H(i) Connected Ditopological Texture Space

In this paper, we introduce the concept of H(i) connected ditopological texture space. We develop some basic properties of bicontinuity and connectedness in term of ditopological texture space which will used inH(i) connected ditopological texture space. We have established some correspondence related to known structure such as bitopological space, fuzzy lattice and topological space.


Introduction
Ditopological texture space may be regarded as a natural combination of texture space, topological space and bitopological space [14] but ditopology corresponds in a natural way to fuzzy topology.The texture is a generalisation of the fuzzy lattice.The notation of texture was introduced by Brown [7] in a point set setting for the study of a fuzzy set.It has been proved useful as a framework to discuss the complement-free mathematical concept.The motivation for the study of texture space is that they allow to represent a classical fuzzy set, L-fuzzy set [12], intuitionistic fuzzy set [1] and intuitionistic set, as a lattice of a crisp subset of some base set.Different fuzzy topological spaces have been studied by Tripathy and Debnath [19] , Tripathy and Ray [21,20] .A detailed analysis of the relation between texture space and the lattice of fuzzy sets of various kind is found in the works due to [1,4,5,6,3].The concept of ditopological texture space is introduced by Brown [15].This paper is totally devoted to the study on bicontinuity [5], connectedness [11] and their applications.In this paper we use the term ω − preserving (a) If X is a set and P (X) the power set of X, then (X, P (X)) is the discrete texture on X.For x ∈ X, P x = {x} and Q x = X − {x}.

bicontinuous if continuous and cocontinuous. or, (another definition)[22] A difunction or an ω−preserving point function between the ditopological texture spaces is called bicontinuous if the inverse image of every open set is open and the inverse image of every closed set is closed.
Definition 2.13.[11] Let (S, Γ) be a texture space and ∅ = Z ⊆ S. {A, B} ⊆ P (S) is said to be a partition of It can be noted that the roles of A and B may interchanged.If {A, B} is a partition of Z, then let B ∩ Z = ∅ and Z A. Let Γ = P (S), then it can be verified that {A, S \B} and {S \A, B} are partition of Z in ordinary meaning.For example, we have Definition 2.14.[11] Let (S, Γ, τ , k) be a ditopological texture space.Z ⊆ S is said to be connected if there exists no partition {G, F } with G ∈ τ and F ∈ k.

Bicontinuitity
In this section, we establish some result on bicontinuity.
Conversely, let V be an open subset of Y .Let P x be a subset of F ← (V ) then f → (P x ) ⊂ V , so that by hypothesis there exists a neighbourhood of We have the following result on the inclusion, composition, restriction of the domain, expanding the range and Local form of the difunctions.
) and coimage set F → (X 1 ), then the difunction (g, G) : X 1 → X 3 obtained by restricting the range of f and corange of F is bicontinuous.If X 3 is a space having X 2 as a subspace, then the difunction (h, H) : X 1 → X 3 obtained by expanding the range of f and corange of F is bicontinuous.
(e) Let X 1 and X 2 be ditopological texture space.The difunction (f, F ) : X 1 → X 2 is bicontinuous if X 1 can be written as join of open set U α and join of closed set V α such that f |U α and F |V α is continuous and cocontinuous for each α.
which is closed in A, (by definition of subspace ditopology texture space, one may refer to [ ]).Hence (j, J) is bicontinuous.
(d) Let the difunction (f, F ) : since X 3 contains the entire image set f → (X 1 ) and and coimage set F → (X 1 ).
Since the difunction Let the difunction (h, H) : X 1 → X 3 be bicontinuous.Let X 2 be a subspace of X 3 and h → (x) = (f oj) → (x) be the composition of the map f : X 1 → X 2 and j : X 2 → X 3 and H → (x) = (F oJ) → (x) is composition of the map F : X 1 → X 2 and J : X 2 → X 3 .
(e) By hypothesis we can write X 1 as the join of the open set U α and also join of closed set V α such that f |U α and F |V α is continuous and cocontinuous for each α.Let U be open set in X 1 and V be closed set in X 1 .Then then both the expressions represent the set of those point x lying in U α for which f (x) ∈ U and y lying in Now we establish a result on Maps into product.
A → X 2 are cocontinuous.The maps f 1 , f 2 are called the coordinate of image set f and F 1 , F 2 are called the coordinate of coimage set F .
Proof.Let the difunction (π 1 , Π 1 ) : X 1 × X 2 → X 1 and (π 2 , Π 2 ) : X 1 × X 2 → X 2 be projections onto the first and second factors respectively.These maps are bicontinuous.For Π Note that for each a ∈ A, If the difunction (f, F ) is bicontinuous, then the coordinate difunctions must be bicontinuous.Conversely, Suppose (f 1 , F 1 ) and (f 2 , F 2 ) are bicontinuous.We show that for each basis element U × E and cobasis element V × F for the ditopological texture space X 1 × X 2 , then their inverse images , where (f 1 , F 1 ) and (f 2 , F 2 ) are bicontinuous functions.So their intersection is also bicontinuous, which implies (f, F ) is bicontinuous.✷

Connectedness
Theorem 4.1.Let {S, Γ, τ , k} be a ditopological texture space on S. P be a connected space and ext(P ) ∩ P = ∅ then P is also connected.
Proof.Let P be not connected then there exists A ∈ τ , B ∈ k such that A ∩ P = B ∩ P , A ∩ P = ∅, P B. Since P is connected .It is obvious that Case (1) Let A ∩ P = ∅, then A ⊆ ext(P ), since ext(P ) ∩ P = ∅ ⇒ A ∩ P = ∅, which is a contradiction.Case (2) Let P ⊆ B ⇒ P ⊆ B B is closed set, and we arrive at a contradiction.Hence P is connected.
Then A is also connected., where a 0 is the smallest element of X if such exists or [a, b 0 ), where b 0 is the largest element of X if such exist.Such type of set act as cobase element which belong to k.It generates a ditopology texture space on X known as order ditopology texture space.Definition 4.4.In order ditopology texture space having X = R and ordered set R has binary operation < is referred as euclidean ditopology texture space.Here we consider A 0 ∈ τ 1 and B 0 ∈ k 1 in the subspace ditopology texture space ([a, b], Γ 1 , τ 1 , k 1 ) which is same as order ditopology texture space.
The set A 0 and B 0 is non-empty because a ∈ A 0 ⊂ B 0 .
Remark Let (X, Γ, τ , k), where X is any set.Suppose {A ∈ τ , B ∈ k} ⊂ Γ be a partition of Z ⊂ X then it must satisfy the condition A ∩ (X − B) = ∅ (where Let c = Sup{A 0 }.We show that P c A 0 ⊂ B 0 , which contradicts to the fact that {A 0 , B 0 } be the partition of [a, b].
Case 1. Suppose P c ⊆ B 0 .Then P c = P b .So neither P c = P a nor P a ⊂ P c ⊂ P b because B 0 is closed in [a, b].Hence the must be some interval of the form [c, e] contained in B 0 .If P c = P a we have a contradiction according to our assumption point c = sup{A 0 }.If P a ⊂ P c ⊂ P b then there exists a point z such that P c ⊂ P z ⊂ P e which is again a contradiction according to our assumption point c = sup{A 0 }.
Case 2. Suppose P c ⊆ A 0 ⇒ P c ⊆ A 0 ⊂ B 0 .So P c ⊆ B 0 which is Case 1, which is a contradiction.
Hence R with usual order ditopology texture space is connected.✷ Definition 4.7.{S, Γ} be texture space non empty set S ⊂ X and S be connected space.A point p of S is called cut point of S means point p separates S provided {A, B} ⊆ Γ is a partition [11] of T = S − p. Otherwise p is a non-cut point of S.
Definition 4.8.let (S, Γ, τ , k) be a ditopological texture space.Let p, q be points of the connected space S. We denote E(p, q), the subset of S consisting of the points p and q together with all cut points of S that separates p and q.
The separation order in E(p, q) is defined as follows: Let x, y be two points in E(p, q), then x precedes y, x < y in E(p, q) if either x = p or if x separates p and y in S. Theorem 4.9.Let (S, Γ, τ , k) be a ditopological texture space.If p, q are two points of connected space S. The separation order in E(p, q) is a simple order.Proof.For each point x in E(p, q), x = p or q then there exist a separation in In the above partition, A r and A s do not contain point r and s respectively but B r and B s contains the point r and s respectively.Let r and s be two points in E(p, q) − p − q.If s is not in B r , then A s contain A r ∪ r and B s contain B r .To see this note that in first case the connected set A r ∪ r contain p but not contain s and so lies entirely in A s .The set B s ∩ {A r ∪ r} = ∅.So B r must lies in B s .The second case is similar.
Let r and s be two points of E(p, q) − p − q.If neither s ∈ A r nor s ∈ B r , then P r ⊂ P s in E(p, q).If neither r ∈ A s nor r ∈ B s , then P s ⊂ P r in E(p, q).Hence any two element in E(p, q) are ordered.Lemma 5.3.For a connected space X, if ξ is a chain of member of the form {A * x such that x ∈ ct(X)} covering X, then for each A * x ∈ ξ, there exists A * y ∈ ξ such that P y = P x and P x ⊆ A * y .
Proof.Suppose that there is some A * x ∈ ξ such that P x A * y for all A * y ∈ ξ, P y = P x .Then A * x A * y for all A * y ∈ ξ, P y = P x .Since ξ is a chain, A * y ⊂ A * x for all A * y ∈ ξ, P y = P x .Thus X = A * x .This implies that X − {x} = A x , which is not possible.Hence the result.✷ Theorem 5.4.Let X be a connected space and x ∈ ctX.Let y be a non-cut point of A * x in A * x and P y = P x .Then y is a non cut point of X.
Proof.Since y is a non-cut point of A * x in A * x , A * x − {y} is connected.Since B * x = B ∩ {x} is connected because B is connected and x is a cut-point.Therefore (A * x − {y}) ∩ B * x is connected.But X − {y} = (A * x − {y}) ∩ B * x , so X − {y} is connected.Thus y is a non-cut point of X. ✷ Theorem 5.5.Let H be a subset of a connected space X.Let P a ⊆ H be such that H − {a} ⊂ ctX.If A * x (a) ⊂ H for every P x ⊆ H − {a}, then H is connected.
Proof.Let W = {A * x (a) such that P x ⊆ H − {a}}.Since W is connected, as join of connected set whose meet is non-empty.For each P x ⊆ H − {a}, P x ⊆ A * x ⊂ W . Thus H ⊂ W .On the other hand, if P x ⊆ W , then P x ⊆ A * y (a) for some P y ⊆ H − {a} and by assumption A * y (a) ⊂ H, so P x ⊆ H. Thus H = W and thus is connected. ✷
are open and closed according to U and V are open and closed in X 1 , Π ← 2 (Y ) = X 1 × Y and π ← 2 (Z) = X 1 × Z are open and closed according to Y and Z are open and closed in X 2 .

Definition 4 . 3 .
Let X be an ordered set and (X, Γ, τ , k) be a ditopological texture space.Assume that X has more then one element.The collection of the sets of the form (a, b) or [a 0 , b), where a 0 is the smallest element of X if it exists or (a, b 0 ], where b 0 is the largest element of X if it exists.Such type of set act as base element which belong to τ .Similarly the set of the form [a, b] or (a 0 , b]

Example 4 . 5 .Theorem 4 . 6 .
The order ditopology texture space on set N of natural number is discrete ditopology texture space where the elements of the base are given by {n} = (n − ε, n + ε); n ∈ N and ε is taken howsoever small real number.Similiarly, cobase are{n} = [n − ε, n + ε].The ditopological texure space (R, Γ, τ , k) with usual order ditopology texture space is connected.Proof.Let {A ∈ τ , B ∈ k} ⊂ Γ be a Partition of R. Take a ∈ A, b ∈ B. Suppose for convenience that a < b ⇒ p a ⊂ p b .The interval [a, b] is contained in R. Hence [a, b] is a partition of A 0 = A ∩ [a, b], B 0 = B ∩ [a, b].It is obvious that A 0 = B 0 [by definition of partition].But according to our assumption [a, b] has partition.So either [a, b] ∩ A 0 = ∅ or [a, b] B 0 .A. K. Saw and B. C. Tripathy For t, t ′ ∈ T and s ∈ S, f Q (s,t) andP (s,t ′ ) F ⇒ P t ′ Q t .) to (T, Γ 2 ) and A ∈ Γ 1 .The image f → A and coimage F → A are defined by, f → A = {Q t |for all s, f Q (s,t) ⇒ A ⊆ Q s }.F → A = {P t |for all s, P (s,t) F ⇒ P s ⊆ A}.
satisfies the following two condition:DF 1.For s, s ′ ∈ S, P s Q s ′ ⇒ there exists t ∈ T