1. Introduction and Preliminaries.

Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.yu/filomat Filomat 22:1 (2008), 97 106 ON δ SETS IN γ SPACES V. Renuka Devi and D. Sivaraj Abstract We consider a collection of subsets of a set X defined in terms of a function on (X), called the γ open sets, which is not a topology but we show that some of the results established for topologies are valid for this collection. In particular, we define δ γ open sets in a γ space and characterize its properties. Also, we discuss the properties of γ rare sets and characterize δ γ open sets in terms of γ rare sets. 1. Introduction and Preliminaries. Let X be a nonempty set and Γ = {γ : (X) (X) γ(a) γ(b) whenever A B}. Also, the subcollections, Γ 1 = {γ Γ γ(x) = X} and Γ 2 = {γ Γ γ(γ(a)) = γ(a) for every subset A of X } of Γ are defined in [3]. If γ Γ, a subset A of X is said to be γ open if A γ(a) [3]. The complement of a γ open set is γ closed. The family of all γ open sets is denoted by µ γ. In [3, Proposition 1.1], it is established that µ γ and arbitrary union of members of µ γ is again in µ γ. Collection of subsets of X satisfying these two conditions is called a generalized topology in [4]. X need not be γ open [3] and so need not be γ closed. X is γ open if γ Γ 1 [3]. The intersection of two γ open sets need not be γ open [3]. The γ interior of A is the largest γ open set contained in A and is denoted by i γ (A). Therefore, A is γ open if and only if A = i γ (A). The smallest γ closed set containing A is called the γ closure of A and is denoted by c γ (A). Therefore, A is γ closed if and only if A = c γ (A). In [3], it is established that c γ Γ 2, i γ Γ 2, i γ c γ = i γ c γ Γ 2, c γ i γ Γ 2 and X i γ (A) = c γ (X A). A subset A of X is said to be γ semiopen [5] if there exists a γ open set G such that G A c γ (G). The complement of a γ semiopen set is said to be γ semiclosed. It is easy to verify that A is γ semiopen if and only if A c γ i γ (A) and A is γ semiclosed if and only if i γ (A) = i γ c γ (A) A. Recall that, a subset A of X is said to be γ dense if X = c γ (A). σ(γ) is the family of all γ semiopen sets, π(γ) = {A X A 2000 Mathematics Subject Classification. 54 A 05, 54 A 10. Keywords and Phrases. γ open, γ interior, γ closure, γ semiopen, γ preopen, γα open, γβ open, γb open sets and generalized topology. Received: November 16, 2007 Communicated by Dragan S. Djordjević

98 V. Renuka Devi and D. Sivaraj i γ c γ (A)} is the family of all γ preopen sets [4], α(γ) = {A X A i γ c γ i γ (A)} is the family of all γα open sets [4], β(γ) = {A X A c γ i γ c γ (A)} is the family of all γβ open sets [4] and b(γ) = {A X A c γ i γ (A) i γ c γ (A)} is the family of all γb open sets [7]. The interior and closure operators of these generalized topologies are respectively denoted by, i σ and c σ, i π and c π, i α and c α, i β and c β and i b and c b. It is clear that µ γ α(γ) σ(γ) π(γ) b(γ) β(γ). In [7], a new family of functions defined on (X), denoted by Γ 4, is introduced. Γ 4 = {γ Γ G γ(a) γ(g A) for every γ open set G and A X}. If γ Γ 4, then the pair (X, µ γ ) is called a γ space. In [7, Example 2.2], it is established that µ γ is not a topology on X even if γ Γ 4 but the intersection of two γ open sets is γ open. It is interesting to note that in a topological space (X, τ), if i is the interior operator, then i Γ 4 and the i space is nothing but the topological space (X, τ). The following lemma will be useful in the sequel. Lemma 1.1. If (X, µ γ ) is a γ space, then the following hold. (a) If A and B are γ open sets, then A B is a γ open set [7, Theorem 2.1]. (b)i γ (A B) = i γ (A) i γ (B) for every subsets A and B of X [7, Theorem 2.3(a)]. (c) c γ (A B) = c γ (A) c γ (B) for every subsets A and B of X [7, Theorem 2.3(b)]. (d) c γ (c σ (A)) = c γ (A) for every subset A of X [7, Theorem 2.5(f)]. (e) i γ c γ (i π (A)) = i γ c γ (c π (A)) = i γ c γ (A) = i π (c γ (A)) for every subset A of X [7, Theorem 2.7(f)]. (f)c γ (i π (A)) = c γ i γ c γ (A) for every subset A of X [7, Theorem 2.7(v)]. (g) If X is a nonempty set, A is a subset of X and γ Γ, then i γ (c σ (A)) = i γ c γ (A) [7, Theorem 2.4(e)]. 2. More results in γ spaces In this section, we establish some of the properties of i γ and c γ in a γ space and also we prove that i γ Γ 4. Also, we characterize γβ open sets, γ locally closed sets and γ preopen sets. Theorem 2.1. If (X, µ γ ) is a γ space, then the following hold. (a) If G is γ open and A X, then G i γ (A) = i γ (G A) and so i γ Γ 4. (b) If G is γ open and A X, then G c γ (A) c γ (G A). (c) i γ (A F ) i γ (A) F where F is γ closed and A X. (d) c γ (A F ) = c γ (A) F where F is γ closed and A X. (e) If G is γ open and D is γ dense, then c γ (G D) = c γ (G). Proof. (a) Let G be γ open and A be any subset of X. Then G i γ (A) is a γ open set by Lemma 1.1(a), such that G i γ (A) G A. Therefore, G i γ (A) i γ (G A) = i γ (G) i γ (A) = G i γ (A), by Lemma 1.1(b). Therefore, G i γ (A) = i γ (G A). Since the set of all i γ open sets coincides with the set of all γ open sets, it follows that i γ Γ 4. (b) Let x G c γ (A) and U be an arbitrary γ open set containing x. Since U G is a γ open set containing x and x c γ (A), (U G) A and so U (G A) which implies that x c γ (G A). Therefore, G c γ (A) c γ (G A). (c) Now X i γ (A F ) = c γ (X (A F )) = c γ ((X A) (X F )) c γ (X A)

On δ sets in γ spaces 99 (X F ), by (b). Therefore, X i γ (A F ) (X i γ (A)) (X F ) = X (i γ (A) F ) and so i γ (A F ) i γ (A) F. (d) Now X c γ (A F ) = i γ (X (A F )) = i γ ((X A) (X F )) = i γ (X A) (X F ) = (X c γ (A)) (X F ) = X (c γ (A) F ) and so c γ (A F ) = c γ (A) F. (e) Since G D G, c γ (G D) c γ (G). By (b), c γ (G D) c γ (D) G = G which implies that c γ (G D) c γ (G) and so c γ (G D) = c γ (G). The following Theorem 2.2 shows that the intersection of two γα open sets is a γα open set and the intersection of a γ semiopen (resp. γ preopen, γβ open, γb open) set with a γα open set is a γ semiopen (resp. γ preopen, γβ open, γb open) set. We will use Lemma 1.1(a), Lemma 1.1(b) and Lemma 1.1(c) in the following Theorem without mentioning them explicitly. Theorem 2.2. If (X, µ γ ) is a γ space, then the following hold. (a)g A is γ semiopen (resp.γ preopen, γβ open, γb open) whenever G is γα open and A is γ semiopen (resp.γ preopen, γβ open, γb open). (b)g A is γα open whenever G and A are γα open. Proof. (a) Suppose G is γα open and A is γ semiopen. Then G A i γ c γ i γ (G) c γ i γ (A) c γ (i γ c γ i γ (G) i γ (A)) = c γ i γ (c γ i γ (G) i γ (A)) c γ i γ c γ (i γ (G) i γ (A)) = c γ i γ c γ i γ (G A) = c γ i γ (G A). Therefore, G A is γ semiopen. Suppose G is γα open and A is γ preopen. Then G A i γ c γ i γ (G) i γ c γ (A) = i γ (c γ i γ (G) i γ c γ (A)) i γ c γ (i γ (G) i γ c γ (A)) = i γ c γ i γ (i γ (G) c γ (A)) i γ c γ i γ c γ (i γ (G) A) i γ c γ (G A) and so G A is γ preopen. Suppose G is γα open and A is γβ open. Then G A i γ c γ i γ (G) c γ i γ c γ (A) c γ (i γ c γ i γ (G) i γ c γ (A)) = c γ i γ (c γ i γ (G) i γ c γ (A)) c γ i γ c γ (i γ (G) i γ c γ (A)) = c γ i γ c γ i γ (i γ (G) c γ (A)) c γ i γ c γ i γ c γ (G A) = c γ i γ c γ (G A) and so G A is γβ open. Suppose G is γα open and A is γb open. Then G A G (c γ i γ (A) i γ c γ (A)) = (G c γ i γ (A)) (G i γ c γ (A)) c γ i γ (G A) i γ c γ (G A) and so G A is γb open. (b) Suppose G and A are γα open. Then G A i γ c γ i γ (G) i γ c γ i γ (A) i γ (c γ i γ (G) i γ c γ i γ (A)) i γ c γ (i γ (G) i γ c γ i γ (A)) = i γ c γ i γ (i γ (G) c γ i γ (A)) i γ c γ i γ c γ (i γ (G) i γ (A)) i γ c γ i γ c γ i γ (G A) = i γ c γ i γ (G A) and so G A is γα open. Theorem 2.3. If (X, µ γ ) is a γ space, G is γ open and A X, then the following hold. (a)g i σ (A) i σ (G A). (b)g i α (A) i α (G A). (c)g i π (A) i π (G A). (d)g i β (A) i β (G A). (e)g i b (A) i b (G A). (f)g c σ (A) c σ (G A). (g)g c α (A) c α (G A). (h)g c π (A) c π (G A). (i)g c β (A) c β (G A). (j)g c b (A) c b (G A).

100 V. Renuka Devi and D. Sivaraj Proof. (a) Let G be γ open and A be a subset of X. Then G i σ (A) is a γ semiopen set by Theorem 2.2(a), such that G i σ (A) G A. Therefore, G i σ (A) i σ (G A). Similarly, we can prove (b), (c) (d) and (e). (f) Let x G c σ (A) and U be an arbitrary σ open set containing x. Since U G is a σ open set containing x and x c σ (A), (U G) A and so U (G A) which implies that x c σ (G A). Therefore, G c σ (A) c σ (G A). Similarly, we can prove (g), (h), (i) and (j). The following Corollary 2.4 shows that if γ Γ 4, then i α Γ 4 and Theorem 2.3(b) above is also true for γα open sets. The proof follows from Theorem 2.2(b) and the fact that the set of all γα open sets coincides with the set of all i α open sets. Corollary 2.4. If (X, µ γ ) is a γ space, G and A are subsets of X, then the following hold. (a) i α (G A) = i α (G) i α (A). (b) If G is γα open, then G i α (A) = i α (G A). (c) i α Γ 4. The following Corollary 2.5 follows from Theorem 2.3. Corollary 2.5. If (X, µ γ ) is a γ space, A X and G is γ open, then the following hold. (a) c σ (G c σ (A)) = c σ (G A). (b) c α (G c α (A)) = c α (G A). (c) c π (G c π (A)) = c π (G A). (d) c β (G c β (A)) = c β (G A). (e) c b (G c b (A)) = c b (G A). Let X be any nonempty set and γ Γ. A subset A of X is said to be γ regular [3] if A = γ(a). The following Theorem 2.6 shows that the intersection of two i γ c γ regular sets is again a i γ c γ regular set and Theorem 2.7 below gives characterizations of γβ open sets in γ spaces. Theorem 2.6. If (X, µ γ ) is a γ space, and A and B are i γ c γ regular sets, then A B is a i γ c γ regular set. Proof. Suppose A and B are i γ c γ regular sets. Now A B = i γ c γ (A) i γ c γ (B) = i γ (c γ (A) c γ (B)) by Lemma 1.1(b) and so i γ c γ (A) i γ c γ (B) i γ c γ (A B). Since the intersection of two γ open set is a γ open set, by Lemma 1.1(a), A B = i γ (A B) i γ c γ (A B). Therefore, A B = i γ c γ (A B) which implies that A B is i γ c γ regular. Theorem 2.7. If (X, µ γ ) is a γ space and A is a subset of X, then the following statements are equivalent. (a) A is γβ open. (b) c γ (A) = c γ i γ c γ (A). (c) c γ (A) is c γ i γ regular. (d) There is a γ preopen set U such that U A c γ (U). (e) c γ (A) is γ semiopen. (f) c σ (A) is γ semiopen.

On δ sets in γ spaces 101 (g) c π (A) is γβ open. Proof. The equivalence of (a) and (b) is clear. (a) (c). If A is γβ open, then c γ (A) = c γ i γ c γ (A) and so c γ (A) is c γ i γ regular. (c) (d). Let U = i π (A). Then U is a γ preopen set such that U A. Now c γ (U) = c γ (i π (A)) = c γ i γ c γ (A), by Lemma 1.1(f). Therefore, c γ (U) = c γ (A) and so U A c γ (U). (d) (a). Suppose U is a γ preopen set such that U A c γ (U). Then c γ (U) = c γ (A). Since U is γ preopen, U i γ c γ (U) and so A c γ (A) = c γ (U) c γ i γ c γ (U) c γ i γ c γ (A) and so A is γβ open. (c) implies (e) is clear. (e) (f). Suppose c γ (A) is γ semiopen. Now i γ c γ (A) = i γ c σ (A), by Lemma 1.1(g) and so i γ c γ (A) c σ (A) c γ (c σ (A)) = c γ (A), by Lemma 1.1(d). Therefore, i γ c γ (A) c σ (A) c γ (A) c γ i γ c γ (A). Since i γ c γ (A) is γ open, c σ (A) is γ semiopen. (f) (a). Suppose c σ (A) is γ semiopen. Then, A c σ (A) c γ i γ (c σ (A)) = c γ i γ c γ (A), by Lemma 1.1(g) and so A is γβ open. (a) (g). Suppose A is γβ open. Since every γ open set is a γ preopen set, c π (A) c γ (A) c γ i γ c γ (A) = c γ i γ c γ (c π (A)), by Lemma 1.1(e) and so (g) follows. (g) (a). Suppose c π (A) is γβ open. Then A c π (A) c γ i γ c γ (c π (A)) = c γ i γ c γ (A), by Lemma 1.1(e). Therefore, A is γβ open. Let X be a nonempty set and γ Γ. A subset A of X is said to be γ locally closed if A = G F where G is γ open and F is γ closed. Since X is γ closed, every γ open set is a γ locally closed set. The following Theorem 2.8 gives a characterization of γ locally closed sets, the proof is similar to the proof of the characterizations of locally closed sets [1] in any topological space and hence is omitted. Theorem 2.9 shows that for γ dense sets, the concepts γ open and γ locally closed on the subsets of X are equivalent. Theorem 2.8. Let X be a nonempty set, γ Γ and A be a subset of X. Then the following statements are equivalent. (a) A is γ locally closed. (b) A = G c γ (A) for some γ open set G. (c) c γ (A) A is γ closed. (d) A (X c γ (A)) is γ open. (e) A i γ (A (X c γ (A))). Theorem 2.9. Let X be a nonempty set, γ Γ and A be a γ dense subset of X. Then the following statements are equivalent. (a) A is γ open. (b) A is γ locally closed. Proof. Enough to prove (b) implies (a). Suppose A is γ dense and γ locally closed. Then A = G c γ (A) for some γ open set G. Therefore, A = G X = G and so A is γ open. The following Theorem 2.10 gives decompositions of γ open sets in γ spaces. Theorem 2.10. Let (X, µ γ ) be a γ space and A be a subset of X. Then the following statements are equivalent. (a) A is γ open.

102 V. Renuka Devi and D. Sivaraj (b) A is γα open and γ locally closed. (c) A is γ preopen and γ locally closed. Proof. It is enough to prove that (c) implies (a). (c) (a). Suppose A is γ preopen and γ locally closed. Since A is γ preopen, A i γ c γ (A). Since A is γ locally closed, A = G c γ (A) for some γ open set G. Now A = A i γ c γ (A) = (G c γ (A)) i γ c γ (A) = G i γ c γ (A) = i γ (G c γ (A)), by Lemma 1.1(b). Therefore, A = i γ (A) which implies that A is γ open. The following Theorem 2.11 gives characterizations of γ preopen sets in a γ space. Theorem 2.11. Let (X, µ γ ) be a γ space and A X. Then the following statements are equivalent. (a) A π(γ). (b) There is an i γ c γ regular set G such that A G and c γ (A) = c γ (G). (c) A = G D where G is a i γ c γ regular set and D is a γ dense set. (d) A = G D where G is a γ open set and D is a γ dense set. Proof. (a) (b). If A π(γ), then A i γ c γ (A) c γ (A) which implies that c γ (A) c γ i γ c γ (A) c γ (A) and so c γ i γ c γ (A) = c γ (A). Let G = i γ c γ (A). Then A G and i γ c γ (G) = i γ c γ i γ c γ (A) = i γ c γ (A) = G which implies that G is i γ c γ regular. Also c γ (G) = c γ i γ c γ (G) = c γ (A). (b) (c). Let G be an i γ c γ regular set such that A G and c γ (A) = c γ (G). Let D = A (X G). Then A = G D where G is i γ c γ regular. Now c γ (D) = c γ (A (X G)) = c γ (A) c γ (X G) = c γ (G) c γ (X G) = c γ (G (X G)) = c γ (X) = X. Hence D is γ dense. (c) (d). The proof follows from the fact that every i γ c γ regular set is a γ open set. (d) (a). Suppose A = G D where G is γ open and D is γ dense. Now G = G X = G c γ (D) c γ (G D) and so G = i γ (G) i γ c γ (G D) = i γ c γ (A) which implies that A i γ c γ (A). Hence A π(γ). 3. δ γ open Sets Let X be a nonempty set, γ Γ and A X. A is said to be δ γ open or A δ γ if and only if i γ c γ (A) c γ i γ (A). In topological spaces, the set of all δ i open sets coincides with the set of all δ sets [2]. The γ boundary of a subset A of X, denoted by bd γ (A), is given by bd γ (A) = c γ (A) i γ (A) = c γ (A) c γ (X A). A subset A of X is said to be µ γ rare if i γ c γ (A) =. In topological spaces, the set of all µ i rare sets coincides with the set of all nowhere dense sets. Every µ γ rare set is a δ γ open set, since i γ c γ (A) = c γ i γ (A). It is easy to show that every γ closed set is a δ γ open set. The following Theorem 3.1 gives some properties of µ γ rare sets. Theorem 3.1. Let X be a nonempty set and γ Γ. Then the following hold. (a) is µ γ rare. (b) Subset of a µ γ rare set is a µ γ rare set. (c) If A is a µ γ rare set, then bd γ (A) is a µ γ rare set.

On δ sets in γ spaces 103 Proof. (a) If M γ = {A A µ γ }, then c γ i γ (X) = c γ (M γ ) = X and so X c γ i γ (X) = which implies that i γ c γ ( ) =. (b) The proof is clear. (c) Since A is µ γ rare, i γ c γ (A) =. Now i γ c γ (bd γ (A)) = i γ c γ (c γ (A) i γ (A)) = i γ c γ (c γ (A) (X i γ (A))) i γ (c γ (A) c γ (X i γ (A))) i γ c γ (A) =. Therefore, bd γ (A) is a µ γ rare set. The following Theorems 3.2, 3.3 and 3.4 deal with µ γ rare sets and γ boundary of subsets of X in a γ space, which are essential to characterize δ γ open sets in Theorem 3.9. Also, in a γ space, one can easily prove the formulas 1 to 15 in [6, Page 56]. Theorem 3.2. Let (X, µ γ ) be a γ space and A and B be subsets X. Then the following hold. (a) If A is γ open, then bd γ (A) = c γ (A) A is µ γ rare. (b)bd γ (A B) bd γ (A) bd γ (B). Proof. (a) i γ c γ (c γ (A) A) = i γ c γ (c γ (A) (X A)) i γ (c γ (A) c γ (X A)) = i γ c γ (A) i γ c γ (X A) = i γ c γ (A) i γ (X A) = i γ c γ (A) (X c γ (A)) =. (b) bd γ (A B) = c γ (A B) c γ (X (A B)) = c γ (A B) (c γ (X A) c γ (X B)) (c γ (A) c γ (B)) (c γ (X A) c γ (X B)) = (c γ (A) (c γ (X A) c γ (X B))) (c γ (B) (c γ (X A) c γ (X B))) (c γ (A) c γ (X A)) (c γ (B) c γ (X B)) = bd γ (A) bd γ (B). Theorem 3.3. Let (X, µ γ ) be a γ space. If A and B are µ γ rare subsets of X, then A B is also a µ γ rare set. Proof. i γ c γ (A B) = i γ (c γ (A) c γ (B)), by Lemma 1.1(c) and so i γ c γ (A B) i γ c γ (A) c γ (B) = c γ (B) by Theorem 2.1(c). Therefore, i γ c γ (A B) i γ c γ (B) = and so A B is µ γ rare. Theorem 3.4. If (X, µ γ ) is a γ space, G is γ open and both A Gand G A are µ γ rare, then B H and H B are µ γ rare, where H = X c γ (G) and B = X A. Proof. Since A c γ (G) A G and A G is µ γ rare, A c γ (G) is µ γ rare. Since c γ (G) A = (G A) ((c γ (G) G) A), by Theorem 3.1(b) and Theorem 3.3, c γ (G) A is µ γ rare. Now B H = B (X c γ (G)) = (X A) c γ (G) = c γ (G) A and H B = (X c γ (G)) B = (X c γ (G)) (X A) = A c γ (G). Therefore, B H and H B are µ γ rare. The following Theorem 3.5 shows that every γ semiopen is a δ γ open set and the complement of a δ γ open set is a δ γ open set. Theorems 3.6 and 3.8 give more properties of δ γ open sets. Theorem 3.5. Let X be a nonempty set and γ Γ. Then the following hold. (a) If A is γ semiopen, then A δ γ. (b) If A δ γ, then X A δ γ. Proof. (a) If A is γ semiopen, then A c γ i γ (A). Now, i γ c γ (A) i γ c γ c γ i γ (A) c γ i γ (A) and so A δ γ. (b) A δ γ implies that i γ c γ (A) c γ i γ (A) and so X c γ i γ (A) X i γ c γ (A) which in turn implies that i γ (X i γ (A)) c γ (X c γ (A)) and so i γ c γ (X A) c γ i γ (X A). Hence X A δ γ. Theorem 3.6. Let (X, µ γ ) be a γ space. If A δ γ and B δ γ, then A B δ γ.

104 V. Renuka Devi and D. Sivaraj Proof. A, B δ γ implies that i γ c γ (A) c γ i γ (A) and i γ c γ (B) c γ i γ (B). Now i γ c γ (A B) i γ (c γ (A) c γ (B)) = i γ c γ (A) i γ c γ (B) by Lemma 1.1(b). Since A δ γ, it follows that i γ c γ (A B) c γ i γ (A) i γ c γ (B) c γ (i γ (A) i γ c γ (B)), by Theorem 2.1(b). Since B δ γ, i γ c γ (A B) c γ (i γ (A) c γ i γ (B)) c γ c γ (i γ (A) i γ (B)) = c γ (i γ (A) i γ (B)) = c γ i γ (A B). Hence i γ c γ (A B) c γ i γ (A B) and so A B δ γ. Corollary 3.7. Let (X, µ γ ) be a γ space. If A δ γ and B δ γ, then A B δ γ. Proof. The proof follows from Theorem 3.5(b) and Theorem 3.6. Theorem 3.8. Let (X, µ γ ) be a γ space and A and B be subsets of X such that A δ γ. Then i γ c γ (A B) = i γ c γ (A) i γ c γ (B). Proof. Since i γ c γ (A)and i γ c γ (B) are γ open sets, i γ c γ (A) i γ c γ (B) is also γ open by Lemma 1.1(a) and so i γ c γ (A) i γ c γ (B) = i γ (i γ c γ (A) i γ c γ (B)) i γ (c γ i γ (A) i γ c γ (B)), since A δ γ. Therefore, i γ c γ (A) i γ c γ (B) i γ c γ (i γ (A) i γ c γ (B)) i γ c γ (i γ (A) c γ (B)) i γ c γ c γ (i γ (A) B) i γ c γ (A B). Also, i γ c γ (A B) i γ (c γ (A) c γ (B)) = i γ c γ (A) i γ c γ (B). Hence i γ c γ (A B) = i γ c γ (A) i γ c γ (B). Theorem 3.9. Let (X, µ γ ) be a γ space and A X. Then the following are equivalent. (a) A δ γ. (b) A is the union of a γ semiopen set and a µ γ rare set. (c) A is the union of a γ open set and a µ γ rare set. (d) bd γ (A) is µ γ rare. (e) There is a γ open set G such that A G and G A are µ γ rare. (f) A = B C where B is γ semiopen and C is γ closed. (g) A = B C where B is γ semiopen and C is γα closed. (h) A = B C where B is γ semiopen and C is γ semiclosed. Proof. (a) (b). A = (A c γ i γ (A)) (A c γ i γ (A)). Let B = A c γ i γ (A) and C = A c γ i γ (A). Then i γ (A) B and B c γ i γ (A) which implies that B c γ i γ (B) and so B is γ semiopen. Now C i γ (A) = (A c γ i γ (A)) i γ (A) = and c γ (C) i γ (A) = c γ (A c γ i γ (A)) i γ (A) (c γ (A) i γ c γ i γ (A)) i γ (A) =. Again, by Lemma 1.1(b), i γ c γ (C) = i γ c γ (A c γ i γ (A)) i γ (c γ (A) i γ c γ i γ (A)) = i γ c γ (A) c γ i γ c γ i γ (A) = i γ c γ (A) c γ i γ (A), since c γ i γ Γ 2. Since A δ γ, i γ c γ (C) c γ i γ (A) c γ i γ (A) = and so C is µ γ rare. (b) (c). Suppose A = B C where B is γ semiopen and C is µ γ rare. Since B is γ semiopen, there exists a γ open set G such that G B c γ (G) and so B = G (B G). Since B G c γ (G) G and c γ (G) G is µ γ rare by Theorem 3.2(a), B G is µ γ rare. Therefore, A = G (B G) C and so (c) follows from Theorem 3.3. (c) (d). Suppose A = G B where G is γ open and B is µ γ rare. Now bd γ (A) = bd γ (G B) bd γ (G) bd γ (B), by Theorem 3.2(b). By Theorem 3.2(a), bd γ (G) is µ γ rare and by Theorem 3.1(c), bd γ (B) is µ γ rare. By Theorem 3.3, bd γ (G) bd γ (B) is µ γ rare and so bd γ (A) is µ γ rare. (d) (e). Suppose G = i γ (A). Then G A = and A G = A i γ (A) c γ (A) i γ (A) = bd γ (A). G is the required γ open set such that G A and A G are µ γ rare. (e) (f). Suppose G is a γ open set such that G A and A G are µ γ rare sets. If H = G c γ (G A), then H is a γ open set such that H A and so H A is

On δ sets in γ spaces 105 µ γ rare. Moreover, A H = A (G c γ (G A)) = (A G) c γ (G A). Since G A and A G are µ γ rare, it follows that A H is µ γ rare. Thus A = H (A H), union of a γ open set and a µ γ rare set which is nothing but (c). If B = X A and K = X c γ (H), then B K and K B are µ γ rare by Theorem 3.4. Thus K is a γ open set such that B K and K B are µ γ rare. Therefore, by (c), B = U R where U is γ open and R is µ γ rare. Hence A = (X U) (X R) where X U is γ closed. Now, c γ i γ (X R) = X i γ c γ (R) = X and so X R is γ semiopen. Therefore, A is the intersection of a γ closed set and a γ semiopen set. (f) (g). The proof follows from the fact that every γ closed set is a γα closed set. (g) (h). The proof follows from the fact that every γα closed set is a γ semiclosed set. (h) (a). Suppose A = B C where B is γ semiopen and C is γ semiclosed. Now i γ c γ (A) = i γ c γ (B C) i γ c γ (c γ i γ (B) C) i γ (c γ i γ (B) c γ (C)) = i γ c γ i γ (B) i γ c γ (C) c γ i γ (B) i γ c γ (C) c γ (i γ (B) i γ c γ (C)) = c γ (i γ (B) i γ (C)), since C is γ semiclosed. Therefore, i γ c γ (A) c γ i γ (B C) = c γ i γ (A). Hence A is δ γ open. REFERENCES [1 ] N. Bourbaki, General Topology (Part I), Addison-Wesley Publishing Company, Inc., 1966. [2 ] C. Chattopadhyay and C. Bandyopadhyay, On structure of δ sets, Bull. Calcutta Math. Soc., 83(1991), 281-290. [3 ] Á. Császár, Generalized Open Sets, Acta Math. Hungar., 75(1-2)(1997), 65-87. [4 ] Á. Császár, Generalized topology, generalized continuity, Acta Math. Hungar., 96(2002), 351-357. [5 ] A. Güldürdek and O. B. Özbakir, On γ semiopen sets, Acta Math. Hungar., 109(4)(2005), 347-355. [6 ] K. Kuratowski, Topology I, Academic Press, New York, 1966. [7 ] P. Sivagami, Remark on γ interior, Acta Math. Hungar., To appear. Address V. Renuka Devi: Department of Mathematics, A. J. College, Sivakasi, Tamil Nadu, INDIA E-mail: renu siva2003@yahoo.com D. Sivaraj:

106 V. Renuka Devi and D. Sivaraj Department of Computer Applications, D. J. Academy for Managerial Excellence, Coimbatore - 641 032, Tamil Nadu, INDIA E-mail: ttn sivaraj@yahoo.com