On new structure of N-topology

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1 PURE MATHEMATICS RESEARCH ARTICLE On new structure of N-topology M. Lellis Thivagar 1 *, V. Ramesh 1 and M. Arockia Dasan 1 Received: 17 February 2016 Accepted: 15 June 2016 First Published: 21 June 2016 *Corresponding author: M. Lellis Thivagar, School of Mathematics, Madurai Kamaraj University, Madurai , Tamil Nadu, India kabilanchelian@gmail.com Reviewing editor: Hari M. Srivastava, University of Victoria, Canada Additional information is available at the end of the article Abstract: In this paper, we propose a new formula to get N-topologies in a non empty set X. Further, we establish its own open sets. We, in addition to it, study its characterizations. Apart from this, we introduce continuous functions on such topological spaces and establish their basic properties and prove the Pasting Lemma. Subjects: Advanced Mathematics; Mathematics & Statistics; Pure Mathematics; Science Keywords: N-topology; Nτ-open sets; Nτ-closed sets; relative topology (Nτ) ; N -continuous functions AMS Subject Classifications: Primary 54A05; Secondary 54A10 1. Introduction The intrinsic nature and beauty of Mathematics is this: One must be in "love" with Mathematics. As a result, the nature of inquisitiveness in a person gets, needless to mention, always enkindled and triggered by the new theorems or axioms or any new findings, even if it is a small in its nature or incredibly big. Indeed, the bitopological space propounded and introduced by Kelly in the year 1963, kept haunting our Mathematical mind. He introduced the bitopological space which is a non empty set X equipped with two arbitrary topologies τ 1 and. In this space, the open sets are called pairwise open sets. In this paper, we establish bitopological space with bitopological axioms and prove the structure of a non M. Lellis Thivagar ABOUT THE AUTHORS M. Lellis Thivagar has published 210 research publications both in national and International journals to his credit. Under his able guidance 15 scholars obtained their doctoral degree. In his collaborative work, he has joined hands with intellectuals of highly reputed persons internationally. He serves as a referee for 12 peerreviewed international journals. At present he is the professor, chairperson, School of Mathematics, Madurai Kamaraj University. V. Ramesh is a research scholar under the guidance of M. Lellis Thivagar at the School of Mathematics, Madurai Kamaraj University, Madurai. Five of his research papers are published/ accepted in the reputed international peerreviewed journals. M. Arockia Dasan is also a research scholar under the guidance of M. Lellis Thivagar at the School of Mathematics, Madurai Kamaraj University, Madurai. Four of his research papers are published/accepted in the reputed international peer-reviewed journals. PUBLIC INTEREST STATEMENT The intrinsic nature and beauty of Mathematics is this: it keeps growing within. It manifests its own manifold beauties, in an exponential quotient, when persons evince keen enthusiasm and starts grappling with and further fathom into its colorful nature and its features. So far, as much we are aware of the publications and other writings in vogue, we may be the first ones, who have tried, herewith, to establish bitopological space with bitopological axioms and proved the structure of non empty set X equipped with more than two topologies. Here, we have defined the structure of N-topology, that is, a non empty set X equipped with N-arbitrary topologies, and which has its own open sets. Further, we introduce continuous functions on N-topological space which in turn has its own impact on the Pasting Lemma The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license. Page 1 of 10

2 empty set X equipped with more than two topologies. Recently many researchers defined various forms of open sets in this space such as τ 1 (Lellis Thivagar, 1991), τ 1,2 (Lellis Thivagar, Ekici, & Ravi, 2008), etc. In addition to our fervent efforts, herein, we have also tried to prove the structure of N- topology, that is, a non empty set X equipped with N-arbitrary topologies τ 1,,, τ N and also established its own open sets. Further, we study its characterizations. Also, we introduce continuous functions on such topological spaces and establish their basic properties and proved the Pasting Lemma. 2. Preliminaries Definition 2.1 (Doitchinov, 1988) A quasi-pseudo-metric on a non empty set X is a function d 1 : X X R + {0} such that (i) d 1 (x, x) =0 for all x X (ii) d 1 (x, z) d 1 (x, y)+d 1 (y, z) for all x, y, z X. where R + is the set of all positive real numbers. Definition 2.2 (Grabiec, Cho & Saadati, 2007) Let d 1 a quasi-pseudo-metric on X, and let a function d 2 : X X R + {0} be defined by d 2 (x, y) =d 1 (y, x) for all x, y X. Trivially d 2 is a quasi-pseudometric defined on X and we say that d 1 and d 2 are conjugate one another. If d 1 is a quasi-pseudo-metric on X, then B d1 (x, k 1 )={y: d 1 (x, y) < k 1 }, the open d 1 -sphere with centre x and radius k 1 > 0. Classically, the collection of all open d 1 -spheres forms a base for a topology, the obtained topology, be denoted by τ 1 and called the quasi-pseudo-metric topology of d 1. Similarly we get a topology for X, due to the quasi-pseudo-metric d 2. Definition 2.3 (Kelly, 1963) A non empty set X equipped with two arbitrary topologies τ 1 and is called a bitopological space and is denoted by (X, τ 1, ). 3. N-topological spaces In this section, we introduce the notion of N-topological spaces and its own open sets. We derive its basic properties. We also define and discuss the relative topology in N-topological spaces. Definition 3.1 Let d 1 and d 2 be conjugate, quasi-pseudo-metrics on X and define a function d 3 :X X R + {0} by d 3 (x, y) = [2d (y, x)+d 1 2 (y, x)] for all x, y X. 3 Then (i) d 3 (x, x) = [2d (x,x)+d (x,x)] (ii) d 3 (x, z) = [2d (z,x)+d (z,x)] = 0 for all x X. [2(d 1 (z,y)+d 1 (y,x))+(d 2 (z,y)+d 2 (y,x))] 3 = d 3 (x, y)+d 3 (y, z) for all x, y, z X. Therefore, d 3 is a quasi-pseudo-metric on X and which is called a Mean Conjugate (simply write M.C) of d 1 and d 1. For each i = 1, 2, 3, the quasi-pseudo metric d i gives a topology τ i whose base is {B di (x, k i )}, where B di (x, k i )={y:d i (x, y) < k i }. Thus we define a non empty set X equipped with three arbitrary topologies τ 1, and τ 3 is called a tritopological space and is denoted by (X,3τ) or (X, τ 1,, τ 3 ). Generally, let d 1,..., d N 1 be quasi-pseudo-metrics on X, d 1 and d 2 be conjugate and d 3, d 4,...; d N 1 be M.C of d 1 and d 1 ; d 1, d 3 and d 1 ;...; d 1,..., d N 2 and d 1, respectively. Define a function d N :X X R + { } by Page 2 of 10

3 d N (x, y) = [d (y, x)+ N 1 d 1 i=1 i (y, x)] N for all x, y X. We can easily verify that d N is a quasi-pseudo-metric on X. Also we note that for each N, d N (x, y) d N (y, x) for all x, y X and d N is called a Mean Conjugate (simply write M.C) of d 1,..., d N 1 and d 1. For each i = 1,, N, the quasi-pseudo metric d i gives a topology τ i whose base is {B di (x, k i )}, where B di (x, k i )={y: d i (x, y) < k i }. Thus we define a non empty set X equipped with N-arbitrary topologies τ 1,,..., and τ N is called a N-topological space and is denoted by (X, Nτ) or (X, τ 1,,, τ N ). Definition 3.2 Let X be a non empty set, τ 1 and be two arbitrary topologies defined on X and the collection 2τ be defined by 2τ ={S X:S =(A 1 A 2 ) (B 1 B 2 ), A 1, B 1 τ 1 and A 2, B 2 } satisfying the following axioms: (i) X, 2τ (ii) i=1 S i 2τ for all S i 2τ (iii) n i=1 S i 2τ for all S i 2τ. Then the pair (X,2τ) is called a bitopological space on X and the elements of the collection 2τ are known as 2τ-open sets on X. We can generalize the above definition as given below: let X be a non empty set, τ 1,,..., τ N be N-arbitrary topologies defined on X and let the collection Nτ be defined by n n Nτ ={S X:S =( A i ) ( B i ), A i, B i τ i }, satisfying the following axioms: (i) X, Nτ i=1 i=1 (ii) i=1 S i Nτ for all S i Nτ (iii) n i=1 S i Nτ for all S i Nτ. Then the pair (X, Nτ) is called a N-topological space on X and the elements of the collection Nτ are known as Nτ-open sets on X. A subset A of X is said to be Nτ-closed on X if the complement of A is Nτopen on X. The set of all Nτ-open sets on X and the set of all Nτ-closed sets on X are, respectively, denoted by NτO(X) and NτC(X). Example 3.3 Let X ={a, b, c, d}. For N = 2, and assume τ 1 O(X) ={X,, {a, b}} and O(X) ={X,, {b, c}}, then 2τO(X) ={X,, {b}, {a, b}, {b, c}, {a, b, c}} and 2τC(X) ={X,, {d}, {a, d}, {c, d}, {a, c, d}}. Therefore, (X,2τ) is a bitopological space on X. For N = 3, and assume τ 1 O(X) ={X,, {a}}, O(X) ={X,, {b, d}} and τ 3 O(X) ={X,, {c, d}}, then 3τO(X) ={X,, {a}, {d}, {a, d}, {b, d}, {c, d}, {a, b, d}, {a, c, d}, {b, c, d}} and 3τC(X) ={X,, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {b, c, d}}. Therefore, (X,3τ) is a tritopological space on X. Remark 3.4 (i) If N = 1, then Nτ = τ 1 = τ. (ii) Intersection of two 2τ is also a 2τ. Intersection of two 3τ is also a 3τ. In general, intersection of two N-topology is again a N-topology. Page 3 of 10

4 (i) is trivial. (ii) Let (Nτ) 1 and (Nτ) 2 be two N-topology defined on X. Clearly, X and are in (Nτ) 1 (Nτ) 2. Let {C i } i I (Nτ) 1 (Nτ) 2, C i I i (Nτ) 1 and C i I i (Nτ) 2 and so in (Nτ) 1 (Nτ) 2. Let {C i } n (Nτ) (Nτ), n C i=1 1 2 i=1 i (Nτ) 1 and n C i=1 i (Nτ) 2 and so in (Nτ) 1 (Nτ) 2. Thus (Nτ) 1 (Nτ) 2 is an N-topology. Remark 3.5 Union of two 2τ need not be a 2τ. Union of two 3τ need not be a 3τ. In general, union of two N-topology need not be a N-topology. Example 3.6 For N = 3, X ={a, b, c, d} and assume τ 1 O(X) ={X,, {a}}, O(X) ={X,, {b, d}} and τ 3 O(X) ={X,, {c, d}}, then (3τ) 1 O(X) ={X,, {a}, {d}, {a, d}, {b, d}, {c, d}, {a, b, d}, {a, c, d}, {b, c, d}}. Also assume τ 1 O(X) ={X,, {b}}, O(X) ={X, } and τ 3 O(X) ={X, }, then (3τ) 2 O(X) ={X,, {b}}. Clearly, (X, (3τ) 1 ) and (X, (3τ) 2 ) are two tritopological spaces on X. Then (3τ) 1 (3τ) 2 ={X,, {a}, {b}, {d}, {a, d}, {b, d}, {c, d}, {a, b, d}, {a, c, d}, {b, c, d}} is not a tritopology, since {a}, {b} (3τ) 1 (3τ) 2 but {a, b} (3τ) 1 (3τ) 2. Definition 3.7 Let X be a non empty set and S be a subset of X. Then (i) (a) The 2τ-interior of S, denoted by 2τ-int(S), and is defined by 2τ-int(S) = {G:G S and G is 2τ-open}. (b) The 3τ-interior of S, denoted by 3τ-int(S), and is defined by 3τ-int(S) = {G:G S and G is 3τ-open}. (c) Generally, the Nτ-interior of S, denoted by Nτ-int(S), and is defined by Nτ-int(S) = {G:G S and G is Nτ-open}. (ii) (a) The 2τ-closure of S, denoted by 2τ-cl(S), and is defined by 2τ-cl(S) = {F:S F and F is 2τ-closed}. (b) The 3τ-closure of S, denoted by 3τ-cl(S), and is defined by 3τ-cl(S) = {F:S F and F is 3τ-closed}. (c) Generally, the Nτ-closure of S, denoted by Nτ-cl(S), and is defined by Nτ-cl(S) = {F:S F and F is Nτ-closed}. Theorem 3.8 Let (X, Nτ) be a N-topological space on X and let A, B X. Then (i) Nτ-cl(A) is the smallest Nτ-closed set which containing A (ii) A is Nτ-closed if and only if Nτ-cl(A) =A. In particular, Nτ-cl( ) = and Nτ-cl(X) =X (iii) A B Nτ-cl(A) Nτ-cl(B) (iv) Nτ-cl(A B) = Nτ-cl(A) Nτ-cl(B) (v) Nτ-cl(A B) Nτ-cl(A) Nτ-cl(B) (vi) Nτ-cl(Nτ-cl(A)) = Nτ-cl(A). Page 4 of 10

5 (i) Since intersection of any collection of Nτ-closed sets is also Nτ-closed, then Nτ-cl(A) is a Nτ-closed set. Trivially A Nτ-cl(A), by the definition of Nτ-closure of A. Now, let B be any Nτ-closed set which containing A. Then Nτ-cl(A) = {F:A F and F is Nτ-closed} B. Therefore, A is the smallest Nτ-closed set which containing A. (ii) Assume A is Nτ-closed, then A is the only smallest Nτ-closed set which containing itself and therefore Nτ-cl(A) =A. Conversely, assume Nτ-cl(A) =A. Then A is the smallest Nτ-closed set containing itself. Therefore, A is Nτ-closed. Particularly, since and X are Nτ-closed sets, then Nτ-cl( ) = and Nτ-cl(X) =X. (iii) Assume A B, and since B Nτ-cl(B), then A Nτ-cl(B). Since Nτ-cl(A) is the smallest Nτ-closed set which containing A. Therefore we have, Nτ-cl(A) Nτ-cl(B). (iv) Since A A B and B A B. Then by (iii), we have Nτ-cl(A) Nτ-cl(B) Nτ-cl(A B). On the other hand, by(i), A B Nτ-cl(A) Nτ-cl(B). Since Nτ-cl(A B) is the smallest Nτ-closed set which containing A B. Then Nτ-cl(A B) Nτ-cl(A) Nτ-cl(B). Therefore we have, Nτcl(A B) = Nτ-cl(A) Nτ-cl(B). (v) Since A B A and A B B, then Nτ-cl(A B) Nτ-cl(A) Nτ-cl(B). (vi) Since Nτ-cl(A) is a Nτ-closed set, then Nτ-cl(Nτ-cl(A)) = Nτ-cl(A). Example 3.9 Let X ={a, b, c, d}. For N = 2, consider τ 1 O(X) ={X,, {a}, {b, d}, {a, b, d}} and O(X) ={X, }, then 2τO(X) ={X,, {a}, {b, d}, {a, b, d}} and also 2τC(X) ={X,, {c}, {a, c}, {b, c, d}}. Let A ={a} and B ={b, c}, then 2τ-cl(A) ={a, c}, 2τ-cl(B) ={b, c, d} and 2τ-cl(A B) =. Therefore, 2τcl(A B) 2τ-cl(A) 2τ-cl(B). That is, equality does not hold in (v) of theorem 3.8. Theorem 3.10 Let (X, Nτ) be a N-topological space on X. Then Nτ-closure satisfies Kuratowski closure axioms given below: (i) Nτ-cl( ) = (ii) A Nτ-cl(A) for each A X (iii) Nτ-cl(A B) = Nτ-cl(A) Nτ-cl(B) for all A, B X (vi) Nτ-cl(Nτ-cl(A)) = Nτ-cl(A) for each A X. is follows from (i), (ii), (iv) and (vi) of theorem 3.8. Theorem 3.11 Let (X, Nτ) be a N-topological space on X and A X. Then x Nτ-cl(A) if and only if G A for every Nτ-open set G containing x. Assume x Nτ-cl(A) and G is a Nτ-open set containing x, then X G is Nτ-closed set and x X G. Suppose that G A =, then A X G. That is, X G is a Nτ-closed set containing A. Since Nτ-cl(A) is the smallest Nτ-closed set which containing A, then Nτ-cl(A) X G. Then x Nτcl(A) X G,which is contradicting to x X G. Hence G A for every Nτ-open set G containing x. Conversely, assume G A for every Nτ-open set G containing x. Suppose that x Nτ-cl(A), then x X Nτ-cl(A), which is a Nτ-open set. By hypothesis, (X Nτ-cl(A)) A. Since X Nτcl(A) X A implies (X Nτ-cl(A)) A (X A) A, then (X A) A, which is a contradiction. Therefore, x Nτ-cl(A). Theorem 3.12 Let (X, Nτ) be a N-topological space X and A X. Then Page 5 of 10

6 (i) Nτ-int(X A) =X Nτ-cl(A) (ii) Nτ-cl(X A) =X Nτ-int(A). (i) Assume x Nτ-int(X A) and suppose x X Nτ-cl(A), then x Nτ-cl(A) implies G A for every Nτ-open set G containing x. Therefore G X A for every Nτ-open set G containing x. Then x Nτ-int(X A), which is a contradiction. Thus x X Nτ-cl(A). On the other hand, let x X Nτ-cl(A), then x Nτ-cl(A). Suppose x Nτ-int(X A), then G X A for every Nτ-open set G containing x. That is, G A for every Nτ-open set G containing x. Then x Nτ-cl(A), which is a contradiction. Thus, x Nτ-int(X A). Therefore, Nτ-int(X A)=X Nτ-cl(A). (ii) Let x X Nτ-int(A). Then x Nτ-int(A) implies G A for every Nτ-open set G containing x. That is, G (X A) for every Nτ-open set G containing x. Then x Nτ-cl(X A). On the other hand, x Nτ-cl(X A), then G (X A) for every Nτ-open set G containing x. That is, G A for every Nτ-open set G containing x. Then x Nτ-int(A) implies x X Nτ-int(A). Therefore, Nτcl(X A) =X Nτ-int(A). Remark 3.13 If we take complement of either side of (i) and (ii) of previous theorem, we get (i) Nτ-cl(A) =X Nτ-int(X A) (ii) Nτ-int(A) =X Nτ-cl(X A). Theorem 3.14 Let (X, Nτ) be a N-topological space X and A, B X. Then (i) Nτ-int(A) is the largest Nτ-open set contained in A (ii) A is Nτ-open set if and only if Nτ-int(A) =A. In particular, Nτ-int( ) = and Nτ-int(X) =X (iii) A B, then Nτ-int(A) Nτ-int(B) (iv) Nτ-int(A B) Nτ-int(A) Nτ-int(B) (v) Nτ-int(A B) = Nτ-int(A) Nτ-int(B) (vi) Nτ-int(Nτ-int(A)) = Nτ-int(A). (i) (ii) Since union of any collection of Nτ-open sets is again a Nτ-open, then Nτ-int(A) is Nτ-open set and by definition of Nτ-interior of A, Nτ-int(A) A. Now, let B be any Nτ-open set which contained in A. Then B {G:G A and G is Nτ-open}=Nτ-int(A) and therefore, Nτ-int(A) is the largest Nτ-open set which contained in A. Assume A is Nτ-open set if and only if X A is Nτ-closed set if and only if Nτ-cl(X A) =X A if and only if X Nτ-cl(X A) =A if and only ifnτ-int(a) =A. In particular, since and X are Nτopen sets, then Nτ-int( ) = and Nτ-int(X) =X. (iii) Assume A B, then X B X A implies Nτ-cl(X A) Nτ-cl(X B) implies Nτint(A) Nτ-int(B). (iv) Assume x Nτ-int(A) Nτ-int(B), then x X (Nτ-int(A) Nτ-int(B)) implies x (Nτcl(X A)) (Nτ-cl(X B)), then x Nτ-cl(X (A B)) implies x Nτ-int(A B). Therefore, Nτint(A B) Nτ-int(A) Nτ-int(B). (v) Assume x Nτ-int(A B), then x X Nτ-int(A B) implies x (Nτ-cl(X A)) (Nτ-cl(X B)), then x (X Nτ-cl(X A)) (X Nτ-cl(X B)). Then x Nτ-int(A) Nτ-int(B). Thus, Nτint(A B) Nτ-int(A) Nτ-int(B). On the other hand, let x Nτ-int(A) Nτ-int(B), then Page 6 of 10

7 x X (Nτ-cl(X A) Nτ-cl(X B)) implies that x Nτ-int(A B). Therefore, Nτint(A B) = Nτ-int(A) Nτ-int(B). (vi) Since Nτ-int(A) is a Nτ-open set, then Nτ-int(Nτ-int(A)) = Nτ-int(A). Example 3.15 Let X ={a, b, c, d, e}. For N = 3, consider τ 1 O(X) ={X,, {a, b}}, O(X) ={X,, {a, b}, {c, d}, {a, b, c, d}} and τ 3 O(X) ={X,, {c}}. Then, we have 3τO(X) ={X,, {c}, {a, b}, {c, d}, {a, b, c}, {a, b, c, d}} and also 3τC(X) ={X,, {e}, {d, e}, {a, b, e}, {c, d, e}, {a, b, d, e}}. Let A ={a, c, d} and B ={b}, then 3τint(A) ={c, d}, 3τ-int(B) =, and 3τ-int(A B) ={a, b, c, d}. Thus, 3τ-int(A B) 3τ-int(A) 3τ-int(B). That is, equality does not hold in (iv) of theorem Theorem 3.16 Let (X, Nτ) be a N-topological space on X and A X. Then (i) Nτ-int(A) τ 1 -int(a) -int(a) τ N -int(a) (ii) Nτ-cl(A) τ 1 -cl(a) -cl(a) τ N -cl(a). (i) Assume x τ 1 -int(a) -int(a) τ N -int(a) and suppose x Nτ-int(A), then G A for every Nτ-open set G containing x. By definition, NτO(X) τ 1 O(X) O(X) τ N O(X), then each G i A for every τ i -open set G i containing x, where i = 1, 2,, N. Then x τ 1 -int(a) - int(a) τ N -int(a), which is a contradiction. Therefore, τ 1 -int(a) -int(a) τ N - int(a) Nτ-int(A). (ii) Since Nτ-int(X A) τ 1 -int(x A) -int(x A) τ N -int(x A), then X Nτ-cl(A) (X τ 1 - cl(a)) (X -cl(a)) (X τ N -cl(a)) which implies Nτ-cl(A) τ 1 -cl(a) -cl(a) τ N - cl(a). Example 3.17 Let X ={a, b, c, d}. For N = 2, consider τ 1 O(X) ={X,, {a, b}} and O(X) ={X,, {a, c}}. Then 2τO(X) ={X,, {a}, {a, b}, {a, c}, {a, b, c}} and also τ 1 C(X) ={X,, {c, d}}, C(X) ={X,, {b, d}}, then 2τC(X) ={X,, {d}, {b, d}, {c, d}, {b, c, d}}. If A ={a} and B ={b}, then we have 2τ-int(A) ={a}, τ 1 - int(a) =, -int(a) =. Therefore, 2τ-int(A) τ 1 -int(a) -int(a). That is, equality does not hold in (i) of theorem If A ={b, c, d}, then we have 2τ-cl(A) ={b, c, d}, τ 1 -cl(a) =X and -cl(a) =X. Therefore, 2τ-cl(A) τ 1 -cl(a) -cl(a). That is, equality does not hold in (ii) of theorem Definition 3.18 (i) Let Y be a non empty subset of a bitopological space (X,2τ). Then the bitopology (2τ) ={Y O:O 2τ} is called the relative (simply induced or subspace) topology on Y for 2τ. The pair (Y, (2τ) ) is called a subspace of (X,2τ). Generally, we can define (ii) Let Y be a non empty subset of a N-topological space (X, Nτ). Then the N-topology (Nτ) ={Y O:O Nτ} is called the relative (simply induced or subspace) topology on Y for Nτ. The pair (Y, (Nτ) ) is called a subspace of (X, Nτ). Example 3.19 Let X ={a, b, c, d, e, f }. For N = 4, consider τ 1 O(X) ={X,, {a}}, O(X) ={X,, {b}}, τ 3 O(X) ={X,, {c}}, and τ 4 O(X) ={X,, {d}}. Then 4τO(X) ={X,, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}}. Let Y ={a, c, e, f } X. Then (4τ) ={Y,, {a}, {c}, {a, c}} is a relative topology for 4τ. Theorem 3.20 Let (Y, (Nτ) ) be a subspace of (X, Nτ) and A Y. Then (i) A is (Nτ) -closed in Y if and only if A = Y F, where F is Nτ-closed in X (ii) If A is (Nτ) -closed in Y and Y is Nτ-closed in X. Then A is Nτ-closed in X. Page 7 of 10

8 (i) A is (Nτ) -closed Y A is (Nτ) -open Y A = Y O for some O Nτ A = Y F where F = X O is Nτ-closed. (ii) Since A is (Nτ) -closed in Y, A = Y F for some Nτ-closed set F in X. Since Y and F are both Nτ closed in X, so is Y F. 4. Continuity in N-topological spaces In this section, we introduce continuous functions in N-topological spaces and discuss the different properties of it. Also, we prove the Pasting Lemma. Throughout this section, the N-topological spaces (X, Nτ) and (Y, Nτ) represented by X and Y, respectively. Definition 4.1 Let X and Y be two N-topological spaces. A function f :X Y is said to be N -continuous on X if the inverse image of every Nτ-open set in Y is a Nτ-open set in X. Example 4.2 For N = 2, let X ={a, b, c, d} and Y ={x, y, z, w}. Consider τ 1 O(X) ={X,, {a, b}}, O(X) ={X,, {a}} and τ 1 O(Y) ={Y,, {x}}, O(Y) ={Y,, {x, y, z}}. Then 2τO(X) ={X,, {a}, {a, b}}, 2τC(X) ={X,, {c, d}, {b, c, d}} and 2τO(Y) ={Y,, {x}, {x, y, z}}, 2τC(Y) ={Y,, {w}, {y, z, w}}. Define f :X Y by f (a) =x, f (b) =y, f (c) =z, f (d) =z. Then f 1 (Y) =X, f 1 ( ) =, f 1 ({x}) = {a}, f 1 ({x, y, z}) = X. That is, the inverse image of every 2τ-open set in Y is a 2τ-open set in X. Therefore, f is 2 -continuous function on X. Theorem 4.3 A function f :X Y is N -continuous on X if and only if the inverse image of every Nτ-closed set in Y is a Nτ-closed set in X. Assume that f :X Y is N -continuous on X and let A be a Nτ-closed set in Y. Then Y A is a Nτopen set in Y. Since f is a N -continuous function on X, then f 1 (Y A) is Nτ-open set in X. Then X f 1 (A) is Nτ-open set in X. Then f 1 (A) is Nτ-closed set in X. Conversely, assume the inverse image of every Nτ -closed set in Y is Nτ-closed set in X and let B be a Nτ-open set in Y. Then Y B is a Nτ-closed set in Y and f 1 (Y B) is a Nτ-closed set in X. Then f 1 (B) is a Nτ-open set in X. Hence f is N -continuous function on X. Theorem 4.4 A function f :X Y is N -continuous on X if and only if f (Nτ-cl(A)) Nτ-cl(f(A)) for every A X. Assume f :X Y be a N -continuous function on X and let A X. Then f (A) Y and Nτ-cl(f(A)) is Nτ-closed set in Y. Since f is N -continuous function on X, then f 1 (Nτ-cl(f(A))) is Nτ-closed set in X. Since f (A) Nτ-cl(f(A)), then A f 1 (Nτ-cl(f(A))). Since Nτ-cl(A) is the smallest Nτ-closed set in X containing A, then Nτ-cl(A) f 1 (Nτ-cl(f(A))). Then f (Nτ-cl(A)) Nτ-cl(f(A)) for every A X. Conversely, assume f (Nτ-cl(A)) Nτ-cl(f(A)) for every A X and let F be a Nτ-closed set in Y. Since f 1 (F) X, then f (Nτcl(f 1 (F))) Nτ-cl(f (f 1 (F))) = Nτ-cl(F). Then Nτ-cl(f 1 (F)) f 1 (Nτ-cl(F)) = f 1 (F). Since F is a Nτ-closed set in Y and also f 1 (F) Nτ-cl(f 1 (F)). Then f 1 (F) =Nτ-cl(f 1 (F)) and also f 1 (F) is Nτ-closed set in X. Therefore, f is N -continuous function on X. Example 4.5 For N = 2. Let X ={a, b, c, d} and Y ={x, y, z, w}. Consider τ 1 O(X) ={X,, {a, c}}, O(X) ={X,, {b, d}} and also τ 1 O(Y) ={Y,, {x, z}, {x, y, z}}, O(Y) ={Y,, {y}}. Then 2τO(X) ={X,, {a, c}, {b, d}}, 2τO(Y) ={Y,, {y}, {x, z}, {x, y, z}} and also 2τC(X) ={X,, {a, c}, {b, d}}, Page 8 of 10

9 2τC(Y) ={Y,, {w}, {y, w}, {x, z, w}}. Define f :X Y by f (a) =y, f (b) =x, f (c) =y and f (d) =x. Clearly, f is 2 -continuous function on X. If A ={a, c} X. Then f (2τ-cl(A)) = f ({a, c}) = {y}. But, 2τ-cl(f (A)) = 2τcl({y}) = {y, w}. Thus, f (2τ-cl(A)) 2τ-cl(f(A)), even though f is 2 -continuous function on X. That is, equality does not hold in the theorem 4.4, even though f is 2 -continuous function on X. Theorem 4.6 A function f :X Y is N -continuous on X if and only if Nτ-cl(f 1 (B)) f 1 (Nτ-cl(B)) for every B Y. Let f :X Y be a N -continuous on X and let B Y. Then Nτ-cl(B) is Nτ-closed set in Y. Since f is N -continuous function on X, then f 1 (Nτ-cl(B)) is Nτ-closed in X. That is, Nτ-cl(f 1 (Nτ-cl(B)))=f 1 (Nτ-cl(B)). Since B Nτ-cl(B), then f 1 (B) f 1 (Nτ-cl(B)). Thus, Nτ-cl(f 1 (B)) Nτ-cl(f 1 (Nτ)-cl(B)))= f 1 (Nτ-cl(B)) for every B Y. Conversely, assume that Nτ-cl(f 1 (B)) f 1 (Nτ-cl(B)) for every B Y and let F be a Nτ-closed set in Y. Then Nτ-cl(F) =F and by assumption, Nτ-cl(f 1 (F)) f 1 (Nτ-cl(F)) = f 1 (F). Since f 1 (F) Nτcl(f 1 (F)), then f 1 (F) =Nτ-cl(f 1 (F)) and f 1 (F) is Nτ-closed set in X. Example 4.7 For N = 2. Let X ={a, b, c, d} and Y ={x, y, z, w}. Consider τ 1 O(X) ={X,, {c}}, O(X) ={X,, {a, d}, {a, c, d}} and τ 1 O(Y) ={Y, }, O(Y) ={Y,, {x, w}}. Then 2τO(X) ={X,, {c}, {a, d}, {a, c, d}}, 2τO(Y) ={Y,, {x, w}} and also 2τC(X) ={X,, {b}, {b, c}, {a, b, d}}, 2τC(Y) ={Y,, {y, z}}. Define f :X Y by f (a) =x, f (b) =y, f (c) =z and f (d) =w. Clearly, f is 2 -continuous function on X. If A ={y} Y. Then f 1 (2τ-cl(A)) = f 1 ({y, z})={b, c}. But, 2τ-cl(f 1 (A)) = 2τ-cl({b}) = {b}. Thus, f 1 (2τ-cl(A)) 2τ-cl(f 1 (A)), even though f is 2 -continuous function on X. That is, equality does not hold in the theorem 4.6, even though f is 2 -continuous function on X. Theorem 4.8 A function f :X Y is N -continuous on X if and only if f 1 (Nτ-int(B)) Nτ-int(f 1 (B)) for every B Y. Let f :X Y be a N -continuous on X and let B Y. Then Nτ-int(B) is Nτ-open set in Y. Since f is N -continuous on X, then f 1 (Nτ-int(B)) is Nτ-open in X. That is, Nτ-int(f 1 (Nτ-int(B))) = f 1 (Nτ-int(B)). Since Nτ-int(B) B, then f 1 (Nτ-int(B)) f 1 (B) which implies Nτ-int(f 1 (Nτ-int(B))) Nτ-int(f 1 (B)). Thus f 1 (Nτ-int(B)) Nτ-int(f 1 (B)) for every B Y. Conversely, assume f 1 (Nτ-int(B)) Nτ-int(f 1 (B)) for every B Y and let G be a Nτ-open set in Y. Then Nτ-int(G) =G and by assumption, f 1 (G) Nτint(f 1 (G)). Also Nτ-int(f 1 (G)) f 1 (G) and hence Nτ-int(f 1 (G)) = f 1 (G) which implies f 1 (G) is Nτ-open in X. Therefore, f is N -continuous function on X. Example 4.9 For N = 2, let X ={a, b, c, d} and Y ={x, y, z, w}. Consider τ 1 O(X) ={X,, {a, b}}, O(X) ={X,, {a}} and τ 1 O(Y) ={Y,, {x}} and O(Y) ={Y,, {x, y, z}}. Then 2τO(X) ={X,, {a}, {a, b}} and 2τC(X) ={X,, {c, d}, {b, c, d}} and also 2τO(Y) ={Y,, {x}, {x, y, z}}, 2τC(Y) ={Y,, {w}, {y, z, w}}. Define f :X Y by f (a) =x, f (b) =y, f (c) =z, f (d) =z. Clearly f is 2 -continuous function on X. If B ={x, y} Y. Then f 1 (2τ-int(B)) = f 1 ({x})={a}. But, 2τ-int(f 1 (B)) = 2τ-int({a, b})) = {a, b}. Thus, f 1 (2τ-int(B)) 2τ-int(f 1 (B)), even though f is 2 -continuous. That is, equality does not hold in the theorem 4.8, when f is 2 -continuous. Theorem 4.10 (The Pasting Lemma) Let X and Y be two N-topological spaces with X = A B, where A and B are Nτ-closed sets in X. Let f :A Y and g:b Y be N -continuous. If f (x) =g(x) for every x A B, then f and g combine to give a N -continuous function h:x Y, defined by setting h(x) =f (x) if x A, and h(x) =g(x) if x B. Page 9 of 10

10 Let F be a Nτ-closed set in Y. Now h 1 (F) =f 1 (F) g 1 (F), by elementary set theory. Since f is N -continuous, f 1 (F) is Nτ -closed in A and therefore, Nτ-closed in X. Similarly, g 1 (F) is Nτ -closed in B and therefore, Nτ-closed in X. Thus their union h 1 (F) is Nτ-closed in X. 5. Conclusion In this paper, we introduce a new venture to establish more topologies on a non empty set. Such efforts prompt us to blissfully convey that these concepts are also applicable in other areas of General topology, Fuzzy topology, intuitionistic topology, ideal topology so on and so forth. The course of human history, unmistakably shown and revealed to us that many great leaps of learning, discoveries, and understanding come from a source not so anticipated, and that in any field of sciences or humanities, and in particular in the field of basic researches often bear fruit well within hundred years, so to say. However, the more we come to grapple with and invest our time and energy to comprehend anything that is new, the better will we be, in order to handle and deal with the challenges and queries that keep facing us in the future, and come up with better results and findings. Funding The authors received no direct funding for this research. Author details M. Lellis Thivagar 1 mlthivagar@yahoo.co.in ORCID ID: V. Ramesh 1 kabilanchelian@gmail.com ORCID ID: M. Arockia Dasan 1 dassfredy@gmail.com ORCID ID: 1 School of Mathematics, Madurai Kamaraj University, Madurai , Tamil Nadu, India. Citation information Cite this article as: On new structure of N-topology, M. Lellis Thivagar, V. Ramesh & M. Arockia Dasan, Cogent Mathematics (2016), 3: References Doitchinov, D. (1988). On completeness in quasi-metric spaces. Topology and its Applications, 30, Grabiec, M. T., Cho, Y. J., & Saadati, R. (2007). Families of quasipseudo-metrics generated by probabilistic quasi-pseudo-metric spaces. Surveys in Mathematics and its Applications, 2, Kelly, J. C. (1963). Bitopological spaces. Proceedings London Mathematical Society, 3, Lellis Thivagar, M. (1991). Generalization of pairwise α-continuous functions. Pure and Applied Mathematicka Sciences, 28, Lellis Thivagar, M., Ekici, E., & Ravi, O. (2008). On (1,2)* sets and bitopological decompositions of (1,2)* continuous mappings. Kochi Journal of Mathematics, (Japan), 3, The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license. You are free to: Share copy and redistribute the material in any medium or format Adapt remix, transform, and build upon the material for any purpose, even commercially. The licensor cannot revoke these freedoms as long as you follow the license terms. Under the following terms: Attribution You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. No additional restrictions You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits. Page 10 of 10