REGULAR GENERALIZED CLOSED SETS IN TOPOLOGICAL SPACES

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1 International Journal of Mathematics and Computing Applications Vol. 3, Nos. -, January-December 0, pp. -5 ISSN: International Science Press REGULAR GENERALIZED CLOSED SES IN OPOLOGICAL SPACES A.I. EL-Maghrabi and A.M. Zahran Abstract: In this paper, we introduce and study the notion of regular generalized γ closed sets. Also, some of their properties are investigated. Further, the notion of regular generalized γ open sets and the notion of regular generalized γ neighbourhood are discussed. Moreover, some of its basic properties are introduced. Finally, we give an applications of rgγ closed sets, say γ regular spaces. Mathematics Subject Classification: 54C08, 54C0, 54C05. Keywords: regular generalized γ closed sets; regular generalized γ open sets; regular generalized γ neighbourhood; regular γ kernel.. INRODUCION AND PRELIMINARIES N. Levine [5] introduced generalized the notion of closed sets in general topology as a generalization of closed sets. his notion was found to be useful and many results in general topology were improved. Many researchers like Balachandran et.al [4], Arockiarani [], Dunham [8], Gnanambal [], Malghan [7], Palaniappan et.al [9], Arya et.al [3] and Keskin et.al [3] have worked on generalized closed sets, their generalizations and related concepts in general topology. he purpose of this paper is to define and study the properties of regular generalized γ closed (breifly, rgγ closed) sets. Also, we define the concept of regular generalized γ open (breifly, rgγ open) sets and obtained some of its properties and results. Further, we define γ regular spaces as the spaces in which every regular generalized γ closed set is γ closed and study its properties. hroughout the paper X and Y denote the topological spaces (X; τ) and (Y, σ) respectively and on which no separation axioms are assumed unless otherwise explicitly stated. For any subset A of a space (X, τ), the closure of A, the interior of A, the gpr closure of A, the gpr interior of A, the w closure of A, the w interior of A, Department of Mathematics, Faculty of Science, Kafr El-Sheikh University, Kafr El-Sheikh, Egypt. aelmaghrabi@yahoo.com Department of Mathematics, Faculty of Science, Al-AZahar University, Assiut, Egypt. amzahranj@yahoo.com

2 International Journal of Mathematics and Computing Applications the gpr closure of A, the gpr interior of A and the complement of A are denoted by cl(a), int(a), γcl(a), γ int(a), ωcl(a), ω int(a), gpr cl(a), gpr int(a) and A c (equivalently, X A or X\A) respectively. (X, τ) will be replaced by X if there is no chance of confusion. Let us recall the following definitions as pre-requesties. Definition.: A subset A of a space X is called: (i) a preopen set [8] if A int(cl(a)) and a preclosed set if cl(int(a)) A, (ii) a semi-open set [4] if A cl(int(a)) and a semi-closed set if int(cl(a)) A, (iii) a γ open [9] (equivlently, b open[ ] or sp open [7 ]) set if A int(cl(a)) cl(int(a)) and a γ closed ( equivlently, b closed or sp closed) set if int(cl(a)) cl(int(a)) A, (iv) a regular open set [] if if A = int(cl(a)) and a regular closed set if cl(int(a)) = A, (v) π open [3] if A is the finite union of regular open sets and the complement of π open set is said to be π closed. he intersection of all γ-open subsets of X containing A is called the γ kernel of A and is denoted by γ ker(a). he intersection of all preclosed ( resp. γ closed ) subsets of X containing A is called the pre-closure (resp. γ closure ) of A and is denoted by pcl(a)) ( resp. γcl(a)). Definition.: A subset A of a space X is called: () generalized closed set (breifly, g closed ) [5] if cl(a) H whenever A H and H is open in X, () γ generalized closed set (breifly, γg closed ) [] if γcl(a) H whenever A H and H is open in X, (3) generalized γ closed set (breifly, gγ closed )[0 ] if γcl(a) H whenever A H and H is γ open in X, (4) generalized preclosed set (breifly, gp closed )[6] if pcl(a) H whenever A H and H is open in X, (5) regular generalized closed set (breifly, rg closed )[9] if cl(a) H whenever A H and H is regular open in X, (6) generalized preregular closed set (breifly, gpr closed )[] if pcl(a) H whenever A H and H is regular open in X,

3 Regular Generalized closed Sets in opological Spaces 3 (7) π generalized closed set (breifly, πg closed )[6] if cl(a) H whenever A H and H is π open in X, (8) weakly closed set, (breifly, ω closed )[0] if cl(a) H whenever A H and H is semi-open in X. he complements of the above mentioned closed sets are their respectively open sets.. BASIC PROPERIES OF REGULAR GENERALIZED CLOSED SES We introduce the following definitions. Definition.: A subset A of a space X is called a regular γ open set ( breifly, rγ open ) if there is a regular open set H such that H A γcl(h). he faimly of all regular γ open sets of X is denoted by RγO(X). Definition.: A subset A of a space X is called a regular generalized γ closed set (breifly, rgγ closed ) if γcl(a) H whenever A H and H is regular γ open in X. he faimly of all regular gγ closed sets of X is denoted by RGγC(X). In the following proposition, we prove that the class of rgγ closed sets has properly lies between the class of gγ closed sets and the class of rg closed sets. Proposition.3: For a space (X, τ), we have (i) Every gγ closed (resp. ω closed, closed, regular closed, π closed ) set is rgγ closed, (ii) Every rgγ closed set is gpr closed ( resp. rg closed ). Proof: he proof follows from definitions and the fact that every regular open sets are rγ open. he converse of the above proposition need not be true, as seen from the following examples. Example.4: Let X = {a, b, c, d, e} with topology τ = {X, φ, {a}, {d}, {e}, {a, d}, {a, e}, {d, e}, {a, d, e}}. hen, () the set A = {a, d, e} is rgγ closed set but not gγ closed set in X, () the set B = {b} is rgγ closed set but not ω closed set in X, (3) the set D = {a, b} is rg closed and gpr closed set but not rgγ closed set in X, Remark.5: From the above discussions and known results we have the following implications, in the following diagram by A B, we mean A implies B but not conversely and A B means A and B are independent of each other.

4 4 International Journal of Mathematics and Computing Applications regular closed π closed πg closed closed ω closed rgγ closed rg closed gpr closed γ closed gγ closed γg closed Remark.6: he union of two rgγ closed subsets of X need not be a rgγ closed subset of X. Let X = {a, b, c, d, e} with a topology τ = {X, φ, {a}, {b}, {a, b}, {a, b, c}, {a, b, c, d}, {a, b, c, e}}. hen two subsets {c} and {d} of X are rgγ closed subsets but their union {c, d} is not rgγ closed subset of X. heorem.7: he arbitrary intersection of any rgγ closed subsets of X is also rgγ closed subset of X. Proof: Let {A i : i I} be any collection of rgγ closed subsets of X such that i A i H and H be rγ open in X. But A i is rgγ closed subset of X, for each i I, then γcl(a i ) H, for each i I. Hence i γcl(a i ) H, for each i I. herefore γcl( i A i ) H. So, i A i a rgγ closed subset of X. heorem.8: A subset A of a space (X, τ) is rgγ closed if and only if for each A H and H is rγ open, there exists a γ closed set F such that A F H. Proof: Suppose that A is a rgγ closed set, A H and H is a rγ open set. hen γcl(a) H. If we put F = γcl(a), hence A F H. Conversely, Assume that A H and H is a rγ open set. hen by hypothesis there exists a γ closed set F such that A F H. So, A γ cl(a) F and hence γcl(a) H. herefore A is rgγ closed. heorem.9: If A is closed and B is a rgγ closed subset of a space X, then A B is also rgγ closed. Proof: Suppose that A B H and H is a rγ open set. hen A H and B H. But B is rgγ closed, then γcl(b) H and hence A B A γcl(b) H. But A γcl(b) a γ closed set, hence there exists a γ closed set A γcl(b) such that A B A γcl(b) H. hus by heorem.8, A B is rgγ closed. heorem.0: If a subset A of X is a rgγ closed set in X, then γcl(a)\a does not contain any non-empty rγ open set in X. Proof: Assume that A is a rgγ closed set in X. Let H be a rγ open set such that H γcl(a)\a and H φ. New H γcl(a)\a. herefore H X\A which implies A X\H. Since H is a rγ open set, then X\H is also rγ open in X. But A is a rgγ closed set in X, then γcl(a) X\H. So, H X\γcl(A). Also, H γcl(a). Hence, H (γcl(a)

5 Regular Generalized closed Sets in opological Spaces 5 (X\γcl(A))) = φ. his shows that, H = φ which is a contradiction. herefore, γcl(a)\a does not contain any non-empty rγ open set in X. he converse of the above theorem need not be true as seen from the following example. Example.: In Example.4, if A = {a, b}, then γcl(a)\a = {a, b, c}\{a, b} = {c} does not contain any non-empty rγ open set in X. But A is not a rgγ closed set in X. Proposition.: If a subset A of X is a rgγ closed set in X, then γcl(a)\a does not contain any regular closed (resp.regular open) set in X. Proof: Follows directly from heorem.0 and the fact that every regular open set is rγ open. heorem.3: For an element p X, the set X /{p} is rgγ closed or rγ open. Proof: Suppose that X /{p} is not a rγ open set. hen X is the only rγ open set containing X /{p}. his implies that γcl(x\{p}) X. Hence X \{p} is a rgγ closed set in X. heorem.4: If A is regular open and rgγ closed, then A is regular closed and hence γ clopen. Proof: Assume that A is regular open and rgγ closed. Since every regular open set is rγ open and A A, we have γcl(a) A. But A γcl(a). herefore A = γcl(a), that is, A is γ closed. Since A is regular open, A is γ open. herefore A is regular closed and γ clopen. heorem.5: If A is a rgγ closed set of X such that A B γcl(a), then B is a rgγ closed set in X. Proof: Let H be a rγ open set of X such that B H. hen A H. But A is a rgγ closed set of X, then γcl(a) H. Now, γcl(b) γcl(γcl(a)) = γcl(a) H. herefore B is a rgγ closed set in X. Remark.6: he converse of the above theorem need not be true in general. In Example.4, if we take a subset A = {b} and B = {b, c}, then A and B are rgγ closed set in (X, τ), but A B is not subset in γcl(a). heorem.7: Let A be a rgγ closed set in X. hen A is γ closed if and only if γcl(a)\a is rγ open. Proof: Suppose that A is γ closed in X. hen γcl(a) = A and so γcl(a)\a = φ, which is rγ open in X. Conversely, Suppose that γcl(a)\a is a rγ open set in X. Since A is rgγ closed, then by heorem.0, γcl(a)\a does not contain any non-empty rγ open set in X. hen γcl(a)\a = φ, hence A is a γ closed set in X.

6 6 International Journal of Mathematics and Computing Applications heorem.8: If A is regular open and rg closed, then A is a rgγ closed set in X. Proof: Let H be any rγ open set in X such that A H. Since A is regular open and rg closed, we have γcl(a) A. hen γcl(a) A H. Hence A is a rgγ closed set in X. heorem.9: If a subset A is both rγ open and rgγ closed in a topological space (X, τ), then A is γ closed. Proof: Assume that A is both rγ open and rgγ closed in a topological space (X, τ). hen γcl(a) A. Hence A is γ closed. heorem.0: If A is both rγ open and rgγ closed subset in X and F is a closed set in X, then A F is a rgγ closed set in X. Proof: Let A be rγ open and rgγ closed subset in X and F be a closed set in X. hen by heorem.9, A is γ closed. So, A F is γ closed. herefore A F is a rgγ closed set in X. heorem.: If A is an open set and H is a γ open set in a topological space (X, τ), then A H is a γ open set in X. heorem.: If A is both open and g closed subset in X, then A is a rgγ closed set in X. Proof: Let A be an open and g closed subset in X and A H, where H is a rγ open set in X. hen by hypothesis γcl(a) A, that is, γcl(a) H. hus A is a rgγ closed set in X. Remark.3: If A is both open and rgγ closed subset in X, then A need not be g closed set in X. In Example.4, the subset {a, d, e} is an open and rgγ closed set but not g closed. heorem.4: For a topological space (X, τ), if RγO(X, φ) = {X, φ}, then every subset of X is a rgγ closed subset in X. Proof: Let (X, τ) be a space and RγO(X, τ) = {X, φ}. Let A be a subset of X. (i) if A = φ, then A is a rgγ closed subset in X. (ii) if A φ, then X is the only a rγ open set in X containing A and so γcl(a) X. Hence A is a rgγ closed subset in X. he converse of heorem.4 need not be true in general as seen from the following example. Example.5: Let X = {a, b, c, d,} with topology τ = {X, φ, {a, b}, {c, d}}. hen every subset of (X, τ) is a rgγ closed subset in X. But RγO(X, τ) = τ. heorem.6: For a topological space (X, τ), then RγO(X, τ) {F X : F is closed} if and only if every subset of X is a rgγ closed subset in X.

7 Regular Generalized closed Sets in opological Spaces 7 Proof: Suppose that RγO(X, τ) {F X : F is closed}. Let A be any subet of X such that A H, where H is a rγ open set in X. hen H RγO(X, τ) {F X : F is closed}. hat is H {F X : F is closed}. hus H is a γ closed set. hen γcl(h) = H. Also, γcl(a) γcl(h) H. Hence A is a rgγ closed subset in X. Conversely, Suppose that every subset of (X, τ) is a rgγ closed subset in X. Let H Rγ (X, τ). Since H H and H is rgγ closed, we have γcl(h) H. hus γcl(h) = H and H {F X : F is closed}. herefore RγO(X, τ) {F X : F is closed}. Definition.7: he intersection of all rγ open subsets of (X, τ) containing A is called the regular γ kernel of A and is denoted by rγ ker(a). Lemma.8: For any subset A of a toplogical space (X, τ), then A rγ ker(a). Proof: Follows directly from Definition.7. Lemma.9: Let (X, τ) be a topological space and A be a subset of X. If A is a rγ open in X, then rγ ker(a) = A. Lemma.30: Let (X, τ) be a topological space and A be a subset of X. hen γ ker(a) rγ ker(a). Proof: Follows from the implication RγO(X, τ) γo(x, τ). 3. REGULAR GENERALIZED OPEN SES In this section, we introduce and study the concept of regular generalized γ open (briefly, rgγ open) sets in topological spaces and obtain some of their properties. Definiton 3.: A subset A of a topological space ( X, τ) is called a regular generalized γ open (breifly, rgγ open) set in X if A c is rgγ closed in X. We denote the family of all rgγ open sets in X by RGγO(X). he following theorem gives equivalent definitions of a rgγ open set. heorem 3.: Let (X, τ) be a topological space and A X. hen the following statements are equivalent: (i) A is a rgγ open set, (ii) for each rγ closed set F contained in A, F γ int(a), (iii) for each rγ closed set F contained in A, there exists a γ open set H such that F H A. Proof: (i) (ii). Let F A and F be a rγ closed set. hen X\A X\F which is rγ open of X, hence γcl(x\a) X\F. So, F γ int(a). (ii) (iii). Suppose that F A and F be a rγ closed set. hen by hypothesis, F γ int(a). Set H = γ int(a), hence there exists a γ open set H such that F H A.

8 8 International Journal of Mathematics and Computing Applications (iii) (i). Assume that X\A V and V is rγ open of X. hen by hypothesis, there exists a γ open set H such that X\V H A, that is, X\A X\H V. herefore, by heorem.8, X\A is rgγ closed in X and hence A is rgγ open in X. heorem 3.3: For a topological space ( X, τ), then every ω open set is rgγ open. Proof: Let A be a ω open set in X. hen A c is ω closed in X. Hence by Proposition.3, A c is rgγ closed in X. herefore, A is rgγ open. he converse of the above theorem need not be true as seen from the following example. Example 3.4: In Example.4, the subset A = {b} is rgγ open but not ω open. heorem 3.5: For a topological space (X, τ), then every open ( resp. regularopen) set is rgγ open but not conversely. Proof: Follows from S. John [] ( resp. Stone [] ) and heorem 3.3. heorem 3.6: Let (X, τ) be a topological space. hen every rgγ open set of X is rg open (resp. gpr open). Proof: (i) Let A be a rgγ open set of X. hen A c a rgγ closed set of X. hen by Proposition.3, A c is a rg closed set of X. herfore A is rg open in X. (ii) For the case of a gpr open set is similar to (i). he converse of the above theorem need not be true as seen from the following example. Example 3.7: In Example.4, the subset {a, b} is rg open (resp. gpr open) but not a rgγ open set in X. heorem 3.8: he arbitrary union of rgγ open sets of X is rgγ open. Proof: Obvious from heorem.7. Remark 3.9: he intersection of two rgγ open subsets of X need not be a rgγ open subset of X. In Remark.6, two subsets {a, b, d, e} and {a, b, c, e} of X are rgγ open subsets but their intersection {a, b, e} is not rgγ open subset of X. heorem 3.0: If A is an open and B is a rgγ open subsets of a space X, then A B is also rgγ open. Proof: Follows from heorem.9. Proposition 3.: If γ int(a) B A and A is a rgγ open set of X, then B is rgγ open. Proposition 3.: Let A be a rγ closed and a rgγ open set of X. hen A is γ open.

9 Regular Generalized closed Sets in opological Spaces 9 Proof: Assume that A is a rγ closed and a rgγ open set of X. hen A γ int(a) and hence A is γ open. heorem 3.3: For a space ( X, τ), if A is a rgγ closed set, then γcl(a)\a is rgγ open. Proof: Suppose that A is a rgγ closed set and F is a rγ closed set contained in γcl(a)\a. hen by heorem.8, F = φ and hence F γ int(γcl(a)\a). herefore, γcl(a)\a is rgγ open. heorem 3.4: If a subset A of a space (X, τ) is rgγ open, then H = X, whenever H is rγ open and int(a) A c H. Proof: Assume that A is rgγ open in X. Let H be rγ open and int(a) A c H. his implies that H c (int(a) A c ) c = (int(a)) c A, that is, H c (int(a)) c \A c. hus H c cl(a) c \A c, since (int(a)) c = cl(a) c. Now H c is rγ open and A c is rgγ closed, hence by heorem.0, it follows that H c = φ. hen H = X. he converse of the above theorem is not true in general as seen from the following example. Example 3.5: Let X = {a, b, c, d} with topology τ = {X, φ, {a}, {b}, {a, b}, {b, c}, {a, b, c}, {a, b, d}}. hen RGγO(X) = {X, φ, {a}, {b}, {c}, {d}, {a, b}, {b, c}, {c, d}, {a, d}, {b, d}, {a, c}, {a, b, c}, {a, b, d}} and RγO(X) = {X, φ, {a}, {b}, {c}, {a, d}, {b, c}, {b, c, d}}. If we take A = {b, c, d}, then A is not rgγ open. However int(a) A c = {b, c} {a} = {a, b, c}. So, for some rγ open H, we have int(a) A c = {a, b, c} H, gives H = X, but A is not rgγ open. heorem 3.6: For a topological space (X, τ), then every singleton of X is eiher rgγ open or rγ open. Proof: Let (X, τ) be a topological space. Let p X. o prove that {p} is eiher rgγ open or rγ open. hat is to prove X\{p} is eiher rgγ closed or rγ open, which follows directly from heorem SOME APPLICAIONS ON rg CLOSED AND rg OPEN SES In this section, we introduce and study the concept of regular generalized γ neighbourhood (briefly, rgγ nbd) in topological spaces by using the concept of rgγ open sets. Further, we prove that every neighbourhood of a point p in X is rgγ nbd of p but conversely not true. Definition 4.: Let (X, τ) be a topological space and let p X. A subset N of X is said to be a regular generalized γ neighbourhood (briefly, rgγ nbd) of p if and only if there exists a rγ open set H such that p H N.

10 0 International Journal of Mathematics and Computing Applications Definition 4.: A subset N of a topological space ( X, τ) is called a regular generalized γ neighbourhood (briefly, rgγ nbd) of a subset A of a space (X, τ) if and only if there exists a rγ open set H such that A H N. Remark 4.3: he rgγ nbd of N of a point p of a space X need not be a rgγ open in X. Example 4.4: Let X = {a, b, c, d} with topology τ = {X, φ, {a}, {b}, {a, b}, {a, b, c}}. hen RGγO(X) = {X, φ, {a}, {b}, {c}, {d}, {a, b}, {c, d}, {a, b, c}, {a, b, d}}. Note that {a, c} is not a rgγ open set, but it is a rgγ nbd of {a}. Since {a} is a rgγ open set such that a {a} {a, c}. heorem 4.5: Every nbd N of p X is a rgγ nbd) of X. Proof: Let N be a nbd of point p. o prove that N is a rgγ nbd of p. hen by definition of nbd, there exists an open set H such that p H N. Since every open set is rgγ open, hence there exists a rgγ open set H such that p H N. herefore N is a rgγ nbd of p. he converse of the above theorem is not true as seen from the following example. Example 4.6: In Example 4.4, the set {a, c} is rgγ nbd of the point c, since {c} is a rgγ open set such that c {c} {a, c}. However the set {a, c} is not a nbd of the point c, since there is no exists an open set H such that c H {a, c}. heorem 4.7: If N is a rgγ open set of a space (X, τ), then N is a rgγ nbd of each of its points. Proof: Assume that N is a rgγ open set. Let p X. We need to prove that N is a rgγ nbd of p. For N is a rgγ open set such that p N N. Since p is an arbitrary point of N, it follows that N is a rgγ nbd of each of its points. he converse of the above theorem is not true in general as seen from the following example. Example 4.8: In Example 4.4, the set {a, d} is a rgγ nbd of the point a, since {a} is the rgγ open set such that a {a} {a, d}. Also, the set {a, d} is a rgγ nbd of the point d, since {d} is the rgγ open set such that d {d} {a, d}, that is {a, d} is a rgγ nbd of each of its points. However the set {a, d} is not a rgγ open set in X. heorem 4.9: If F is a rgγ closed set of a space (X, τ) and p F c, then there exists a rgγ nbd N of p such that N F = φ. Proof: Let F be a rgγ closed subset of a space (X, τ) and p F c. hen F c is a rgγ open set of X. So, by heorem 4.7, F c contains a rgγ nbd of each of its points. Hence, there exists a rgγ nbd N of p such that N F c, that is, N F = φ. Definition 4.0: Let p be a point in a space X. he set of all a rgγ nbd of p is called the rgγ nbd system at p and is denoted by rgγ N(p).

11 Regular Generalized closed Sets in opological Spaces heorem 4.: Let (X, τ) be a topological space and for each p X. Let rgγ N(p) be the collection of all rgγ nbds of p. hen the following statements are hold. () rgγ N(p) φ, for every p X, () If N rgγ N(p), then p N, (3) If N rgγ N(p) and N M, then M rgγ N(p), (4) If N rgγ N(p) and M rgγ N(p), then N M rgγ N(p), (5) If N rgγ N(p), then there exists M rgγ N(p) such that M N and M rgγ N(q), for every q M. Proof: () Since X is a rgγ open set, then X is a rgγ nbd for every p X. Hence there exists at least one rgγ nbd (namely, X), for each p X. herefore rgγ N(p) φ, for every p X. () If N rgγ N(p), then N is a rgγ nbd of p. Hence by definition of rgγ nbd, p N. (3) Let N rgγ N(p) and N M. hen there is a rgγ open set H such that p H N. Since N M, then p H M and so M is a rgγ nbd of p. Hence M rgγ N(p). (4) Let N rgγ N(p) and M rgγ N(p). hen by definition of rgγ nbd, there exist rgγ open sets H and H such that p H N and p H M. Hence p H H N M... (i). Since H H is a rgγ open set, (being the intersection of two rgγ open sets), it follows from (i) that N M is a rgγ nbd of p. So, N M rgγ N(p). (5) If N rgγ N(p), then there exists a rgγ open set M such that p M N. Since M is a rgγ open set, then M is a rgγ nbd of each of its points. herefore M rgγ N(q), for every q M. heorem 4.: For each p X, let rgγ N(p) be a non empty collection of subsets of X satisfying the following conditions. () If N rgγ N(p) implies that p N, () If N rgγ N(p) and M rgγ N(p) implies that N M rgγ N(p). Let τ consists of the empty set φ and all those non-empty subsets H of X having the property that p H implies that there exists an N rgγ N(p) such that p H N. hen τ is a topology on X. Proof: () Since φ τ (given), now we show that X τ. Let p be an arbitrary element of X. Since rgγ N(p) is non empty, then there is an N rgγ N(p) and so by (), p N. Since N is a subset of X, then we have p N X. Hence X τ. () Let H τ and H τ. If p H H, then p H and p H. Since H

12 International Journal of Mathematics and Computing Applications τ and H τ, hence there exists N rgγ N(p) and M rgγ N(p) such that p N H and p M H. hen p N M H H. But by (), N M rgγ N(p). Hence H H τ. (3) Let H i τ for every i I. If p {H i : i I}, then p H i p, for some i p I. Since H ip τ, then there exists an N rgγ N(p) such that p N H and consequently, p N {H i : i I}. Hence {H i : i I} τ. herefore τ is a topology on X. In the following, we introduce the notion of γ regular spaces. Definition 4.3: A space (X, τ) is called a γ regular space if every rgγ closed set is γ closed. Example 4.4: Any set with an indiscrete topology is an example for a γ regular space. Remark 4.5: he notions of γ regular and spaces are independent of each other. Example 4.6: Let X = {a, b, c} with topologies τ = {X, φ, {a}, {b}, {a, b}} and σ = {X, φ, {c}, {a, b}}. hen (X, τ) is but not γ regular where as (X, σ) is γ regular but not. heorem 4.7: For a topological space ( X, τ), the following conditions are equivalent: () X is γ regular, () Every singleton of X is either regular closed or γ open. Proof: () (). Let p X and assume that {p} is not regular closed. hen X\{p} is not regular open and hence X\{p} is rgγ closed. Hence, by hypothesis, X\{p} is γ closed and thus {p} is γ open. () (). Let A X be rgγ closed and p γcl(a). We will show that p A. For consider the following two cases: Case (). he set {p} is regular closed. hen, if p A, then there exists a regular closed set in γcl(a)\a. Hence by Proposition., p A. Case (). he set {p} is γ open. Since p γcl(a), then {p} γcl(a) φ. hus p A. So, in both cases, p A. his shows that γcl(a) A or equivalently A is γ closed. Definition 4.8: A space (X, τ) is called: ip

13 Regular Generalized closed Sets in opological Spaces 3 () locally indiscrete (equivalently, partition space)[5] if every closed set is regular closed, () γ space [] if every γg closed set is γ closed. heorem 4.9: For a topological space ( X, τ), the following conditions are equivalent: () X is γ regular, () X is locally indiscrete and γ. Proof: () (). Let ( X, τ) be a γ regular space. hen by heorem 4.7, every singleton of (X, τ) is closed or γ open. his implies that (X, τ) is a γ space. () (). Since the space X is γ, then every singleton of (X, τ) is closed or γ open. But (X, τ) is locally indiscrete, hence all closed singletons are regular closed. So, by theorem 4.7, the space X is γ regular. Remark 4.0: Every γ regular space is γ. But the converse is not true as seen by the following example. Example 4.: Let X = {p, q, r} and τ = {φ, X, {p}, {q}, {p, q}}. hen every singleton of (X, τ) is closed or γ open. herefore (X, τ) is γ, but not γ regular, since the rgγ closed set {p, q} is not γ closed. Proposition 4.: For a space (X, τ), we have γ (X, τ) RGγO(X, τ). Proof: Let A be a γ open set. hen A c is γ closed and so rgγ closed. his implies that A is rgγ open. Hence γo(x, τ) RGγO(X, τ). heorem 4.3: For a topological space ( X, τ), the following conditions are equivalent: () X is γ regular, () γo(x, τ) = RGγO(X, τ). Proof: () (). Let (X, τ) be a γ regular space and let A RGγO(X, τ). hen A c is rgγ closed. By hypothesis, A c is γ closed and thus A is γ open this implies that A γo(x, τ). Hence, γo(x, τ) = RGγO(X, τ). () (). Let γo(x, τ) = RGγO(X, τ) and let A be a rgγ closed set. hen A c is rgγ open. Hence A c γo(x, τ). hus A is γ closed thereby implying that (X, τ) is γ regular.

14 4 International Journal of Mathematics and Computing Applications heorem 4.4: For a topological space ( X, τ), the following conditions are equivalent: () X is γ regular, () Every nowhere dense singleton of X is regular closed, (3) he only nowhere dense subset of X is the empty set, (4) Every subset of X is γ open. Proof: () (). By heore m 4.7, every singleton is γ open or regular closed. Since non-empty nowhere dense set can not be γ open at the same time, then every nowhere dense singleton of X is regular closed. () (). Since every singleton is either γ open or nowhere dense [0], then by hypothesis, every singleton of X is either γ open or regular closed. Hence, by heorem 4.7, X is γ regular. () (3). Let A X be a non-empty nowhere dense set. hen, there exists a nowhere dense singleton in X and hence by hypothesis, this singleton is regular closed. If the nowhere dense singleton {p} is regular closed, then {p} = cl(int{p}) = cl(φ) = φ which is the contradiction with this assumption. (3) (4). By hypothesis and the fact that singletons are either γ open or nowhere dense, it follows that every singleton of X is γ open. Since the union of γ open sets is γ open. Hence, every subset of X is γ open. (4) (). From the definition of a γ regular space and by hypothesis, every subset of X is γ open and hence γ closed. REFERENCES [] D. Andrijevic, On b-open Sets, Mat.Vesnik, 48(), (996), [] I. Arockiarani, Studies on Generalizations of Generalized Closed Sets and Maps in opological Spaces, Ph.D. hesis, Bharathiar University, Coimbatore, (997). [3] S.P. Arya, and R. Gupta, On Strongly Continuous Mappings, Kyungpook Math. J.,, (974), [4] K. Balachandran, P. Sundaram, and H. Maki, On Generalized Continuous Maps in opological Spaces, Mem. I ac Sci.Kochi Univ. Math., (99), 5-3. [5] D.E. Cameron, Properties of S closed Spaces, Proc.Amer.Math.Soc., 7, (978), [6] J. Dontchev, and. Noiri, Quasi-normal Spaces and πg closed Sets, Acta Math.Hungar., 89(3), (000), -9. [7] J. Dontchev and M. Przemki, On the Various Decompositions of Continuous and Some Weakly Continuous Functions, Acta Math.Hungar.,7(-), (996), 09-0.

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