr − R0. 3. Λr − R1 Spaces Definition 2. A space (X ,τ) is Λr − R1 if for each x , y ∈ X with Λr − cl({x}) 6= Λr − cl({y}), ∃Λr -open sets U and V such that Λr − cl({x}) ⊆ U, Λr − cl({y}) ⊆ V and U ∩ V = ;. Proposition 1. If (X ,τ) is Λr − R1, then (X ,τ) is Λr − R0. Proof. Let (X ,τ) be Λr − R1. Let U be Λr -open in X and x ∈ U . For each y ∈ X − U , Λr − cl({x}) 6= Λr − cl({y}). Then ∃ disjoint Λr -open sets Uy and Vy such that Λr − cl({x}) ⊆ Uy and Λr − cl({y}) ⊆ Vy . Take V = ∪{Vy/y ∈ X − U}. Then V is Λr -open, X − U ⊆ V and x /∈ V . Therefore Λr − cl({x}) ⊆ X − V ⊆ U and hence (X ,τ) is Λr − R0. Theorem 6. If (X ,τ) is Λr − T2, then (X ,τ) is Λr − R1. Proof. Let x , y ∈ X such that x 6= y and Λr − cl({x}) 6= Λr − cl({y}). Since (X ,τ) is Λr − T2, ∃Λr -open sets U and V such that x ∈ U , y ∈ V and U ∩ V = ;. That is, {x} ⊆ U and {y} ⊆ V . Since (X ,τ) is Λr − T2, it is Λr − T1. Therefore for every x ∈ X , {x} = Λr − cl({x}). Thus Λr − cl({x}) ⊆ U , Λr − cl({y}) ⊆ V and U ∩ V = ;. Hence (X ,τ) is Λr − R1. Remark 1. The converse of the above theorem need not be true. For example, let X = {a, b, c, d} and τ = {X ,;, {a}, {b, c}, {a, b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {b, c}, {a, b, c}{b, c, d}, {a, d}}. Here (X ,τ) is Λr − R1 but not Λr − T2. Note that Λr − T0 and Λr − R1 are independent as in the following examples 3 and 4. S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 362 Example 3. Let X = {a, b, c} and τ = {X ,;, {c}, {a, c}, {b, c}}. Then ΛrO(X ,τ) = {X ,;, {c}, {a, c}, {b, c}}. Here (X ,τ) is Λr − T0 but it is not Λr − R1. Example 4. Let X = {a, b, c, d} and τ = {X ,;, {a}, {b, c}, {a, b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {a, d}, {b, c}, {a, b, c}, {b, c, d}}. Here (X ,τ) is Λr − R1 but it is not Λr − T0. Theorem 7. For a space (X ,τ), the following statements are equivalent: (1) (X ,τ) is Λr − R1, (2) If x , y ∈ X such that Λr − cl({x}) 6= Λr − cl({y}), then ∃Λr -closed sets F1 and F2 such that x ∈ F1, y /∈ F1, x /∈ F2, y ∈ F2 and X = F1 ∪ F2. Proof. (1)⇒ (2) Let x , y ∈ X such that Λr−cl({x}) 6= Λr−cl({y}). Then by (1), ∃ disjoint Λr -open sets U and V such that Λr − cl({x})⊆ U and Λr − cl({y}) ⊆ V . Take F1 = X − V and F2 = X − U . Then F1 and F2 are Λr -closed sets such that x ∈ F1, y /∈ F1, x /∈ F2, y ∈ F2 and X = F1 ∪ F2. (2)→ (1) Let x , y ∈ X such that Λr − cl({x}) 6= Λr − cl({y}). Then by (2) ∃Λr -closed sets F1 and F2 such that x ∈ F1, y /∈ F1, x /∈ F2, y ∈ F2 and X = F1 ∪ F2. Take U = X − F2 and V = X − F1. Then U and V are Λr -open sets, x ∈ U , y ∈ V and U ∩ V = ;. Therefore (X ,τ) is Λr − T2 and hence (X ,τ) is Λr − R1. 4. Λr − Dk Spaces Definition 3. Let (X ,τ) be a topological space and A be a subset of X . Then A is called Λr - difference set (shortly Λr -D set) if ∃U , V ∈ ΛrO(X ,τ) such that U 6= X and A = U − V . The collection of all Λr -difference sets of (X ,τ) is denoted by Λr D(X ,τ). Remark 2. Every Λr -open set A different from X is a Λr -D set if U = A and V = ;. But the con- verse need not be true. For example, let X = {a, b, c, d} and τ = {X ,;, {b, d}, {b, c, d}, {a, b, d}}. Then ΛrO(X ,τ) = {X ,;, {b, d}, {b, c, d}, {a, b, d}} and Λr D(X ,τ) = {;, {b, d}, {b, c, d}, {a, b, d}, {c}, {a}}. Here {a} is a Λr -D set but not Λr -open set. Definition 4. A space (X ,τ) is called (1) Λr − D0 if for x , y ∈ X , x 6= y, ∃ a Λr -D set containing one of x and y but not the other (2) Λr − D1 if for x , y ∈ X , x 6= y, ∃Λr -D sets U and V in X such that x ∈ U, y /∈ U and y ∈ V , x /∈ V (3) Λr − D2 if for x , y ∈ X , x 6= y, ∃Λr − D sets U and V in X such that x ∈ U, y ∈ V and U ∩ V = ;. Theorem 8. A space (X ,τ) is Λr − D0 iff it is Λr − T0. S. Missier, M. Jeyanthi, A. Kılıçman / Eur. J. Pure Appl. Math, 5 (2012), 357-364 363 Proof. Suppose (X ,τ) is Λr − D0. Let x , y ∈ X such that x 6= y. Then ∃ an Λr -D set A containing one of x and y but not the other, say x ∈ A but y /∈ A. Since A is a Λr -D set, A= U − V where U 6= X and U , V ∈ ΛrO(X ,τ). Since x ∈ A, x ∈ U and x /∈ V . For y /∈ A, we have two cases (a) y /∈ U (b) y ∈ U and y ∈ V . In case (a), x ∈ U but y /∈ U . In case (b), y ∈ V but x /∈ V . Hence (X ,τ) is Λr − T0. Conversely, suppose (X ,τ) is Λr − T0. Let x , y ∈ X such that x 6= y. Then ∃ an Λr -open set U containing one of x and y but not the other, say x ∈ U but y /∈ U . Then U 6= X and hence U is a Λr -D set. Therefore U is a Λr -D set containing x but not y. Hence (X ,τ) is Λr − D0. The following examples 5 and 6 show that Λr − R0 and Λr − D0 are independent. Example 5. Let X = {a, b, c} and τ = {X ,;, {b}, {a, c}}. Then ΛrO(X ,τ) = {X ,;, {b}, {a, c}}. Here (X ,τ) is Λr − R0 but it is not Λr − D0. Example 6. Let X = {a, b} and τ = {X ,;, {a}}. Then ΛrO(X ,τ) = {X ,;, {a}}. Here (X ,τ) is Λr − D0 but it is not Λr − R0. Remark 3. Examples 7 and 8 below show that Λr − R1 and Λr − D0 are independent. Example 7. Let X = {a, b, c, d} and τ= {X ,;, {a}, {b, c, d}}. Then ΛrO(X ,τ) = {X ,;, {a}, {b, c, d}}. Here (X ,τ) is Λr − R1 but it is not Λr − D0. Example 8. Let X = {a, b, c} and τ = {X ,;, {b}, {a, b}}. Then ΛrO(X ,τ) = {X ,;, {b}, {a, b}}. Here (X ,τ) is Λr − D0 but it is not Λr − R1. Theorem 9. A space (X ,τ) is Λr − D1 iff it is Λr − D2. Proof. Suppose (X ,τ) is Λr − D1. Then for each pair of distinct points x , y ∈ X , we have Λr -D sets A and B such that x ∈ A, y /∈ A and y ∈ B, x /∈ B. Let A= U1 − V1 and B = U2 − V2. Then U1, V1, U2 and V2 are Λr -open sets, U1 6= X and U2 6= X . For x /∈ B, we have two cases (i) x /∈ U2 (ii) x ∈ U2 and x ∈ V2. Case(i) x /∈ U2 then since y /∈ A, either y /∈ U1 or ( y ∈ U1 and y ∈ V1 ). If y /∈ U1, from y ∈ B = U2− V2, it follows that y ∈ U2− (V2 ∪U1). From x ∈ A= U1− V1 and x /∈ U2, x ∈ U1− (V1 ∪U2). Also (U2− (V2∪U1))∩ (U1− (V1∪U2)) = ;. If y ∈ U1 and y ∈ V1, then x ∈ U1 − V1. That implies (U1 − V1)∩ V1 = ;. Case(ii) x ∈ U2 and x ∈ V2 then we have y ∈ B = U2 − V2, x ∈ V2 and (U2 − V2) ∩ V2 = ;. Hence (X ,τ) is Λr − D2. Conversely, suppose (X ,τ) is Λr − D2. Let x , y ∈ X such that x 6= y. Then ∃Λr − D sets A and B such that x ∈ A, y ∈ B and A∩ B = ;. Therefore x ∈ A, y /∈ A and y ∈ B, x /∈ B. Hence (X ,τ) is Λr − D1. REFERENCES 364 Corollary 2. If (X ,τ) is Λr − D1, then it is Λr − T0. Remark 4. The converse of the above corollary need not be true. For example, let X = {a, b, c} and τ = {X ,;, {c}, {a, c}}. Then ΛrO(X ,τ) = {X ,;, {c}, {a, c}} and Λr D(X ,τ) = {;, {a}, {c}, {a, c}}. Here (X ,τ) is Λr − T0 but not Λr − D1. Remark 5. Examples 9 and 10 below show that Λr − R0 and Λr − D2 are independent. Example 9. Let X = {a, b, c, d} and τ = {X ,;, {a}, {b, c}, {a, b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {a, d}, {b, c}, {a, b, c}, {b, c, d}}. Here (X ,τ) is Λr − R0 but it is not Λr − D2. Example 10. Let X = {a, b, c} and τ = {X ,;, {b}, {c}, {b, c}, {a, c}}. Then ΛrO(X ,τ) = {X ,;, {b}, {c}, {b, c}, {a, c}}. Here (X ,τ) is Λr − D2 but it is not Λr − R1. Similar to the previous cases the examples 11 and 12 show that Λr − R1 and Λr − D1 are independent. Example 11. Let X = {a, b, c, d} and τ = {X ,;, {a}, {b, c}, {a, b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {a, d}, {b, c}, {a, b, c}, {b, c, d}}. Here (X ,τ) is Λr − R1 but it is not Λr − D1. Example 12. Let X = {a, b, c} and τ = {X ,;, {a}, {c}, {a, c}, {b, c}}. Then ΛrO(X ,τ) = {X ,;, {a}, {c}, {a, c}, {b, c}}. Here (X ,τ) is Λr − D1 but it is not Λr − R1. References [1] M. Caldas and S. Jafar. On some low separation axioms in topological space. Houston Journal of Math, 29:93–104, 2003.