CHAPTER I INTRODUCTION & PRELIMINARIES

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1 CHAPTER I INTRODUCTION & PRELIMINARIES 1.1 INTRODUCTION In 1965, O.Njastad [44] defined and studied the concept of α - sets. Later, these are called as α-open sets in topology. In 1983, A. S. Mashhour et al [32], have defined the concepts like α-closed sets, α - closure of a set, α-openness and α - closedness in topology. Later, many topologists have studied these concepts in detail. G.B. Navalagi [35] has given some properties on α- neighbourhoods. In this paper we give some more properties of α-open sets. Throughout this paper, (X,τ), (Y,σ), and so forth (or simply X, Y, etc.) will always denote topological spaces. 1.2 PRELIMINARIES DEFINITION 1.2.1: Let X be a topological space and A X. Then A is called a) an α-open [44] if A int(cl(int(a))) b) a Semi-open [24] if A cl(int(a)) c) a Pre-open [33] if A int(cl(a)) d) a Regular open [28] if A = int cl(a) The family of all α open (resp. semi-open, pre-open) sets in a space X is denoted by αo(x) or α τ (resp. SO(X) PO(X.) The complement of an α-open (resp. semi-open, pre-open) set is said to be α- 2

2 closed (resp. semi-closed, pre-closed), denoted by αf(x) (resp. SF(X) PF(X.). Clearly, every open set is α -open. Also every closed set is α - closed (resp. semi-closed, pre-closed). But converse need not be true. DEFINITION :Let X be a topological space and A X. Then A is called a α- neighbourhood [35], denoted by α-nhd, of a point x in X, if there exist a α-open set U in X such that x U A. The α-nhd system of a point x X is denoted by α-n(x). NOTE 1.2.3: Clearly every α -open set of A is α -nhd of the points of A. DEFINITION : The union of all semi-open (preopen, α-open) sets contained in A is called the semi-interior of A, denoted by sint(a) [14] (resp. pre interior of A, pinta [34], α-interior of A, I α (A) [32]. DEFINITION 1.2.5: The intersection of all semi-closed ( preclosed and α-closed ) sets containing A is called the semi-closure of A, denoted by scl(a) [14 & 11 ] (resp.preclosure of A, pcl A [ 18 & 34 ], α-closure of A, C α (A) [35]). DEFINITION 1.2.6: The union of all α-open sets contained in the complement of A is called the α-exterior of A, denoted by Ext α (A) [7] DEFINITION : A point x X is said to be a α-limit point [35] of A X iff U αo(x) implies U (A-{x}). DEFINITION : The set of all α-limit points of A X is called the α-derived set [35] of A and is denoted by D α (A). DEFINITION : The set C α - I α is called the α -frontier [35] of A is denoted by F α (A). 3

3 DEFINITION : The set b α = A- I α (A) [or b α = A\I α (A)] is said to α-boundary of A or α-border of A [7]. DEFINITION : A subset A of a space (X,τ) is called a α - generalized closed (or αg-closed) [31] (generalized preclosed or gpclosed [3],) if C α (A) U (pcla U ) whenever A U and U is an open set. DEFINITION : A subset A of X is called a αg open set (gp-open set) if its complement is αg-closed set (gp-closed set) in X. The family of all αg-open sets (gp-open sets) in X is denoted by αgo(x) (GPO(X)) and the family of all αg -closed sets (gp-closed sets) of X is denoted by αgf(x) (GPC(X)). DEFINITION :Let X be a topological space and A X. Then A is called a αg- neighbourhood, denoted by αg-nhd, of a point x in X, if there exist a αg-open set U in X such that x U A. The αg-nhd system of a point x X is denoted by αg-n(x). DEFINITION : A point x A is said to be an αg-interior point of A if A is αg-nhd of x. In other words, it means that there exists a αg-open set G containing x such that G A The set of all αg-interior points of A is said to be αg-interior of A and is denoted by I αg (A). DEFINITION :[42] A subset A of a topological space X is called γ g open set iff A B GPO(X) for every B GPO(X). 4

4 The family of all γ g open subset of X is denoted by γ g O(X). The complement of a γ g open set is called γ g closed set. The family of γ g - closed subsets of X is denoted by γ g C(X). DEFINITION : [19] A subset A of a topological space X is called γ gαg -open set iff A B GPO(X) for every B αgo(x). The family of all γ gαg -open subset of X is denoted by γ gαg O(X). The complement of a γ gαg -open set is called γ gαg -closed set. The family of γ gαg - closed subsets of X is denoted by γ gαg C(X). DEFINITION : A function f: X Y is said to be α-continuous [32] (resp. pre-continuous [33], semi-continuous[24]) if f -1 (V) is α - open (resp. pre-open, semi-open) in X for every open set V of Y. DEFINITION : A function f: X Y is said to be strongly precontinuous [33] (strongly semi-continuous [1] if f -1 (V) is open in X for every pre-open (semi-open) set V of Y. DEFINITION : A function f: X Y is said to be preirresolute [47] (resp. irresolute [13], and α-irresolute [27] ) if f -1 (V) is preopen (resp. semiopen and α-open) set in X for each preopen (resp. semiopen and α-open ) set of Y. DEFINITION : A function f :X Y is said to be strongly - irresolute[16] if for each x X and each V SO(Y) containing f(x), there exists U SO(X) containing x such that f(sclu) V. DEFINITION : A function f:x Y is said to be contra α- continuous [20] (resp. contra pre-continuous [22] ) if f -1 (V) is α closed (resp. pre-closed) in X for every open set V of Y. 5

5 DEFINITION :A space X is said to be α-t 0 [35] ) if for each pair of distinct points in X, there exists a α-open set of X containing one point but not the other. DEFINITION : A space X is said to be α -T 1 [35] if for each pair of distinct points x and y of X, there exist α-open sets U and V containing x and y resp. such that y U and x V. DEFINITION : A space X is said to be α-t 2 [35] if for each pair of distinct points x and y in X, there exist disjoint α-open sets U and V in X such that x U and y V. DEFINITION [30]:A space X is said to be αt 1 if (X, τ α ) is a T 1. DEFINITION [ 7 ] : A space X is said to be an αt b if every αg-closed set is closed. DEFINITION : A function f : X Y is said to be preopen [33] (resp. semiopen [5], α-open [32]) if f(u) is preopen (resp. semiopen, α-open) in Y for each open set U in X. DEFINITION : A function f :X Y is said to be preclosed [18] (resp. semiclosed [6], α-closed [32]) if f(f) is preclosed (resp. semiclosed, α-closed) set in Y for each closed set F in X. DEFINITION [48] : A function f:x Y is called (i) strongly α- open (resp. strongly semiopen, strongly preopen ) if the image of each α-open (resp. semiopen, preopen ) set in X is a α-open (resp. semiopen, preopen ) set in Y. 6

6 CHAPTER 2 BASIC PROPERTIES OF α-open SETS αg- OPEN SETS AND γ gαg OPEN SETS 2.1 INTRODUCTION G.B.Navalagi[35], has given some properties on α- neighbourhoods. Latter many topologists have studied the properties of α- neighbourhoods of a point. Levine[25] introduced the class of generalized closed sets (g-closed sets) and studied their most basic properties. The concepts of gα-closed and αg-closed sets were introduced and all these notions were defined through α-open sets. In this chapter we study some more properties α- neighbourhoods of a point,, α-interior of a set, α - closure and α- derived sets, αg neighbourhoods of a point, αg- interior of a set, αg - closure and αg- derived sets. In 1987, Andrijevic [3] introduced a new class of sets called γ-open sets. In this chapter we introduce a new set which is analogous to γ-open sets but whose concept is defined via αg-open sets and GP-open sets. We shall call this new class of sets as γ gαg open sets. 2.2 SOME MORE PROPERTIES OF α- NEIGHBOURHOODS OF A POINT, α-interior OF A SET, α- CLOSURE AND α DERIVATIVE SETS LEMMA 2.2.1:[35] Any arbitrary union of α-nhds of a point x is again a α-nhd of x. LEMMA 2.2.2: Finite intersection of α-nhds of a point x is again a α- nhd of x. 7

7 THEOREM 2.2.3: Every nhd of x in X is α-nhd of x. PROOF: Easy to prove Converse of the theorem is not true in general, For, EXAMPLE 2.2.4: Let X = {a,b,c} & τ= {X,Ø,{a},{a,b}} be a topology on X. αo(x) = {X,Ø,{a},{a,b},{a,c}}. Then {a,c} is α-nhd of c but not nhd of c. THEOREM 2.2.5:[35] A subset A is α-open iff it is α-nhd of each of its points. THEOREM 2.2.6: The α-nhd system α-n(x) of a point x X satisfies the following properties. i) α-n(x) x X ii) [35] if N α-n(x) then x N iii) [35] N α-n(x) & N M M α-n(x) iv) N α-n(x) & M α-n(x) N M α-n(x) v) [35] N α-n(x) M α-n(x) such that M N and M α- N(y) y M. THEOREM 2.2.7: For each x X, let α-n(x) satisfies the following conditions [P1] N α-n(x) x N [P2] N α -N(x) M α-n(x) N M α -N(x) Let τ consists of the empty set and all the non-empty subsets A of X 1 having the property that x A there exists an N α-n(x) such that x N A, then τ is a topology for X. 1 8

8 THEOREM 2.2.8: Let A be a subset of X then α-interior of A is the union of all α-nhd subsets of A. i.e., I α (A) = {U α-n(x): U A }. THEOREM 2.2.9: Let A be a subset of X. Let P(X) denote the collection of all possible subsets of a non empty set X, then a closure operator is a mapping C α of P(X) into itself satisfies the following. [i] C α ( ) = [ii] A C α (A) [iii] A B C α (A) C α (B) [iv] [35] C α (A B) C α (A) C α (B) [v] [35] C α (C α (A))=C α (A) [vi] [35] C α (A B) C α (A) C α (B) THEOREM : Let A and B be subsets of X then the derived sets D α (A) & D α (B) have the following properties. (i) D α (Ø) = Ø. (ii) [35] A B D α (A) D α (B) (iii) x D α (A) x D α [A {x}] (iv) [35] D α (A) D α (B) D α (A B) (v) [35] D α (A B) D α (A) D α (B) THEOREM : I α (A), Ext α (A) are disjoint where A X and hence for x A, U α -N(x) & Ext (A) are disjoint α THEOREM : If A is a subset of X and A Fr (A) = Ø then A is α-nhd of its points. Converse of theorem is not true in general, for, EXAMPLE : Let X = {1,2,3,4,5} α 9

9 τ = { Ø,X,{2},{3,4},{2,3,4},{1,3,4},{1,2,3,4}} A = {1,2,4}. Then A is α-nhd of 2. But A Fr (A) = {1, 2,4} {1,3,4,5} Ø. α 2.3 SOME MORE PROPERTIES OF αg- NEIGHBOURHOODS OF A POINT & αg-interior OF A SET LEMMA 2.3.1: Any arbitrary union of αg-nhds of a point x is again a αg-nhd of x. NOTE :Similarly finite intersection of αg-nhd of a point is also a αg-nhd of that point. THEOREM 2.3.3: Every nhd of x in X is αg-nhd of x. PROOF: Easy proof Converse of the theorem is not true in general, For, EXAMPLE 2.3.4: Let X = {a,b,c} & τ= {X,Ø,{a},{a,b}} be a topology on X. αgo(x) = {X,Ø,{a},{a,b},{a,c}}. Then {a,c} is αg-nhd of c but not nhd of c. LEMMA 2.3.5: Let {B i i I} be a collection of αg -open sets in X. Then B αgo(x) i I i THEOREM 2.3.6: A subset A is αg-open iff it is αg-nhd of each of its points. THEOREM 2.3.7: The αg-nhd system αg-n(x) of a point x X satisfies the following properties. i) αg-n(x) x X ii) if N αg-n(x) then x N 10

10 iii) N αg-n(x) & N M M αg-n(x) iv) N αg-n(x) & M αg-n(x) N M αg-n(x) v) N αg-n(x) M αg-n(x) such that M N and M αg-n(y) y M THEOREM 2.3.8: For each x X, let αg-n(x) satisfies the following conditions [P1] N αg-n(x) x N [P2] N αg -N(x) M α g-n(x) N M α g-n(x) Let τ consists of the empty set and all the non-empty subsets A of X 1 having the property that x A there exists an N αg-n(x) such that x N A, then τ is a topology for X. 1 EXAMPLE 2.3.9: Let X = {a,b,c}, τ = { Ø, {a},{b},{a,b},x}. Then αgo(x)={φ,{a},{b},{a,b},{a,c},{b,c},x}. Let α -N(a) ={a}, α -N(b) = {{b},{a,b}}, α -N(c) = {{a,c},{b,c}, X} be any collection of subsets of X. We observe that α -N(a), α -N(b), α-n(c) satisfy [P1] & [P2] for, Consider α-n(b), {a,b} α-n(b) b {a,b} which is [P1]. Also {a,b} {b}={b} α -N(b) which is [P2]. Similarly we can show that α-n(a) & α-n(c) satisfies [P1] & [P2]. Now we find the members of τ 1 as follows. (1) By definition Ø τ. Also X τ because a X there 1 1 exists X α -N(c) such that that a X X, b X X α -N(c), such that b X X, c X X α-n(c) such that c X X. Now (i) (a,c) τ 1 because a {a,c} {a} α-n(a) such that a {a} {a,c} c {a,c} {a,c} α-n(c) such that c {a,c} {a,c}. (ii){c} τ because c {c} 1 {a,c} α -N(c) such that c {a,c} {c}. (iii){a,b} τ because a {a,b}, 1 11

11 there exist {a} in α -N(a) which contains a and is contained in {a,b}. Similarly b {a, b}, there exist {b} in α-n(b) which contains b and is contained in {a,b}. (iv) {b,c} τ because for b {b,c} there exist 1 {a,b} α-n(b) such that {b} {a,b} {a,c},x} which is a topology. {b,c}. Hence τ ={Ø,{a},{b},{a,b}, 1 COROLLARY : If A is a αg-closed subset of X and x X A, then there exists a αg-nhd N of x such that N A =. NOTE : Since every α-open set is αg-open set, every α-interior point of a set A X is αg-interior point of A. Thus I α (A) I αg (A). In general I αg (A) I α (A) which is shown in the following example EXAMPLE : Let X = {a, b, c,d} and τ ={ Ø, {a},{a,b},{a,b,d},x} be a topology. Here αox = { Ø, {a},{a, b},{a, b, c}, {a, b, d}, X} and αf(x) = { Ø,{c},{d}, {c,d}, {b,c,d}, X}. Let A 1 = {a,c,d}. Then A 1 is αg-closed sets since C α (A 1 ) = X X. So complement of A 1 ={b} is αgopen in X. Now let A = {b}. Then I αg (A) = {b}. But I α (A) ={ Ø }. So I αg (A 1 ) I α (A 1 ). THEOREM : A subset A of X is αg-open iff A =I αg (A). LEMMA : If A and B are subsets of X and A B then I αg (A) I αg (B). NOTE : I αg (A) = I αg (B) does not imply that A = B. This is shown in the following example. EXAMPLE : Let X = {a, b, c} and τ = { Ø, {a, b}, X} be a topology. Let A ={a} and B ={a,c}, then I αg (A) = I αg (B) but A B. 12

12 THEOREM : Let A and B be subsets of X. Then (i) I αg (A) I αg (B) I αg (A B) (ii) I αg (A B) I αg (A) I αg (B) THEOREM : Let A be a subset of X then αg-interior of A is the union of all αg-nhd subsets of A. i.e., I αg (A) = {U αg -N(x): U A } PROPERTIES OF αg- CLOSURE, αg DERIVATIVE SETS αg-exterior, αg-frontier, αg- BOUNDARY SET THEOREM : Let A be a subset of X. Let P(X) denote the collection of all possible subsets of a non empty set X, then a closure operator is a mapping C αg of P(X) into itself satisfies the following. [C1] C αg ( ) = [C2] A C αg (A) [C3] A B C αg (A) C αg (B) [C4] C αg (A B) C αg (A) C αg (B) [C5] [ C αg (A B) C αg (A) C αg (B) [C6] C αg (C αg (A)) = C αg (A) NOTE 2.4.2: Equality does not hold in Theorem [C4], which is shown by the following. EXAMPLE 2.4.3: X= {a,b,c,d} τ = {, {a},{a,d},{b,c},{a,b,c},{b,c,d}, X} be a topology of X. Now αgc(x) = {, {a}, {b}, {c}, {d}, {a,d}, {b,c},{b,d},{b,c,d},{a,c,d},x}. Take A = {a} and B ={b} then A B = {a,b}. Now, we obtain that C αg (A) ={a}, C αg (B) = {b} and C αg (A B) = X. It follows that C αg (A) C αg (B) C αg (A B). 13

13 THEOREM 2.4.4: Let C αg be a closure operator defined on X satisfying the properties of Theorem Let F be the family of all subsets F of X for which C αg (F) = F and τ be the family of all complements of members of F, then τ is topology for X such that for any arbitrary subset A of X, A=C αg (A). DEFINITION 2.4.5: Let (X, τ ) be a topological space and A be a subset of X. Then a point x X is called a αg-limit point of A if and only if every αg-nhd. of x contains a point of A distinct from x. i.e., (N -{x}) A αg-nhd. N of x. Alternatively, we can say that a point x X is called a αg-limit point of A if and only if every αg-open set G containing x contains a point of A other than x. The set of all αg-limit points of A is called a αg-derived set of A and is denoted by D αg (A). EXAMPLE 2.4.6: Let (X, D) be any discrete topological space and let A be any subset of X. Then no point x X can be a αg-limit point of A since {x} is a αg-open set which contains no points of A other than (possibly) x. Thus D αg (A) =. EXAMPLE 2.4.7: Consider any indiscrete topological space (X,I). Let A be a subset of X containing two or more points of X. Then every point x X is a αg-limit point of A since the only αg-open set containing x is X which contains all the points of A and must therefore contains a point of A other than x (Since A contains more than one point). Hence D αg (A)=X. If A = {p] consisting of a single point p. Then all the points of X other than p are αg-limit points of A, for x p X, then the only αg- 14

14 open set containing x is X which contains the point p of A which is different from x. Hence D αg (A) = X-{p}. If A =, then evidently no point of X can be a αg-limit point of A and so D αg (A) = THEOREM 2.4.8: Let A and B be subsets of X then the derived sets D αg (A). & D αg (B) have the following properties. (1) D αg (Ø) = Ø. (2) A B D αg (A) D αg (B) (3) x D αg (A) x D αg [A {x}] (4) D αg (A) D αg (B) D αg (A B) (5) D αg (A B) D αg (A) D αg (B). NOTE 2.4.9: Equality does not hold in Theorem 2.4.8(4), which are shown by the following. EXAMPLE : X= {a,b,c,d} τ = {, {a}, {b}, {a,b}, {b,c}, {a,b,c}, X} be a topology of X. Then αgo(x) ={,{a},{b},{c}, {a,b}, {a,c}, {a,d},{b,c},{b,d},{a,b,c},{a,b,d},{b,c,d},x}. Take A = {a,c,d} and B ={d} then A B = X. Now, we obtain that D αg (A) = {d}, D αg (B) = { } and D αg (A B) = X. It follows that {d}= D αg (A) D αg (B) D αg (A B) = X. NOTE : D αg (A)=D αg (B) does not imply A = B. For, EXAMPLE : Consider X= {a,b,c,d} τ = {,{a},{b},{a,b},{b,c}, {a,b,c},x} be a topology of X. If A = {c,d} and B = {d}. Then D αg (A) = D αg (B) =. Hence the example. THEOREM : Let A be a subset of X. Then A is αg-closed if and only it contains the set of its αg-limit points. PROOF: By definition, X-A is αg-open, if A is αg-closed set. Thus, A is αg-closed if and only if for every point in X-A has a αg-nhd., contained 15

15 in X-A i.e., no point of X-A is αg-limit point of A, which is equivalently saying that A contains each of its αg-limit points. THEOREM : Let (X, τ ) be a topological space and A be a subset of X. Then A is αg-closed if and only if D αg (A) A. THEOREM : Let A be any subset of a topological space (X, τ ). Then A D αg (A) is αg-closed set. THEOREM : In any topological space (X, τ ), every D αg (A) is a αg-closed set. THEOREM : I αg (A), Ext αg (A) are disjoint where A X and hence for x A, U α g-n(x) & Ext (A) are disjoint. αg THEOREM : If A is a subset of X and A Fr (A) = Ø then A is αg αg-nhd of its points. Converse of theorem is not true in general, for, EXAMPLE : Let X = {1,2,3,4,5} τ = { Ø,X,{2},{3,4},{2,3,4},{1,3,4},{1,2,3,4}} A = {1,2,4}. Then A is αg-nhd of 2. But A Fr (A) = {1,2,4} {1,3, 4,5} Ø. αg DEFINITION : (i) The set b αg = A- I αg (A) is said to αg-boundary of A (ii) The set C αg - I αg is called the αg -frontier of A is denoted by F αg (A). THEOREM : For a subset A of a space X, the following statements hold: (1) b αg (A) b(a) where b(a) denotes the boundary of A (2) A = I αg (A) b αg (A) (3) I αg (A) b αg (A) = (4) A is an αg open set if and only if b αg (A) = (5) b αg ( I αg (A)) = (6) I αg ( b αg (A)) = (7) b αg (b αg (A)) = b αg (A) (8) b αg (A)) = A Cl αg (X-{A} (9) b αg (A) = D αg (X-A) 16

16 The converse of the theorem (1) is not true in general EXAMPLE : X={a,b,c} τ= { Ø, {a}, {a,b}, X}. If A = {a,c}, then b αg (A) = Ø and b(a) = {c}. Hence b(a) b αg (A) THEOREM : For a subset A of a space X, the following statements hold. 1) Fr αg (A) Fr(A) where Fr(A) denotes the frontier of A 2) C αg (A) = I αg (A) Fr αg (A) 3) I αg (A) Fr αg (A) = 4) b αg (A) Fr αg (A) 5) Fr αg (A) = b αg (A) D αg (A) 6) A is an αg open set if and only if Fr αg (A) = D αg (A) 7) Fr αg (A) = C αg (A) C αg (X-A) 8) Fr αg (A)= Fr αg (X-A) 9) Fr αg (A) is αg-closed 10) Fr αg (Fr αg (A)) Fr αg (A); 11) Fr αg (I α (A)) Fr αg (A); 12) Fr αg (C αg (A)) Fr αg (A); 13) I αg (A) = A -Fr αg (A); The converse of (1) and (4) of Theorem are not true in general, as shown by the following example EXAMPLE : X = {a,b,c} τ = { Ø, {a}, {a,b}, X}. Then αgo(x) = {Ø,{a},{a,b},{a,c},X} & αgc(x) ={Ø,{b},{c},{b,c},{a,c},X}. If A = {b}, B = {a, b}, then Fr(A) = {b, c} {b} = Fr αg (A), Fr αg (B) ={c} b αg (B) =. 17

17 2.5 γ gαg - OPEN SETS DEFINITION 2.5.1: A subset A of a topological space X is called γ gαg - open set iff A B GPO(X) for every B αgo(x). The family of all γ gαg -open subset of X is denoted by γ gαg O(X). The complement of a γ gαg -open set is called γ gαg -closed set. The family of γ gαg - closed subsets of X is denoted by γ gαg C(X). THEOREM 2.5.2: αgo(x) γ gαg O(X) PROOF: We know that αgo(x) GPO(X). Now for all A αgo(x), A X=A αgo(x) GPO(X) A X=A GPO(X) A γ gαg O(X). REMARK 2.5.3: The following two example illustrates that αgo(x) γ gαg O(X) GPO(X). EXAMPLE 2.5.4: (i) Let X = {a,b,c,d}, τ ={Ø,{a,b},{a,b,c},X} αo(x) = {Ø,{a,b},{a,b,c},{a,b,d}, X}. αc(x) = {Ø,{c,d},{d},{c},X}. αgc(x) ={Ø,{b,c,d},{a,c,d},{a,b,d},{c,d},{a,d},{b,d},{d},{c},X}. αgo(x) ={Ø,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},{a,b,d},X}. PO(X) ={Ø,{a},{b}{a,b},{a,c},{a,d},{b,c},{b,d},{a,b,c},{a,b,d},{a,c,d}, {b,c,d},x}. PC(X) ={Ø,{a},{b},{c},{d},{a,c},{a,d},{c,d},{b,c},{b,d},{a,c,d}, {b,c,d},x}. GPC(X) ={Ø,{a},{b},{c},{d},{a,c},{a,d},{c,d},{b,c},{b,d},{a,b,d}, {a,c,d},{b,c,d},x}. GPO(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{b,c},{b,d},{a,b,c}, {a,b,d},{a,c,d},{b,c,d},x}. 18

18 γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{b,c},{b,d},{a,b,c}, {a,b,d},{a,c,d},{b,c,d},x}. So αgo(x) γ gαg O(X). (ii) Let X = {a,b,c,d}, τ ={Ø,{a,d},{a,b,c},X} αo(x) = {Ø,{a,d},{a,b,c},{a,b,d},{a,c,d},X}. αc(x) = {Ø,{b,c},{d},{c},{b},X}. αgc(x) ={Ø,{b,c,d},{a,c,d},{a,b,d},{c,d},{b,c},{b,d},{b},{c},{d},X}. αgo(x)={ø,{a},{b},{c},{a,b},{a,c},{b,d},{a,b,c},{a,b,d},{a,c,d},x}. PO(X) ={Ø,{a},{a,b},{a,c},{a,d},{c,d},{b,d},{a,b,c},{a,b,d},{a,c,d}, {b,c,d},x}. PC(X) ={Ø,{a},{b},{c},{d},{a,b},{a,c},{c,d},{b,c},{b,d},{b,c,d},X}. GPC(X) ={Ø,{a},{b},{c},{d},{a,b},{a,c},{c,d},{b,c},{b,d},{a,b,d}, {a,c,d},{b,c,d},x}. GPO(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{c,d},{b,d},{a,b,c},{a,b,d}, {a,c,d},{b,c,d},x}. γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{a,b,c},{a,b,d},{a,c,d}, {b,c,d},x}. So γ gαg O(X) GPO(X). THEOREM 2.5.5: γ g O(X) γ gαg O(X) PROOF: We have γ g O(X) GPO(X). Let A γ g O(X). Then A GPO(X). Also for all B αgo(x) GPO(X), A B GPO(X) A γ gαg O(X). Hence the result. REMARK 2.5.6: The following example illustrates that both αgo(x) & γ g O(X) are the subsets of γ gαg O(X). 19

19 EXAMPLE 2.5.7: Let X= {a,b,c,d}, τ ={Ø,{a},{a,b},{a,b,c},X} αgo(x) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{b,c},{a,b,c},{a,b,d},{a,c,d}, X}. GPO(X) ={Ø, {a}, {b}, {c}, {a,b},{a,c}, {a,d}, {b,c},{b,d}, {a,b,c}, {a,b,d}, {a,c,d}, X}. γ g O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},{a,b,d},{a,c,d}, X}. γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d}, {b,c}, {a,b,c}, {a,b,d}, {a,c,d}, X}. αgo(x) γ gαg O(X) & γ g O(X) γ gαg O(X). REMARK 2.5.8: The following diagram establishes the relationships between γ gαg closed sets and some other sets. A B(resp A B. A B) represents A implies B but not conversely (resp. A need not imply B, A and B are independent of each other) gp closed γ g -closed closed α-closed γ gαg -closed αg-closed g-closed semi-closed g#-closed g*-closed Semi preclosed g#s-closed g*p-closed pre semi preclosed 20

20 THEOREM 2.5.9: For a topological space X, every singleton subset of X is αg-open if it is γ gαg -open. PROOF: Suppose every singleton subset of X is αg-open. Then for x X and any subset B of X, {x} B is either {x} or Ø, both which are gpopen. Hence {x} is in γ gαg O(X). REMARK : For a topological space X if a singleton subset of X is γ gαg -closed then it is not necessary that it is αg-closed. This is illustrated in the following example. EXAMPLE : Let X = {a,b,c,d}, τ ={Ø,{a},{a,d},{a,b,c},X} αgc(x) ={Ø,{b,c,d},{a,c,d},{a,b,d},{c,d},{b,d},{b,c},{d},{c},{b}, X}. γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{a,b,c},{a,b,d},{a,c,d}, X}. Here {a} is γ gαg -closed set but it is not αg-closed set. 2.6 PROPERTIES OF γ gαg -NEIGHBOURHOODS DEFINITION 2.6.1: Let X be a topological space and A X. Then A is called a γ gαg - neighbourhood, denoted by γ gαg -nhd, of a point x in X, if there exist a γ gαg -open set U in X such that x U A. The γ gαg -nhd system of a point x X is denoted by γ gαg -N(x). NOTE 2.6.2: Clearly every γ gαg -open set A is γ gαg -nhd of the points of A. LEMMA 2.6.3: Any arbitrary union of γ gαg -nhds of a point x is again a γ gαg -nhd of x. PROOF: Let { A } be the arbitrary collection of γ λ λ I gαg-nhds of a point x X. We have to prove that A for λ I also a γ gαg -nhd of x. Now λ 21

21 there exist γ gαg -open set M x such that x M A A for λ I. i.e., x M x A λ for x λ I. Therefore A for λ I is a γ gαg -nhd of x. i.e., arbitrary union of γ gαg -nhd of a point x is again a γ gαg -nhd of x. λ λ λ NOTE 2.6.4: Similarly finite intersection of γ gαg -nhd of a point is also a γ gαg -nhd of that point. THEOREM 2.6.5: Every nhd of x in X is γ gαg -nhd of x. Converse of the theorem is not true in general, For, EXAMPLE 2.6.6: Let X = {a,b,c} & τ= {X,Ø,{a},{a,c}} be a topology on X. Then γ gαg O(X) ={X,Ø,{a},{c},{a,b},{a,c}} Then {a,b} is γ gαg -nhd of b but not nhd of b. LEMMA 2.6.7: Let {B i i I} be a collection of γ gαg -open sets in X. Then U B γ gαg O(X).... i I i PROOF: Easy to prove THEOREM 2.6.8: A subset A is γ gαg -open iff it is γ gαg -nhd of each of its points. PROOF: Let A X be a γ gαg -open set. Then for each x A, x A A and A is γ gαg -open A is γ gαg -nhd of x. Conversely suppose U is γ gαg - nhd of each of its points. Then for each x U, there exists N x γ gαg O(X) such that N x U. Then U = { N x X}. Since each N x is γ gαg -open it follows that U is γ gαg -open by lemma x THEOREM 2.6.9: The γ gαg -nhd system γ gαg -N(x) of a point x X satisfies the following properties. 22

22 i) γ gαg -N(x) Ø x X ii) if N γ gαg -N(x) then x N iii) N γ gαg -N(x) & N M M γ gαg -N(x) iv) N γ gαg -N(x), M γ gαg -N(x) N M γ gαg -N(x) v) N γ gαg -N(x) M γ gαg -N(x) such that M N and M γ gαg -N(y) y M PROOF: (i) is trivial (ii) Let N γ gαg -N(x) N is the γ gαg -nhd of x x N. (iii) Let N γ gαg -N(x) and N M. Therefore there exists U γ gαg O(X) such that x U N M M is γ gαg -nhd of x M γ gαg -N(x) (iv) N γ gαg -N(x) & M γ gαg -N(x) γ gαg -open sets N x & M x such that x N N & x 2.6.4, the result follows x M M. Now x N M N M. By note x x x (v) Let N γ gαg -N(x). Therefore there exists M γ gαg O(X) such that x M N. Since every γ gαg -open is γ gαg -nhd, M is γ gαg -nhd of its points y M, M is γ gαg -nhd of y M γ gαg -N(y) y M. COROLLARY : If A is a γ gαg -closed subset of X and x X A, then there exists a γ gαg -nhd N of x such that N A=. PROOF: If A is a γ gαg -closed set in X, then X-A is γ gαg -open set. By theorem 2.6.8, X-A contains a γ gαg -nhd. of each of its points. This implies 23

23 that there exists a γ gαg -nhd. N of x such that N X-A. It is clear that no point of N belongs to A and hence N A =. THEOREM : For each x X, let γ gαg -N(x) satisfies the following conditions [P1] N γ gαg -N(x) x N [P2] N γ gαg -N(x), M γ gαg -N(x) N M γ gαg -N(x). Let τ consists of the empty set and all the non-empty subsets of A of X 1 having the property that x A there exists an N γ gαg -N(x) such that x N A, then τ is a topology for X. 1 PROOF: 1) By definition Ø τ. Let x X, then γ 1 gαg -N(x) being nonempty, there exists an N γ gαg -N(x) so that by [P1] we have x N. Also N is a subset of X so that x N X and therefore X τ 1.. 2) Let A τ 1 and B τ 1. We will show that A B τ. Now choose x to 1 be an arbitrary element of A B x A and x B. But as A τ and B τ, there exists N γ gαg -N(x) and M γ gαg -N(x) such that 1 1 x N A and x M B or x N M A B. But N M γ gαg -N(x) by [P2], so that A B τ. 1 3) Let {A : λ Λ } be an arbitrary collection of τ open sets and we will λ 1 show that {A : λ Λ } τ. Let x be an arbitrary element of λ 1 { A : λ α Λ } so that x A for at least one λ Λ. Now x and λ A λ τ, therefore there exist some N γ 1 gαg -N(x) such that x N A λ. Or x N { A : λ Λ } { A : λ Λ} τ. Thus the collection τ is a λ λ 1 1 topology for X. A λ 24

24 EXAMPLE : Let X ={a,b,c} & τ= {X,Ø,{a},{a,c}} be a topology on X. Then γ gαg O(X)={X,Ø,{a},{c},{a,b},{a,c}}. Let γ gαg -N(a) ={{a},x} γ gαg -N(b) ={{a,b}}, γ gαg -N(c) ={{c},{a,c}} be any collection of subsets of X. We observe that γ gαg -N(a), γ gαg -N(b), γ gαg -N(c) satisfy [P1] & [P2] Now we find the members of τ as follows: (i) By definition Ø τ. 1 1 Also X τ because a X there exists X γ 1 gαg -N(a) such that that a X X, b X X γ gαg -N(a), such that b X X, c X X γ gαg -N(a) such that c X X. (ii){c} τ because c {c}, but there exist 1 {c} in γ gαg -N(c) containing c and contained in {c}.(iii) Similarly {a} τ. 1 (iv) {b} τ 1 since for b {b} there exist {a,b} in γ gαg -N(b) containing b but not contained in {b}. (v) (b,c) τ because b {b,c}, but there does 1 not exist any set in γ gαg -N(b) containing b and contained in {b,c} (vi) {a,c} τ because a {a,c} and there exist {a} in γ 1 gαg -N(a) which contains a and is contained in {a,c}. (vi) Similarly {a,b} τ. 1 Hence τ ={ Ø,{a},{c},{a,b},{a,c},X} which is a topology. 1 DEFINITION :The intersection of all γ gαg -closed sets containing a set A is called the γ gαg -closure of A and is denoted by γ gαg Cl(A). Thus γ gαg Cl(A) = {F A F and F γ gαg Cl(X)}. REMARK : For a subset A of X γ gαg Cl(A) is not necessarily γ gαg - closed set as shown in the following example. EXAMPLE : Let X = {a,b,c,d}, τ ={Ø,{a},{a,d},{a,b,c},X} γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{a,b,c},{a,b,d},{a,c,d}, X}. γ gαg C(X) ={Ø,{b},{c},{d},{c,d},{b,d},{b,c},{b,c,d},{a,b,d},{a,c,d}, X}. Now γ gαg Cl({a}) = {a,b,d} {a,c,d} X = {a,d} which is not γ gαg -closed. 25

25 THEOREM : Let A be a subset of X. Let P(X) denote the collection of all possible subsets of a non empty set X, then a closure operator is a mapping γ gαg Cl of P(X) into itself satisfies the following. [C1] γ gαg Cl (Ø) = Ø [C2] A γ gαg Cl (A) [C3] A B γ gαg Cl (A) γ gαg Cl (B) [C4] γ gαg Cl (A B) γ gαg Cl (A) γ gαg Cl (B) [C5] γ gαg Cl (γ gαg Cl (A)) = γ gαg Cl (A). PROOF: Easy to prove. THEOREM : Let γ gαg Cl be a closure operator defined on X satisfying the properties of Theorem Let F be the family of all subsets F of X for which γ gαg Cl = F and τ be the family of all complements of members of F, then τ is topology for X such that for any arbitrary subset A of X, A = γ gαg Cl(A) PROOF: Proof is similar to the corresponding result under therorem REMARK : γ gαg Cl(A)= γ gαg Cl(B) does not necessarily imply that A = B. This is shown in the following example. EXAMPLE : Let X = {a,b,c,d}, τ ={Ø,{a},{a,d},{a,b,c},X} γ gαg C(X) ={Ø,{b},{c},{d},{c,d},{b,d},{b,c},{b,c,d},{a,b,d},{a,c,d}, X}. Now γ gαg Cl({a,b}) = X and γ gαg Cl({a,b,c}) =X. But {a,b} {a,b,c}. 26

26 DEFINITION : Let (X, τ ) be a topological space and A be a subset of X. Then a point x X is called a γ gαg -limit point of A if and only if every γ gαg -nhd. of x contains a point of A distinct from x. i.e., (N {x}) A Ø γ gαg -nhd N of x. Alternatively, we can say that if and only if every γ gαg -open set G containing x contains a point of A other than x. The set of all γ gαg -limit points of A is called a γ gαg -derived set of A and is denoted by γ gαg D(A). EXAMPLE : Let (X, D) be any discrete topological space and let A be any subset of X. Then no point x X can be a γ gαg -limit point of A since {x} is a γ gαg -open set which contains no points of A other than (possibly) x. Thus γ gαg D(A) = Ø. EXAMPLE : Consider any indiscrete topological space (X,I). Let A be a subset of X containing two or more points of X. Then every point x X is a γ gαg -limit point of A since the only γ gαg -open set containing x is X which contains all the points of A and must therefore contains a point of A other than x. (Since A contains more than one point). Hence γ gαg D(A) =X. If A = {p} consisting of a single point p. Then all the points of X other than p are γ gαg -limit points of A, for x p X, then the only γ gαg - open set containing x is X which contains the point p of A which is different from x. Hence γ gαg D(A) = X-{p}. If A = Ø., then evidently no point of X can be a γ gαg limit point of A and so γ gαg D(A) = Ø. 27

27 THEOREM : Let (X, τ ) be a topological space and A be a subset of X. Then A is γ gαg -closed if and only if γ gαg D(A) A. PROOF: ( ) Suppose A is γ gαg -closed. i.e., X-A is γ gαg -open. Now we show that γ gαg D(A) A. Let x γ gαg D(A) implies x is a γ gαg -limit point of A. i.e., every γ gαg -nhd. of x contains a point of A different from x. Now suppose x A so that x X-A, which is γ gαg -open and by definition of γ gαg -open sets there exists a γ gαg -nhd. N of x such that N X-A. From this we conclude that N contains no point of A, which is a contradiction. Therefore x A and hence γ gαg D(A) A. ( ) Assume γ gαg D(A) A. We show that A is a γ gαg -closed set in (X, τ ). Or we show that there exists γ gαg -nhd.n of x for each x X-A. Let x be an arbitrary point of X-A so that x A. Since γ gαg D(A) A, x A implies x γ gαg D(A). i.e., there exists a γ gαg -nhd.n of x, which does not contain any point of A. i.e., there exists a γ gαg -nhd.n of x which consists of only points of X-A. This means that X-A is γ gαg -open and hence A is γ gαg -closed. THEOREM : Let A is any subset of a topological space (X, τ ). Then A γ gαg D(A) is αg-closed set. PROOF :( )A γ gαg D(A) is γ gαg -closed set in X if X [A γ gαg D(A)] is a γ gαg -open set in X. But X [A γ gαg D(A)] = [X A] [X-γ gαg D(A)]. Thus we prove that [X A] [X-γ gαg D(A)] is γ gαg -open set in X i.e., it contains a γ gαg -nhd. of each of its points. Let x [X A] [X-γ gαg D(A)] x [X A] and x [X-γ gαg D(A)] or x A and x γ gαg D(A). Now x γ gαg D(A) x is not a γ gαg -limit point of A, it follows that there exist γ gαg -nhd N of x which contains no points of A other than possibly x. But x A so that N contains no point of A and so N X- A. Again N is 28

28 γ gαg -open and is a γ gαg -nhd of each of its points. Therefore no point of N can be a limit point of A. i.e., no point of N can belong to γ gαg D(A). Thus N X - γ gαg D(A). Thus we have proved that for all x [X A] [Xγ gαg D(A)] that there exist γ gαg -open neighbourhood N of x such that N [X- A] [X-γ gαg D(A)] which implies [X- A] [X-γ gαg D(A)] is γ gαg -open set in X. Hence the proof. THEOREM : In any topological space (X, τ ), every γ gαg D(A) is a γ gαg -closed set. PROOF : Let A be a set of X and γ gαg D(A) is γ gαg -derived set of A. Then by theorem , A is γ gαg -closed set if and only if γ gαg D(A) A. Hence γ gαg D(A) is γ gαg -closed if and only if γ gαg D(γ gαg D(A) γ gαg D(A. i.e., every γ gαg -limit point of γ gαg D(A) belongs to γ gαg D(A). Let x be a γ gαg -limit point of γ gαg D(A). i.e., x γ gαg D(γ gαg D(A)) so that there exists a γ gαg -open set U containing x such that (U {x}) (γ gαg D(A)) Ø (U {x}) A Ø since every γ gαg -nhd of an element of γ gαg D(A) has at least one point of A. Therefore, x is a γ gαg - limit point of A. i.e., x γ gαg D(A). Thus x γ gαg D(γ gαg D(A) x γ gαg D(A). Therefore γ gαg D(A) is γ gαg -closed set in X. DEFINITION : The union of all γ gαg -open sets which are contained in A is called the γ gαg -interior of A and is denoted by γ gαg Int(A). i.e., γ gαg Int(A) = { U U A and U γ gαg O(X)} DEFINITION : The union of all γ gαg -open sets contained in the complement of A is called the γ gαg -exterior of A, denoted by γ gαg Ext(A). DEFINITION : The set γ gαg Cl(A) γ gαg Int(A) is called the γ gαg - frontier of A is denoted by γ gαg Fr(A). 29

29 DEFINITION : The set γ gαg b(a) = A - γ gαg Int(A) is said to γ gαg - Boundary of A. THEOREM :Let A be a subset of X then γ gαg -interior of A is the union of all γ gαg -nhd subsets of A. i.e., γ gαg Int(A) = {U γ gαg -N(x): U A}. PROOF : Let x γ gαg Int(A) then γ gαg -open set U of A such that x U A. Since every γ gαg -open set is a γ gαg -nhd, U γ g g -N(x) such that x U A. and so x {U γ g g -N(x) : U A}. Hence γ gαg Int(A) α α {U γ -N(x) : U A}. Now let x {U γ g g g g -N(x) : U A} x U α α γ -N(x), which is contained in A for some U there exist a γ g g gαg - α open set V in X such that x V U A x γ gαg Int(A). {U γ -N(x) : U A} I gαg α (A). Hence the result. NOTE : Since every γ g -open set is γ gαg -open set, every γ g -interior point of a set A X is γ gαg -interior point of A. Thus γ g Int(A) γ gαg Int(A). In general γ gαg Int(A) γ g Int(A) which is shown in the following example EXAMPLE :Let X = {a,b,c,d}, τ ={Ø,{a},{a,b},{a,b,c},X} γ g O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},{a,b,d},{a,c,d}, X}. γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d}, {b,c}, {a,b,c}, {a,b,d}, {a,c,d}, X}. Let A = {a,d}. Then γ g Int(A) ={a} but γ gαg Int(A) ={a,d} γ g Int(A) γ gαg Int(A). THEOREM : A subset A of X is γ gαg -open iff A = γ gαg Int(A). PROOF: ( ) Let A be an γ gαg -open set. Now A is the largest γ gαg -open 30

30 set contained in A. But γ gαg Int(A) is the largest γ gαg -open set of A. Hence A = γ gαg Int(A). ( ) Let A = γ gαg Int(A). Then by definition γ gαg Int(A) is an γ gαg -open set. This implies that A is also γ gαg -open set. Hence the result. LEMMA : If A and B are subsets of X and A B then γ gαg Int(A) γ gαg Int(B). PROOF: Easy proof. NOTE : γ gαg Int(A) = γ gαg Int(B) does not imply that A = B. This is shown in the following example. EXAMPLE : Let X = {a,b,c,d}, τ ={Ø,{a},{a,d},{a,b,c},X} γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{a,b,c},{a,b,d},{a,c,d}, X}. Now γ gαg Int({b,c}) = {b} and γ gαg Int({b,c,d}) ={b}. But {b,c} {b,c,d}. THEOREM : Let A and B be subsets of X. Then (i) γ gαg Int(A) γ gαg Int(B) γ gαg Int(A B) (ii) γ gαg Int(A B) γ gαg Int(A) γ gαg Int(B) PROOF: Proof follows from lemma THEOREM : γ Int(A), γ Ext(A) are disjoint where A X and g g g g α α hence for x A, U γ g g -N(x) & γ Ext(A) are disjoint α gαg PROOF: γ Cl(A) A and Ext(A) c c γ = γ Int(A ) A g g g g gαg α α γ g α g c Int(A) γ g g Ext(A) = γ Int(A) γ Int(A ) = γ Int(A A c ) = α gαg gαg gαg γ Int(Ø) = Ø. Thus γ Int(A) & γ Ext(A) are disjoint. gαg gαg gαg 31

31 Now by Theorem γ Int(A) = gαg U {U γ -N(x): U A} U and g α g γ Ext(A) are disjoint. Hence the result. gαg THEOREM : For a subset A of a space X, the following statements hold: 1) γ gαg b(a) b(a) where b(a) denotes the boundary of A; 2) A = γ gαg Int(A) b α (A); 3) γ gαg Int(A) γ gαg b(a) = Ø; 4) A is an γ gαg open set if and only if γ gαg b(a) = Ø; 5) γ gαg b(γ gαg Int(A)) = Ø; 6) γ gαg Int (γ gαg b(a)) = Ø; 7) γ gαg b(γ gαg b(a)) = γ gαg b(a); 8) γ gαg b(a)) = A γ gαg Cl(X-{A} 9) γ gαg b(a) = γ gαg D(X-A) Proof: Easy to prove. The converse of the theorem (1) is not true in general EXAMPLE : X = {a,b,c} τ = { Ø, {a}, {a,b}, {a,c}, X}. If A = {a,b}, then γ gαg b(a) = Ø and b(a) = {b}. Hence b(a) γ gαg b(a) THEOREM : For a subset A of a space X, the following statements hold. 1) γ gαg Fr(A) Fr(A) where Fr(A) denotes the frontier of A; 2) γ gαg Cl(A) =γ gαg Int(A) γ gαg Fr(A); 3) γ gαg Int(A) γ gαg Fr(A) = Ø; 4) γ gαg b(a) γ gαg Fr(A) 5) γ gαg Fr(A) = γ gαg b(a) γ gαg D(A); 6) A is an γ gαg open set if and only if γ gαg Fr(A) = γ gαg D(A); 7) γ gαg Fr(A) = γ gαg Cl(A) γ gαg Cl(X-A); 8) γ gαg Fr(A)= γ gαg Fr(X-A); 32

32 9) γ gαg Fr(A) is γ gαg -closed; 10) γ gαg Fr(γ gαg Fr(A)) γ gαg Fr(A); 11) γ gαg Fr(γ gαg Int(A)) γ gαg Fr(A); 12) γ gαg Fr(γ gαg Cl(A)) γ gαg Fr(A); 13) γ gαg Int(A) = A -γ gαg Fr(A); PROOF: Proof is straight forward The converse of (1) and (4) of Theorem are not true in general, as shown by the following example. EXAMPLE : Let X = {a,b,c} & τ= {X,Ø,{a},{a,c}} be a topology on X. Then γ gαg O(X) ={X,Ø,{a},{c},{a,b},{a,c}}. If A = {c}, B = {b,c}, then Fr(A) = {b, c} {c} = γ gαg Fr(A), γ gαg Fr(B) = {b,c} γ gαg b (B)={c}. THEOREM : Let (X,τ) be a topological space and A be the subset of X. Then A is an γ gαg -open in X if and only if and only if F γ gαg - Int(A), where F is γ gαg -closed and F A. PROOF: Proof is easy. 33

33 CHAPTER 3 CONTINOUS FUNCTIONS 3.1. INTRODUCTION Mashour et.al [32] in 1983, introduced and investigated, α -continuity in topological space. In this chapter we introduce and investigate the concept of (p,α)-continuity, (s,α)-continuity, αg-continuity, (α,αg)- continuity, contra αg-continuity. Some characterizations and basic properties of the new type of functions are obtained. 3.2 γ gαg CONTINUOUS FUNCTIONS, (α,γ gαg )-CONTINUOUS & γ gαg -IRRESOLUTE FUNCTIONS DEFINITION 3.2.1: A function f:(x,τ) (Y,σ) is said to be γ gαg - continuous if the inverse image of every open set in Y is γ gαg -open in X. EXAMPLE 3.2.2: Let X = Y = {a,b,c}. τ = {,{a},{b},{a,b},x}, σ = {, {a},{a,b},y}. Then γ gαg C(X) ={,{c},{b,c},{a,c},x}, Let f be the identity map then f is γ gαg -continuous. THEOREM 3.2.3: A function f: (X,τ) (Y,σ) is γ gαg -continuous if and only if f -1 (V) is γ gαg -closed in X for every closed set in Y. PROOF: Let f: (X,τ) (Y,σ) is γ gαg -continuous and V be an closed set in Y. Then V c is open in Y and since f is γ gαg -continuous, f -1 (V c ) is γ gαg - open set in X. But f -1 (V c ) = [f -1 (V)] c. So f -1 (V) is γ gαg -closed in X. Conversely, assume that f -1 (V) is γ gαg -close in X for each closed set V in Y. Let F be a open set in Y. Then F c is closed in Y and by assumption 34

34 f -1 (F c ) is γ gαg -closed in X. Since f -1 (F c ) = [f -1 (F)] c, we have f -1 (F) is γ gαg - open set in X and so f is γ gαg -continuous. THEOREM 3.2.4: If f: X Y is γ gαg -continuous and g: Y Z is continuous then their composition gof : X Z is γ gαg -continuous. PROOF: Let F be any open set in Z. Since g is continuous, g -1 (F) is open set in Y. Since f is γ gαg -continuous and g -1 (F) is open in Y f -1 (g -1 (F)) = (gof) -1 (F) is γ gαg -open set in X and so gof is γ gαg -continuous. DEFINITION 3.2.5: A function f: X Y is called (α,γ gαg )-continuous if f -1 (V) is γ gαg -closed set in X for every α-closed set V of Y. EXAMPLE 3.2.6: Let X = Y = {a,b,c}. τ = {,{a,b},x}, σ = {, {a}, {b},{a,b},y}. Then γ gαg C(X)={,{a},{b},{c},{b,c},{a,c},X} and σc(x) ={,{c},{b,c},{a,c},x}. Let f be the identity map then f is (α,γ gαg )- continuous. THEOREM 3.2.7: If f: (X,τ) (Y,σ) is α-irresolute then it is (α,γ gαg )- continuous but not conversely. PROOF: Let V be any α-closed in Y. since f is α-irresolute, f -1 (V) is α- closed set in X. Every α-closed is γ gαg -closed in X. So f -1 (V) is γ gαg -closed set in X. So f is (α,γ gαg )-continuous. However the converse of the above theorem need not be true. EXAMPLE 3.2.8: Let X=Y={a,b,c}. τ ={,{a,b},x}, σ = {, {a}, {b}, {a,b},{a,c}y}. Then αc(x)={,{c},x} & γ gαg C(X) ={,{a}, {b}, {c}, {b,c},{a,c},x} and σc(x) ={,{b}, {c},{b,c}, {a,c}, X}. Let f be the 35

35 identity map then f is (α,γ gαg )-continuous but not α-irresolute because the inverse image of α-closed set {b} in Y is {b}, which is not α-closed in X. REMARK 3.2.9: The composition of two (α,γ gαg )-continuous functions need not be (α,γ gαg )-continuous. EXAMPLE : Let X = Y =Z = {a,b,c}, τ ={,{a},{b},{a,b},{a,c}, X}, σ = {, {a}, {b}, {a,b},y} and η ={,{a},z}. Let f: (X,τ) (Y,σ) be an identity map and g:(y,σ) (Z,η) defined by g(a)=c, g(b)= b, g(c)=a. Then f and g are (α,γ gαg )-continuous but their composition function gof: (X,τ) (Z,η) is not (α,γ gαg )-continuous because V ={c} is α-closed set in Z, but (gof) -1 (V) = f -1 g -1 (V) = f -1 (g -1 ({c}) = f -1 ({a}) = {a} is not γ gαg - closed in X. DEFINITION : A function f:x Y is called γ gαg -irresolute function if the inverse image of every γ gαg -closed set in Y is γ gαg -closed set in X. REMARK : The following example shows that the notions of irresolute functions and γ gαg -irresolute functions are independent. EXAMPLE : Let X = Y = {a,b,c}. τ = {, {a}, {a,b}, X}, σ ={, {a},{b},{a,b},y}. Then γ gαg C(X) = {, {b}, {c}, {b,c}, {a,c}, X} and γ gαg C(Y) ={,{c},{b,c},{a,c},y}. Then f : X Y defined by f(a) = a, f(b) = b, f(c) =c is γ gαg -irresolute function, but it is not irresolute. Since {b.c} is semi-open and γ gαg -closed in Y but f -1 {b,c} ={b,c} is not semiopen in X where as {b,c} is γ gαg -closed in X. 36

36 EXAMPLE : Let X = Y = {a,b,c}. τ ={, {a},{b},{a,b},x}, σ = {,{a,b},y}. Then γ gαg C(X) ={,{c},{b,c},{a,c}, X}. γ gαg C(Y) ={,{a}, {b},{c},{b,c},{a,c},y} and Then f : X Y defined by f(a) = a, f(b) = b, f(c) =c is not γ gαg -irresolute function, but it is irresolute. THEOREM : A function f: (X,τ) (Y,σ) is γ gαg -irresolute function if and only the inverse image of every γ gαg -open set in Y is γ gαg - open set in X. PROOF: Let f: (X,τ) (Y,σ) be a γ gαg -irresolute function and U be an γ gαg -open set in Y. Then U c is γ gαg -closed set in Y and since f is γ gαg - irresolute, f -1 (U c ) is γ gαg -closed set in X. But f -1 (U c ) = [f -1 (U c )] c and so f -1 (U) is γ gαg -open in X. Conversely, assume that f -1 (U) is γ gαg -open in X for each γ gαg - open set U in Y. Let F be γ gαg -closed set in Y. Then F c is γ gαg -open in Y and by assumption f -1 (F c ) is γ gαg -open in X. Since f -1 (F c ) = [f -1 (F)] c, we have f -1 (F) is γ gαg -closed set in X and so f is γ gαg -irresolute. THEOREM : If a function f: (X,τ) (Y,σ) is γ gαg -irresolute then it is γ gαg -continuous. PROOF: Proof is trivial. REMARK : The converse of the theorem need not be true, which is illustrated in the following theorem. EXAMPLE : Let X = Y = {a,b,c}. τ ={,{a},{b},{a,b},{a,c}, X}, σ = {,{a},{a,c},y}. Then γ gαg C(X) ={,{b},{c},{a,c},{b,c},x}, αc(y) ={, {b,c},{b}, X} and γ gαg C(Y) ={X,,{b},{c},{a,b},{b,c}}. Then the identity function f:x Y is γ gαg -continuous, since inverse image of every 37

37 closed set in Y is γ gαg -closed set in X. But f is not γ gαg -irresolute since the inverse image of the γ gαg -closed set {a,b} is {a,b}, which is not a γ gαg - closed set. THEOREM : Let X, Y, Z be topological spaces and f:(x,τ) (Y,σ) and g: (Y,σ) (Z,η) be two functions such that (i) f is γ gαg -irresolute and g is γ gαg -continuous function or (ii) f is γ gαg -continuous and g is continuous function Then their composition gof: (X,τ) (Z,η) is γ gαg -continuous function. PROOF: (i) Let V be an open set in Z. Then g -1 (V) is γ gαg -open. Since f is γ gαg -irresolute, f -1 [g -1 (V)] is γ gαg -open. But f -1 [g -1 (V)] =(gof) -1 (V). So gof is γ gαg -continuous. Let U be a closed set in Z. Then g -1 (U) be a closed set in Y because g is continuous. Since f is γ gαg -continuous, f -1 [g -1 (U)] is γ gαg -clossed set in X. But f -1 [g -1 (U)] = (gof) -1 (U). So gof is γ gαg -continuous. THEOREM : Let X, Y, Z be topological spaces and f:(x,τ) (Y,σ) and g: (Y,σ) (Z,η) be two γ gαg -irresolute functions. Then their composition gof: (X,τ) (Z,η) is γ gαg -irresolute function. PROOF: Proof is trevial. 38

38 CHAPTER 4 SEPARATION AXIOMS 4.1 INTRODUCTION α-separation axioms play a very important role in general topology. Indeed there are some research papers which deal with different α-separation axioms and also many topologists worldwide are doing research in this area. It is the aim of this part to offer some new types of α-separation axioms by using α-open sets. Separation axioms on the new set γ gαg -open sets are also introduced. As α-separation axioms play a very important role in general topology, γ gαg separation axioms can also give their supporting contribution. It is the aim of this part to offer some new types of γ gαg -separation axioms by using γ gαg -open sets. 4.2 SOME BASIC CHARACTERIZATIONS OF γ gαg -SEPARATION AXIOMS DEFINITION 4.2.1:A space X is said to be γ gαg -T 0 if for each pair of distinct points x & y in X, there exists a γ gαg -open set of X containing one point but not the other. THEOREM 4.2.2: A topological space X is γ gαg -T 0 space if and only if γ gαg -closure of distinct points are distinct. PROOF: Let x, y X with x y and (X,τ) is γ gαg -T 0 space. We will show that γ gαg Cl({x}) γ gαg Cl({y}). Since (X,τ) is γ gαg -T 0, there exists γ gαg -open set G such that x G but y G. Also x X-G and y X-G, 39

39 where X-G is γ gαg -closed set in (X,τ). Now {y} is the intersection of all γ gαg -closed sets which contain y. Hence y γ gαg Cl({y}) but x γ gαg Cl({y}) as x X-G. Therefore that γ gαg Cl({x}) γ gαg Cl({y}). Conversely, for any pair of distinct points x,y X and γ gαg Cl({x}) γ gαg Cl({y}). Then there exists at least one point z X such that z γ gαg Cl({x}) but z γ gαg Cl({y}). We claim that x γ gαg Cl({y}) because if x γ gαg Cl({y}), then {x} γ gαg Cl({y}) γ gαg Cl({x}) γ gαg Cl({y}) (Using theorem 2.6.4). So z γ gαg Cl({y}) which is contradiction. Hence x γ gαg Cl({y}). Now x γ gαg Cl({y}) x X- γ gαg Cl({y}), which is an γ gαg -open set in (X,τ) containing x but not y. Hence (X,τ) is γ gαg -T 0 space. THEOREM 4.2.3: Every subspace of γ gαg -T 0 space is γ gαg -T 0 space. In other words the property of being a γ gαg -T 0 space is a hereditary property. PROOF : Let (X,τ) be a topological space. (Y,τ*) be a subspace of (X,τ), where τ* is a relative topology. Let y 1 & y 2 be two distinct points of Y and Y X, therefore these two are distinct points of X. Since (X,τ) is a γ gαg -T 0 space, there exists γ gαg open set G such that y 1 G and y 2 G. Then by definition, G Y is γ gαg open set in (Y, τ*) which contains y 1 but not y 2. hence (Y,τ*) is a γ gαg -T 0 space. DEFINITION : A space X is said to be γ gαg -T 1 if for each pair of distinct points x and y of X, there exist γ gαg -open sets U and V containing x and y respectively such that y U and x V. EXAMPLE 4.2.5: (1) Let X = {a,b,c,d}, τ ={Ø,{a,d},{a,b,c},X} γ gαg O(X) ={Ø,{a},{b},{c},{a,b},{a,c},{a,d},{a,b,c},{a,b,d},{a,c,d}, {b,c,d},x}. Then (X,τ) is γ gαg -T 1. 40

40 (2) Let X={a,b,c} τ ={Ø,{a},{b},{a,b},X}. Then γ gαg O(X) ={Ø,{a},{b}, {a,b}, X}. Then (X,τ) is not γ gαg -T 1 since for the elements a and c of X, there does not exist γ gαg -open sets U and V containing a and c respectively such that c U and a V. REMARK 4.2.6:(i) Every α-t 1 space is γ gαg -T 1. Since every α-open set is γ gαg -open set. (ii) Every γ gαg -T 1 space is γ gαg -T 0. The converse of the above remark need not be true as seen from following examples. EXAMPLE 4.2.7: (1) Let X = {a,b,c,d}, τ ={Ø,{a,d},{a,b,c},X}. Then (X,τ) is γ gαg -T 1 but not α-t 1 space. (2) Let X={a,b,c} τ ={Ø,{a},{a,b},{a,c},X}. Then γ gαg O(X) = {Ø, {a}, {a,b},{a,c}, X}. Then (X,τ) is γ gαg -T 0 but not γ gαg -T 1 since for the elements a and c of X, there does not exist γ gαg -open sets U and V containing a and c respectively such that c U and a V. PROPOSITION 4.2.8: If A is γ gαg closed, then C γgα g(a)-a does not contain non-empty closed set. PROOF: Let F be a closed subset of C γgα g(a)-a. Then C γgαg(a) F c. We have F C γ g α g c (A) C γgα g(a) = Ø and hence F = Ø PROPOSITION 4.2.9:(i) For each x X, {x} is closed or its complement X-{x} is γ gαg closed in (X, τ ) (ii) For each x X, {x} is α -closed or its complement X-{x} is γ gαg closed in (X, τ ) 41