# Double-Connected Intuitionistic Space in Double Intuitionistic Topological Spaces

Authors: Asmaa Ghassob Raoof; Taha H. Jassim; Ngiste Amare
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

The structures of the families of fuzzy sets that arise out of various notions of openness and closeness in a double fuzzy topological space are explored. This research is to present a new portion of space (Double-connected intuitionistic space) in Double intuitionistic topological spaces. Through these concepts, we advance some of their characteristics and relate to themselves.


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## Body

## 1. Introduction

Ajmal and Kohli [1] explain connectedness in fuzzy topological spaces. Yasry [2] investigated lectures in advanced topology. El-Hamed et al. [3] presented Double connected spaces. Kandil et al. [4] investigated on flour (intuitionistic) topological spaces. Kendal et al. [5] studied on flour (intuitionistic) compact space. The concept was used to define intuitionistic sets and the intuitionistic gradation of openness by Coker [6–8]. Atanassov and Stoeva [9] describe Intuitionistic fuzzy sets. Atanassov [10] studied more intuitionistic fuzzy sets. Ozcelik and Narli [11] introduced the concept of submaximal intuitionistic topological spaces. Tantawy et al. [12] researched soft connections with double spaces. Selma and Coker [13] examined the concept of connectedness in intuitionistic fuzzy special topological spaces. Levine [14, 15] introduced generalized closed sets in topology. The concept of a fuzzy set was presented by Zadeh in his classic paper from 1965 and 2002 [14, 16]. After that, we introduce a new class of sets in DITS, namely the Double connected set, the separated Double I sets, the strongly Double connected, Double CO connected I space, and the Double I component. We also presented several examples of each type and concluded that there are relationships between them that were presented through theorems.


## 2. Preliminaries

We remind the following definitions, which are needed in our effort:Letx≠∅, as well as ζ and L be IS having the form ζ=x.ζ1.ζ2.L=x.L1.L2, respectively. Also, ζi:i∈I be an arbitrary family of I S in X, where ζi=x.ζi1.ζi2, afterward:(1)
∅˜=x.∅.x; x˜=x.x.∅.(2)
ζ⊆Liffζ1⊆L1andζ2⊇L2.(3)
ζc=x.ζ2.ζ1.(4)
∪ζi=x.∪ζi1.∩ζi2.∩ζi=x.∩ζi1.∪ζi2 [7]. Let x be a nonempty set, an intuitionistic set B (I S, for short) is an object having g the form B=x.B1.B2, where B1andB2 are disjoint subset of x. The set B1 is called set of members of B, while B2 is called set of nonmembers of B [7]. An intuitionistic topology (IT, for short) on a nonempty set x is a family T of IS in X containing ∅˜.x˜ and closed under arbitrary unions and finitely intersections. The pair x.T is called ITS [11]. Let x≠∅.(1)
A Double-set (D-set, for short) U is an ordered pair U1,U2 ∈ p (x) × p (x), such that U1 ⊆ U2.(2)
D (x) = {U1,U2 ∈ p (x) × p (x), U1 ⊆ U2} is the family of all D-sets on x.(3)
TheD-set x˜ = (x, x) is called the universal D-set, and the D-set ∅˜ = (∅, ∅) is called the empty D-set.(4)
LetU = (U1, U2); ϑ = (ϑ1; ϑ2) ∈ D (x):(1)
Uc=U2c,U1c, where Uc is the complement of U.(2)
U−ϑ=U1−ϑ2,U2−ϑ2 [4]. Let x be a nonempty set. The family η of D-sets in x is called a double topology on x if it satisfies the following axioms:(a)
∅,x ∈ η.(b)
IfU, ϑ ∈ η, then U ∩ ϑ ∈ η,(c)
If {Uz: z ∈ Ζ} ⊆ η, then ∪z∈ΖUz ∈ η. The pair (x, η) is called a DTS. Each element of η is called an open D-set in x. The complement of open D-set is called closed D-set [4]. Let x be a nonempty set:(1)
IN (x) = {∅, x} is a DTS, which is called indiscrete DTS.(2)
dis (x) = px) × px is a DTS, which is called discrete DTS [4]. Let (X, η) be a DTS and ∂ ∈ D (X). The double closure of U, denoted by cl η (U) defined by cl η (U) = ∩ {ϑ: ϑ ∈ ηc and U ⊆ ϑ} [2, 4]. Let X nonempty set, w∈X a fixed element in X, and let M=x.M1.M2 be an intuitionistic set (IS, for short). The IS w˙ defined by w˙ = x.w.wc is called an intuitionistic point Iwforshort in X. The IS w¨ =  x.Ø.wc is called a vanishing I point (VIw, for short) in X. The IS w˙ is said to be contained in Mw˙∈M.forshort only if w∈M1, and similarly IS w¨ contained in Mw¨∈Mforshort only if w∉M2. For a given IS w in X, we may write M = ∪w:w˙∈M∪∪w¨:w¨∈M, and whenever M is not a proper IS ( i.e., if M is not of the form M = x.M1.M2 where M1∪M2≠X), then M = ∪ {w˙:w˙∈M hold. In general, any IS M in X can be written in the form M = M˙∪M¨ where M˙ =  ∪w˙:w˙∈M}, and M¨ =  ∪w¨:w¨∈M [6]. A topological space X is connected if it cannot be written as X = X1∪X2 , where X1 and X2 are both open and X1∩X2 =  ∅, otherwise X is called disconnected [2]. Let (X, η) be a DTS and Ƴ be a nonempty subset of X. Then, ηƳ = {q∩Ƴ: q∈M and Ƴ = (Ƴ, Ƴ)} is a double topology on Ƴ. The DTS (Ƴ, ηƳ) is called a double topological subspace of (X, η) (DT-subspace, for short) [4]. Let (X, η) be a DTS and let ɣ, ɧ∈D (X): ɣ, ɧ are said to be separated double sets (separated D-sets, for short) if cl η (ɣ) ∩ɧ = ∅ and cl η (ɧ) ∩ɣ = ∅ [3]. Let (X, η) be a DTS, and let be a nonempty subset of X. If there exist two nonempty separated D-sets ɣ, ɧ∈D(X) such that ɣ∪ɧ = Ɲ, then the D-sets ɣ and ɧ form a D-separation of Ɲ and it is said to be double disconnected set (D-disconnected set, for short). Otherwise, Ɲ is said to be double connected set (d-connected set, for short) [3]. The DTS (X, Π) is said to be (1) C5- disconnected, if (X, Π) has a proper open and closed D-set in Π. (2) C5-connected, if (X, Π) is not C5-disconnected [13]. An intuitionistic fuzzy special topological space (X, Π) is said to be strongly connected if there exist nonempty IFSC, SS ɣ, and ɧ in X such that ɧ∩ɣ = ∅ [13]. Let (X, Π) be a DTS and ƴ ⊆ X with Πh ∈ (ƴ˜). The double component of ƴ with respect to Πh is the maximal of all D-connected subsets of (ƴ, Π) containing the D-point Πh and denoted by C (ƴ˜ , Πh), i.e., C (ƴ, Πh). = ∪ {Ƶ˜ ⊆ ƴ˜: Πh ∈ Ƶ˜, Ƶ˜ is a D-connected set} [12].


## 3. Double-Connected Intuitionistic Space in DITS

In this section, we define new kinds ofx is called Double connected I space, separated Double I sets, strongly Double connected, Double CO connected I space, and Double I component in Double intuitionistic topological spaces, and joined to other kinds of sets that are defined in this work.We start this section by the following definitions:Definition 1.
Letx be a nonempty set.(1)
A Double intuitionistic set (DoubleI set, for short) is an ordered pair Q=x.Q1.Q2.x.D1.D2∈px×pxs.tQ⊆D.(2)
DI (x) = {(Q, D) ∈ p (x) × p (x), Q ⊆ D} is the family of all Double I sets on x.(3)
The DoubleI set 〈x.x.∅, x.x.∅ = (x˜, x˜) is called the universal Double I set, and the Double I set ∅˜,∅˜ = (x.∅.x.x.∅.x is called the empty Double I set.(4)
Let (Q, D), (C, G) ∈ Double I (x): 1) (Q, D)c =  Dc,Qc =  x.D1.D2c.x.Q1.Q2c=x.D2.D1.x.Q2.Q1.2Q.D/Q.DC,=Q/G,D/C=x.Q1.Q2.x.D1.D2/x.C1.C2.x.G1.G2=x.Q1.Q2/x.G1.G2,x.D1.D2.x.C1.C2
Each element of Ѡ is called a DIOS inx. The complement of DIOS is called DICS.
Now, we want to introduce the important theorem to construct the Double intuitionistic topological spaces.Theorem 1.
Letx ≠ ∅. The family T of all a Double intuitionistic open sets in x is Double intuitionistic topological spaces (DITS).Proof.
(1)
Letx.T be intuitionistic topological spaces (ITS), then ∅˜=x.∅.x , x˜ = x.x.∅ ∈ IT⟶∅˜,∅˜,x˜,x˜∈DITS(2)
Let (τ.σ), (ϕ.ϵ) ∈ DIT ⟶ τ, σ, φ, ε ∈ IT. Since IT is an intuitionistic topology; then, τ ∩ ∈ IT and φ ∩ ε ∈ IT. Now, let K=τ.σ and W=ϕ.ϵ ⟶ (K.W) = ((τ, ϕ), (σ.ϵ)) ∈ DITS.(3)
Letθr.ϕr) be a family of IS and r ∈ R and θr.ϕr) ∈ DIT ⟶ θr.ϕr ∈ ITS, since IT is intuitionistic topology, then ∪r∈Rϴr ∈ IT and ∪r∈Rϕr ∈ IT. Thus, ∪r∈Rϴr.ϕr ∈ DIT. Therefore, DITS is Double intuitionistic topological spaces.Definition 2.
Letx nonempty set, (b˜, b˜) ∈xa fixed element in x, and let Q.D=x.Q1.Q2.x.D1.D2 be an Double intuitionistic set (DIS).The DIS (b˜, b˜) defined by (b˜, b˜) = x.b.bc.x.b.bc is called an Double intuitionistic point DIbforshort in X. The DIS b˜.b˜is said to be contained in Q.D, if and only if b˜.b˜∈Q1.D1.Definition 3.
Let (x, Ѡ) be a DITS, let (ℒ, Ӻ), (ℐ, Ω) ∈ DI (x); then, (ℒ, Ӻ), (ℐ, Ω) are said to be separated Double intuitionistic sets (separated Double I sets, for short) if cl (ℒ, Ӻ) ∩ (ℐ, Ω) = ∅˜,∅˜; and (ℒ, Ӻ) ∩ cl(ℐ, Ω) = ∅˜,∅˜. Or (cl (ℒ, Ӻ) ∩ (ℐ, Ω)) ∪ ((ℒ, Ӻ) ∩ cl (ℐ, Ω)) =  ∅˜,∅˜.Proposition 1.
Let (x, Ѡ) be a DITS and let (ℒ, Ӻ), (ℐ, Ω) ∈ DI (x), if (ℒ, Ӻ), (ℐ, Ω) are separated Double I sets, then (ℒ, Ӻ) ∩ (ℐ, Ω) =  ∅˜,∅˜.Proof.
Let (ℒ, Ӻ), (ℐ, Ω) are separated DoubleI sets. By definition if cl (ℒ, Ӻ) ∩ (ℐ, Ω) =  ∅˜,∅˜ and (ℒ, Ӻ) ∩ cl (ℐ, Ω) = ∅˜,∅˜. But (ℒ, Ӻ) ⊆ cl (ℒ, Ӻ).So, (ℒ, Ӻ) ∩ (ℐ, Ω) =  ∅˜,∅˜.
The following example shows that the converse is not true:Example 1.
Letx = {γ, β, ε, Λ}: Ѡ = {∅˜,∅˜, X˜,X˜, (x1.x2),(x3.x2), (x4.x2), (x4.x5), (x4.x3), x2.x2)}, where x1.x2 = (x.ε.β.Λ, x.γ.β.ε.Λ), (x3.x2) = (x.γ.β.ε.Λ.x.γ.β.ε.Λ, x4.x2) =  x.∅.β.ε.Λ.x.γ.β.ε.Λ.x4.x5=x.∅.β.ε.Λ.x.βε.Λ, x4.x3=x.∅.β.ε.Λ, x.γ.β.ε.Λandx2.x2=x.γ.β.ε.Λ, x.γ.β.ε.Λ. Ѡc = {∅˜,∅˜, X˜,X˜, (x2c.x1c), x2c.x3c), (x2c.x4c), (x5c.x4c), (x3c.x4c),(x2c.x2c)}, where(x2c.x1c) = (x.Λ.γ.β.ε, x.β.Λ.ε), x2c.x3c = x.Λ.γ.β.ε.x.ε.Λ.γ.β. (x2c.x4c=x.Λ.γ.β.ε. x.β.ε.Λ.∅.x5c.x4c = x.ε.Λ.β.x.β.ε.Λ.∅, (x3c.x4c=x.ε.Λ.γ.β, x.β.ε.Λ.∅andx2c.x2c=x.Λ.γ.β.ε, x.Λ.γ.β.ε. Let x6.x5c =  x.Λ.γ.β.x.ε.Λ.β and x7.x7 = (x.β.γ.ε.Λ.x.β.γ.ε.Λsoclx6.x5c = (x3c.x4c. Hence x3c.x4c∩x7.x7 = ∅˜,∅˜, but cl x6.x5c∩x7.x7≠∅˜,∅˜. Therefore, x6.x5c, x7.x7 is not separated Double I sets.Proposition 2.
Any two DoubleI closed (Double I open) subset say v,u, s,l of DITS, (x, Ѡ) are separated Double I sets if and only if they are two disjoint.Proof.
Since separated DoubleI sets are already two disjoint; then, we only need to show that the Double I sets v,u, s,l are separated Double I sets, they are Double I closed and two disjoint, v,u ∩ s,l = ∅˜,∅˜. Since cl v,u = v,u and cl s,l = s,l ⟶ cl v,u ∩ s,l = and v,u ∩ cl s,l = ∅˜,∅˜⟶v,u,s,l are separated Double I sets.
Now, takev,u, s,l are Double I open and two disjoint ⟶ v,u ∩ s,l = ∅˜,∅˜. Since v,u and s,l are Double I open ⟶ v,uc and s,lc are Double I closed ⟶ cl v,uc = v,uc, cl s,lc = s,lc [v,u ⊆ s,lc, s,l ⊆ v,uc] ⟶ cl v,u ⊆ cl s,lc = s,lc ⟶ cl v,u ⊆ s,lc and cl s,l ⊆ cl v,uc = v,uc ⟶ cl s,l ⊆ v,uc. Hence, cl v,u ∩ s,l = ∅˜,∅˜ and v,u ∩ cl s,l = ∅˜,∅˜. So, v,u, s,l are separated Double I sets.
The following definition of Double-connected intuitionistic sets in DITS:Definition 4.
Let (x, Ѡ) be a DITS; let (k, Z) be a nonempty subset of X. If there exist two nonempty separated Double I sets (ℒ, Ӻ), (ℐ, Ω) ∈ DI (X) such that (ℒ, Ӻ) ∪ (ℐ, Ω) = (k, Z), then the Double I sets (ℒ, Ӻ) and (ℐ, Ω) form a Double separation of (k, Z) and it is said to be Double disconnected intuitionistic sets (Double disconnected I sets, for short) Otherwise, (k, Z) is said to be Double-connected intuitionistic sets (Double connected I sets, for short).
The following two examples one of them is Double connectedI sets, and the other is not:Example 2.
LetX = {l.q}; Ѡ = {∅˜,∅˜, (X˜,X˜), (∅˜.ξ1), (∅˜.ξ2), (∅˜.ξ3), (∅˜.ξ3c ), ∅˜.ξ1c, ∅˜.ξ2c, (∅˜.ξ4), (ξ1.X˜), (ξ3.X˜), (ξ2.X˜), (ξ3c.X˜), (ξ1c.X˜), (ξ2c.X˜), (ξ4.X˜), ξ3,ξ1.ξ1,ξ1.ξ3,ξ3.ξ2,ξ2.ξ3c.ξ3c.ξ1c.ξ1c.ξ2c.ξ2c.ξ4,ξ4.ξ3c,ξ2.ξ1c,ξ1.ξ1c,ξ2.ξ1c.ξ3c.ξ1c,ξ4.ξ2c,ξ1.ξ2c,ξ3.ξ2c,ξ2.ξ2c,ξ4.ξ4,ξ1.ξ4,ξ2.∅˜.X˜ where.(∅˜.ξ1) =  x.∅.X.x.l.∅. (∅˜.ξ2) =  x.∅.X.x.q.∅. (∅˜.ξ3) =  x.∅.X.x.l.q. (∅˜.ξ3c) = x.∅.X.x.q.l.∅˜.ξ1c=x.∅.X.x.∅.l.∅˜.ξ2c =  x.∅.X.x.∅.q.∅˜.ξ4 =  x.∅.X.x.∅.∅.ξ1.X˜=x.l.∅.x.X.∅.ξ3.X˜=x.l.q.x.X.∅, ξ2.X˜=x.q.∅.x.X.∅, (ξ3c.X˜) =  x.q.l.x.X.∅.ξ1c.X˜=x.∅.l.x.X.∅, ξ2cX˜=x.∅.q.x.X.∅, (ξ4.X˜) =  x.∅.∅.x.X.∅.ξ3,ξ1=x.l.q.x.l.∅.ξ1,ξ1=x.l.∅.x.l.∅, ξ3,ξ3=x.l.q.x.l.q, ξ2,ξ2=x.q.∅.x.q.∅.ξ3c.ξ3c=x.q.l.x.q.l,ξ1c.ξ1c=x.∅.l.x.∅.l, ξ2c,ξ2c=x.∅.q.x.∅.q, ξ4,ξ4=x.∅.∅.x.∅.∅.ξ3c,ξ2=x.q.l.x.q.∅.ξ1c,ξ1=x.∅.l.x.l.∅.ξ1c,ξ2=x.∅.l.x.q.∅, ξ1c.ξ3c=x.∅.l.x.q.lξ1c,ξ4=x.∅.l.x.∅.∅, ξ2c,ξ1 =  x.∅.q.x.l.∅.ξ2c,ξ3=x.∅.q.x.l.q, ξ2c,ξ2 =  x.∅.q.x.q.∅, ξ2c,ξ4=x.∅.q.x.∅.∅, ξ4,ξ1=x.∅.∅.x.l.∅.ξ4,ξ2=x.∅.∅.x.q.∅, and∅˜.X˜=x.∅.X.x.X.∅. Let ξ3,ξ3 = cl ξ3,ξ3∩∅˜.ξ1c =  ∅˜,∅˜ and ξ3,ξ3∩cl∅˜.ξ1c = ∅˜.ξ1c⟶ξ3,ξ3∩∅˜.ξ1c = ∅˜,∅˜, such that ξ3,ξ3∪∅˜.ξ1c = ξ3,ξ1. Therefore, ξ3,ξ3.∅˜.ξ1c are not Double connected I sets.Example 3.
LetX = {ɕ, ɗ, ɤ}; Ѡ = {(∅˜,∅˜), (X˜,X˜), (ß1.ß2), (ß2.ß3), (ß4.ß5), ∅˜.ß4)} where ß1.ß2 = (x.ɤ.ɕ.ɗ, x.ɕ.ɤ.ɗ), (ß2.ß3) = (x.ɕ.ɤ.ɗ.x.ɕ.ɤ.∅.ß4.ß5=x.ɕ.ɗ.ɤ.x.ɕ.ɤ, and ∅.˜ß4 =  〈x.∅.X>.x.ɕ.ɗ.ɤ). Ѡc = {(∅˜,∅˜,(X˜,X˜), ß2c.ß1c.ß3c.ß2c.ß5c.ß4c.ß4c.X˜ where ß2c.ß1c = (x.ɗ.ɕ.ɤ,x.ɕ.ɗ.ɤ), ß3c.ß2c = (x.∅.ɕ.ɤ, x.ɗ.ɕ.ɤ), ß5c.ß4c = (x.ɤ.ɕ,x.ɗ.ɤ.ɕ), and ß4c.X˜ = x.ɗ.ɤ.ɕ.<x, X, ∅>). Let ß1.ß2 and ß2.ß3. Then, ß4c.X˜ = cl ß1.ß2∩ß2.ß3 = ß1.ß3 and ß1.ß2∩clß2.ß3 = (X˜,X˜)) ⟶ß1.ß2∩ (X˜,X˜) = ß1.ß2 and ß1.ß2∪ß2.ß3 = ß2.ß3. Hence, ß1ß2.ß2.ß3 are Double connected I sets.Definition 5.
Let (x, Ѡ) be a DITS; If there exist two nonempty separated Double I sets (ℒ, Ӻ), (ℐ, Ω) ∈ DI (x) such that (ℒ, Ӻ) ∪ I,Ω = (X˜,X˜); then, (ℒ, Ӻ) and (ℐ, Ω) are said to be Double I division for DITS (x, Ѡ). (x, Ѡ) is said to be Double disconnected intuitionistic space (Double disconnected I space, for short), if (x, Ѡ) has a Double I division. Otherwise, (x, Ѡ) is said to be Double-connected intuitionistic space (Double connected I space, for short).Example 4.
Letx = {40, 41, 42}; Ѡ = {(∅˜,∅˜), (X˜,X˜), (ƣ1.ƣ1), (ƣ2.ƣ3), (ƣ4.ƣ5), ƣ1c.ƣ1c} where ƣ1.ƣ1 = (x.40.42.41, x.40.42.41), (ƣ2.ƣ3) =  x.41.42.x.40.41.∅.ƣ4.ƣ5=x.∅.41.42.x.40.41 and ƣ1c.ƣ1c =x.41.40.42>.x.41.40.42. ɸc = {(∅˜,∅˜), (X˜,X˜), ƣ1c.ƣ1c.ƣ3c.ƣ2c.ƣ5c.ƣ4c.ƣ1.ƣ1 where ƣ1c.ƣ1c =(x414042,x.41.40.42), ƣ3c.ƣ2c =(x.∅.40.41,x.42.41), ƣ5c.ƣ4c = (x.41.40, x.41.42.∅) and ƣ1.ƣ1 = 〈x.40.42.41, x.40.42.41). Let ƣ1.ƣ1 = cl ƣ4.ƣ5∩ƣ4.ƣ5 =  ƣ4.ƣ5 and ƣ4.ƣ5∩clƣ4.ƣ5 = ƣ1.ƣ1 and ƣ4.ƣ5∪ƣ4.ƣ5≠X˜,X˜. So, ƣ4.ƣ5 is Double connected I spaces.Remark 1.
(1)
(x, IN) is Double connected I space always since the only Double I open sets are ∅˜,∅˜), (X˜,X˜, and this I sets not make the I space is Double disconnected.(2)
(X, dis) is Double disconnected I space if x contains more than two element, since there exist (ℒ,Ӻ), ∅˜≠ (ℒ,Ӻ) ⊆ x⟶X˜,X˜)= (ℒ,Ӻ) ∪ (ℒ,Ӻ) c, [(ℒ,Ӻ),(ℒ,Ӻ) c] ∈dis (ℒ,Ӻ) ∩ (ℒ,Ӻ) c =(∅˜,∅˜), (ℒ,Ӻ) ≠∅˜,∅˜, and (ℒ,Ӻ) c≠∅˜,∅˜).Theorem 2.
Let (x,Ѡ) is Double connected I space, if f (X˜,X˜) cannot be written as the union of two disjoint nonempty Double I closed sets.Proof.
(⟹) Suppose that (x,Ѡ) is Double connected to prove that (X˜,X˜) ≠ (Ω, Υ) ∪ (Υ, Π), (Ω, Υ) and (Υ, Π) are Double I closed set, cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩ cl (Υ, Π) = (∅˜,∅˜). Let (X˜,X˜) = (Ω, Υ) ∪ (Υ, Π), (Ω, Υ) and (Υ, Π) are Double I closed set, cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩ cl (Υ, Π) = (∅˜,∅˜) ⟶ cl (Ω, Υ) ⊆ (Υ, Π)c = X˜,X˜\(Υ, Π) ⊆ (Υ, Π), but (Ω, Υ) ⊆ cl (Ω, Υ), so that (Ω, Υ) = (Υ, Π)c ∧ (Υ, Π) = (Ω, Υ)c ⟶ (Ω, Υ) ∈Ѡ ∧ (Υ, Π) ∈Ѡ (since (Ω, Υ) = (Υ, Π)c ∧ (Υ, Π) is Double I closed set and (Υ, Π) = (Ω, Υ)c ∧ (Ω, Υ) is Double I closed set) ⟶ (X˜, X˜) = (Ω, Υ) ∪ (Υ, Π) [(Ω, Υ), (Υ, Π) ∈Ѡ], cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩ cl (Υ, Π) = (∅˜, ∅˜) ⟶ (x,Ѡ) is Double disconnected, which a contradiction (since (x,Ѡ) is Double connected) ⟶ (X˜,X˜) ≠ (Ω, Υ) ∪ (Υ, Π), (Ω, Υ) and (Υ, Π) are Double I closed set, cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩ cl (Υ, Π) = (∅˜,∅˜). Hence, X˜,X˜ cannot be written as the union of two disjoint nonempty Double I closed sets (⇐). Suppose that X˜,X˜≠ (Ω, Υ) ∪ (Υ, Π), (Ω, Υ), and (Υ, Π) are Double I closed set, cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩cl (Υ, Π) = (∅˜,∅˜) to prove that (x,Ѡ) is Double connected I space. Let (x,Ѡ) is Double disconnected I space ⟶X˜,X˜= (Ω, Υ) ∪ (Υ, Π) [(Ω, Υ), (Υ, Π) ∈Ѡ], cl (Ω, Υ) ∩ (Υ, Π) = (∅˜,∅˜) and (Ω, Υ) ∩ cl (Υ, Π) = ∅˜,∅˜⟶ (Ω, Υ) = (Υ, Π)c ∧ (Υ, Π) = (Ω, Υ)c ⟶ (Ω, Υ) and (Υ, Π)are Double I closed set, which is a contradiction. Since the complement of every one of them is Double I open set and this contradiction with hypotheses. Therefore, (x, Ψ) is Double connected I space.Theorem 3.
Let (x, Ѡ) be a DITS and let Ұ be a nonempty subset of x. Then, if (Ω, Υ) and (Υ, Π) are Double I closed sets in Ұ; then, (Ω, Υ) and (Υ, Π) are separated Double I sets in Ұ if and only if (Ω, Υ) and (Υ, Π) are separated Double I sets in x.Proof.
cl (Ω, Υ)∩ (Υ, Π) = Ұ˜,Ұ˜∩ cl (Ω, Υ) ∩ (Υ, Π); (Υ, Π) ⊆ Ұ˜,Ұ˜ = Ұ˜,Ұ˜ ∩ (Υ, Π) ∩ cl (Ω, Υ) = (Υ, Π) ∩Ұ˜.Ұ˜∩ cl (Ω, Υ) = (Υ, Π) ∩ clѠҰ (Ω, Υ) = (∅˜,∅˜). Similarly, we have cl (Υ, Π) ∩ (Ω, Υ) = Ұ˜,Ұ˜ ∩ cl (Υ, Π) ∩ (Ω, Υ); (Ω, Υ) ⊆ Ұ˜,Ұ˜ = Ұ˜,Ұ˜ ∩ (Ω, Υ) ∩ cl (Υ, Π) = (Ω, Υ) ∩ Ұ˜,Ұ˜ ∩ cl (Υ, Π) = (Ω, Υ) ∩ clѠҰ (Υ, Π) = (∅˜,∅˜).
Conversely, clѠҰ (Ω, Υ) ∩ (Υ, Π) = Ұ˜,Ұ˜∩ cl (Ω, Υ) ∩(Υ, Π) = Ұ˜,Ұ˜∩[cl (Ω, Υ) ∩ (Υ, Π)] = Ұ˜,Ұ˜∩(∅˜,∅˜) =(∅˜,∅˜). Also, clѠҰ (Υ, Π) ∩ (Ω, Υ) = Ұ˜,Ұ˜ ∩ cl (Υ, Π) ∩(Ω, Υ) = Ұ˜,Ұ˜ ∩ [cl (Υ, Π) ∩ (Ω, Υ)] = Ұ˜,Ұ˜ ∩ (∅˜,∅˜) =(∅˜,∅˜).Theorem 4.
(x,Ѡ) is Double disconnected I space if f there exist a nonempty proper subset of x, which are both Double I open and Double I closed sets in x.Proof.
Suppose that (ƪ, ƭ),(ƙ, ƞ) are a nonempty proper subset of x, which are both Double I open and Double I closed sets to prove x is Double disconnected. Let (ƙ, ƞ)c = (ƪ, ƭ)c; then, (ƙ, ƞ) ≠ (∅˜,∅˜) and (ƪ, ƭ) ≠ (∅˜,∅˜), Moreover, (ƪ, ƭ) ∪ (ƙ, ƞ) = X˜,X˜ and(1)ƪ,ƭ∩ƙ,ƞ=∅˜,∅˜.
Since (ƪ, ƭ) is DoubleI closed as well as Double I open, then (ƙ, ƞ) is also Double I open and Double I closed subset of x, so (ƪ, ƭ) = cl (ƪ, ƭ) and (ƙ, ƞ) = cl (ƙ, ƞ) ⟶ cl (ƪ, ƭ) ∩ (ƙ, ƞ) = (∅˜, ∅˜) (from (1)) and cl (ƙ, ƞ) ∩ (ƪ, ƭ) =(∅˜,∅˜). Hence, X is Double disconnected I spaces.
Conversely: suppose thatx is Double disconnected, then there exist a nonempty subset (ƪ, ƭ), (ƙ, ƞ) of x such that (ƪ, ƭ) ∪ (ƙ, ƞ) = X˜,X˜; (ƪ, ƭ) ≠∅˜,∅˜, (ƙ, ƞ)≠∅˜,∅˜⟶ cl (ƪ, ƭ) ∩ (ƙ, ƞ) = ∅˜,∅˜ and (ƪ, ƭ) ∩ cl (ƙ, ƞ) = ∅˜,∅˜.Since (ƪ, ƭ) ⊆ cl (ƪ, ƭ) ⟶ cl (ƪ, ƭ) ∩ (ƙ, ƞ) = ∅˜,∅˜⟶ (ƪ, ƭ) ∩ (ƙ, ƞ) = ∅˜,∅˜. Hence, (ƪ, ƭ) =(ƙ, ƞ)c ( since (ƪ, ƭ)⊆ (ƙ, ƞ)c = X˜,X˜\(ƙ, ƞ)) = X˜\ ƞ). (X˜\ ƙ)) = (ƙ, ƞ)c) and (ƙ, ƞ)∪(ƙ, ƞ)c = X˜,X˜, (ƙ, ƞ) ⊆ X˜,X˜\ (ƪ, ƭ)) = X˜\ƭ), (X˜\ ƪ)) = (ƪ, ƭ)c ⟶ (ƙ, ƞ) = (ƪ, ƭ)c is a proper subset of x. Now (ƪ, ƭ) ∪ cl (ƙ, ƞ) = X˜,X˜. Also, (ƪ, ƭ) ∩ cl (ƙ, ƞ) = ∅˜,∅˜⟶(ƙ, ƞ) = (cl (ƙ, ƞ))c. Similarly, cl (ƪ, ƭ) ∩ (ƙ, ƞ) = ∅˜,∅˜⟶(ƪ, ƭ) = (cl (ƪ, ƭ))c.Theorem 5.
Let (x, Ѡ) be a DITS and let ℋ be a nonempty subset of x such that (ℋ, Ѡℋ) is Double connected, if (Ω, Υ) and (Υ, Π) are separated Double I sets such that H˜,H˜ ⊆ (Ω, Υ) ∪ (Υ, Π), then H˜,H˜ ⊆ (Ω, Υ)or H˜,H˜ ⊆ (Υ, Π).Proof.
SinceH˜,H˜ ⊆ (Ω, Υ) ∪ (Υ, Π), we have cl (Ω, Υ) ∩ (Υ, Π) = ∅˜,∅˜, and (Ω, Υ) ∩ cl (Υ, Π) = ∅˜,∅˜, then H˜,H˜ = H˜,H˜∩ [(Ω, Υ)∪(Υ, Π)] = [H˜,H˜∩ (Ω,Υ)]∪H˜,H˜∩ (Υ, Π)] ((by Theorem 4)) H˜,H˜∩ (Ω, Υ) and H˜,H˜∩ (Υ, Π) are separated Double I sets of ℋ. Let H˜,H˜∩ (Ω, Υ) and H˜,H˜∩ (Υ, Π) are nonempty⟶H˜,H˜∩(Ω, Υ)≠∅˜,∅˜ and H˜,H˜∩(Υ, Π)≠∅˜,∅˜⟶[H˜,H˜∩(Ω, Υ)] ∩clH˜,H˜∩ (Υ, Π)] = [H˜,H˜∩(Ω, Υ)]∩clH˜,H˜∩cl (Υ, Π)] = H˜,H˜∩clH˜,H˜]∩[(Ω, Υ)∩cl (Υ, Π)] = H˜,H˜∩clH˜,H˜∩∅˜,∅˜ = ∅˜,∅˜. Similarly, cl [H˜,H˜∩(Ω, Υ)] ∩H˜,H˜∩(Υ, Π)] = ∅˜,∅˜. i. e, H˜,H˜∩ (Ω, Υ) and H˜,H˜∩ (Υ, Π) are separated Double I sets. So (ℋ, Ѡℋ) is Double disconnected, which is a contradiction. Let H˜,H˜∩(Ω)=∅˜,∅˜⟶H˜,H˜ = H˜,H˜∩(Υ, Π)⟶H˜,H˜⊆(Υ, Π). Let H˜,H˜∩(Υ, Π)=∅˜,∅˜⟶H˜,H˜ = H˜,H˜∩(Ω, Υ)⟶H˜,H˜⊆(Ω, Υ). Therefore, H˜,H˜⊆(Ω, Υ) or H˜,H˜⊆(Υ, Π).Theorem 6.
Let (x, Ѡ) be a DITS and let ℋ be a nonempty subset of x such that (ℋ, Ѡℋ) is Double connected and ℜ be a subset of x, such that H˜,H˜⊆R˜,R˜⊆clH˜,H˜, then (ℜ, Ѡℜ) is Double connected subspace of (x, Ѡ). In particular, (cl H˜,H˜, ѠclH˜,H˜) is Double connected subspace of (x, Ѡ), Υ) ƙ, ƞ).Proof.
Let (ℜ, Ѡℜ)is Double disconnected subspace of (x, Ѡ), then R˜,R˜ has a Double separation (Ω, Υ)and (Υ, Π) such that cl (Ω, Υ)∩(Υ, Π) = ∅,˜∅˜ and (Ω, Υ) ∩ cl (Υ, Π) = ∅,˜∅˜ and (Ω, Υ)∪(Υ, Π) = R˜,R˜, we have H˜,H˜⊆R˜,R˜⟶H˜,H˜⊆(Ω, Υ)∪(Υ, Π)⟶H˜,H˜⊆(Ω, Υ) or H˜,H˜ ⊆ (Υ, Π) (by Theorem 5). Let H˜,H˜ ⊆ (Ω, Υ) ⟶clH˜,H˜⊆cl (Ω, Υ) ⟶clH˜,H˜∩(Υ, Π)⊆cl (Ω, Υ) ∩ (Υ, Π) ⟶clH˜,H˜∩ (Υ, Π) ⊆ ∅,˜∅˜, but ∅,˜∅˜ ⊆clH˜,H˜∩ (Υ,Π).i.e., (2)clH˜,H˜∙∩Υ,Π∅,˜∅˜.
AgainR˜,R˜ ⊆ cl H˜,H˜⟶ (Ω,Υ) ∪ (Υ,Π) ⊆ cl H˜,H˜⟶ (Υ,Π) ⊆ cl.(3)H˜,H˜⟶Υ.Π∩clH˜,H˜=Υ,Π.
From (2) and (3), we have (Υ,Π) = ∅,˜∅˜, which is a contradiction. So, (ℜ,Ѡℜ) is Double connected. Similarly, cl H˜<i></i>H˜ is Double connected.Theorem 7.
Let (x, Ѡ) be a DITS, if Yβ.Vβ.ѠYβ.Vβ:β∈J is a family of nonempty Double connected subspace of X, ∩β∈JYβ.Vβ is nonempty; then, (Z1.Z2 =  ∪β∈JYβ.Vβ.Ѡ∪β∈JYβ.Vβ is Double connected subspace of (x, Ѡ).Proof.
LetYβ.Vβ.ѠYβ.Vβ:β∈J is Double connected subspace of (x, Ѡ), ∩β∈JYβ.Vβ≠∅,˜∅˜ to show that Z1.Z2=∪β∈JYβ.Vβ is Double connected, if possible suppose that (Z1.Z2 is Double disconnected there exist two nonempty disjoint Double I open sets (π, μ), (ζ, ρ) such that (π, μ) ∩ (ζ, ρ) = ∅,˜∅˜, (ζ, ρ) ∩ Z1.Z2 = ∅,˜∅˜⟶ [(π, μ) ∩ Z1.Z2] ∩ [(ζ, ρ) ∩ Z1.Z2] = ∅,˜∅˜ and [(π, μ) ∩ Z1.Z2] ∪ [(ζ, ρ) ∩ Z1.Z2]= (Z1.Z2⟶ = [(π, μ) ∪ (ζ, ρ)] ∩ (Z1.Z2⟶ (Z1.Z2 ⊆ [(π, μ) ∪ (ζ, ρ)], which is a contradiction, now ∩ Yβ.Vβ≠ .Choose a Double I point b˜.b˜∈∩β∈JYβ.Vβ⟶b˜.b˜∈Yβ.Vβ⟶b˜.b˜∈∪β∈JYβ.Vβ⟶b˜.b˜∈ (Z1.Z2⟶b˜.b˜∈ [(π, μ) ∪ (ζ, ρ)] ⟶b˜.b˜∈ (π, μ) and b˜.b˜∈ (ζ, ρ). Let b˜.b˜∈ (π, μ) also Yβ.Vβ ⊆ ∪β∈JYβ.Vβ⟶ (Z1.Z2 ⊆ [(π, μ) ∪ (ζ, ρ)] ⟶Yβ.Vβ ⊆ [(π, μ) ∪ (ζ, ρ)] ⟶Yβ.Vβ ∩ [(π, μ) ∪ (ζ, ρ)] =  Yβ.Vβ⟶ [Yβ.Vβ ∩ (π, μ)] ∪ [Yβ.Vβ∩ (ζ, ρ)] =  Yβ.Vβ⟶ [Yβ.Vβ ∩ (π, μ)] ∩ [Yβ.Vβ ∩ (ζ, ρ)] = ∅,˜∅˜.Also, Yβ.Vβ∩ (π, μ), Yβ.Vβ ∩ (ζ, ρ) are nonempty two disjoint Double I open set of Yβ.Vβ⟶Yβ.Vβ ⊆ [(π, μ) ∪ (ζ, ρ)], so Yβ.Vβ is Double connected either Yβ.Vβ ⊆ (π, μ) or Yβ.Vβ ⊆ (ζ, ρ) ⟶b˜.b˜∈ (π, μ), b˜.b˜∈Yβ.Vβ⟶Yβ.Vβ ⊆ (π, μ) ⟶∪β∈JYβ.Vβ ⊆ (π, μ)) ⟶ (Z1.Z2 ⊆ (π, μ), (π, μ) ∩(ζ, ρ) = ∅,˜∅˜⟶ (Z1.Z2∩ (ζ, ρ) =  ∅,˜∅˜, which is a contradiction to the fact that (Z1.Z2 ∩ (ζ, ρ) ≠∅,˜∅˜ and b˜.b˜∈ (ζ, ρ), which a contradiction ⟶ (Z1.Z2 is Double connected subspace of (x,Ѡ).Theorem 8.
Let (x, ΨѠ) be a DITS, then, if (x, Ѡ) is Double disconnected I space, then (x, (δ1.δ2 are disconnected I space.Proof.
Suppose that (x, Ѡ) is Double disconnected, then there exist [(q1.q2, (J1.J2]∈ Ψ such that (q1.q2 ∩ (J1.J2 = ∅,˜∅˜ and (q1.q2∪ (J1.J2 =  X˜, X˜). So, q1 ∩J1 = ∅˜ , q2 ∩ J2 = ∅˜ and q1∪J1 =  X˜, q2∪J2 =  X˜ .Hence, (q1.q2 ∩ (J1.J2 =  ∅,˜∅˜ and (q1.q2∪ (J1.J2 =  X˜, X˜),[(q1.q2, (J1.J2]∈ (δ1.δ2.Therefore, (x, δ1) and (x, δ2 ) are disconnected I space.
The following example shows that the converse is not true in general:Example 5.
LetX = {i.j.h}; δ1 = {∅˜,X˜,M1.M2}, where M1=x.i.j.h, M2 = x.h.i.j, and δ2=∅˜.X˜.M3.M4, where M3 =  x.i.h.j and M4 = x.j.i.h. Then, (x, δ1) and (x, δ2) are intuitionistic topological spaces and disconnected I space. Since Ѡ = (δ1.δ2= {(∅˜, ∅˜), (X˜, X˜), ∅˜.M3, ∅˜.M4, ∅˜.X˜, M1.X˜, M2.X˜), (M2.M3)}, where ∅˜.M3=〈x.∅.X>.x.i.h.j), ∅˜.M4 =  x.∅.X>.x.j.i.h, (∅˜.X˜ =  〈x.∅.X>.< x,X, ∅>), M1.X˜= (x.i.j.h.< x,X, ∅>), M2.X˜)=(x.h.i.j.<x,X,∅>),and(M2.M3)= (x.h.i.j.x.i.h.j. Ѡc =  δ2c.δ1c = {(∅˜, ∅˜),(X˜, X˜), M3c.X˜.M4c.X˜.∅˜.X˜.∅˜.M1c.∅˜.M2c.M3c.M2c, where M3c.X˜= (x.j.i.h.< x,X, ∅>), M4c.X˜=x.i.h.j, < x,X, ∅>), ∅˜.X˜=〈x.∅.X>.< x,X, ∅>), ∅˜.M1c=x.∅.X>.x.h.i.j, ∅˜.M2c=〈x.∅.X>.x.i.j.h, and M3c.M2c =  〈x.j.i.h, x.i.j.h. Therefore, (x, Ѡ) is not Double disconnected I space.Definition 6.
Let (x, Ѡ) be a DITS is said to be the following:(1)
Double CO disconnected, if (x, Ѡ) has a proper Double I open and Double I closed sets in x.(2)
Double CO connected, if (x, Ѡ) is not Double CO disconnected.Definition 7.
The DITS (x, Ѡ) is said to be the following:(1)
Strongly Double connectedI space, if there exist no nonempty Double I closed sets [(ν, μ), (η, ϵ)] ∈ x. Such that (ν, μ) ∩ (η, ϵ) =  ∅˜∅˜).(2)
Strongly Double disconnectedI space, if (x, Ѡ) is not strongly Double connected I space.Proposition 3.
(x, Ѡ) is strongly Double connected, if and only, if there exist no Double I open sets (ν, μ),(η, ϵ) in x such that (ν, μ) ≠ (X˜, X˜) ≠ (η, ϵ)and (ν, μ) ∪ (η, ϵ) = (X˜, X˜).Proof.
Let (ν, μ), (η, ϵ) be Double I open sets in x such that (ν, μ) ≠ (X˜, X˜) ≠ (η, ϵ). If we take (k, t)= (ν, μ)c and (n, m) = (η, ϵ)c, then (k, t) and (n, m) become Double I closed sets in x and (k, t) ≠∅˜, ∅˜) ≠ (n, m) and (k, t) ∩ (n, m) =  ∅˜, ∅˜), which is a contradiction.
Conversely, it is obvious.Remark 2.
Strongly Double connectedness does not imply Double CO connectedness, and Double CO connectedness does not imply Strongly Double connectedness.Example 6.
Letx = {Ω.τ.μ}; Ѡ = {(∅˜, ∅˜),(X˜, X˜),(ϕ1.ϕ2),(ϕ3.ϕ3), ∅˜.ϕ4),(ϕ1.ϕ1}where (ϕ1.ϕ2)=(x.Ω.τ.μ, x.Ω.τ.∅),(ϕ3.ϕ3)=(x.τ.μ.Ω.x.τ.μ.Ω.∅˜.ϕ4=x.∅.X>.x.τ.Ω and (ϕ1.ϕ1 = x.Ω.τ.μ.x.Ω.τ.μ. Ѡc = {(∅˜, ∅˜), (X˜ , X˜), ϕ2c.ϕ1c.ϕ1.ϕ1ϕ4c.X˜, ϕ3.ϕ3,whereϕ2c.ϕ1c = (x.∅.Ω.τ, x.τ.μ.Ω), ϕ1.ϕ1=x.Ω.τ.μ.x.Ω.τ.μ.ϕ4c.X˜= (x.Ω.τ, < x,X, ∅>), and (ϕ3.ϕ3)=(x.τ.μ.Ω.x.τ.μ.Ω. Therefore, (x, Ѡ) is strongly Double connected, but is not Double CO connected, for there exist (ϕ1.ϕ1, ϕ3.ϕ3 are both Double I open and Double I closed sets.Example 7.
Letx = {a.ƀ.Ƈ.Ɖ}; Ѡ = {(∅˜, ∅˜), (X˜, X˜),(Ɣ1.Ɣ4), Ɣ2.Ɣ3), Ɣ3.X˜.Ɣ2.Ɣ4), Ɣ5.Ɣ1), Ɣ4.X˜, Ɣ5.Ɣ2), Ɣ4.Ɣ4), Ɣ5.Ɣ4),(Ɣ2.X˜, Ɣ2.Ɣ2), Ɣ5.Ɣ5), Ɣ3.Ɣ3)}, where Ɣ1.Ɣ4=x.ƀ.Ƈ.Ɖ.x.a.ƀ.Ƈ.∅, Ɣ2.Ɣ3=x.a.Ƈ.x.a.Ɖ.Ƈ, Ɣ3.X˜ =(x.a.Ɖ.Ƈ, < x,X, ∅>), Ɣ2.Ɣ4)=(x.a.Ƈ.x.a.ƀ.Ƈ.∅, Ɣ5.Ɣ1=x.∅.Ƈ.Ɖ.x.ƀ.Ƈ.Ɖ, Ɣ4.X˜ =(x.a.ƀ.Ƈ.∅.< x,X, ∅>), Ɣ5.Ɣ2)=(x.∅.Ƈ.Ɖ.x.a.Ƈ, Ɣ4.Ɣ4 =(x.a.ƀ.Ƈ.∅.x.a.ƀ.Ƈ.∅, Ɣ5.Ɣ4)=(x.∅.Ƈ.Ɖ.x.a.ƀ.Ƈ.∅,(Ɣ2.X˜ =(x.a.Ƈ.< x,X, ∅>), Ɣ2.Ɣ2)= (x.a.Ƈ.x.a.Ƈ, Ɣ5.Ɣ5)= (x.∅.Ƈ.Ɖ.x.∅.Ƈ.Ɖ), and Ɣ3.Ɣ3 =(x.a.Ɖ.Ƈ.x.a.Ɖ.Ƈ). Ѡc = {(∅˜, ∅˜),(X˜ , X˜), Ɣ4c.Ɣ1c.Ɣ3c.Ɣ2c.Ɣ4c.Ɣ2c.Ɣ1c.Ɣ5c. (∅˜.Ɣ4c.Ɣ2c.Ɣ5c.Ɣ4c.Ɣ4c.Ɣ4c.Ɣ5c. (∅˜.Ɣ2c, Ɣ2c.Ɣ2c.Ɣ5c.Ɣ5c.Ɣ3c.Ɣ3c. (∅˜.Ɣ3c}, where Ɣ4c.Ɣ1c = x.∅.a.ƀ.Ƈ.x.Ɖ.ƀ.Ƈ, Ɣ3c.Ɣ2c=x.Ƈ.a.Ɖ.x.Ƈ.a , Ɣ4c.Ɣ2c=x.∅.a.ƀ.Ƈ.x.Ƈ.a, Ɣ1c.Ɣ5c=x.Ɖ.ƀ.Ƈ.x.Ƈ.Ɖ.∅, (∅˜.Ɣ4c=x.∅.X>.x.∅.a.ƀ.Ƈ, Ɣ2c.Ɣ5c=x.Ƈ.a.x.Ƈ.Ɖ.∅, Ɣ4c.Ɣ4c=x.∅.a.ƀ.Ƈ.x.∅.a.ƀ.Ƈ, Ɣ4c.Ɣ5c=x.∅.a.ƀ.Ƈ.x.∅.Ƈ.Ɖ, ∅˜.Ɣ2c = x.∅.X>.x.Ƈ.a, Ɣ2c.Ɣ2c=x.Ƈ.a.x.Ƈ.a, Ɣ5c.Ɣ5c=x.Ƈ.Ɖ.∅.x.Ƈ.Ɖ.∅, Ɣ3c.Ɣ3c = (x.Ƈ.a.Ɖ.x.Ƈ.a.Ɖ, and ∅˜.Ɣ3c = x.∅.X>.x.Ƈ.a.Ɖ. (x, Ѡ) is Double CO connected, but it is not strongly Double connected, for there exist [(Ɣ1.Ɣ4), Ɣ3.Ɣ3] ∈ Ψ, and (Ɣ1.Ɣ4) ∪Ɣ3.Ɣ3 = (X˜, X˜).Definition 8.
Let (x, Ѡ) be a DITS and (y˜, y˜) ⊆ (X˜, X˜) with b˜.b˜=x.b.bc.x.b.bc ∈ DI b (y). The union of all Double connected subsets of (y˜, y˜) containing the Double I point b˜.b˜ is called Double I component of y with respect to b˜.b˜, denoted by c1.c2 [(y˜, y˜), b˜.b˜. i.e., c1.c2 [(y˜, y˜), b˜.b˜ =  ∪ {(ν, μ) ⊆(y˜, y˜): b˜.b˜ ∈ (ν, μ), and (ν, μ) is Double connected I set }.Example 8.
Letx = {Ƹ.ƿ.Ʊ}; Ѡ = {(∅˜, ∅˜), (X˜, X˜), (Ʈ1.Ʈ1), Ʈ1c.Ʈ1c}, where Ʈ1.Ʈ1x.Ƹ.ƿ.Ʊ.x.Ƹ.ƿ.Ʊ , Ʈ1c.Ʈ1c=x.ƿ.Ʊ.Ƹ.x.ƿ.Ʊ.Ƹ. Let Ʈ1.Ʈ1=Ƹ˜.Ƹ˜ = (x.Ƹ.ƿ.Ʊ.x.Ƹ.ƿ.Ʊ ⊆ x⟶Ʈ1.Ʈ1 is Double connected. Ʈ2.Ʈ2=ƿ˜.ƿ˜ = (x.ƿ.Ƹ.Ʊ.x.ƿ.Ƹ.Ʊ ⊆ x⟶Ʈ2.Ʈ2 is do double connected. Ʈ3.Ʈ3=Ʊ˜.Ʊ˜ = (x.Ʊ.Ƹ.ƿ.x.Ʊ.Ƹ.ƿ ⊆ x⟶Ʈ3.Ʈ3 is Double connected. Ʈ4.Ʈ4=ƿ.Ʊ˜.ƿ.Ʊ˜ = (x.ƿ.Ʊ.Ƹ.x.ƿ.Ʊ.Ƹ, and subE; x⟶Ʈ4.Ʈ4 is Double connected. Ʈ5.Ʈ5=Ƹ.ƿ˜.Ƹ.ƿ˜ = (x.Ƹ.ƿ.Ʊ.x.Ƹ.ƿ.Ʊ ⊆ x⟶Ʈ5.Ʈ5 is not Double connected. Ʈ6.Ʈ6=Ƹ.Ʊ˜.Ƹ.Ʊ˜ = (x.Ƹ.Ʊ.ƿ.x.Ƹ.Ʊ.ƿ ⊆ x⟶Ʈ6.Ʈ6 is not Double connected.
Note that,Ʈ1.Ʈ1, Ʈ2.Ʈ2, Ʈ3.Ʈ3, and Ʈ4.Ʈ4 are Double connected to find the Double I component of any Double I point in x, so Ʈ1.Ʈ1 ⊆ Ʈ1.Ʈ1 only. Hence, Ʈ1.Ʈ1 is Double I component.
Ʈ2.Ʈ2 ⊆ Ʈ2.Ʈ2 and Ʈ4.Ʈ4, so Ʈ2.Ʈ2 is not Double I component Ʈ3.Ʈ3 ⊆ Ʈ3.Ʈ3 and Ʈ4.Ʈ4, so Ʈ3.Ʈ3 is not Double I component, Ʈ4.Ʈ4 ⊆ Ʈ4.Ʈ4 only. Hence, Ʈ4.Ʈ4 is Double I component. Therefore, Ʈ1.Ʈ1 , Ʈ4.Ʈ4 are Double I component ⟶Ʈ1.Ʈ1∪Ʈ4.Ʈ4 = (X˜, X˜).Theorem 9.
Every DoubleI component of a DITS is Double I closed set.Proof.
Let (x, Ѡ)be a DITS and let c1.c2 [(y˜, y˜), b˜.b˜ be a Double I component of the DITS (x, Ѡ) with respect to an arbitrary Double I point b˜.b˜ ∈ DI b (x).Then, c1.c2 [(y˜, y˜), b˜.b˜ is Double connected subset of x (by Theorem 6 and Theorem 7), cl [c1.c2 [(y˜, y˜), b˜.b˜ is Double connected subset of (X˜, X˜) containing b˜.b˜, then cl [c1.c2 [(y˜, y˜), b˜.b˜ ⊆c1.c2 [(y˜, y˜), b˜.b˜.But (c1.c2 [(y˜, y˜), b˜.b˜ ⊆ cl [c1.c2 [(y˜, y˜), b˜.b˜.Therefore, (c1.c2 [(y˜, y˜), b˜.b˜ = cl [(c1.c2 [(y˜, y˜), b˜.b˜]]. This indicates the Double I component c1.c2 [(y˜, y˜), b˜.b˜ is Double I closed set.


## 4. Conclusions

In this paper, we got the following next results:(1)
We have presented a new set of the following concepts: Double intuitionistic set (DIS) (resp., Double intuitionistic topological spaces (DITS)), DoubleI point, Double connected ΨI set, separated Double I sets, strongly Double connected, Double CO connected I space, and Double I component Ψ in DITS.(2)
Study the basic characteristics and qualities related to these types and the relationships between them; giving examples is incorrect.



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