ON INTUITIONISTIC FUZZY SUBSPACES

Commun. Korean Math. Soc. 24 (2009), No. 3, pp. 433 450 DOI 10.4134/CKMS.2009.24.3.433 ON INTUITIONISTIC FUZZY SUBSPACES Ahmed Abd El-Kader Ramadan and Ahmed Aref Abd El-Latif Abstract. We introduce a new concept of intuitionistic fuzzy topological subspace, which coincides with the usual concept of intuitionistic fuzzy topological subspace due to Samanta and Mondal [18] in the case that µ = χ A for A X. Also, we introduce and study some concepts such as continuity, separation axioms, compactness and connectedness in this sense. 1. Introduction and preliminaries Šostak [19], introduce the fundamental concept of a fuzzy topological structure as an extension of both crisp topology and Chang s fuzzy topology [4], in the sense that not only the object were fuzzified, but also the axiomatics. In [20, 21] Šostak gave some rules and showed how such an extension can be realized. Chattopadhyay et al. [5, 6] have redefined the similar concept. In [16] Ramadan gave a similar definition namely Smooth fuzzy topology for lattice L = [0, 1], it has been developed in many direction [9-11]. As a generalization of fuzzy sets, the notion of intuitionistic fuzzy sets was introduced by Atanassov [1-3]. By using intuitionistic fuzzy sets, Çoker and his colleague [7, 8] introduced the topology of intuitionistic fuzzy sets. Samanta and Mondal [17, 18] introduced the notion of intuitionistic fuzzy topology which is a generalization of the concepts of fuzzy topology and the topology of intuitionistic fuzzy sets. Recently, much work has been done with this concept [12-14]. Throughout this paper, let X be a nonempty set I = [0, 1], I 0 = (0, 1], I 1 = [0, 1) and I X denote the set of all fuzzy subsets of X. For µ I X, we call A µ = {ν I X : ν µ}. IF stand for intuitionistic fuzzy. For α I, α(x) = α for all x X. A fuzzy point x t for t I 0 is an element of I X such that, for y X, { t if y = x, x t (y) = 0 if y x. Received December 30, 2005; Revised April 16, 2009. 2000 Mathematics Subject Classification. 54A40. Key words and phrases. intuitionistic fuzzy subspace, intuitionistic fuzzy µ (continuity, separation axioms, compactness and connectedness). 433 c 2009 The Korean Mathematical Society

434 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF The set of all fuzzy points in X is denoted by P t(x). A fuzzy set λ is quasicoincident with a fuzzy set µ, denoted by λqµ, if there exists x X such that λ(x) + µ(x) > 1. Otherwise λ qµ [15]. Lemma 1.1 ([22]). If f(x t )qλ[], then x t qf 1 (λ)[µ], where x t qλ[µ] means t + λ(x) > µ(x). Definition 1.1 ([18]). An intuitionistic gradation of openness (IGO, for short) on X is an ordered pair (τ, τ ) of mappings τ, τ : I X I satisfies the following conditions: (IGO1) τ(λ) + τ (λ) 1 for each λ I X ; (IGO2) τ(0) = τ(1) = 1, τ (0) = τ (1) = 0; (IGO3) τ(λ 1 λ 2 ) τ(λ 1 ) τ(λ 2 ) and τ (λ 1 λ 2 ) τ (λ 1 ) τ (λ 2 ) for any λ 1, λ 2 I X ; (IGO4) τ( i Γ λ i) i Γ τ(λ i) and τ ( i Γ λ i) i Γ τ (λ i ) for any {λ i : i Γ} I X. The triplet (X, τ, τ ) is called an intuitionistic fuzzy topological space (briefly, IFTS). τ and τ may be interpreted as gradation of openness and gradation of nonopenness, respectively. Definition 1.2 ([18]). Let f : (X, τ, τ ) (Y, σ, σ ) be a mapping from an IFTS (X, τ, τ ) to another IFTS (Y, σ, σ ). Then f is said to be IF -continuous if for each ν I Y, σ(ν) τ(f 1 (ν)) and σ (ν) τ (f 1 (ν)). 2. Intuitionistic fuzzy subspaces Definition 2.1. Let (X, τ, τ ) be an IFTS and µ I X. The pair of mappings (τ µ, τ µ) : A µ I defined by: τ µ (ν) = {τ(λ) : λ I X, λ µ = ν} τ µ(ν) = {τ (λ) : λ I X, λ µ = ν} is an intuitionistic fuzzy µ-topology induced over µ by (τ, τ ). For any ν A µ, the number τ µ (ν) is called the µ-openness degree of ν, while τ µ(ν) is called µ-nonopenness degree of ν. Remark 2.1. If A X and µ = χ A, we have just the usual concept of intuitionistic fuzzy subspace due to Samanta and Mondal [18]. Given (τ µ, τ µ) and ν A µ we can define ((τ µ ) ν, (τ µ) ν ), the intuitionistic fuzzy ν-topology induced over ν by (τ µ, τ µ). We have trivially τ ν = (τ µ ) ν and τ ν = (τ µ) ν, that is, an intuitionistic fuzzy subspace of an intuitionistic fuzzy subspace is also an intuitionistic fuzzy subspace. Theorem 2.1. Let (X, τ, τ ) be an IFTS and µ I X. Then (τ µ, τ µ) verifies the following properties: (µigo1) τ µ (ν) + τ µ(ν) 1 for each ν A µ.

ON INTUITIONISTIC FUZZY SUBSPACES 435 (µigo2) τ µ (0) = τ µ (µ) = 1, τ µ(µ) = τ µ(0) = 0. (µigo3) τ µ (ν 1 ν 2 ) τ µ (ν 1 ) τ µ (ν 2 ) and τ µ(ν 1 ν 2 ) τ µ(ν 1 ) τ µ(ν 2 ) for each ν 1, ν 2 A µ. (µigo4) τ µ ( i J ν i) i J τ µ(ν i ) and τ µ( i J ν i) i J τ µ(ν i ) for each {ν i : i J} A µ. Proof. (µigo1) and (µigo2) are clear. (µigo3) Suppose that there exist ν 1, ν 2 A µ such that Then there exists s (0, 1) such that τ µ(ν 1 ν 2 ) τ µ(ν 1 ) τ µ(ν 2 ). τ µ(ν 1 ν 2 ) > s τ µ(ν 1 ) τ µ(ν 2 ). Since τ µ(ν 1 ) s and τ µ(ν 2 ) s, there exist λ 1, λ 2 I X with τ (λ 1 ) s and τ (λ 2 ) s such that ν 1 = λ 1 µ and ν 2 = λ 2 µ and hence ν 1 ν 2 = (λ 1 λ 2 ) µ. Since τ (λ 1 λ 2 ) τ (λ 1 ) τ (λ 2 ) s, τ µ(ν 1 ν 2 ) s. It is a contradiction. Hence, τ µ(ν 1 ν 2 ) τ µ(ν 1 ) τ µ(ν 2 ) for each ν 1, ν 2 A µ. Similarly, we can show τ µ (ν 1 ν 2 ) τ µ (ν 1 ) τ µ (ν 2 ) for each ν 1, ν 2 A µ. (µigo4) Suppose that there exist a family {ν i : i J} A µ such that τ µ( i J Then there exists s (0, 1) such that τ µ( i J ν i ) i J ν i ) > s i J τ µ(ν i ). τ µ(ν i ). Since τµ(ν i ) s for all i J, there exists λ i I X with τ (λ i ) s such that ν i = λ i µ. Thus i J ν i = ( i J λ i) µ. Since τ ( i J λ i) i J τ (λ i ) s, τµ( i J ν i) s. It is a contradiction. Hence τµ( i J ν i) i J τ µ(ν i ) for each {ν i : i J} A µ. Similarly we can show τ µ ( i J ν i) i J τ µ(ν i ) for each {ν i : i J} A µ. Theorem 2.2. Let (X, τ, τ ) be an IFTS and µ I X. Define the mappings F τµ, Fτ : A µ µ I by: F τµ (ν) = τ µ (µ ν) and Fτ (ν) = τ µ µ(µ ν) for each ν A µ. Then (F τµ, Fτ ) satisfies the following properties: µ (µgic1) F τµ (ν) + Fτ (ν) 1 for each ν A µ. µ (µgic2) F τµ (0) = F τµ (µ) = 1, Fτ (µ) = F µ τ (0) = 0. µ (µgic3) F τµ (ν 1 ν 2 ) F τµ (ν 1 ) F τµ (ν 2 ) and Fτ (ν µ 1 ν 2 ) Fτ (ν 1) µ Fτ (ν 2) for each ν µ 1, ν 2 A µ. (µgic4) F τµ ( i J ν i) i J F τ µ (ν i ) and Fτ ( µ i J ν i) i J F τ (ν i) for µ each {ν i : i J} A µ. Proof. It is clear.

436 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF Definition 2.2. Let (X, τ, τ ) be an IFTS, µ I X and x t µ. Then for r I 0, s I 1 with r + s 1. We say that ν A µ is (r, s)-if µ-q-neighborhood of x t, if there is η A µ with τ µ (η) r and τ µ(η) s such that x t qη[µ] and η ν. We denote the family of all (r, s)-if µ-q-neighborhoods of x t by Q τµ,τ µ (x t, r, s). Theorem 2.3. Let (X, τ, τ ) be an IFTS, µ I X and x t µ. Then for r I 0, s I 1 with r + s 1, Q τµ,τ µ (x t, r, s) = {λ µ : λ Q τ,τ (x t, r, s)}. Proof. Let λ Q τ,τ (x t, r, s). Then there is ξ I X with τ(ξ) r and τ (ξ) s such that x t qξ and ξ λ. Then, ξ µ λ µ. Put η = ξ µ. Then τ µ (η) r and τµ(η) s. Since x t qξ, t+ξ(x) > 1 which implies that t+(ξ µ)(x) > µ(x). Then x t qη[µ]. Thus there is η = ξ µ A µ with τ µ (η) r,τµ(η) s, x t qη[µ] and η λ µ. Hence λ µ Q τµ,τµ (x t, r, s). Theorem 2.4. Let (X, τ, τ ) be an IFTS, µ I X and x t µ. Then for r I 0, s I 1 with r + s 1, Q τµ,τµ (x t, r, s) satisfies the following: (µq1) If ν Q τµ,τµ (x t, r, s), then x t qν[µ]. (µq2) If ν 1, ν 2 Q τµ,τµ (x t, r, s), then ν 1 ν 2 Q τµ,τµ (x t, r, s). (µq3) If ν Q τµ,τµ (x t, r, s) and ν A µ such that ν ν, then ν Q τµ,τµ (x t, r, s). (µq4) If ν Q τµ,τµ (x t, r, s), there is ν Q τµ,τµ (x t, r, s) such that ν Q τµ,τµ (y m, r, s) for each y m qν [µ]. Proof. It is clear. Theorem 2.5. Let (X, τ, τ ) be an IFTS, µ I X. Then for each ν A µ and each r I 0, s I 1 with r+s 1, we define the operator C τµ,τµ : A µ I 0 I 1 A µ as follows: C τµ,τµ (ν, r, s) = {η A µ : η ν, τ µ (µ η) r, τµ(µ η) s}. For each ν, ν 1, ν 2 A µ and r I 0, s I 1 with r +s 1, the operator C τµ,τµ satisfies the following: (µc1) C τµ,τµ (0, r, s) = 0. (µc2) ν C τµ,τµ (ν, r, s). (µc3) C τµ,τµ (ν 1 ν 2, r, s) = C τµ,τµ (ν 1, r, s) C τµ,τµ (ν 2, r, s). (µc4) C τµ,τµ (ν, r, s) C τ µ,τµ (ν, m, n) if r m, s n and m + n 1. (µc5) C τµ,τµ (C τ µ,τµ (ν, r, s), r, s) = C τ µ,τµ (ν, r, s). Proof. It is straightforward. Theorem 2.6. Let (X, τ, τ ) be an IFTS, µ I X. Then for each ν A µ and each r I 0, s I 1 with r +s 1, we define the operator I τµ,τµ : A µ I 0 I 1

ON INTUITIONISTIC FUZZY SUBSPACES 437 A µ as follows: I τµ,τ µ (ν, r, s) = {η A µ : η ν, τ µ (η) r, τ µ(η) s}. For each ν, ν 1, ν 2 A µ and r I 0, s I 1 with r + s 1, the operator I τµ,τµ satisfies the following: (µi1) I τµ,τµ (µ, r, s) = µ. (µi2) I τµ,τµ (ν, r, s) ν. (µi3) I τµ,τµ (ν 1 ν 2, r, s) = I τµ,τµ (ν 1, r, s) I τµ,τµ (ν 2, r, s). (µc4) I τµ,τµ (ν, r, s) I τ µ,τµ (ν, m, n) if r m, s n and m + n 1. (µc5) I τµ,τµ (I τ µ,τµ (ν, r, s), r, s) = I τ µ,τµ (ν, r, s). Proof. It is straightforward. Theorem 2.7. Let (X, τ, τ ) be an IFTS and µ I X. For each ν A µ and each r I 0, s I 1 with r + s 1, we have (i) I τµ,τµ (µ ν, r, s) = µ C τ µ,τµ (ν, r, s). (ii) C τµ,τµ (µ ν, r, s) = µ I τ µ,τµ (ν, r, s). Proof. (i) For each ν A µ and each r I 0, s I 1 with r + s 1, we have the following: µ C τµ,τ µ (ν, r, s) = µ {η A µ : ν η, τ µ (µ η) r, τ µ(µ η) s} = {µ η : µ η µ ν, τ µ (µ η) r, τ µ(µ η) s} = I τµ,τµ (µ ν, r, s). (ii) It is similar to (i). Theorem 2.8. Let (X, τ, τ ) be an IFTS, µ I X and x t µ. For ν A µ and r I 0, s I 1 with r + s 1, x t C τµ,τµ (ν, r, s) if and only if for each η A µ with τ µ (η) r, τµ(η) s and x t qη[µ] we have νqη[µ]. Proof. Let x t C τµ,τµ (ν, r, s), η A µ with τ µ (η) r, τµ(η) s and x t qη[µ]. Suppose that ν qη[µ], then ν µ η. Since x t qη[µ], t + η(x) > µ(x). This implies x t µ η. Since ν µ η, τ µ (µ (µ η)) = τ µ (η) r and τµ(µ (µ η)) = τµ(η) s, we have x t C τµ,τµ (ν, r, s). It is a contradiction. Conversely, let η A µ with τ µ (η) r, τµ(η) s, x t qη[µ] and νqη[µ]. Suppose that x t C τµ,τµ (ν, r, s). Then there exists ξ A µ with τ µ (µ ξ) r, τµ(µ ξ) s, ν ξ and x t ξ. Then ξ(x) < t which implies (µ ξ)(x) + t > µ(x). Thus x x q(µ ξ)[µ]. Then from our hypothesis νq(µ ξ)[µ]. Thus there is y X such that ν(y) + (µ ξ)(y) > µ(y). Thus ν(y) > ξ(y) which is a contradiction with ν ξ. Hence x t C τµ,τµ (ν, r, s).

438 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF 3. IF µ-continuity Definition 3.1. Let (X, τ, τ ) and (Y, σ, σ ) be two IFTSs and µ I X. Then the mapping f : (X, τ, τ ) (Y, σ, σ ) is called IF µ-continuous if τ µ (f 1 (ν) µ) σ (ν) and τ µ(f 1 (ν) µ) σ (ν) for each ν A. Remark 3.1. Every IF -continuous mapping is IF µ-continuous for all µ but the converse is not true in general as the following example shows. Example 3.1. Let X = I. We define the IGO(τ, τ ) and IGO(σ, σ ) on X as follows: for each λ I X 1, if λ = 0, 1 0, if λ = 0, 1 τ(λ) = 0.5, if λ = t, t 0.6 τ (λ) = 0.4, if λ = t, t 0.6 0, otherwise. 1, otherwise. 1, if λ = 0, 1 0, if λ = 0, 1 0.2, if λ = t, t 0.6 σ(λ) = σ 0.7, if λ = t, t 0.6 (λ) = 0.4, if λ = 0.7 0.5, if λ = 0.7 0, otherwise. 1, otherwise. Let µ = 0.5, then 1, if ν = 0, µ τ µ (ν) = 0.5, if 0 < ν < 0.5 0, otherwise. 1, if ν = 0, σ (ν) = 0.2, if 0 < ν < 0.5 0, otherwise. 0, if λ = 0, µ τµ(ν) = 0.4, if 0 < ν < 0.5 1, otherwise. 0, if λ = 0, σ (ν) = 0.7, if 0 < ν < 0.5 1, otherwise. Then the identity mapping id X : (X, τ, τ ) (X, σ, σ ) is an IF µ-continuous but not IF -continuous. Theorem 3.1. Let (X, τ, τ ) and (Y, σ, σ ) be two IFTSs, f : X Y be a mapping and {µ i : i J} I X such that i J µ i = 1. Then f is IF µ i - continuous for each i J if and only if f is IF -continuous. Proof. Due to Remark 3.1, it suffices to show that if f is IF µ i -continuous for each i J, then f is IF -continuous. Suppose there exists λ I Y such that Then there exists s (0, 1) such that τ (f 1 (λ)) σ (λ). τ (f 1 (λ)) > s σ (λ). Since σ (λ) s, σ f(µ i ) (λ f(µ i)) s for each i J. Since f is an IF µ i - continuous for each i J we have τ µ i (f 1 (λ f(µ i )) µ i ) σ f(µ i) (λ f(µ i)) s

ON INTUITIONISTIC FUZZY SUBSPACES 439 but f 1 (λ f(µ i )) µ i = f 1 (λ) f 1 (f(µ i )) µ i = f 1 (λ) µ i. Then τ µ i (f 1 (λ) µ i ) s for each i J. Then for each i J there exists ν i I X with τ (ν i ) s such that f 1 (λ) µ i = ν i µ i. This implies that i J (f 1 (λ) µ i ) = i J (ν i µ i ), thus f 1 (λ) ( i J µ i ) = ( i J ν i ) ( i J µ i ). Since i J µ i = 1, f 1 (λ) = i J ν i. Then τ (f 1 (λ)) = τ ( ν i ) τ (ν i ) s. i J i J It is a contradiction. Thus τ (f 1 (λ)) σ (λ) for each λ I Y. Similarly, we can show τ(f 1 (λ)) σ(λ) for each λ I Y. Thus f is an IF -continuous. Theorem 3.2. Let (X, τ, τ ) and (Y, σ, σ ) be two IFTSs, µ I X and f : X Y be an injective mapping. For r I 0, s I 1 with r + s 1, the following statements are equivalent: (i) f is IF µ-continuous. (ii) F τµ (f 1 (λ) µ) F σ (λ) and F τ µ (f 1 (λ) µ) F σ (λ) for each λ A. (iii) f(c τµ,τ µ (ν, r, s)) C σ,σ (f(ν), r, s) for each ν A µ. (iv) C τµ,τµ (f 1 (λ) µ, r, s) f 1 (C σ,σ (λ, r, s)) µ for each λ A. (v) f 1 (I σ,σ (λ, r, s)) µ I τ µ,τµ (f 1 (λ) µ, r, s) for each λ A. (vi) For each x t µ and λ A with σ (λ) r, σ (λ) s and f(x t )qλ[], there is ν A µ with τ µ (ν) r, τµ(ν) s such that x t qν[µ] and f(ν) λ. (vii) For each x t µ and λ Q σ,σ (f(x t), r, s), there exists ν Q τµ,τµ (x t, r, s) such that f(ν) λ. (viii) For each x t µ and each λ Q σ,σ (f(x t), r, s), f 1 (λ) Q τµ,τ µ (x t, r, s). Proof. (i) (ii) For each λ A, we have F σ (λ) = σ ( λ) τ µ(f 1 ( λ) µ) = τ µ(f 1 () f 1 (λ)) µ) = τ µ(µ f 1 (λ) µ) = F τ µ (f 1 (λ) µ). Similarly, we can show F σ (λ) F τµ (f 1 (λ) µ) for each λ A.

440 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF (ii) (iii) For each ν A µ and r I 0, s I 1 with r + s 1, we have f 1 (C σ,σ (f(ν), r, s)) = f 1 ( {λ A : f(ν) λ, σ ( λ) r, σ ( λ) s}) = {f 1 (λ) : ν f 1 (λ), F σ (λ) r, F σ (λ) s} {f 1 (λ) µ A µ : ν f 1 (λ) µ, F τµ (f 1 (λ) µ) r, F τ µ (f 1 (λ) µ) s} {f 1 (λ) µ A µ : ν f 1 (λ) µ, τ µ (µ (f 1 (λ) µ)) r, = C τµ,τµ (ν, r, s). τ µ(µ (f 1 (λ) µ)) s} This implies that f(c τµ,τµ (ν, r, s)) C σ,σ (f(ν), r, s). (iii) (iv) For each λ A and r I 0, s I 1 with r + s 1, we have f(c τµ,τ µ (f 1 (λ) µ, r, s)) C σ,σ (f(f 1 (λ) µ), r, s) C σ,σ (λ, r, s). Thus C τµ,τµ (f 1 (λ) µ, r, s) f 1 (C σ,σ (λ, r, s)) µ. (iv) (v) It is clear from Theorem 2.7. (v) (i) Suppose that there exist λ A and r 0 I 0, s 0 I 1 with r 0 + s 0 1 such that τ µ (f 1 (λ) µ) < r 0 σ (λ) and τ µ(f 1 (λ) µ) > s 0 σ (λ). Since σ (λ) r 0 and σ (λ) s 0, λ = I σ,σ (λ, r 0, s 0 ). By (v) we have Thus f 1 (λ) µ = f 1 (I σ,σ (λ, r 0, s 0 )) µ I τµ,τ µ (f 1 (λ) µ, r 0, s 0 ). f 1 (λ) µ = I τµ,τ µ (f 1 (λ) µ, r 0, s 0 ). This meaning, τ µ (f 1 (λ) µ) r 0 and τµ(f 1 (λ) µ) s 0. It is a contradiction. Then τ µ (f 1 (λ) µ) σ (λ) and τµ(f 1 (λ) µ) σ (λ) for each λ A. Hence f is IF µ-continuous. (i) (vi) Let x t µ, λ A and r I 0, s I 1 with r + s 1 such that σ (λ) r, σ (λ) s and f(x t)qλ[]. Since f is IF µ-continuous we have τ µ (f 1 (λ) µ) σ (λ) r and τµ(f 1 (λ) µ) σ (λ) s.

ON INTUITIONISTIC FUZZY SUBSPACES 441 Since f(x t )qλ[] and by Lemma 1.1, we have x t qf 1 (λ)[µ]. Since f is injective, f 1 (λ) µ and hence x t qf 1 (λ) µ[µ]. Then there exists ν = f 1 (λ) µ A µ with τ µ (ν) r, τµ(ν) s and x t qν[µ]. Also, f(ν) = f(f 1 (λ) µ) f(f 1 (λ)) λ = λ. (vi) (iii) Let ν A µ, x t C τµ,τµ (ν, r, s) and λ A with σ (λ) r and σ (λ) s such that f(x t)qλ[]. By (vi), there exists η A µ with τ µ (η) r, τµ(η) s such that x t qη[µ] and f(η) λ. Since x t C τµ,τµ (ν, r, s), τ µ (η) r, τµ(η) s and x t qη[µ], then by using Theorem 2.8, we have νqη[µ] which implies that f(ν)qf(η)[] and hence f(ν)qλ[]. Thus f(x t ) C σ,σ (f(ν), r, s). This implies that x t f 1 (C σ,σ (f(ν), r, s)). Then C τµ,τµ (ν, r, s) f 1 (C σ,σ (f(ν), r, s)). Hence f(c τµ,τµ (ν, r, s)) C σ,σ (f(ν), r, s). (vi) (vii) Let x t µ and λ Q σ,σ (f(x t), r, s). Then there exists η A with σ (η) r, σ (η) s such that f(x t)qη[] and η λ. By (vi), there exists ν A µ with τ µ (ν) r, τµ(ν) s such that x t qν[µ] and f(ν) η λ. Hence ν Q τµ,τµ (x t, r, s) and f(ν) λ. (vii) (viii) Let x t µ and λ Q σ,σ (f(x t), r, s). By (vii), there exists ν Q τµ,τµ (x t, r, s) such that f(ν) λ. So, there is ξ A µ with τ µ (ξ) r, τµ(ξ) s such that x t qξ[µ] and ξ ν f 1 (λ). Hence f 1 (λ) Q τµ,τµ (x t, r, s). (viii) (vi) Let x t µ and λ A with σ (λ) r, σ (λ) s and f(x t )qλ[]. Then λ Q σ,σ (f(x t), r, s). By (viii), we have f 1 (λ) Q τµ,τµ (x t, r, s) and hence there is ν A µ with τ µ (ν) r, τµ(ν) s such that x t qν[µ] and ν f 1 (λ), so f(ν) λ. Theorem 3.3. Let (X, τ, τ ), (Y, σ, σ ) and (Z, δ, δ ) be IFTSs, µ I X, f : X Y and g : Y Z. If f is IF µ-continuous and g is IF -continuous, then g f is IF µ-continuous. Proof. For each λ A (g f)(µ) we have δ g() (λ) σ (g 1 (λ) ) τ µ(f 1 (g 1 (λ) ) µ) = τ µ((g f) 1 (λ) f 1 () µ) = τ µ((g f) 1 (λ) µ). Similarly, δ g() (λ) τ µ ((g f) 1 (λ) µ). Thus g f is IF µ-continuous. Definition 3.2. Let (X, τ, τ ) be an IFTS and µ I X. For r I 0, s I 1 with r + s 1, ν A µ is called:

442 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF (i) (r, s)-if µ-regular open set if I τµ,τµ (C τ µ,τµ (ν, r, s), r, s) = ν. (ii) (r, s)-if µ-regular closed set if C τµ,τµ (I τ µ,τµ (ν, r, s), r, s) = ν. Definition 3.3. Let (X, τ, τ ), (Y, σ, σ ) be two IFTSs and µ I X. Then the mapping f : X Y is called IF µ-almost continuous if τ µ (f 1 (ν) µ) r and τ µ(f 1 (ν) µ) s for each (r, s)-if µ-regular open set ν in A. Remark 3.2. Every IF µ-continuous mapping is also IF µ-almost continuous but the converse is not true in general, as the following example shows. Example 3.2. Consider Example 3.1, and put 1, if λ = 0, 1 0, if λ = 0, 1 0.5, if λ = 0.5 σ(λ) = σ 0.5, if λ = 0.5 (λ) = 0.6, if λ = 2.5 0.2, if λ = 2.5 0, otherwise. 1, otherwise. Since µ = 0.5, we have 1, if ν = 0, σ (ν) = 0.6, if ν = 2.5 0, otherwise. 0, if λ = 0, σ (ν) = 0.2, if ν = 2.5 1, otherwise. Then, the identity mapping id X : (X, τ, τ ) (X, σ, σ ) is IF µ-almost continuous but not IF µ-continuous. Theorem 3.4. Let (X, τ, τ ) and (Y, σ, σ ) be two IFTSs, µ I X and f : X Y be an injective mapping. For r I 0, s I 1 such that r + s 1 the following statements are equivalent: (i) f is IF µ-almost continuous. (ii) F τµ (f 1 (λ) µ) r and Fτ (f 1 (λ) µ) s for each (r, s)-if µ- µ regular closed set λ in A. (iii) f 1 (λ) µ I τµ,τµ (f 1 (I σ,σ (C σ,σ (λ, r, s), r, s) µ), r, s) for each λ A with σ (λ) r and σ (λ) s. (iv) f 1 (λ) µ C τµ,τµ (f 1 (C σ,σ (I σ,σ (λ, r, s), r, s) µ), r, s) for each λ A with σ ( λ) r and σ ( λ) s. (v) For each x t µ and λ A with σ (λ) r, σ (λ) s and f(x t )qλ[], there is ν A µ with τ µ (ν) r, τµ(ν) s such that f(ν) I σ,σ (C σ,σ (λ, r, s), r, s). (vi) For each x t µ and λ Q σ,σ (f(x t), r, s), there exists ν Q τµ,τµ (x t, r, s) such that f(ν) I σ,σ (C σ,σ (λ, r, s), r, s). (vii) For each x t µ and each λ Q σ,σ (f(x t), r, s) we have f 1 (I σ,σ (C σ,σ (λ, r, s), r, s)) Q τ µ,τ µ (x t, r, s).

ON INTUITIONISTIC FUZZY SUBSPACES 443 Proof. (i) (ii) Let λ be (r, s)-if µ-regular closed set in A. Then λ is (r, s)-if µ-regular open set. Since f is IF µ-almost continuous we have τ µ (f 1 ( λ) µ) r and τ µ(f 1 ( λ) µ) s. Since f is injective, τ µ ((µ f 1 (λ)) µ) r and τ µ((µ f 1 (λ)) µ) s. Let ν = µ (f 1 (λ) µ). Then ν(x) = µ(x) (f 1 (λ) µ)(x) = µ(x) min{f 1 (λ)(x), µ(x)}. If µ(x) f 1 (λ)(x), then ν(x) = µ(x) µ(x) = 0. Then and F τµ (f 1 (λ) µ) = τ µ (µ (f 1 (λ) µ)) = τ µ (ν) = τ µ (0) = 1 r. F τ µ (f 1 (λ) µ) = τ µ(µ (f 1 (λ) µ)) = τ µ(ν) = τ µ(0) = 0 s. If µ(x) > f 1 (λ)(x), then Then ν = µ f 1 (λ) = (µ f 1 (λ)) µ. τ µ (ν) = τ µ ((µ f 1 (λ)) µ) r and τ µ(ν) = τ µ((µ f 1 (λ)) µ) s. Thus F τµ (f 1 (λ) µ) = τ µ (ν) r and F τ µ (f 1 (λ) µ) = τ µ(ν) s. (ii) (i) It is clear. (i) (iii) Let λ A and r I 0, s I 1 with r + s 1 such that σ (λ) r, σ (λ) s. Then λ I σ,σ (C σ,σ (λ, r, s), r, s). Since I σ,σ (C σ,σ (λ, r, s), r, s) is (r, s)-if µ-regular open and f is IF µ-almost continuous, and So, τ µ (f 1 (I σ,σ (C σ,σ (λ, r, s), r, s)) µ) r τµ(f 1 (I σ,σ (C σ,σ (λ, r, s), r, s)) µ) s. f 1 (λ) µ f 1 (I σ,σ (C σ,σ (λ, r, s), r, s)) µ = I τµ,τµ (f 1 (I σ,σ (C σ,σ (λ, r, s), r, s)) µ, r, s). (iii) (iv) It follows from Theorem 2.7, (iv) (ii) Let λ be (r, s)-if -regular closed. Then σ ( λ) r ( λ) s. By (iv) we have, σ f 1 (λ) µ C τµ,τµ (f 1 (C σ,σ (I σ,σ (λ, r, s), r, s) µ), r, s) = C τµ,τ µ (f 1 (λ) µ, r, s).

444 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF Thus f 1 (λ) µ = C τµ,τ µ (f 1 (λ) µ, r, s). By Theorem 2.5, we have and F τµ (f 1 (λ) µ) = τ µ (µ f 1 (λ) µ) r F τ µ (f 1 (λ) µ) = τ µ(µ f 1 (λ) µ) s. (i) (v) (iii) and (v) (vi) (vii) (v) are similar to that of Theorem 3.2. 4. IF µ-separation axioms Definition 4.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. µ is said to be (r, s)-if µt 2 -space if for each x α, y β (x y) µ, there are ν 1, ν 2 A µ with τ µ (ν 1 ) r, τ µ (ν 2 ) r, τ µ(ν 1 ) s and τ µ(ν 2 ) s such that x α ν 1, y β ν 2 and ν 1 qν 2 [µ]. Theorem 4.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. µ is (r, s)-if µt 2 -space if and only if for each x α, y β (x y) µ, we have y β {C τµ,τµ (ν, r, s) : τ µ(ν) r, τµ(ν) s, x α ν}. Proof. Let x α, y β (x y) µ and m = µ(y) β. Then x α, y m (x y) µ. Since µ is (r, s)-if µt 2 -space, there are ν 1, ν 2 A µ with τ µ (ν 1 ) r, τ µ (ν 2 ) r, τ µ(ν 1 ) s and τ µ(ν 2 ) s such that x α ν 1, y m ν 2 and ν 1 qν 2 [µ]. Thus ν 1 µ ν 2, τ(µ (µ ν 2 )) r and τ (µ (µ ν 2 )) s which implies, C τµ,τ µ (ν 1, r, s) µ ν 2. Since y m ν 2, β = µ(y) m > µ(y) ν 2 (y) (C τµ,τ µ (ν 1, r, s))(y) and hence y β C τµ,τµ (ν 1, r, s). Thus y β {C τµ,τµ (ν, r, s) : τ µ(ν) r, τµ(ν) s, x α ν}. Conversely, let x α, y β (x y) µ. Then, x α, y µ(y) β (x y) µ. By hypothesis, there is ν 1 A µ with τ µ (ν 1 ) r, τµ(ν 1 ) s such that x α ν 1 and y µ(y) β C τµ,τµ (ν 1, r, s) and hence, µ(y) β > (C τµ,τµ (ν 1, r, s))(y) which implies y β µ C τµ,τµ (ν 1, r, s) = ν 2 and τ µ (ν 2 ) r, τµ(ν 2 ) s. Since ν 2 = µ C τµ,τµ (ν 1, r, s) µ ν 1, ν 1 qν 2 [µ]. Hence µ is (r, s)-if µt 2 -space. Definition 4.2. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. µ is said to be (r, s)-if µ-regular space if for each ξ A µ with τ µ (µ ξ) r, τ µ(µ ξ) s and for each fuzzy point x t µ with x t qξ[µ], there are ν, η A µ with τ µ (ν) r, τ µ (η) r, τ µ(ν) s and τ µ(η) s such that x t ν, ξ η and ν qη[µ]. Theorem 4.2. Let (X, τ, τ ) be an IFTS, µ I X r + s 1. Then the following are equivalent: (i) µ is (r, s)-if µ-regular. and r I 0, s I 1 with

ON INTUITIONISTIC FUZZY SUBSPACES 445 (ii) For each fuzzy point x t µ and each ν A µ with x t ν, τ µ (ν) r, τµ(ν) s there is η A µ with τ µ (η) r, τµ(η) s such that x t η C τµ,τµ (η, r, s) ν. (iii) For each fuzzy point x t µ and each ξ A µ with τ µ (µ ξ) r and τµ(µ ξ) s such that x t qξ[µ], there are ν, η A µ with τ µ (ν) r, τ µ (η) r, τµ(ν) s and τµ(η) s such that x t ν, ξ η and (ν, r, s) qη[µ]. C τµ,τ µ Proof. (i) (ii) Let x t µ be a fuzzy point, r I 0, s I 1 with r + s 1 and ν A µ with τ µ (ν) r, τµ(ν) s and x t ν. Then τ µ (µ (µ ν)) r, τµ(µ (µ ν)) s and x t q(µ ν)[µ]. By (i), there are η, ν 1 A µ with τ µ (η) r, τ µ (ν 1 ) r, τµ(η) s and τµ(ν 1 ) s such that x t η, µ ν ν 1 and η qν 1 [µ]. Since η qν 1 [µ], η µ ν 1 ν and hence C τµ,τµ (η, r, s) µ ν 1 ν. Thus x t η C τµ,τµ (η, r, s) ν. (ii) (iii) Let x t µ be a fuzzy point, r I 0, s I 1 with r + s 1 and ξ A µ with τ µ (µ ξ) r, τµ(µ ξ) s and x t qξ[µ]. Then x t µ ξ. By (ii), there is η 1 A µ with τ µ (η 1 ) r and τµ(η 1 ) s such that x t η 1 C τµ,τµ (η 1, r, s) µ ξ. By (ii) again, there is ν A µ with τ µ (ν) r and τµ(ν) s such that x t ν C τµ,τ µ (ν, r, s) η 1 C τµ,τ µ (η 1, r, s) µ ξ. Put η = µ C τµ,τµ (η 1, r, s). Hence there are ν, η A µ with τ µ (ν) r, τ µ (η) r, τµ(ν) s and τµ(η) s such that x t ν, ξ η and C τµ,τµ (ν, r, s) qη[µ]. (iii) (i) It is clear. 5. IF µ-compactness Definition 5.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. µ is said to be (r, s)-if µ-compact if for every family {ν i : i J} in {ν : ν A µ, τ µ (ν) r, τ µ(ν) s} such that ( i J ν i)(x) = µ(x) for each x X, there exists a finite subset J 0 of J such that ( i J 0 ν i )(x) = µ(x). Definition 5.2. Let X be a non-empty set and µ I X. A collection β A µ is said to form a fuzzy µ-filterbasis if for each finite subcollection {ν 1, ν 2,..., ν n } of β, ( n i=1 ν i)(x) > 0 for some x X. Theorem 5.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. µ is (r, s)-if µ-compact if and only if for each fuzzy µ-filterbasis β A µ, ( ν β C τ µ,τµ (ν, r, s))(x) > 0 for some x X. Proof. Let {ν i : i J} be a family in Λ = {ν : ν A µ, τ µ (ν) r, τµ(ν) s} such that ( i J ν i)(x) = µ(x). Suppose that there is no finite subset J 0 of J such that ( i J 0 ν i )(x) = µ(x). Then for each finite subcollection {ν 1, ν 2,..., ν n } of Λ, there exists x X such that ν i (x) < µ(x) for each i = 1, 2,..., n. Then µ(x) ν i (x) > 0 for each i = 1, 2,..., n. So, n i=1 (µ ν i )(x) > 0 and hence β = {µ ν i : ν i Λ, i J} forms a fuzzy µ-filterbasis.

446 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF Since ( i J ν i)(x) = µ(x) for each x X and τ µ (ν i ) r, τ µ(ν i ) s for each i J we have ( C τµ,τ (µ ν µ i, r, s))(x) = ( i J i J (µ ν i ))(x) = (µ i J ν i )(x) = 0 for each x X. It is a contradiction. Thus there exists a finite subset J 0 of J such that ( i J 0 ν i )(x) = µ(x) for all x X. Thus µ is (r, s)-if µ-compact. Conversely, Suppose that there exists fuzzy µ-filterbasis β such that for each x X. Then ( C τµ,τ (ν, r, s))(x) = 0 µ ν β ( ν β(µ C τµ,τµ (ν, r, s))(x) = µ(x) for each x X. Since τ µ (µ C τµ,τµ (ν, r, s)) r and τ µ(µ C τµ,τµ (ν, r, s)) s and µ is (r, s)- IF µ-compact, there exists a finite subcollection {µ C τµ,τµ (ν i, r, s)) : i = 1, 2,..., n} of {µ C τµ,τµ (ν, r, s)) : ν β} such that n ( (µ C τµ,τµ (ν i, r, s)))(x) = µ(x) for each x X. i=1 Since ν i (x) C τµ,τµ (ν i, r, s)(x) for each x X we have n n ( (µ ν i ))(x) ( (µ C τµ,τµ (ν i, r, s)))(x) = µ(x) for each x X. i=1 i=1 Then ( n i=1 (µ ν i))(x) = µ(x) for each x X. Thus ( n i=1 ν i)(x) = 0 for each x X. It is a contradiction. Hence ( ν β C τ µ,τµ (ν, r, s))(x) > 0 for some x X. Theorem 5.2. Let (X, τ, τ ), (Y, σ, σ ) be two IFTSs, f : X Y be an IF µ-continuous bijective mapping. For r I 0, s I 1 with r + s 1, if µ is (r, s)-if µ-compact, then is (r, s)-if -compact. Proof. Let {λ i : i J} be a family in {λ : λ A, σ (λ) r, σ (λ) s} such that ( i J λ i)(y) = ()(y) for all y Y. Since f is IF µ-continuous for each i J we have, Since f is injective, τ µ (f 1 (λ i ) µ) σ (λ i ) r, τ µ(f 1 (λ i ) µ) σ (λ i) s. i J(f 1 (λ i ) µ) = f 1 ( i J λ i ) µ = f 1 () µ = µ. Since µ is (r, s)-if µ-compact, there exists a finite subset J 0 of J such that ( i J 0 (f 1 (λ i ) µ))(x) = µ(x) for all x X. This implies that

ON INTUITIONISTIC FUZZY SUBSPACES 447 f( i J 0 (f 1 (λ i ) µ)) = and hence i J 0 (f(f 1 (λ i ))) =. Since f is surjective, i J 0 λ i =. Thus ( i J 0 λ i )(y) = (y) for all y Y. Hence is (r, s)-if -compact. 6. IF µ-connected sets Definition 6.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. Then ν, η A µ are said to be (r, s)-if µ-separated if ν qc τµ,τµ (η, r, s)[µ] and η qc τµ,τµ (ν, r, s)[µ]. Theorem 6.1. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. Then for ν, η A µ, (i) If ν and η are (r, s)-if µ-separated and ν 1, η 1 A µ such that 0 ν 1 ν, 0 η 1 η, then ν 1 and η 1 are (r, s)-if µ-separated. (ii) If ν qη[µ] and either τ µ (ν) r, τ µ (η) r, τ µ(ν) s and τ µ(η) s or τ µ (µ ν) r, τ µ (µ η) r, τ µ(µ ν) s and τ µ(µ η) s, then ν and η are (r, s)-if µ-separated. (iii) If τ µ (ν) r, τ µ (η) r, τ µ(ν) s and τ µ(η) s or τ µ (µ ν) r, τ µ (µ η) r, τ µ(µ ν) s and τ µ(µ η) s, then ν (µ η) and η (µ ν) are (r, s)-if µ-separated. Proof. (i) Since ν 1 ν, C τµ,τµ (ν 1, r, s) C τµ,τµ (ν, r, s). Since ν, η are (r, s)- IF µ-separated, η qc τµ,τµ (ν, r, s)[µ]. Thus η 1 η µ C τµ,τ µ (ν, r, s) µ C τ µ,τ µ (ν 1, r, s). Then η 1 qc τµ,τ µ (ν 1, r, s)[µ]. Similarly, ν 1 qc τµ,τ µ (η 1, r, s)[µ]. Hence ν 1 and η 1 are (r, s)-if µ-separated. (ii) Let ν qη[µ], τ µ (ν) r, τ µ (η) r, τ µ(ν) s and τ µ(η) s. Since ν qη[µ], ν µ η. Thus C τµ,τµ (ν, r, s) C τ µ,τµ (µ η, r, s) = µ η. Then C τµ,τµ (ν, r, s) qη[µ]. Similarly, C τ µ,τµ (η, r, s) qν[µ]. Then ν and η are (r, s)-if µ-separated. Let τ µ (µ ν) r, τ µ (µ η) r, τµ(µ ν) s and τµ(µ η) s. By Theorem 2.5, ν = C τµ,τµ (ν, r, s) and η = C τ µ,τµ (η, r, s). Since ν qη[µ] we have, C τµ,τµ (ν, r, s) qη[µ] and C τ µ,τµ (η, r, s) qν[µ]. Then ν and η are (r, s)-if µ-separated. (iii) Let τ µ (ν) r, τ µ (η) r, τµ(ν) s and τµ(η) s. Since ν (µ η) µ η, C τµ,τµ (ν (µ η), r, s) C τ µ,τµ ((µ η), r, s) = µ η. Then Since η (µ ν) η, η µ C τµ,τµ (ν (µ η), r, s). η (µ ν) µ C τµ,τµ (ν (µ η), r, s).

448 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF Thus C τµ,τµ (ν (µ η), r, s) q(η (µ ν))[µ]. Similarly, we can show C τ µ,τµ (η (µ ν), r, s) q(ν (µ η))[µ]. Hence, ν (µ η) and η (µ ν) are (r, s)-if µ- separated. Let τ µ (µ ν) r, τ µ (µ η) r, τµ(µ ν) s and τµ(µ η) s. By Theorem 2.5, we have ν = C τµ,τµ (ν, r, s) and η = C τ µ,τµ (η, r, s). Then ν (µ η) µ η = µ C τµ,τµ (η, r, s). Thus C τµ,τµ (η, r, s) µ (ν (µ η)). Since η (µ ν) η, C τµ,τµ ((η (µ ν)), r, s) C τ µ,τµ (η, r, s) µ (ν (µ η)). Thus C τµ,τµ (η (µ ν), r, s) q(ν (µ η))[µ]. Similarly, C τµ,τµ (ν (µ η), r, s) q(η (µ ν))[µ]. Hence, ν (µ η) and η (µ ν) are (r, s)-if µ-separated. Theorem 6.2. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. Then ν, η A µ are (r, s)-if µ-separated if and only if there exist ν 1, η 1 A µ with τ µ (ν 1 ) r, τ µ (η 1 ) r, τµ(ν 1 ) s and τµ(η 1 ) s such that ν ν 1, η η 1, ν qη 1 [µ] and η qν 1 [µ]. Proof. Let ν, η A µ are (r, s)-if µ-separated. Then ν µ C τµ,τ µ (η, r, s) = ν 1(say) and η µ C τµ,τ µ (ν, r, s) = η 1(say). Then τ µ (ν 1 ) r, τ µ (η 1 ) r, τ µ(ν 1 ) s and τ µ(η 1 ) s. Since ν 1 = µ C τµ,τ µ (η, r, s) µ η and η 1 = µ C τµ,τ µ (ν, r, s) µ ν, we have ν 1 qη[µ] and η 1 qν[µ]. Conversely, let ν 1, η 1 A µ with τ µ (ν 1 ) r, τ µ (η 1 ) r, τ µ(ν 1 ) s and τ µ(η 1 ) s such that ν ν 1, η η 1, ν qη 1 [µ] and η qν 1 [µ]. Then and C τµ,τ µ (ν, r, s) C τ µ,τ µ (µ η 1, r, s) = µ η 1 µ η C τµ,τ µ (η, r, s) C τ µ,τ µ (µ ν 1, r, s) = µ ν 1 µ ν. Thus C τµ,τµ (ν, r, s) qη[µ] and C τ µ,τµ (η, r, s) qν[µ]. Hence ν, η are (r, s)-if µ- separated. Definition 6.2. Let (X, τ, τ ) be an IFTS, µ I X and r I 0, s I 1 with r + s 1. Then ν A µ is said to be (r, s)-if µ-connected if it can t be expressed as the union of two (r, s)-if µ-separated sets. Theorem 6.3. Let (X, τ, τ ) and (Y, σ, σ ) be IFTS, µ I X and f : X Y be IF µ-continuous injective mapping. For r I 0, s I 1 with r + s 1, if λ is (r, s)-if µ-connected, then f(λ) is (r, s)-if -connected. Proof. Suppose that f(λ) is not (r, s)-if -connected. Then there exist two (r, s)-if -separated sets ν, η A such that f(λ) = ν η. By Theorem 6.2, there exist ν 1, η 1 A with σ (ν 1 ) r, σ (η 1 ) r, σ (ν 1) s and σ (η 1) s such that ν ν 1, η η 1, ν qη 1 [] and

ON INTUITIONISTIC FUZZY SUBSPACES 449 η qν 1 []. Since f is injective and ν 1, f 1 (ν 1 ) f 1 () = µ and hence f 1 (ν) f 1 (ν 1 ) µ. Similarly, f 1 (η) f 1 (η 1 ) µ. Since f is IF µ-continuous, we have τ µ (f 1 (ν 1 ) µ) σ (ν 1 ) r and τ µ(f 1 (ν 1 ) µ) σ (ν 1) s. Similarly, τ µ (f 1 (η 1 ) µ) r, τ µ(f 1 (η 1 ) µ) s. Since f is injective, f 1 (ν)(x) + (f 1 (η 1 ) µ)(x) = f 1 (ν)(x) + f 1 (η 1 )(x) = ν(f(x)) + η 1 (f(x)) = ν(y) + η 1 (y) (y) = µ(x) and hence f 1 (ν) q(f 1 (η 1 ) µ)[µ]. Similarly, f 1 (η) q(f 1 (ν 1 ) µ)[µ]. Then by Theorem 6.2, f 1 (ν) and f 1 (η) are (r, s)-if µ-separated. Since f is injective, λ = f 1 (f(λ)) = f 1 (ν η) = f 1 (ν) f 1 (η). It is a contradiction with λ is (r, s)-if µ-connected. Hence f(λ) is (r, s)-if - connected. References [1] K. Atanassov, Intuitionistic fuzzy sets, VII ITKR s Session, Sofia (September, 1983) (in Bularian). [2], Intuitionistic fuzzy sets, Fuzzy Sets and Systems 20 (1986), no. 1, 87 96. [3], New operators defined over the intuitionistic fuzzy sets, Fuzzy Sets and Systems 61 (1993), no. 2, 131 142. [4] C. L. Chang, Fuzzy topological spaces, J. Math. Anall. Apll. 24 (1968), 182 190. [5] K. C. Chattopadhyay, R. N. Hazra, and S. K. Samanta, Gradation of openness: fuzzy topology, Fuzzy Sets and Systems 94 (1992), 237 242. [6] K. C. Chattopadhyay and S. K. Samanta, Fuzzy topology: fuzzy closure operator, fuzzy compactness and fuzzy connectedness, Fuzzy Sets and Systems 54 (1993), 207 212. [7] D. Çoker, An introduction to intuitionistic fuzzy topological spaces, Fuzzy Sets and Systems 88 (1997), 81 89. [8] D. Çoker and A. H. Es, On fuzzy compactness in intuitionistic fuzzy topological spaces, J. Fuzzy Mathematics 3 (1995), no. 4, 899 909. [9] M. Demirci, Neighborhood structures in smooth topological spaces, Fuzzy Sets and Systems 92 (1997), 123 128. [10] U. Höhle and A. P. Sostak, A general theory of fuzzy topological spaces, Fuzzy Sets and Systems 73 (1995), 131 149. [11] Y. C. Kim, Initial L-fuzzy closure spaces, Fuzzy Sets and Systems 133 (2003), 277 297. [12] Y. C. Kim and S. E. Abbas, Connectedness in intuitionistic fuzzy topological spaces, Commun. Korean Math. Soc. 20 (2005), no. 1, 117 134. [13] E. P. Lee and Y. B. Im, Mated fuzzy topological spaces, Int. Journal of Fuzzy Logic and Intelligent Systems 11 (2001), no. 2, 161 165. [14] W. K. Min and C. K. Park, Some results on intuitionistic fuzzy topological spaces defined by intuitionistic gradation of openness, Commun. Korean Math. Soc. 20 (2005), no. 4, 791 801. [15] P. M. Pu and Y. M. Liu, Fuzzy topology I: Neighborhood structure of a fuzzy point and Moor-Smith convergence, J. Math. Anal. Appl. 76 (1980), 571 599. [16] A. A. Ramadan, Smooth topological spaces, Fuzzy Sets and Systems 48 (1992), 371 375.

450 AHMED ABD EL-KADER RAMADAN AND AHMED AREF ABD EL-LATIF [17] S. K. Samanta and T. K. Mondal. Intuitionistic gradation of openness: intuitionistic fuzzy topology, Busefal 73 (1997), 8 17. [18], On intuitionistic gradation of openness, Fuzzy Sets and Systems 31 (2002), 323 336. [19] A. P. Šostak, On a fuzzy topological structure, Supp. Rend. Circ. Math. Palermo (Ser.II) 11 (1985), 89 103. [20], On the neighbourhood structure of a fuzzy topological space, Zb. Rodova Univ. Nis, ser Math. 4 (1990), 7 14. [21], Basic structure of fuzzy topology, J. of Math. Sciences 78 (1996), no. 6, 662 701. [22] A. M. Zahran, On fuzzy subspaces, Kyungpook Math. J. 41 (2001), 361 369. Ahmed Abd El-Kader Ramadan Department of Mathematics Faculty of Science Beni-Suef University, Egypt E-mail address: aramadan58@yahoo.com Ahmed Aref Abd El-Latif Department of Mathematics Faculty of Science Beni-Suef University, Egypt E-mail address: ahmeda73@yahoo.com