\mu - k - Connectedness in GTS

Bol. Soc. Paran. Mat. c SPM –ISSN-2175-1188 on line SPM: www.spm.uem.br/bspm (3s.) v. 33 2 (2015): 161–165. ISSN-00378712 in press doi:10.5269/bspm.v33i2.23419 µ - k - Connectedness in GTS Shyamapada Modak and Takashi Noiri abstract: Császár [4] introduced µ - semi - open sets, µ - preopen sets, µ - α open sets and µ - β - open sets in a GTS (X, µ). By using the µ - σ - closure, µ - π - closure, µ - α - closure and µ - β - closure in (X, µ), we introduce and investigate the notions µ - k - separated sets and µ - k - connected sets in (X, µ). Key Words: µ - k - separated, µ - k - connected, µ - k - component. Contents 1 Introduction and preliminaries 161 2 µ - k - separated sets 162 1. Introduction and preliminaries Let X be a set and exp X denote the power set of X. We call a class µ ⊂exp X a generalized topology [2] (briefly GT) if ∅ ∈ µ and any union of elements of µ belongs to µ. A set with GT is called a generalized topological space (briefly GTS). For a GTS (X, µ), the elements of µ are called µ - open sets and the complements of µ - open sets are called µ - closed sets. For A ⊂ X, we denote by cµ (A) the intersection of all µ - closed sets containing A and iµ (A) the union of all µ - open sets contained in A. Then we recall that iµ (iµ (A)) = iµ (A), cµ (cµ (A)) = cµ (A) and iµ (A) = X −cµ (X −A). Also, we consider by [1] that iµ (cµ (iµ (cµ (A)))) = iµ (cµ (A)) and cµ (iµ (cµ (iµ (A)))) = cµ (iµ (A)). A set A ⊂ X is said to be be µ - semi open (resp. µ - preopen, µ - α - open, µ - β - open) [4] if A ⊂ cµ (iµ (A)) (resp. A ⊂ iµ (cµ (A)), A ⊂ iµ (cµ (iµ (A))), A ⊂ cµ (iµ (cµ (A)))). We denote by σ(µ) (resp. π(µ), α(µ), β(µ)) the class of all µ - semi - open sets (resp. µ - preopen sets, µ α - open sets, µ - β - open sets). The complement of a µ - α - open (resp. µ - σ open, µ - π - open, µ - β - open) set is said to be µ - α - closed (resp. µ - σ - closed, µ - π - closed, µ - β - closed) [5]. iα (A) (resp. iσ (A), iπ (A), iβ (A)) denotes the union of µ - α - open (resp. µ - σ - open, µ - π - open, µ - β - open) sets included in A, and cα (A) (resp. cσ (A), cπ (A), cβ (A)) [5] denotes the intersection of µ - α - closed (resp. µ - σ - closed, µ - π - closed, µ - β - closed) sets including A. Obviously µ ⊂ α(µ) ⊂ σ(µ) ⊂ β(µ) and α(µ) ⊂ π(µ) ⊂ β(µ). Given U, V ⊂ X, let us say U and V are γ - separated [3] if cµ (U ) ∩ V = cµ (V ) ∩ U = ∅. Let us say that a set S ⊂ X is γ - connected if S = U ∪ V, U and V are γ separated imply U = ∅ or V = ∅. The space X is said to be γ - connected if it is a γ - connected subset of itself (here space X means GTS (X, µ)). 2000 Mathematics Subject Classification: 54A05, 54D05 161 Typeset by BSP style. M c Soc. Paran. de Mat. 162 Shyamapada Modak and Takashi Noiri The purpose of this paper is to introduce and investigate the notions of µ - k connected sets by using cµ (A), cα (A), cσ (A), cπ (A) and cβ (A) in GTS (X, µ). 2. µ - k - separated sets Definition 2.1. Let (X, µ) be a GTS. Two nonempty subsets U, V of X are said to be µ - k - separated if cµ (U ) ∩ ck (V ) = ∅ = ck (U ) ∩ cµ (V ), where k = α, σ, π or β. If we assign the values k = σ, π, α, β, then we get different types µ - k separated sets. Observe that two µ - k - separated sets are disjoint. Moreover, if U and V are µ - k - separated, U ′ ⊂ U, V ′ ⊂ V , then U ′ and V ′ are µ - k - separated as well. Again every µ - k - separated sets is a γ - separated set. From the above definition we obtain the following diagram: DIAGRAM I γ-separated ks µ-σ-separated ks µ-α-separated  µ-β-separated ks  µ-π-separated Definition 2.2. A subset A of a GTS X is said to be µ - k - connected if A is not the union of two µ - k - separated sets in X. From the above definition for a subset of a GTS the following diagram holds: DIAGRAM II γ-connected +3 µ-β-connected +3 µ-π-connected  µ-σ-connected  +3 µ-α-connected In the sequel, a GTS is briefly called a space. Theorem 2.3. Let X be a space. If A is a µ - k - connected subset of X and H, G are µ - k - separated subsets of X with A ⊂ H ∪ G, then either A ⊂ H or A ⊂ G. Proof: Let A be a µ - k - connected set. Let A ⊂ H∪G. Since A = (A∩H)∪(A∩G), then ck (A ∩ G) ∩ cµ (A ∩ H) ⊂ ck (G) ∩ cµ (H) = ∅ and cµ (A ∩ G) ∩ ck (A ∩ H) ⊂ cµ (G) ∩ ck (H) = ∅. Suppose A ∩ H and A ∩ G are nonempty. Then A is not µ - k - connected. This is a contradiction. Thus, either A ∩ H = ∅ or A ∩ G = ∅. This implies that A ⊂ H or A ⊂ G. ✷ µ - k - Connectedness in GTS 163 Theorem 2.4. If A and B are µ - k - connected sets of a space X such that A and B are not µ - k - separated, then A ∪ B is µ - k - connected. Proof: Let A and B be µ - k - connected sets in X. Suppose A ∪ B is not µ k - connected. Then, there exist two nonempty µ - k - separated sets G and H such that A ∪ B = G ∪ H. Since A and B are µ - k - connected, by Theorem 2.3, either A ⊂ G and B ⊂ H or B ⊂ G and A ⊂ H. Now if A ⊂ G and B ⊂ H, then cµ (A) ∩ ck (B) ⊂ cµ (G) ∩ ck (H) = ∅ and ck (A) ∩ cµ (B) ⊂ ck (G) ∩ cµ (H) = ∅. Thus, A and B are µ - k - separated, which is a contradiction. In case B ⊂ G and A ⊂ H a contradiction is similarly shown. Hence, A ∪ B is µ - k - connected. ✷ Theorem 2.5. If T {Mi : i ∈ I} is aSnonempty family of µ - k - connected sets of a space X, with i∈I Mi 6= ∅, then i∈I Mi is µ - k - connected. S S Proof: Suppose i∈I Mi is not µ - k - connected. Then we have T i∈I Mi = H ∪G, where H and G are nonempty µ - kS- separated sets in X. Since i∈I Mi 6= ∅, we have a point x ∈ ∩i∈I Mi . Since x ∈ i∈I Mi , either x ∈ H or x ∈ G. Suppose that x ∈ H. Since x ∈ Mi for each i ∈ I, then Mi and H intersect for each i ∈ I. By Theorem 2.3, S Mi ⊂ H or Mi ⊂ G. Since H and G are disjoint, Mi ⊂ H for all i ∈ I and hence i∈I Mi ⊂ H. This implies that G is empty. This is a contradiction. Suppose that x ∈ G.S By the similar way, we have that H is empty. This is a ✷ contradiction. Thus, i∈I Mi is µ - k - connected. Theorem 2.6. Let X be a space, {Aα : α ∈ △} be a family of µ - k - connected sets and A be a µ - k - connected set. If A ∩ Aα 6= ∅ for every α ∈ △, then A ∪ (∪α∈△ Aα ) is µ - k - connected. Proof: Since A∩Aα 6= ∅ for each α ∈ △, by Theorem 2.5, A∪Aα is µ - k - connected for each α ∈ △. Moreover, A ∪ (∪Aα ) = ∪(A ∪ Aα ) and ∩(A ∪ Aα ) ⊃ A 6= ∅. Thus ✷ by Theorem 2.5, A ∪ (∪Aα ) is µ - k - connected. Theorem 2.7. If A is a µ - k - connected subset of a space X and A ⊂ B ⊂ ck (A), then B is also a µ - k - connected subset of X. Proof: Suppose B is not a µ - k - connected subset of X then there exist µ k - separated sets H and G such that B = H ∪ G. This implies that H and G are nonempty and ck (G) ∩ cµ (H) = ∅ = cµ (G) ∩ ck (H). By Theorem 2.3, we have that either A ⊂ H or A ⊂ G. Suppose that A ⊂ H. Then ck (A) ⊂ ck (H) and cµ (G) ∩ ck (A) ⊂ cµ (G) ∩ ck (H) = ∅. This implies that G ⊂ B ⊂ ck (A) and G = ck (A) ∩ G ⊂ ck (A) ∩ cµ (G) = ∅. Thus G is an empty set. Since G is nonempty, this is a contradiction. Hence, B is µ - k - connected. ✷ Corollary 2.8. If A is a µ - k - c (...truncated)