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Int Jr. of Mathematical Sciences & Applications Vol.4, No.1, January-June 2014 Copyright Mind Reader Publications ISSN No: 2230-9888 www.journalshub.com SEPARATION AXIOMS ON g̃*-OPEN SETS IN INTUITIONISTIC TOPOLOGICAL SPACES Dr.G.Vasuki email: [email protected] & Dr.B.Amudhambigai email: [email protected] Department of Mathematics Sri Sarada College for Women, Salem-16 Tamil Nadu, India Abstract The purpose of this paper is to introduce the concepts intuitionistic g̃*-regular space and intuitionistic g̃*-normal space, intuitionistic g̃*-To space, intuitionistic g̃*-Ro space, intuitionistic T2 spaces are introduced. Some interesting properties and characterizations are introduced and discussed. Keywords: Intuitionistic g̃-open set, Intuitionistic g̃-closed set, Intuitionistic g̃*open set, Intuitionistic g̃*-closed set, Intuitionistic g̃*-regular space, Intuitionistic g̃*-normal space, Intuitionistic g̃*-To space, Intuitionistic g̃*-Ro space. 2010 AMS Subject classification primary: 54A05, 54A10, 54A20. 1 INTRODUCTION The concept of intuitionistic sets in topological spaces was introduced by Coker in 1996. The concept of g̃-closed set was introduced by Jafarietal. This notion was further studied by Rajesh and Rkici. The purpose of this paper is to introduce the concepts intuitionistic g̃-closed set, intuitionistic g̃*-closed set, intuitionistic g̃*-regular space, 1 81 intuitionistic g̃*-normal space, intuitionistic g̃*-To space, intuitionistic g̃*-Ro space, intuitionistic T2 spaces are introduced. Some interesting properties and characterizations are introduced and discussed. 2 Preliminaries Definition 2.1. [1] Let X be a nonempty set. An intuitionistic set (IS for short) A is an object having the form A = < X, A1 , A2 > where A1 and A2 are subsets of X satisfying A1 ∩ A2 = φ. The set A1 is called the set of members of A, where A2 is called the set of non-members of A. Every crisp set A on a nonempty set X is obviously an IS having the form < X, A1 , A2 >. Definition 2.2. [1] Let X be a nonempty set. Let A = < X, A1 , A2 > and B = < X, B 1 , B 2 > be an IS’s on X and let {Ai : i ∈ J} be an arbitrary family of IS’s in X, where Ai = < X, A1i , A2i > then 1. A ⊆ B iff A1 ⊆ A2 and B 2 ⊆ A2 2. A = B iff A ⊆ B and B ⊆ A 3. A ⊂ B iff A1 ∪ A2 ⊇ B 1 ∪ B 2 4. A = < X, A2 , A1 > 5. ∪Ai = < X, ∪A1i , ∩A2i > 6. ∩Ai = < X, A1i , A2i > 7. A−B = A ∩ B c 8. [ ]A = < X, A1 , (A1 ) > c 9. < >A = < X, (A2 ) , A2 > 2 82 10. φ∼ = < X, φ, X > and X∼ = < X, X, φ >. Definition 2.3. [1] Let f be a function from a set X to a set Y. Let A = < X, A1 , A2 > be an IS in X and B = < Y, B 1 , B 2 > an IS in Y. Then the pre image f −1 (B) is an IS in X defined by f −1 (B) = < X, f −1 (B 1 ), f −1 (B 2 ) > and the image f(A) is an IS in c c Y defined by f(A) = < Y, f (A1 ), f− (A2 ) > where f− (A2 ) = (f (A2 ) ) . Definition 2.4. [2] An intuitionistic topology (IT f orshort) on a nonempty set X is a family τ of IS’s in X satisfying the following axioms: 1. φ∼ , X∼ ∈ T 2. G1 ∩ G2 ∈ T for any G1 , G2 ∈ T 3. ∪Gi ∈ T for any arbitrary family {Gi |i ∈ J} ⊆ T where (X,T) is called an intuitionistic topological space (ITS for short) any intuitionistic set in T is called an intuitionistic open set (IOS for short) in X. The Complement A of an IOS A is called an intuitionistic closed set (ICS for short)in X. Definition 2.5. [2] Let X be a nonempty set and p X a fixed element in X and let c A = < X, A1 , A2 > be an IS. Then the IS p∼ defined by p∼ =< X, {p}, {p} > is called an intuitionistic point (IP for short in X). Definition 2.6. [2] Let (X, T) and (X, T 0 ) be any two ITS’s and let f : X→Y be a function. Then f is said to be intuitionistic open iff the image of IS in T is IS in T 0 . Definition 2.7. [3] A subset A of a space (X,T) is called 1. ĝ-closed if cl(A) ⊆ U whenever A ⊆ U and U is semi-open in (X,T). The complement of ĝ-closed set is said to be ĝ-open. 2. *g-closed set if cl(A) ⊆ U whenever A ⊆ U and U is ĝ-open in (X,T). The complement of *g-closed set is said to be *g-open. 3 83 3. ] g semi-closed (briefly, ] gs-closed) set if Scl(A) ⊆ U whenever A ⊆ U and U is ] g-open in (X,T). The complement of ] gs-closed set is said to be ] gs-open. 4. g̃-closed set if cl(A) ⊆ U whenever A ⊆ U and U is an ] gs-open in (X,T). The complement of g̃-closed set is said to be g̃-open. Definition 2.8. [7] Let (X, T) be a topological space and E ⊆ X. Then g̃-closure of E (briefly, g̃-cl(E)) is define to be the intersection of all g̃-closed set containing E. Definition 2.9. [10] For a subset A of a topological space (X,T), clθ (A) = {x X : cl(U) ∩ A 6= φ, U T and x U} Definition 2.10. [6] A subset B of a topological space X is called g-closed in X if cl(B) ⊂ G whenever B ⊂ G and G is open in X. The complement of g-closed is g-open in X. Definition 2.11. A function f : (X,T)→(Y,σ) is called 1. a g̃-continuous [8] if f −1 (V ) is g̃-closed in (X,T) for every closed set V in (Y,σ). 2. a g̃-irresolute [9] if f −1 (V ) is g̃-closed in (X,T) for every g̃-closed set V in (Y,σ). 3. a ] g-semi irresolute [11] (briefly ] gs-irresolute) if f −1 (V ) is ] gs-closed in (X,T) for each ] gs-closed set V in (Y,σ). 4. a g̃-closed [4] if the image of every closed set in (X,T) is g̃-closed in (Y,σ). 5. Weakly Continuous [5] if for each point x ∈ X and each open set V if (Y,σ) containing f(x), there exists an open set U containing x such that f(U) ⊆ cl(V). 4 84 3 INTUITIONISTIC g̃-REGULAR SPACES Definition 3.1. Let (X,T) be an intuitionistic topological space and A = < X, A1 , A2 >, B = < X, B 1 , B 2 > be an ICs in X. Then A is said to be 1. Iĝ-closed if cl(A) ⊆ B whenever A ⊆ B and B is an intuitionistic semi open. The complement of Iĝ-closed set is said to be Iĝ-open. 2. I*g-closed if cl(A) ⊆ B whenever A ⊆ B and B is an Iĝ-open. The complement of I*g-closed set is said to be I*g-open. 3. I ] g-semi closed if scl(A) ⊆ B whenever A ⊆ B and B is an I*g-open. The complement of I ] g-semi closed set is said to be I ] g-semi open. 4. I g̃-closed if cl(A) ⊆ B whenever A ⊆ B and B is an I ] gs-open. The complement of I g̃-closed set is said to be I g̃-open. Definition 3.2. Let (X,T) be an ITS and let p∼ be an IP in X and S be a subset of X. Then S is called an intuitionistic neighbourhood of p∼ in X if there exists an intuitionistic open set R such that p∼ ∈ R ⊆ S. Definition 3.3. Let (X,T) be an ITS and let P = < X, P 1 , P 2 > be any IS in X. Then an intuitionistic g̃-closure of P (briefly, I g̃cl(P)) is defined by I g̃cl(P) = ∩ {R :R is an I g̃CS in X and P ⊆ R}. Definition 3.4. Let (X,T) be an ITS and let A = < X, A1 , A2 > be any IS in X. Then an intuitionistic g̃ of A (briefly, I g̃int(A)) is defined by I g̃int(A) = ∪ {B:B is an I g̃OS in X and B ⊆ A}. Definition 3.5. Let (X,T) be an ITS and let A = < X, A1 , A2 >, B = < X, B 1 , B 2 > and C = < X, C 1 , C 2 > be any ISs in X. Then A is said to be an I g̃*-open set (briefly, I g̃*OS) if A = B ∩ C where B is an I g̃-open and C is an intuitionistic open set. 5 85 Definition 3.6. Let (X,T) be an ITS and let p∼ be an Ip in X and V be an subset of X. Then V is called an intuitionistic g̃*-neighbourhood of p∼ (briefly, I g̃*nbd) in (X,T) if there exists an I g̃*OS U of (X,T) such that p∼ ∈ U ⊆ V. Definition 3.7. Let (X,T) be an ITS and let E = < X, E 1 , E 2 > be an IS in X. Then an intuitionistic g̃*-closure of E (briefly, I g̃*cl(E)) is defined by I g̃*cl(E) = ∩ {B :B is I g̃*CS in X and E ⊆ B} Definition 3.8. Let (X,T) be an ITS and let P = < X, P 1 , P 2 > be an IS in X. Then an intuitionistic g̃*-interior of R (briefly, I g̃*int(R)) is defined by I g̃*int(R) = ∪ {R :R is I g̃*OS in X and R ⊆ P}. Definition 3.9. An ITS (X,T) is said to be I g̃*-regular space (briefly, I g̃*RS) if for every I g̃* closed set F and each point p∼ in X where p∼ ∈ / F, there exists disjoint IOSs U and V such that F ⊆ U and p∼ ∈ V. Proposition 3.1. Let (X,T) be an ITS. Then the following statements are equivalent: 1. (X,T) is an I g̃*RS. 2. For each p∼ ∈ X and I g̃*nbd W of p∼ there exists an IOS V of p∼ such that cl(v) ⊆ w. Proof (i)⇒(ii) Let W be any I g̃*nbd of p∼ . Then there exists an I g̃*OS G such that p∼ ∈ G ⊆ w. Since G is an I g̃*CS and p∼ ∈ / G by hypothesis there exists IOSs U and V such that G ⊆ U, p∼ ∈ V where U ∩ V = φ and and so V ⊆ U . Now cl(V) ⊆ cl(U ) = U and hence G ⊆ U implies U ⊆ G ⊆ W. Therefore cl(V) ⊆ W. (ii)⇒(i) Let F be any I g̃*CS and p∼ ∈ / F. Then p∼ ∈ F and F be an I g̃*OS and so F is 6 86 an I g̃*nbd of p∼ . By hypothesis, there exists an IOnbd V of p∼ such that p∼ ∈ V and cl(V) ⊆ F which implies F ⊆ cl(V ). Then cl(V ) is an IOS containing F and V ∩ cl(V ) = φ. Therefore (X,T) is an I g̃*RS. Proposition 3.2. Suppose that B ⊆ A ⊆ X, B is an intuitionistic I g̃*CS in X. Then B is I g̃*CS in X. Proposition 3.3. If (X,T) is an I g̃*RS and Y is an intuitionistic open and I g̃*CS of X, then the subspace Y is an I g̃*RS. Proof Let F be any I g̃*CS of Y and y ∈ F . Then by Proposition 3.2, F is an I g̃*CS in X. Since (X,T) is I g̃*RS, there exists disjoint IOSs U and V of (X,T) such that y∼ ∈ U and F ⊆ V. Therefore, U ∩ Y and V ∩ Y are disjoint IOSs of the subspace Y such that y∼ ∈ U ∈ Y and F ⊆ V ∩ Y. Hence the subspace Y is an I g̃*RS. Proposition 3.4. An ITS (X,T) is an I g̃*RS iff for each I g̃*CS F of (X,T) and p∼ ∈ F there exist IOSs U and V of (X,T) such that p∼ in U, F ⊆ V and cl(U) ∩ cl(V) = φ∼ . Proof Let F be an I g̃*CS of (X,T) with p∼ ∈ / F. Then there exist IOSs Uo and V of (X,T) such that p∼ ∈ Uo , F ⊆ V where Uo ∩ V = φ∼ which implies Uo ∩ cl(V) = φ∼ . Since cl(V) is an ICS, it is an I g̃*CS and p∼ ∈ / cl(V). Since (X,T) is an I g̃*RS there exist IOSs G and H of (X,T) such that p ∈ G, cl(V) ⊆ H where G ∩ H = φ∼ which implies cl(G) ∩ H = φ∼ . Let U = Uo ∩ G, then U and V are IOSs of (X,T) such that p∼ ∈ U, F ⊆ V and cl(U) ∩ cl(V) = φ∼ . The converse part is trivial. Proposition 3.5. Let (X,T) be an ITS. Then the following statements are equivalent: 1. (X,T) is an I g̃*RS. 7 87 2. For each point p∼ ∈ X and for each I g̃*nbd W of p∼ , there exists an IOnbd V of p∼ such that cl(U) ⊆ W. 3. For each point p∼ ∈ X and for each I g̃*CS F not containing p∼ , there exists an IOnbd V of p∼ such that cl(V) ∩ F = φ∼ . Proof (i)⇒(ii) Follows from Proposition 3.1 (ii)⇒(iii) Let p∼ ∈ X and F be an I g̃*CS such that p∼ ∈ / F. Then F is an I g̃*nbd of p∼ and by hypothesis, there exists an IOnbd V of p∼ such that cl(V) ⊆ F and hence cl (V) ∩ F = φ∼ . (iii)⇒(ii) Let p∼ ∈ X and W be an I g̃*nbd of p∼ . Then there exists an I g̃*OS G such that p∼ ∈ G ⊆ W. Since G is I g̃*CS and p ∈ / G by hypothesis there exists an IOnbd U of p∼ such that cl(U) ∩ G = φ. Therefore, cl(U) ⊆ G ⊆ W. Definition 3.10. For an intuitionistic subset A of an ITS (X,T) Iclθ∗ (A) = {p∼ ∈ X : cl(U) ∩ A 6= φ∼ , U ∈ T and p∼ ∈ U } Proposition 3.6. Let (X,T) be an ITS. Then the following statements are equivalents 1. (X,T) is I g̃*RS. 2. Iclθ∗ (A) = I g̃*-cl(A) for each intuitionistic subset A of X and 3. Iclθ∗ (A) = A for each I g̃*CS of X. Proof (i)⇒(ii) For any intuitionistic subset A ⊆ X, A ⊆ I g̃*cl(A) ⊆ Iclθ∗ (A). Let p∼ ∈ I g̃*cl(A). 8 88 Then there exists an I g̃*CS F such that p∼ ∈ F and A ⊆ F. By assumption, there exist disjoint IOSs U and V such that p∼ ∈ U and F ⊆ V. Now, p∼ ∈ U ⊆ cl(V) ⊆ V ⊆ F ⊆ A and therefore cl(U) ∩ A = φ∼ . Thus p∼ ∈ Iclθ∗ (A) and hence Iclθ∗ (A) = I g̃*cl(A). (ii)⇒(iii) Proof is trivial. (iii)⇒(i) Let F be any I g̃*CS with p∼ ∈ F . Since F is I g̃*CS, by assumption p∼ ∈ Iclθ∗ (F ) and so there exists an IOS U such that p∼ ∈ U and cl(U) ∩ F = φ∼ . Then F ⊆ cl(U ). Let V = cl(U ). Then V is an IOS such that F ⊆ V. Also the IOSs U and V are disjoint and hence (X,T) are I g̃*RS. Definition 3.11. An intuitionistic subset B of an ITS (X,T) is called an intuitionistic g-closed set (briefly, IGCS) in (X,T) if cl(B) ⊆ G whenever B ⊆ G and G is an IOS inX. The complement of an IGCS is an intuitionistic g-open set (briefly, IGOS). Definition 3.12. Let (X,T) and (Y,σ) be any two ITSs. A function f :(X,T)→ (Y, σ) is called: 1. an I g̃*-continuous function (briefly, I g̃*CF) if f −1 (v) is an I g̃*CS in (X,T) for every ICS V in (Y,σ). 2. an I g̃*-irresolute function (briefly, I g̃*IRF) if f −1 (v) is an I g̃*CS in (X,T) for every I g̃*CS V in (Y,σ). 3. an I ] g-semi irresolute (briefly, I ] gS-irresolute) if f −1 (v) is I ] gS-closed in (X,T) for each I ] gS-closed set V of (Y,σ). 4. an I g̃ -closed if the image of every closed set in (X,T) is an IgCS in (Y,σ). 5. an intuitionistic weakly continuous function (briefly, IWCF) if for each point p∼ ∈ X and IOS V if (Y,σ) containing f(p∼ ). There exists an IOSU containing p∼ such that f(U) ⊆ cl(V). 9 89 Proposition 3.7. Let f :(X,T)→ (Y, σ) is an I ] gS-irresolute I g̃-closed and A is an I g̃-closed set of (X,T) then f(A) is I g̃-closed. Proof Let U be an I ] gS-open set in (Y,σ) such that f(A) ⊂ U. Since f is an (Y,σ)Sirresolute f −1 (U) is a I ] gS-open set containing A. Hence cl(A) ⊂ f −1 (U) as A is an I g̃-closed in (X,T). Since f is an I g̃-closed, f(cl(A)) is an I g̃-closed set contained in the I ] gS-open set U which implies that cl(f(cl(A))) ⊆ U and hence cl(f(A)) ⊆ U. Therefore, f(A) is an I g̃-closed set. Proposition 3.8. Let f :(X,T)→ (Y, σ) is an I ] gS-irresolute, I g̃-closed, intuitionistic continuous, injection and (Y,σ) is Ig ∗ -regular, then (X,T) is an I g̃*-regular. Proof Let F be any I g̃*CS of (X,T) and p∼ ∈ F. Since f is an I ] gS-irresolute, I g̃-closed by Proposition 3.2, f(F) is I g̃-closed in (Y,σ) and f (p∼ ) ∈ / f(F). Since (Y,σ) is I g̃*-regular, there exists disjoint IOSs U and V in(Y,σ) such that f (p∼ ) ∈ U and f(F) ⊆ V. That is, p∼ ∈ f −1 (U), F ⊆ f −1 (V) and f −1 (U) ∩ f −1 (V) = φ. Therefore,(X,T) is an I g̃*-regular space. Proposition 3.9. Let (X,T) and (Y,σ) be any two ITSs. Then for any function f:(X,T)→(Y,σ), the following statements are equivalent: 1. f is an intuitionistic weakly continuous function. 2. f −1 (V) ⊆ intf −1 (cl(V )) for every V ∈ (Y,σ). 3. cl(f −1 (V)) ⊆ f −1 (cl(V )) for every V ∈ (Y,σ). Proof (i)⇒(ii) Suppose that V ∈ (Y,σ) is an IOS and let p∼ ∈ f −1 (V). From (i), f(U) ∈ cl(V) for some IOS U in (X,T). Hence, U ⊆ f −1 (cl(V )) and p∼ ∈ U ⊆ intf −1 (cl(V )). 10 90 Therefore, f −1 (V ) ⊆ intf −1 (cl(V )). (ii)⇒(iii) Suppose that V ∈ (Y,σ) is an IOS and p∼ ∈ / f −1 (cl(V )). Then,f(p∼ ) ∈ / cl(V). There exists an IOS G in (Y,σ) such that G ∩ V = φ∼ . Since V ∈ (Y,σ) is an IOS, cl(G) ∩ V = φ∼ and hence int(f −1 (cl(G))) ∩ f −1 (V) = φ∼ . By (ii), p∼ ∈ f −1 (G) ⊆ int(f −1 (cl(G))). Therefore, p∼ ∈ / cl(f −1 (V )). Hence, cl(f −1 (V )) ⊂ f −1 (cl(V )). (iii)⇒(i) Let p∼ ∈ X and V ∈ (Y,σ) is an IOS. Then, p∼ ∈ / f −1 (cl(Y − cl(V ))). Since Y-cl(V) ∈ / V(Y,σ) is an IOS and by (iii), p∼ ∈ / cl(f −1 (Y − cl(V ))). Hence, there exists an IOS U ⊆ X such that U ∩ f −1 (Y − cl(V )) = φ. Therefore, f(U) ∩ (Y-cl(V)) = φ. Hence, f(U) ⊂ cl(V). This shows that f is intuitionistic weakly continuous. Proposition 3.10. If f :(X,T)→ (Y, σ) is IWCF I g̃*-closed injection and (Y,σ) is I g̃*-regular, then (X,T) ia an I g̃*-regular. Proof Let F be any I g̃*CS of (X,T) and p∼ ∈ / F. Since f is I g̃-closed, f(F) is an I g̃-closed in (Y,σ) and f (p∼ ) ∈ / f(F). Since (Y,σ) is I g̃*-regular, there exists IOSs U and V such that f (p∼ ) ∈ U, f(F) ⊆ V and cl(U) ∩ cl(V) = φ. Since f is IWCF it follows that p∼ ∈ f −1 (U) ⊆ int(f −1 (cl(U))), F ⊆ f −1 (V) ⊆ int(f −1 (cl(V))) and int(f −1 (cl(U))) ∩ int(f −1 (cl(V))) = φ∼ . Therefore, (X,T) is an I g̃*-regular space. 4 I g̃*-NORMAL SPACE Definition 4.1. An intuitionistic topological space (X,T) is said to be an I g̃*-normal space (briefly, I g̃*NS) if for any pair of disjoint I g̃*CSs A and B, there exist disjoint IOSs U and V such that A ⊆ U and B ⊆ V. 11 91 Proposition 4.1. If (X,T) is an I g̃*NS and Y is an IO and I g̃*CS of (X,T), then the intuitionistic subspace is an I g̃*NS. Proof Let A and B be any two disjoint I g̃*CSs of Y, then A and B are I g̃*CS in (X,T). Since (X,T) is I g̃*NS, there exist disjoint IOSs U and V of (X,T) such that A ⊆ U and B ⊆ V. Then A ⊆ U ∩ Y and B ⊆ V ∩ Y. So the intuitionistic subspace is I g̃*NS. Proposition 4.2. Let (X,T) be an ITS. Then the following statements are equivalent 1. (X,T) is I g̃*NS; 2. For each I g̃*CS F and each I g̃*OS U containing F, there exists an IOS V containing F such that cl(V) ⊆ U; 3. For each pair of disjoint I g̃*CS A and B in (X,T), there exists an IOSs U containing A such that cl(U) ⊆ B = φ∼ ; 4. For each pair of disjoint I g̃*CS A and B in (X,T), there exists an IOSs U containing A and V containing B such that cl(U) ∩ cl(V) = φ; Proof (i)⇒(ii) Let F be an I g̃*CS and U be an I g̃*OS such that F ⊆ U. The F ∩ U = φ∼ . By assumption, there exist IOSs V and W such that F ⊆ V, U ⊆ W and V ∩ W = φ∼ which implies cl(V) ∩ W = φ∼ . Now, cl(V) ∩ U ⊆ cl(V) ∩ W = φ and so cl(V) ⊆ U. (ii)⇒(iii) Let A and B be disjoint I g̃*CSs of (X,T). Since A ∩ B = φ∼ , A ⊆ B and B is an I g̃*OS. By assumption, there exists an open set U containing A such that cl(U) ⊆ B and so cl(U) ∩ B = φ∼ . (iii)⇒(iv) Let A and B be any two disjoint I g̃*CS of (X,T). Then by assumption, there exists 12 92 an IOS U containing A such that cl(U) ∩ B = φ∼ . Since cl(U) is an ICS, it is I g̃*CS and so B and cl(U) are disjoint I g̃*CSs in (X,T). Therefore by assumption, there exists an IOS V containing B such that cl(U) ∩ cl(V) = φ∼ . (iv)⇒(i) Let A and B be any two disjoint I g̃*CS of (X,T). By assumption, there exists an IOS U contains A and V containing B such that cl(U) ∩ cl(V) = φ∼ . We have U ∩ V φ∼ and therefore (X,T) is an I g̃*NS. Proposition 4.3. If f:(X,T)→(Y,σ) is I g̃*CS, intuitionistic continuous injection and (Y,σ) is I g̃*-normal,then (X,T) is I g̃*-normal. Proof Let A and B be any two disjoint I g̃*CSs of (X,T). Since f is I g̃*CS, f(A) and f(B) are disjoint I g̃*CSs of (Y,σ). Since (Y,σ) is I g̃*-normal,there exist disjoint IOSs U and v such that f(A) ⊆ U and f(B) ⊆ V. ie, A ⊆ f −1 (V) and f −1 (U) ∩ f −1 (V) = φ. Since f is an intuitionistic continuous, f −1 (U) and f −1 (V) are IOSs is (X,T). Therefore (X,T) is an I g̃*NS. Proposition 4.4. If f:(X,T)→(Y,σ) is an intuitionistic weakly continuous, I g̃*-closed, injection and (Y,σ) is I g̃*-normal then (X,T) is normal. Proof Let A and B be any two disjoint closed sets of (X,T). Since f is injective and I g̃*closed, f(A) and f(B) are disjoint I g̃*-closed sets of (Y,σ). Since (Y,σ) is an I g̃*NS by Proposition 4.2, there exist intuitionistic open sets U and V such that f(A) ⊆ U, f(B) ⊆ V and cl(U) ∩ cl(V) = φ. Since f is an intuitionistic weakly continuous, it follows that from Proposition 3.9, A ⊆ f −1 (U ) ⊆ int(f −1 (cl(U ))), B ⊆ f −1 (V ) ⊆ int(f −1 (cl(V ))) and int(f −1 (cl(U ))) ∩ int(f −1 (cl(V ))) = φ. Therefore, (X,T) is I g̃*NS. 13 93 5 OTHER SEPARATION AXIOMS Definition 5.1. An intuitionistic topological space (X,T) is called an 1. I g̃*-To space if for each pair of distinct intuitionistic points there exists an I g̃*OS containing one point but not the other. 2. I g̃*-Ro space if cl{p∼ } ⊆ U whenever U is an I g̃*OS and p∼ ∈ U. Definition 5.2. An intuitionistic topological space (X,T) is called an I g̃*-T2 space if for each pair of distinct intuitionistic points p∼ and q∼ in X, there exist disjoint I g̃*OSs U and V in X such that p∼ ∈ U and q∼ ∈ V. Definition 5.3. An intuitionistic topological space (X,T) is called an IT2 space if for each pair of distinct intuitionistic points p∼ and q∼ in X, there exist disjoint IOSs U and V in X such that p∼ ∈ U and q∼ ∈ V. Definition 5.4. Let (X,T) be an ITS and let A = < X, A1 , A2 > be an intuitionistic set in X is said to be an intuitionistic regular set if A is both intuitionistic open and intuitionistic closed. Proposition 5.1. For any ITS (X,T) the following statements are equivalent: 1. (X,T) is an I g̃*RS. 2. Every I g̃*OS U is the union of intuitionistic regular set. 3. Every I g̃*CS A is the intersection of intuitionistic regular set. Proof (i)⇒(ii) Let U be an I g̃*OS and let p∼ ∈ U. If A = X−U the I g̃*CS. By assumption, there exist disjoint IOSs W1 and W2 of X such that p∼ ∈ W1 and A ⊆ W2 . If V = cl(W1 ), then V is ICS and V ∩ A ⊆ V ∩ W2 = φ. p∼ ∈ V and V ⊆ U. Thus U is a union of 14 94 an intuitionistic regular sets. (ii)⇒(iii) Is obvious (iii)⇒(i) Let A be an I g̃*CS and let p∼ ∈ / A. By assumption there exists an intuitionistic regular set V such that A ⊆ V and p∼ ∈ / V. If U = X−V then U is an intuitionistic open set containing p∼ and U ∩ V = φ. Thus (X,T) is I g̃*RS. Proposition 5.2. Every I g̃*RS (X,T) is both IT2 and I g̃Ro *. Proof Let (X,T) be an I g̃*RS and let p∼ , q∼ ∈ X be two distinct intuitionistic points.[{p∼ } is either IO and IC since every space is T2 ]. If {p∼ } is intuitionistic open, hence I g̃*OS, then by Proposition 5.1, it is an intuitionistic regular. Thus {p∼ } and X−{p} are separating IOSs. If {p∼ } is intuitionistic closed, X−{p∼ } is intuitionistic open and since every intuitionistic open set is an I g̃*OS and by Proposition 5.1 X−{p∼ } is the union of intuitionistic regular sets. Hence there is an intuitionistic regular set V ⊆ X−{p∼ } containing q∼ . Hence (X,T) is IT2 . Again by Proposition 5.1 it follows immediately that (X,T) also I g̃ ∗ Ro . Definition 5.5. An ITS (X,T) is called an ITg̃∗ -space if every I g̃*CS of (X,T) is closed in (X,T). Definition 5.6. A map f:(X,T)→(Y,σ) is said to be I g̃*-closed map if the image of every closed set in (X,T) is I g̃*CS in (Y,σ). Definition 5.7. A subset B of an intuitionistic topological space (X,T) is called intuitionistic g-closed (briefly, Ig-closed) in X if cl(B) ⊂ G whenever B ⊂ G and G is an intuitionistic open in X. 15 95 Proposition 5.3. Let f:(X,T)→(Y,S) be a I g̃*-continuous and (Y,S) is an ITg̃∗ space and g:(Y,S)→(Z,R) is I g̃*-irresolute then their composition g ◦ f:(X,T)→(Z,R) is I g̃*irresolute. Proof Let A be an I g̃*CS in (Z,R). Since g is an I g̃*-irresolute then g −1 (A) is I g̃*CS in (Y,S). Since (Y,S) is an ITg̃∗ space g −1 (A) is closed in (Y,S). Since f is I g̃*-continuous f −1 (g −1 (A)) = (g ◦ f)−1 (A) is I g̃*CS in (X,T). Hence g ◦ f is I g̃*-irresolute. Proposition 5.4. Let f:(X,T)→(Y,S) be I g̃*-irresolute, g:(Y,S)→(Z,R) be I g̃*-continuous mapping. 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