weak separation axioms in generalized topological ... - Springer Link

Acta Math. Hungar., 131 (1–2) (2011), 110–121 DOI: 10.1007/s10474-010-0017-7 First published online November 3, 2010

WEAK SEPARATION AXIOMS IN GENERALIZED TOPOLOGICAL SPACES M. S. SARSAK Department of Mathematics, The Hashemite University, P.O. Box 150459, Zarqa 13115 Jordan e-mail: [email protected] (Received March 19, 2010; accepted April 20, 2010)

Abstract. We study a unified approach of weak separation axioms in generalized topological spaces.

1. Introduction and preliminaries A generalized topology (breifly GT) [1] μ on a nonempty set X is a collection of subsets of X such that ∅ ∈ μ and μ is closed under arbitrary unions. Elements of μ will be called μ-open sets, and a subset A of (X, μ) will be called a μ-neighborhood of a point x of X if there is U ∈ μ such that x ∈ U ⊂ A. A subset A of X will be called μ-closed if X\A is μ-open. The pair (X, μ) will be called generalized topological space (breifly GTS). By a space X or (X, μ), we will always mean a GTS. A space (X, μ) is said to be a quasi-topological space [3] if μ is closed under finite intersections. Clearly, every topological space is a quasi-topological space and every quasi-topological space is a GTS. If A is a subset of a space (X, μ), then the μ-closure of A [2], cμ (A), is the intersection of all μ-closed sets containing A and the μ-interior of A [2], iμ (A), is the union of all μ-open sets contained in A. It was pointed out in [2] that each of the operators cμ and iμ are monotonic [4], i.e. if A ⊂ B ⊂ X, then cμ (A) ⊂ cμ (B) and iμ(A) ⊂ iμ (B), idempotent [4], i.e. if A ⊂ X,   then cμ cμ (A) = cμ (A) and iμ iμ (A) = iμ (A), cμ is enlarging [4], i.e. if A ⊂ X, then cμ (A) ⊃ A, iμ is restricting [4], i.e. if A ⊂ X, then iμ (A) ⊂ A, A is μ-open if and only if A = iμ (A), and cμ (A) = X\iμ (X\A). Clearly, A is μ-closed if and only if A = cμ (A), cμ (A) is the smallest μclosed set containing A and iμ (A) is the largest μ-open set contained in A, and x ∈ cμ (A) if and only if any μ-open set containing x intersects A. Key words and phrases: μ-open, μ-closed, generalized topology, μ-T0 , μ-T1 , μ-T2 , μ-D0 , μ-D1 , μ-D2 . 2000 Mathematics Subject Classification: 54A05, 54A10, 54D10. c 2010 Akad´ 0236-5294/$ 20.00  emiai Kiad´ o, Budapest, Hungary

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If A is a subset of a space (X, μ), Ac stands for X\A. For the concepts and terminology not defined here, we refer the reader to [5]. 2. Dμ -sets and associated separation axioms Definition 2.1. A subset A of a space (X, μ) is called a Dμ -set if there are two U, V ∈ μ such that U = X and A = U \V . Remark 2.2. Letting A = U and V = ∅ in the above definition, it is easy to see that every proper μ-open set U is a Dμ -set. Definition 2.3. A space (X, μ) is called μ-D0 if for any pair of distinct points x and y of X there exists a Dμ -set of X containing x but not y or a Dμ -set of X containing y but not x. Definition 2.4. A space (X, μ) is called μ-D1 if for any pair of distinct points x and y of X there exists a Dμ -set of X containing x but not y and a Dμ -set of X containing y but not x. Definition 2.5. A space (X, μ) is called μ-D2 if for any pair of distinct points x and y of X there exist disjoint Dμ -sets G and E of X containing x and y, respectively. Definition 2.6. A space (X, μ) is called μ-T0 if for any pair of distinct points of X, there is a μ-open set containing one of the points but not the other. Definition 2.7. A space (X, μ) is called μ-T1 if for any pair of distinct points x and y of X, there exists a μ-open set U of X containing x but not y and a μ-open set V of X containing y but not x. Definition 2.8. A space (X, μ) is called μ-T2 if for any pair of distinct points x and y of X, there exist disjoint μ-open sets U and V of X containing x and y, respectively. The following remark follows immediately from the definitions and Remark 2.2. Remark 2.9. (1) If (X, μ) is μ-Ti , then it is μ-Ti−1 , i = 1, 2. (2) If (X, μ) is μ-Ti , then (X, μ) is μ-Di , i = 0, 1, 2. (3) If (X, μ) is μ-Di , then it is μ-Di−1 , i = 1, 2. Theorem 2.10. For a space (X, μ), the following statements hold: (1) (X, μ) is μ-D0 if and only if it is μ-T0 . (2) (X, μ) is μ-D1 if and only if it is μ-D2 . Acta Mathematica Hungarica 131, 2011

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Proof. (1) Necessity. Let (X, μ) be μ-D0 . Then for all distinct points x, y ∈ X, at least one of x, y, say x, belongs to a Dμ -set G but y ∈ G. Suppose that G = U1 \U2 where U1 = X and U1 , U2 ∈ μ. Then x ∈ U1 , and for y ∈ G we have two cases: (a) y ∈ U1 ; (b) y ∈ U1 and y ∈ U2 . In case (a), U1 contains x but does not contain y. In case (b), U2 contains y but does not contain x. Hence, X is μ-T0 . Sufficiency. Follows from Remark 2.9 (2). (2) Necessity. Let X be μ-D1 . Then for all distinct points x, y ∈ X, we have Dμ -sets G1 , G2 such that x ∈ G1 , y ∈ G1 ; y ∈ G2 , x ∈ G2 . Let G1 = U1 \U2 , G2 = U3 \U4 . From x ∈ G2 , we have either x ∈ U3 or x ∈ U3 and x ∈ U4 . We discuss the two cases separately. (1) x ∈ U3 . From y ∈ G1 , we obtain the following two subcases: (a) y ∈ U1 . From x ∈ U1 \U2 we have x ∈ U1 \(U2 ∪ U3 ) and from  y ∈ U3 \U4 we have y ∈ U3 \(U1 ∪ U4 ). It is easy to see that U1 \(U2 ∪ U3 ) ∩ (U3 \(U1 ∪ U4 ) = ∅. (b) y ∈ U1 and y ∈ U2 . We have x ∈ U1 \U2 , y ∈ U2 and (U1 \U2 ) ∩ U2 = ∅. (2) x ∈ U3 and x ∈ U4 . We have y ∈ U3 \U4 , x ∈ U4 and (U3 \U4 ) ∩ U4 = ∅. Hence, X is μ-D2 . Sufficiency. Follows from Remark 2.9 (3).  Corollary 2.11. If (X, μ) is μ-D1 , then it is μ-T0 . Proof. Follows from Remark 2.9 (3) and Theorem 2.10 (1).  The following diagram summarizes the implications among the introduced concepts. μ-T2

μ-D2

μ-T1

μ-D1

μ-T0

μ-D0 



Example 2.12. Let X = {a, b, c} and μ = X, ∅, {a, b}, {a, c}, {b, c} . Then (X, μ) is μ-T1 but not μ-T2 (observe that a and b cannot be separated by disjoint μ-open sets). Thus, (X, μ) is μ-D2 but not μ-T2 . 



Example 2.13. Let X = {a, b, c} and μ = ∅, {a}, {a, b} . Then (X, μ) is μ-T0 but not μ-D1 (observe that the Dμ subsets of X are: ∅, {a}, {a, b}, {b}). Thus, (X, μ) is μ-T0 (μ-D0 ) but not μ-T1 (μ-D1 ). Acta Mathematica Hungarica 131, 2011

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Example 2.14. Let X = {a, b, c} and μ = X, ∅, {a}, {a, b}, {b, c} . Then (X, μ) is μ-D1 (observe that the Dμ subsets of X are: ∅, {a}, {a, b}, {b, c}, {b}, {c}). However, (X, μ) is not μ-T1 . Thus, (X, μ) is also μ-D2 (μ-T0 ) but not μ-T2 (μ-T1 ). Theorem 2.15. A space (X,μ) is μ  -T0 if and only if for each pair of distinct points x, y of X , cμ {x} = cμ {y} . Proof. Necessity. Let (X, μ) be a μ-T0 space and x, y be any two distinct points of X. Then there exists a μ-open set G containing x, say c y but not x. but not y, and  therefore G is a μ-closed set which contains  μ-closed set containing y, c {y} ⊂ Gc , and Since cμ {y}  is the smallest μ     so x ∈ cμ {y} . Thus, cμ {x} = cμ {y} . that x, y ∈ X and x = y. Then by  Sufficiency.   Suppose   assumption,  of X such that z ∈ c {x} and z ∈ cμ  {x} = cμ {y} . Let z be a point μ       cμ {y} , say. We claim that x ∈ cμ {y} . For, if x ∈ cμ {y} , then cμ {x}       c ⊂ cμ {y} , a contradiction with z ∈ cμ {y} . Thus, x ∈ (cμ {y} ) , but   c (cμ {y} ) is a μ-open set that does not contain y. Hence, (X, μ) is μ-T0 .  Theorem 2.16. A space (X, μ) is μ-T1 if and only if the singletons of X are μ-closed. Proof. Necessity. Suppose that (X, μ) is μ-T1 and x is any point of X. Let y ∈ {x}c . Then x = y, and so there exists a μ-open set Uy such that y ∈Uy but x ∈ Uy . Consequently y ∈ Uy ⊂ {x}c , i.e. {x}c = ∪ Uy : y ∈ {x}c which is μ-open. Sufficiency. Suppose that {p} is μ-closed for every p ∈ X. Let x, y ∈ X with x = y. Then by assumption, {x}c is a μ-open set containing y but not x. Similarly {y}c is a μ-open set containing x but not y. Hence, X is μ-T1 .  Definition 2.17. A point x of a space (X, μ) which has X as the only μ-neighborhood is called a μ-neat point. Remark 2.18. It is clear that if a μ-T0 space (X, μ) has a μ-neat point, then it is unique, because if x and y are distinct μ-neat points in X, then at least one of them, say x, has a μ-neighborhood U containing x but not y. Thus, U = X, a contradiction. Theorem 2.19. A space (X, μ) is μ-D1 if and only if (X, μ) is μ-T0 and has no μ-neat points. Proof. Necessity. Since (X, μ) is μ-D1 , each point x of X is contained in a Dμ -set O = U \V and thus in U . By definition U = X. Hence, x is not a μ-neat point. (X, μ) being μ-T0 follows from Corollary 2.11. Acta Mathematica Hungarica 131, 2011

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Sufficiency. Since (X, μ) is μ-T0 , for each pair of distinct points x, y ∈ X, there exists a μ-open set U containing x, say but not y. Thus by Remark 2.2, U is a Dμ -set, but X has no μ-neat point, so y is not a μ-neat point. Thus, there exists a μ-open set V containing y such that V = X, and therefore, y ∈ V \U, x ∈ V \U and V \U is a Dμ -set. Hence, X is μ-D1 .  Corollary 2.20. For a μ-T0 space (X, μ), the following are equivalent: (1) (X, μ) is μ-D1 ; (2) (X, μ) has no μ-neat points. Definition 2.21. A function f : (X, μ) → (Y, κ) is called (μ, κ)-continuous if the inverse image of each κ-open set is μ-open. Theorem 2.22. If f : (X, μ) → (Y, κ) is a (μ, κ)-continuous surjective function and E is a Dκ -set in Y , then the inverse image of E is a Dμ -set in X . Proof. Let E be a Dκ -set in Y . Then there are κ-open sets U1 and U2 in Y such that E = U1 \U2 and U1 = Y . By the (μ, κ)-continuity of f , f −1 (U1 ) and f −1 (U2 ) are μ-open in X. Since U1 = Y , f −1 (U1 ) = X. Hence, f −1 (E) = f −1 (U1 )\f −1 (U2 ) is a Dμ -set.  Theorem 2.23. If (Y, κ) is κ-D1 and f : (X, μ) → (Y, κ) is (μ, κ)-continuous and bijective, then (X, μ) is μ-D1 . Proof. Suppose that Y is a κ-D1 space. Let x and y be any pair of distinct points of X. Since f is injective and Y is κ-D1 , there exist Dκ -sets Gx and Gy of Y containing f (x) and f (y), respectively, such that f (y) ∈ Gx and f (x) ∈ Gy . By Theorem 2.22, f −1 (Gx ) and f −1 (Gy ) are Dμ -sets in X containing x and y, respectively, and y ∈ f −1 (Gx ), x ∈ f −1 (Gy ). Hence, X is μ-D1 .  Theorem 2.24. A space (X, μ) is μ-D1 if and only if for each pair of distinct points x, y ∈ X , there exists a (μ, κ)-continuous surjective function f : (X, μ) → (Y, κ), where Y is a κ-D1 space such that f (x) and f (y) are distinct. Proof. Necessity. For every pair of distinct points of X, it suffices to take the identity function on X. Sufficiency. Let x and y be any pair of distinct points of X. By hypothesis, there exists a (μ, κ)-continuous, surjective function f from X onto a κ-D1 space Y such that f (x) = f (y). Thus by Theorem 2.10 (2), there exist disjoint Dκ -sets Gx and Gy in Y such that f (x) ∈ Gx and f (y) ∈ Gy . Since f is (μ, κ)-continuous and surjective, by Theorem 2.22, f −1 (Gx ) and f −1 (Gy ) are disjoint Dμ -sets in X containing x and y, respectively. Hence, X is μ-D1 .  Acta Mathematica Hungarica 131, 2011

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3. μ-R0 spaces and μ-R1 spaces Definition 3.1. A space (X, μ) is said to be a μ-R0 space if every μopen set contains the μ-closure of each of its singletons. Definition 3.2.    A space (X, μ) is said to be μ-R1 if for x, y in X with c {x} =  c {y} , there exist disjoint μ-open sets U and V such that μ μ    cμ {x} ⊂ U and cμ {y} ⊂ V . 



Example 3.3. Let X = {a, b,c} and μ= X, ∅, {a, b}, {c} . Then    (X, μ) is μ-R1 (observe that cμ {a} = cμ {b} = {a, b} and cμ {c} = {c}). Theorem 3.4. If (X, μ) is μ-R1 , then (X, μ) is μ-R0 . 



Proof. Let U be a μ-open set and x ∈ U . If y ∈ U , then x ∈ cμ {y} ,  c {y} . Since (X, μ) is μ-R1 , there exists a μ-open and thus, cμ {x} =  μ   that cμ {y} ⊂ Vy and x ∈ Vy . Thus, y ∈ cμ {x} . Therefore, set Vy such  cμ {x} ⊂ U , and hence, (X, μ) is μ-R0 .  The converse of Theorem 3.4 need not be true as will be seen in Remark 3.12. Definition 3.5. A space (X, μ) is said to be μ-symmetric if for each  x, y ∈ X, x ∈ cμ {y} implies y ∈ cμ {x} . Theorem 3.6. A space (X, μ) is μ-R0 if and only if (X, μ) is μsymmetric. 



Proof. Necessity. Assume that X is μ-R0 . Let x ∈ cμ {y} and U be any μ-open set such that y ∈ U . Then by hypothesis, x ∈U . Therefore,  every μ-open set which contains y contains x. Hence, y ∈ cμ {x} .   {y} , Sufficiency. Let U be a μ-open set and x ∈ U . If y ∈  U , then x ∈  c μ     and thus by assumption, y ∈ cμ {x} . Therefore, cμ {x} ⊂ U , and hence, (X, μ) is μ-R0 .  Theorem 3.7. A space (X, μ) is μ-T1 if and only if (X, μ) is μ-T0 and μ-R0 . Proof. Necessity. Follows from Remark 2.9 (1) and Theorem 2.16. Sufficiency. Let x, y ∈ X and x = y. Since X is μ-T0 , we may assume c without μ-open set G. Thus,  loss  of generality that x ∈ G ⊂ {y} for some   x ∈ cμ {y} , and so by Theorem 3.6, y ∈ cμ {x} . Therefore, X\cμ {x} is a μ-open set containing y but not x. Hence, (X, μ) is μ-T1 .  Remark 3.8. It is clear from Theorem 3.7 that if (X, μ) is μ-T0 but not μ-T1 , then (X, μ) is not μ-R0 (see Example 2.13, Example 2.14). Acta Mathematica Hungarica 131, 2011

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Corollary 3.9. For a μ-R0 space (X, μ), the following are equivalent: (1) (X, μ) is μ-T0 ; (2) (X, μ) is μ-T1 ; (3) (X, μ) is μ-D1 . Proof. (1) → (2). Follows from Theorem 3.7. (2) → (3). Follows from Remark 2.9 (2). (3) → (1). Follows from Corollary 2.11.  Theorem 3.10. For a space (X, μ), the following are equivalent: (1) (X, μ) is μ-T2 ; (2) (X, μ) is μ-T1 and μ-R1 ; (3) (X, μ) is μ-T0 and μ-R1 . Proof. (1) → (2). Follows from Remark 2.9(1) and Theorem 2.16. (2) → (3). Follows from Remark 2.9(1). (3) → (1). It follows from Theorem 3.4 and Theorem 3.7 that (X, μ) is μ-T1 , but (X, μ) is μ-R1 , thus again by Theorem 2.16, (X, μ) is μ-T2 .  Corollary 3.11. For a μ-R1 space (X, μ), the following are equivalent: (1) (X, μ) is μ-T2 ; (2) (X, μ) is μ-T1 ; (3) (X, μ) is μ-T0 . Remark 3.12. It is clear from Theorem 3.7 and Theorem 3.10 that any space that is μ-T1 but not μ-T2 , is μ-R0 but not μ-R1 (see Example 2.12). Definition 3.13. Let A be a subset of a space (X, μ). The μ-kernel of A, denoted by μ- ker (A), is the set ∩{U ∈ μ : A ⊂ U }. Note that if there is no μ-open set containing A, then μ- ker (A) = X. Lemma 3.14. Let (X, μ) be a space and A ⊂ X . Then 



μ- ker (A) = {x ∈ X : cμ {x} ∩ A = ∅} 





 c



 c

Proof. Let cμ {x} ∩ A = ∅. Then x ∈ (cμ {x} ) , but (cμ {x} ) is a μ-open set containing A, so x ∈ μ- ker (A). Next suppose that x ∈ μ- ker (A). set U containing A and x ∈ U .   Then there exists a μ-open  Then cμ {x} ∩ U = ∅, thus cμ {x} ∩ A = ∅.  Lemma 3.15. Let (X, μ) be a space and x ∈ X . Then 

y ∈ μ- ker {x}



if and only if 







x ∈ cμ {y} .

exists a μ-open set V Proof. Suppose that y ∈ μ- ker {x} . Then there  containing x such that y ∈ V . Therefore, x ∈  cμ {y} . The converse is similarly shown.  Acta Mathematica Hungarica 131, 2011

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Lemma 3.16. Let x and y be any points of a space (X, μ). Then 





μ- ker {x} = μ- ker {y}



if and only if 









cμ {x} = cμ {y} . 





= μ- ker {y} .  Then Proof. Necessity. Suppose that μ- ker {x}    and z ∈ μ- ker {y} , there exists a point z in X such that z ∈ μ- ker {x}    that {x} ∩ cμ {z} = ∅ which implies say. From z ∈ μ- ker {x} , it follows     {y} , we have {y} ∩ cμ {z} = ∅. Since that x∈ cμ {z} . By z ∈ μ- ker     that c {x} ⊂ c {z} , and thus, {y} ∩ cμ {x} x ∈ cμ {z} , it follows μ μ    = ∅. Hence, cμ {x} = cμ {y} .    . Then there exists a point Sufficiency. Suppose that cμ {x} = cμ {y}  exists a μz in X such that z ∈ cμ {x} and z ∈ cμ {y} , say. Thus there   {x} . Hence, open set containing z and therefore x but not y, so y ∈  μker     μ- ker {x} = μ- ker {y} .  Corollary3.17.  A space (X, μ)  is μ-R1 if and only if for each x, y ∈ X such that μ- ker {x} = μ- ker {y} , there exist disjoint μ-open sets U and  V such that cμ {x} ⊂ U and cμ {y} ⊂ V . Proof. Follows from Lemma 3.16.  ifand only if for any x Theorem 3.18.   A space  (X,  μ) is a μ-R0 space  and y in X , cμ {x} = cμ {y} implies cμ {x} ∩ cμ {y} = ∅.

Necessity. Assume that (X, μ) isμ-R0 and x, y ∈ X such  Proof.     that  cμ {x} = cμ {y} . Then there exists z ∈ c {x} such that z ∈  c {y} , μ μ   say. Since z ∈ cμ {y} , there exists V ∈ μ such that y ∈ V and z ∈ V ; but       c z ∈ cμ {x} , so x ∈ V and thus x ∈ cμ {y} . Therefore, x ∈ (cμ {y} )     c ∈ μ, but (X, μ) is μ-R0 , so cμ {x} ⊂ (cμ {y} ) .   Sufficiency. Let V ∈ μ and let x ∈ V .  We will show that cμ {x}   ⊂ V . Suppose that y ∈ V . Then x ∈ cμ {y} . Thus, cμ {x} = cμ {y} .   By assumption, cμ {x} ∩ cμ {y} = ∅, and thus, y ∈ cμ {x} . Therefore,  cμ {x} ⊂ V , and hence, (X, μ) is μ-R0 .  Theorem 3.19. A space (X,  μ) is a μ-R0 space  if and only if for  x and y in X , μ ker {x} =  μ ker {y} implies μ- ker {x} any points   ∩ μ- ker {y} = ∅. Proof. Necessity. Suppose that (X, μ) is a μ-R Then by  0 space.    Lemma 3.16, for any points x and y in X, if μker {x} =  μker {y} ,       = c {y} . Assume that z ∈ μ- ker {x} ∩μ- ker {y} . By then cμ {x}   μ  z ∈ μ- ker {x} and Lemma 3.15, it follows that x ∈ cμ {z} . Thus by Acta Mathematica Hungarica 131, 2011

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Theorem 3.18, cμ {x} = cμ {z} . Similarly, we have cμ {y} = cμ {z}      = cμ {x} , a contradiction. Hence, μ- ker {x} ∩ μ- ker {y} = ∅. Sufficiency. Let x  (X, μ) be a space such that  for any points   and =  μker {y} implies μker {x} ∩ μker {y} = ∅. y of X, μ- ker {x}       {x} =  c {y} . Then by Lemma 3.16, μker {x} = Assume that c μ μ       therefore by assumption, μ- ker {x} ∩ μ- ker {y} = ∅. μ- ker {y} , and   Now if z ∈cμ {x} , then by Lemma 3.15, x ∈ μ- ker  {z} , and therefore,   =  ∅. By hypothesis, μker {x} = μker {z} . μ- ker {x} ∩ μ- ker {z}   Thus z ∈ cμ {x} ∩ cμ {y} implies that 











μ- ker {x} = μ- ker {z} = μ- ker {y} , 











a contradiction. Therefore, cμ {x} = cμ {y} implies that cμ {x} ∩  cμ {y} = ∅, and thus by Theorem 3.18, (X, μ) is μ-R0 .  Theorem 3.20. For a space (X, μ), the following properties are equivalent: (1) (X, μ) is a μ-R0 space. (2) For any nonempty subset A of X and G ∈ μ such that A ∩ G = ∅, there exists a μ-closed set F such that A ∩ F = ∅ and F ⊂ G. (3) For any G ∈ μ, G = ∪{F : F is μ-closed, F ⊂ G}. (4) For any μ-closed set  F ,F = μ- ker  (F ). (5) For any x ∈ X , cμ {x} ⊂ μ- ker {x} . Proof. (1) → (2). Let A be a nonempty subset of X and G ∈ μ such  {x} that A ∩ G = ∅. Then there exists x ∈ A ∩ G. Since x ∈ G ∈ μ, c μ  ⊂ G. Set F = cμ {x} . Then F is μ-closed, F ⊂ G and A ∩ F = ∅. (2) → (3). Let G ∈ μ. Then G ⊃ ∪{F : F is μ-closed, F ⊂ G}. Let x be any point of G. Then there exists a μ-closed set F such that x ∈ F and F ⊂ G. Therefore, x ∈ F ⊂ ∪{F : F is μ-closed, F ⊂ G}, and hence G = ∪{F : F is μ-closed, F ⊂ G}. (3) → (4). Clear.   (4) → (5). Let x be any point of X and y ∈ μ- ker  {x} . Then there exists V ∈μ such that x ∈ V and y ∈ V ; hence cμ {y} ∩ V = ∅. By (4),  (μ- ker(cμ {y} )) ∩ V = ∅, and thus there exists G ∈ μ such that x ∈G and cμ {y} ⊂ G. Therefore, cμ {x} ∩ G = ∅, and thus y ∈ cμ (cμ {x} )       = cμ {x} . Hence, cμ {x} ⊂ μ- ker {x} . (5) → (1). Clear.  Theorem 3.21. For a space (X, μ), the following properties are equivalent: (1) (X, μ) is a μ-R0 space   (2) If F is μ-closed and x ∈ F, thenμ- ker {x} ⊂ F.  (3) If x ∈ X , then μ- ker {x} ⊂ cμ {x} . Acta Mathematica Hungarica 131, 2011

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Proof. (1) → (2). Let F be μ-closed and x ∈ F . Then μ- ker {x} ⊂ μ- ker (F  ). By (1), it follows from Theorem 3.20 that μ- ker (F ) = F . Thus, μ- ker {x} ⊂ F .     (2) → (3). Since x ∈ cμ {x} and cμ {x} is μ-closed, by (2), 







μ- ker {x} ⊂ cμ {x} . 







by Lemma 3.15, y ∈ μ- ker {x} . By (3) → (1). Let x ∈ cμ {y} . Then      (3), y ∈ cμ {x} . Therefore, x ∈ cμ {y} implies that y ∈ cμ {x} . Hence by Theorem 3.6, (X, μ) is μ-R0 .  



Corollary 3.22. A space (X, μ) is μ-R0 if and only if cμ {x} =   μ- ker {x} for all x ∈ X . Proof. Follows from Theorems 3.20 and 3.21.  Definition 3.23. A net {xα }α∈Λ in a space (X, μ) is called μ-convergent to a point x in X, if for any μ-open subset U of X containing x, there exists α0 ∈ Λ such that xα ∈ U for each α  α0 . Lemma 3.24. Let (X, μ) be a space and let x and y be any two points of X such that every net in X μ-converging to y μ-converges to x. Then x ∈ cμ {y} . Proof. Suppose that xn = y for each n ∈ N. Then {xn }n∈N is a net in X that μ-converges   to y. Thus by assumption, {xn }n∈N μ-converges to x. Hence, x ∈ cμ {y} .  Theorem 3.25. For a space (X, μ), the following statements are equivalent: (1) (X, μ) is a μ-R0 space.   (2) If x, y ∈ X , then y ∈ cμ {x} if and only if every net in X μconverging to y μ-converges to x. 



Proof. (1) → (2). Let x, y ∈ X be such that y ∈ cμ {x} . Let {xα}α∈Λ be a net in X such that{xα }α∈Λ μ-converges to y. Since y ∈ cμ {x} , by   Theorem 3.6, x ∈ cμ {y} . Since {xα }α∈Λ μ-converges to y and x ∈ cμ {y} , {xα }α∈Λ μ-converges to x. Conversely, let x, y ∈ X such that every net in X μ-converging to y μ-converges to x. Then x ∈ cμ {y} by Lemma 3.24.   By Theorem 3.6, y ∈ cμ {x} . (2) → (1). Assume that x and y are any two points of X such that  y ∈ cμ {x} . Suppose that xn = y for each n ∈ N. Then {xn }n∈N is a net  in X that μ-converges to y. Since y ∈ cμ {x} and {xn }n∈N μ-converges   to y, it follows from (2) that {xn }n∈N μ-converges to x. Thus, x ∈ cμ {y} . Hence by Theorem 3.6, (X, μ) is μ-R0 .  Acta Mathematica Hungarica 131, 2011

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M. S. SARSAK

4. Weakly μ-D1 spaces Definition 4.1. A space (X, μ) is said to be weakly μ-D1 if 





cμ {x} = ∅.

x∈X

Example 4.2. Let (X, μ) be the space of Example 3.3. Then (X, μ) is clearly weakly μ-D1 . 



Example 4.3. Let X = {a, b, c} and μ =  ∅, {a, b} . Then (X, μ) is not     weakly μ-D1 (observe that cμ {a} = cμ {b} = X and cμ {c} = {c}). Theorem 4.4. A space (X, μ) is weakly μ-D1 if and only if X has no μ-neat points. Proof. Necessity. Suppose  that  (X, μ) is weakly μ-D1 and that X has a μ-neat point y. Then y ∈ cμ {x} for each x ∈ X, a contradiction. Sufficiency. Suppose that (X, μ)has no μ-neat  points and that X is not weakly μ-D1 . Then there exists y ∈ x∈X cμ {x} , and thus, any μ-open set containing y must be X, that is, y is a μ-neat point of X, a contradiction.  Corollary 4.5. A space (X, μ) is μ-D1 if and only if (X, μ) is μ-T0 and weakly μ-D1 . Proof. Follows from Theorem 2.19 and Theorem 4.4.  Example 4.6. It follows from Corollary 4.5, that Example 4.2 is a weakly μ-D1 space that is not μ-D1 as it is not μ-T0 . 

Theorem 4.7. A space (X, μ) is weakly μ-D1 if and only if μ- ker {x} = X for every x ∈ X .



Proof. Necessity. Suppose that (X,  μ) is weakly μ-D1 . Assume that {y} = X. Thus by Lemma 3.14, there is a point y in X such that μker    y ∈ x∈X cμ {x} , a contradiction.   Sufficiency. Assume that μ- ker {x} = X for every x ∈ X. If X is not weakly   μ-D1 , then by Theorem 4.4, X has a μ-neat point y. Hence, μ- ker {y} = X, a contradiction.  Definition 4.8. A function f : (X, μ) → (Y, κ) is called (μ, κ)-closed if the image of every μ-closed subset of X is κ-closed in Y . The proof of the following lemma is straightforward, and thus omitted. f : (X, μ) → (Y, κ) is (μ, κ)-closed if and only Lemma   4.9. A function  if cκ f (A) ⊂ f cμ (A) for each subset A of X . Acta Mathematica Hungarica 131, 2011

WEAK SEPARATION AXIOMS IN GENERALIZED TOPOLOGICAL SPACES

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Theorem 4.10. If f : (X, μ) → (Y, κ) is a (μ, κ)-closed injection and X is weakly μ-D1 , then Y is weakly μ-D1 . Proof. Since X is weakly μ-D1 , tion, so ∅=f

 





cμ {x}

x∈X



x∈X cμ



=







{x} = ∅, but f is an injec



f (cμ {x}

)

x∈X

Since f is a (μ, κ)-closed function, it follows from Lemma 4.9 that 



y∈Y

Hence,

 y∈Y





cκ {y} ⊂



cκ ({f (x)}) ⊂

x∈X







f (cμ {x}

)

x∈X



cκ {y} = ∅, that is, Y is weakly μ-D1 .  References

´ Cs´ [1] A. asz´ ar, Generalized topology, generalized continuity, Acta Math. Hungar., 96 (2002), 351–357. ´ Cs´ [2] A. asz´ ar, Generalized open sets in generalized topologies, Acta Math. Hungar., 106 (2005), 53–66. ´ Cs´ [3] A. asz´ ar, Further remarks on the formula for γ-interior, Acta Math. Hungar., 113 (2006), 325–332. ´ Cs´ [4] A. asz´ ar, Remarks on quasi topologies, Acta Math. Hungar., 119 (2008), 197–200. [5] R. Engelking, General Topology, Second edition, Sigma Series in Pure Mathematics 6, Heldermann Verlag (Berlin, 1989).

Acta Mathematica Hungarica 131, 2011