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## Original Article

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Int. J. Fuzzy Log. Intell. Syst. 2015; 15(1): 79-86

Published online March 31, 2015

https://doi.org/10.5391/IJFIS.2015.15.1.79

© The Korean Institute of Intelligent Systems

## Intuitionistic Fuzzy Topology and Intuitionistic Fuzzy Preorder

Sang Min Yun, and Seok Jong Lee

Department of Mathematics, Chungbuk National University, Cheongju, Korea

Correspondence to :
Seok Jong Lee (sjl@cbnu.ac.kr)

Received: December 24, 2014; Revised: January 20, 2015; Accepted: March 14, 2015

This paper is devoted to finding relationship between intuitionistic fuzzy preorders and intuitionistic fuzzy topologies. For any intuitionistic fuzzy preordered space, an intuitionistic fuzzy topology will be constructed. Conversely, for any intuitionistic fuzzy topological space, we obtain an intuitionistic fuzzy preorder on the set. Moreover, we will show that the family of all intuitionistic fuzzy preorders on an underlying set has a very close link to the family of all intuitionistic fuzzy topologies on the set satisfying some extra condition.

Keywords: Intuitionistic fuzzy topology, Intuitionistic fuzzy preorder

The theory of rough sets was introduced by Pawlak [1]. It is an extension of set theory for the research of intelligent systems characterized by insufficient and incomplete information. The relations between rough sets and topological spaces have been studied in some papers [2?4]. It is proved that the pair of upper and lower approximation operators is a pair of closure and interior of a topological space under a crisp reflexive and transitive relation.

The concept of fuzzy rough sets was proposed by replacing crisp binary relations with fuzzy relations in [5]. Furthermore in [6], the axiomatic approach for fuzzy rough sets were provided. In [7], the authors presented a general framework for the research of fuzzy rough sets in which both constructive and axiomatic approaches are used.

In [8], the authors showed that there is one to one correspondence between the family of fuzzy preorders on a nonempty set and the family of fuzzy topologies on this set satisfying certain extra conditions, and hence they are essentially equivalent.

This paper is devoted to finding relationship between intuitionistic fuzzy preorders and intuitionistic fuzzy topologies. For any intuitionistic fuzzy preordered space, an intuitionistic fuzzy topology will be constructed. Conversely, for any intuitionistic fuzzy topological space, we obtain an intuitionistic fuzzy preorder on the set. Moreover, we will show that the family of all intuitionistic fuzzy preorders on an underlying set has a very close link to the family of all intuitionistic fuzzy topologies on the set satisfying some extra condition.

Let X be a nonempty set. An intuitionistic fuzzy set A is an ordered pair

A=(μA,νA).

where the functions μA : XI and νA : XI denote the degree of membership and the degree of nonmembership, respectively and μA + νA ≤ 1 (See [9]). Obviously every fuzzy set μ in X is an intuitionistic fuzzy set of the form (μ, 1? ? μ).

Throughout this paper, I ? I denotes the family of all intuitionistic fuzzy numbers (a, b) such that a, b ∈ [0, 1] and a + b ≤ 1, with the order relation defined by

(a,b)(c,d)?iff?ac?and?bd.

And IF(X) denotes the family of all intuitionistic fuzzy sets in X, and ‘IF’ stands for ‘intuitionistic fuzzy.’

Any IF set A = (μA, νA) on X can be naturally written as a function A : XI ? I defined by A(x) = (μA(x), νA(x)) for any xX.

For any IF set A = (μA, νA) of X, the value

πA(x)=1-μA(x)-νA(x)

is called an indeterminancy degree (or hesitancy degree) of x to A (See [9]). Szmidt and Kacprzyk [10] call πA(x) an intuitionistic index of x in A. Obviously

0πA(x)1,?????????xX.

Note πA(x) = 0 iff νA(x) = 1 ? μA(x). Hence any fuzzy set μA can be regarded as an IF set (μA, νA) with πA = 0.

### Definition 2.1

([11]). An IF set R on X × X is called an IF relation on X. Moreover, R is called

• (i) reflexive if R(x, x) = (1, 0) for all xX,

• (ii) symmetric if R(x, y) = R(y, x) for all x, yX,

• (iii) transitive if R(x, y) ? R(y, z) ≤ R(x, z) for all x, y, zX.

A reflexive and transitive IF relation is called an IF preorder. A symmetric IF preorder is called an IF equivalence. An IF preorder on X is called an IF partial order if for any x, yX, R(x, y) = R(y, x) = (1, 0) implies that x = y. In this case, (X, R) is called an IF partially ordered space. An IF preorder R is called an IF equality if R is both an IF equivalence and an IF partial order.

### Remark 2.2

R? 1 is called the inverse of R if R? 1(x, y) = R(y, x) for any x, yX. If R is an IF preorder, so is R? 1. RC is called the complement of R if RC(x, y) = (νR(x,y), μR(x,y)) where R(x, y) = (μR(x,y), νR(x,y)). It is obvious that R?1 ≠ = RC.

### Definition 2.3

([12]). Let R be an IF relation on X. Then the two functions R?, R : IF(X) → IF(X), defined by

R?(A)(x)=?yX(R(x,y)A(y)),R_(A(x))=?yX(RC(x,y)A(y)),

are called the upper approximation operator and the lower approximation operator on X, respectively. Moreover, (X, R) is called an IF approximation space.

For any IF number (a, b) ∈ I ? I, (?, b?) is an IF set which has the membership value a constant “a” and the nonmember-ship value a constant “b” for all xX.

### Proposition 2.4

([12, 13]). Let (X, R) be an IF approximation space. Let A, B ∈ IF(X), {Aj| jJ} ⊆ IF(X) and (a, b) ∈ I ? I. Then we have

• R((1?, 0?)) = (1?, 0?),

R?((0?, 1?)) = (0?, 1?),

• ABR(A) ⊆ R(B),

R?(A) ⊆ R?(B),

• R(AC) = (R?(A))C,

R?(AC) = (R(A))C,

• R(AB) = R(A) ∩ R(B),

R?(AB) = R?(A) ∪ R?(B),

• R(?jJAj) = ?jJ (R(Aj)),

R?(?jJAj) = ?jJ (R?(Aj)),

• R((?,b?) ∪ A) = (?,b?) ∪ R(A),

R?((?,b?) ∩ A) = (?,b?) ∩ R?(A).

### Theorem 2.5

([12, 13]). Let (X, R) be an IF approximation space. Then

• R is reflexive AIF(X),R_(A)AAIF(X),AR?(A).

• R is transitive AIF(X),R_(A)R_(R_(A))AIF(X),R?(R?(A))R?(A).

### Theorem 2.6

Let (X, R) be a reflexive IF approximation space. If R(Aj) = Aj for each jJ, then R(?jJAj) = ?jJAj.

Proof

By the reflexivity of R and Theorem 2.5, R(?jJAj) ⊆ ?jJAj. By Proposition 2.4,

R_(?jJAj)?jJR_(Aj)=?jJAj.

Thus R(?jJAj) = ?jJAj.

### Example 2.7

Let X = {x1, x2}. Let R = {?(x1, x1), 1, 0?, ?(x1, x2), 0.2, 0.5?, ?(x2, x1), 0.4, 0.3?, ?(x2, x2), 1, 0?}. Then (X, R) is an IF approximation space. Let A = {?x1, 0.6, 0.3?, ?x2, 0.5, 0.4?} be an IF set on X, then

?yX(μR(x1,y)μA(y))=(10.6)(0.20.5)=0.6,νR?(A)(x1)=?yX(νR(x1,y)νA(y))=(00.2)(0.50.4)=0.2.

Similarly, we obtain

μR?(A)(x2)=0.5,νR?(A)(x2)=0.3.

Hence,

R?(A)={?x1,0.6,0.2?,?x2,0.5,0.3?}.

Similarly, we have

R_(A)={?x1,0.5,0.3?,?x2,0.5,0.4?}.

### Proposition 2.8

([12]). For an IF relation R on X and A ∈ IF(X), the pair R? and R are “dual”, i.e.,

R?(A)=(R_(AC))C,R_(A)=(R?(AC))C,

where AC is the complement of A.

### Definition 2.9

([13, 14]). An IF topology on X in the sense of Lowen [15] is a family of IF sets in X that is closed under arbitrary suprema and finite infima, and contains all constant IF sets. The IF sets in are called open, and their complements, closed.

### Definition 2.10

([8]). A Kuratowski IF closure operator on X is a function k : IF(X) → IF(X) satisfying for (a, b) ∈ I ? I, A, B ∈ IF(X),

• (i) k((?,b?)) = (?,b?),

• (ii) Ak(A),

• (iii) k(AB) = k(A) ∨ k(B),

• (iv) k(k(A)) = k(A).

A Kuratowski IF closure operator k on X is called saturated if for all Aj ∈ IF(X), jJ,

k(?{Aj?jJ})=?{k(Aj)?jJ}.

Furthermore, an IF topology is called saturated if it has a saturated IF closure operator.

### Remark 2.11

([13]). Every Kuratowski IF closure operator k on X gives rise to an IF topology on X in which an IF set B is closed iff k(B) = B.

### 3. Intuitionistic Fuzzy Implication Operator

Generally, for (a1, a2), (b1, b2) ∈ I ? I, the implication operator (or residual implicator) [16, 17] is defined as follows;

(a1,a2)(b1,b2)=sup{(d1,d2)I?I?(a1,a2)(d1,d2)(b1,b2)}.

If given IF numbers (a1, a2), (b1, b2) are comparable, then the IF implication operator is clearly given by

(a1,a2)(b1,b2)={(1,0)if(a1,a2)(b1,b2),(b1,b2)if(a1,a2)>(b1,b2),

for all (a1, a2), (b1, b2) ∈ I ? I.

But it is not always able to compare given IF numbers. Nevertheless, in many papers the IF implication operator is studied where given IF numbers are comparable. In this paper, we consider the IF implication operator to the extend that the given IF numbers are not comparable with some restrictions.

### Remark 3.1

The IF residual implicator is clearly given by

(a1,a2)(b1,b2)={(1,0)if(a1,a2)(b1,b2),(b1,b2)if(a1,a2)>(b1,b2),(1-b2,b2)ifa1<b1,a2b2,(b1,1-b1)ifa1>b1,a2b2,

for all (a1, a2), (b1, b2) ∈ I ? I.

### Definition 3.2

(1) For (a1, a2) ∈ I ? I, A ∈ IF(X) and xX, the map ((a1, a2) ∧ A) : I ? I × IF(X) → IF(X) is defined by ((a1,a2)A)(x)=(a1,a2)A(x).

(2) For (a1, a2) ∈ I ? I, A ∈ IF(X) and xX, the map ((a1, a2) → A) : I ? I × IF(X) → IF(X) is defined by ((a1,a2)A)(x)=(a1,a2)A(x).

(3) For A, B ∈ IF(X), the map (AB) : IF(X)×IF(X) → IF(X) is defined by (AB)(x)=A(x)B(x),?????????xX.

### Remark 3.3

(1) We denote by [(a1, a2), (1, 0)] the rectangular plane which represents [a1, 1]× [0, a2]. For a set AX, an IF set χA : XI ? I is a map defined by

χA(x)={(1,0)ifxA,(0,1)ifx?A.

From the above, if (a1, a2) ∈ I ? I and A ∈ IF(X) are comparable, then we have that

(a1,a2)A=χA-1[(a1,a2),(1,0)]A.

(2) If (a1, a2), (b1, b2), (c1, c2) ∈ I ? I are all comparable, then

(a1,a2)(b1,b2)(c1,c2)iff?(b1,b2)(a1,a2)(c1,c2).

Clearly the following holds;

((a1,a2)(b1,b2))((b1,b2)(c1,c2))((a1,a2)(c1,c2)).

### Theorem 3.4

The implication operator ”→” is an IF preorder on I ? I.

Proof

By the above property and the fact

(a1,a2)(a1,a2)=(1,0),

it follows.

### 4. Intuitionistic Fuzzy Preorder and Intuitionistic Fuzzy Topology

Let (X, ≤) be a preordered space and A ⊆ X. Let ↑ A = {yX | yx, for some xA}. If ↑ A = A, then A is called an upper set. Dually if B =↓ B = {yX | yx, for some xB}, then B is called a lower set. The family of all the upper sets of X is clearly a topology on X, which is called the Alexandrov topology (See [18]) on X, and denoted Γ(≤). We write simply Γ(X) for the topological space (X, Γ(≤)).

On the other hand, for a topological space (X, ) and x, yX, let xy if xU implies yU for any open set U of X, or equivalently, x{y}?. Then ≤ is a preorder on X, called the specialization order (See [18]) on X. Denote this preorder by Ω( ). We also write simply Ω(X) for (X, Ω( )).

A function f : (X, ≤1) → (Y, ≤2) between two preordered sets is called order-preserving if x1y implies f(x) ≤2f(y).

From now on we are going to enlarge the above ideas to the IF theories in a natural way.

### Definition 4.1

Let (X, R) be an IF preordered space. Then A ∈ IF(X) is called an IF upper set in (X, R) if

A(x)R(x,y)A(y),???x,yX.

Dually, A is called an IF lower set if A(y) ∧ R(x, y) ≤ A(x) for all x, yX.

Let R be an IF preorder on X. For x, yX, the real number R(x, y) can be interpreted as the degree to which x is less than or equal to y. The condition A(x) ∧ R(x, y) ≤ A(y) can be interpreted as the statement that if x is in A and xy then y is in A. Particularly, if R is an IF equivalence relation, then an IF set A is an upper set in (X, R) if and only if it is a lower set in (X, R).

The classical preorder relation xy can be naturally extended to R(x, y) = (1, 0) in IF preorder relation. Since (1, 0) = R(x, y) ≤ A(x) → A(y), A(x) ≤ A(y) for any IF upper set A. That is, xy means A(x) ≤ A(y). Obviously, the notion of IF upper sets and IF lower sets agrees with that of upper sets and lower sets in classical preordered space.

### Definition 4.2

A function f : (X, R1) → (Y, R2) between IF preordered spaces is called order-preserving if

R1(x,y)R2(f(x),f(y)),???x,yX.

### Definition 4.3

Let X be a totally ordered set. An IF set A on X is called

• (i) increasing if for all x, yX with x < y, A(x) ≤ A(y),

• (ii) decreasing if for all x, yX with x < y, A(x) ≥ A(y),

• (iii) monotone if A is increasing or decreasing.

### Definition 4.4

Let A = (μA, νA) be an IF set. A is called simple if μA or νA is a constant function.

### Remark 4.5

(1) If indeterminancy degree πA is a constant function t, then for any two different elements x, yX, πA(x) = πA(y) = t. So, if μA(x) ≤ μA(y), then νA(x) ≥ νA(y). Hence A(x) ≤ A(y). Therefore A(x) and A(y) are comparable for any x, yX provided that the hesitancy degree of an IF set A is a constant function.

(2) Suppose that the universal set X is a totally ordered set. If A is monotone or simple, then A(x) and A(y) are comparable for any different element x, yX.

From now on, in order to avoid getting imprecise value in acting with the implication operator, we will consider only the IF set A : XI ? I such that A(x) and A(y) are comparable for any x, yX. By Remark 4.5, any function which is monotone or simple or of constant hesitancy degree is an example of such function.

### Lemma 4.6

For a given IF preordered space (X, R), an IF set B : XI ? I is an IF upper set of (X, R) if and only if B : (X, R) → (I ? I, →) is an order-preserving function.

Proof

For any B ∈ IF(X), we have the following relations.

B is an IF upper set in (X, R)

B(x)R(x,y)B(y)?for?all?x,y,XR(x,y)B(x)B(y)?for?all?x,yXThe?map?B:(X,R)(I?I,)?defined?by?x?B(x)=(μB(x),νB(x))?preserves?order.

### Theorem 4.7

If (X, R) is an IF preordered space, then the family of all the upper sets in X satisfies the following conditions, and hence it is an IF topology on X.

For any IF sets Aj, A;

• (i) ∀(a, b) ∈ I ? I, (?,b?) ∈ ,

• (ii) ?jJAj,

• (iii) ?jJAj,

• (iv) ∀(a, b) ∈ I ? I, ((a, b) ∧ A) ∈ ,

• (v) Suppose that for any A and x, yX and (a, b) ∈ I ? I, A(x) and A(y) and (a, b) are all comparable. Then ((a, b) → A) ∈ .

Proof

(i) Let (a, b) ∈ I ? I, then (?, b?)(x) ∧ R(x, y) = (a, b) ∧ R(x, y) ≤ (a, b) = (?, b?) (y). So (?, b?) ∈ .

(ii) (?Aj)(x)∧R(x, y) = ?(Aj(x)∧R(x, y)) ≤ ?(Aj(y)) = (?Aj)(y).

(iii) (?Aj)(x)∧R(x, y) = ?(Aj(x)∧R(x, y)) ≤ ?(Aj(y)) = (?Aj)(y).

(iv) If A is an upper set, then A(x) ∧ R(x, y) ≤ A(y) for any x, yX. So, we obtain that

((a,b)A(x))R(x,y)((a,b)A(y)),

for all (a, b) ∈ I ? I. This means that ((a, b) ∧ A) ∈ .

(v) By Remark 3.3, ((a, b) → A(x)) ∧ (A(x) → A(y)) ≤ ((a, b) → A(y)). Thus ((a, b) → A(x)) → ((a, b) → A(y)) ≥ (A(x) → A(y)) ≥ R(x, y). Hence ((a, b) → A(x)) ∧ R(x, y) ≤ ((a, b) → A(y)).

### Definition 4.8

For an IF preordered space (X, R), let

TR?={AIF(X)?A=R?(A)},TR_={AIF(X)?A=R_(A)}.

If is a family of IF sets in X which satisfies conditions (i)?(v) of Theorem 4.7, then there exists an IF preorder R on X such that consists of all the upper sets of the IF preordered space (X, R). It will be shown in the following theorem.

### Lemma 4.9

Let Λ be a subfamily of IF(X) such that for any A ∈ Λ and for any x, yX, A(x) and A(y) are comparable. Let R(x, y) = ?B∈Λ(B(x) → B(y)). Then R is an IF preorder.

Proof

(i) R(x, x) = ?B∈Λ (B(x) → B(x)) = (1, 0).

(ii) R(x, y)∧R(y, z) = (?B∈Λ(B(x) → B(y)))∧(?D∈Λ(D(y) → D(z))) ≤ ?B∈Λ ((B(x) → B(y)) ∧ (B(y) → B(z))) ≤ ?B∈Λ(B(x) → B(z)) = R(x, z).

### Theorem 4.10

Let Λ be a subfamily of IF(X) satisfying (i)?(v) of Theorem 4.7 such that for any A ∈ Λ and for any x, yX, A(x) and A(y) are comparable. Then there exists an IF preorder R such that Λ is the family of all upper sets in X with respect to R.

Proof

Suppose that Λ ⊆ IF(X) satisfy the conditions (i)?(v) of Theorem 4.7. Define R(x, y) =?B∈Λ(B(x) → B(y)). By the above lemma, R is an IF preorder on X. Let B ∈ Λ. Since B(x) and B(y) are comparable for any x, yX, we have R(x, y) ≤ B(x) → B(y). Hence B(x) ∧ R(x, y) ≤ B(y), i.e. B is an IF upper set. Thus Λ ⊆ .

What remains is to show that ⊆ Λ. Take D. For a given xX, define mx : XI ? I by mx(z) = D(x) ∧ R(x, z) for all zX. Then mx(x) = D(x) and mx(z) ≤ D(z) for all zX. Thus D = ?xXmx.

For each B ∈ Λ and previously given x, define gB : XI ? I by gB(z) = B(x) → B(z) for all zX. By (v), gB ∈ Λ. By (iii), we obtain that ?B∈ΛgB ∈ Λ. Since

mx(z)=D(x)R(x,z)=D(x)(?BΛ(B(x)B(z))=D(x)(?BΛgB(z))),

we have mx ∈ Λ. Note that D = ?xXmx. Therefore D ∈ Λ by (ii).

### Example 4.11

Let X = [0, 1] be the universal set. Let Λ be the family of all constant IF sets on X and the IF set A = (μA, νA), where μA(x) = 1 ? x, νA(x) = x. Take Γ by arbitrary suprema and arbitrary infima with members of Λ. Then clearly Γ ⊆ IF(X) and it satisfies (i)?(v) of Theorem 4.7. We can define the order R on X by

R(x,y)=?BT(B(x)B(y))?for?all?x,yX.

Then clearly R is reflexive. Take x, yX such that xy, then R(x, y) = A(y) and R(y, x) = (1, 0). So we know that R is transitive. Therefore R is a preorder on X. Consider A(x) ∧ R(x, y) when xy. Then R(x, y) = A(y), and hence A(x) ∧ R(x, y) = A(x) ∧ A(y) ≤ A(y). Consequently we know that Γ is the family of all upper sets in X with respect to R.

The following result relates lower sets and upper sets in an IF preordered space (X, R) with the upper approximation operator R? and R-1?, respectively.

### Proposition 4.12

Let (X, R) be an IF preordered space and A ∈ IF(X). Then A is a lower set iff A = R?(A).

Proof

A is a lower set

iff?A(y)R(x,y)A(x),x,yXiff??{A(y)R(x,y)?yX}A(x),xXiff?R?(A)(x)A(x),xXiff?R?(A)Aiff?R?(A)=A.

### Proposition 4.13

Let (X, R) be an IF preordered space and A ∈ IF(X). Then A is an upper set iff A=R-1?(A).

Proof

A is an upper set

iff?A(x)R(x,y)A(y),x,yXiff??{A(x)R(x,y)?xX}A(y),yXiff??{A(x)R-1(y,x)?xX}A(y),yXiff?R-1?(A)(y)A(y),yXiff?R-1?(A)Aiff?R-1?(A)=A.

### Remark 4.14

Let be the family of all upper sets of an IF preordered space (X, R). By Theorem 4.7, is an Alexandrov IF topology on X. Furthermore T=TR-1? by the above proposition.

### Proposition 4.15

Let (X, R) be an IF preordered space. Then = {A ∈ IF(X) | A = R(A)} is an Alexandrov IF topology on X.

Proof

(1) Take (a, b) ∈ I ? I, then we know that R((?,b?))(x) = ?yX((?,b?)(y) ∨ RC(x, y)) ≥ ?yX(?,b?)(y) = (a, b) = (?,b?)(x). So R((?,b?)) ≥ (?,b?). Clearly R((?,b?) ≤ (?,b?). Hence (?,b?) = R((?,b?)). Therefore (?,b?) ∈ for any (a, b) ∈ I ? I.

(2) Take Aj. Then Aj = R(Aj) for each jJ. So (?jJAj) (x) = (?jJR(Aj)) (x) = ?jJR(Aj) (x) = ?J (?yX(Aj(y) ∨ RC(x, y))) = ?yX ((?jJAj)(y)∨ RC(x, y) = R(?jJAj)(x). Thus ?jJAj.

(3) Take Aj. Then Aj = R(Aj) for each jJ. So (?jJAj) (x) = (?jJR(Aj)) (x) = ?jJR(Aj) (x) = ?jJ (?yX (Aj(y) ∨ RC(x, y))) = ?yX ((?jJAj)(y) ∨ RC(x, y)) = R(?jJAj)(x). Thus ?jJAj.

### Remark 4.16

The IF topology is dual to the IF topology . It follows from the fact that for A ∈ IF(X), AA = R(A) ⇔ A = (R?(AC))CAC = R?(AC) ⇔ AC. In addition, the IF topologies and are Alexandrov IF topologies.

### Proposition 4.17

Let (X, R) be an IF preordered space and A ∈ IF(X). Then

(i) R is the IF interior operator for the IF topology ,

(ii) R is the IF closure operator for the IF topology .

Proof

(i) We will show that int(A) = R(A) for any A ∈ IF(X). Since int(A) = ?{B | B, BA}, int(A) = ?{R(B) | R(B) ≤ A}. Since ?{R(B) | R(B) ≤ A} ≤ A, R(?{R(B) | R(B) ≤ A}) ≤ R(A). By Theorem 2.6, R(?{R(B) | R(B) ≤ A}) = ?{R(B) | R(B) ≤ A}. So ?{R(B) | R(B) ≤ A} ≤ R(A). On the other hand, by R(A) ≤ A, we obtain R(A) ≤ ?{R(B) | R(B) ≤ A}. Hence ?{R(B) | R(B) ≤ A} = R(A). Therefore int(A) = R(A).

(ii) We will show that cl(A) = R? (A) for any A ∈ IF(X). Since cl(A) = ?{BC | B, BCA}, cl(A) = ?{R(B)C | R(B)CA}. By the duality, R?(A) = (R(AC))C = (?{R(B) | R(B) ≤ AC})C = ?{R(B)C | R(B)CA}. Hence cl(A) = R?(A).

### Proposition 4.18

Let k be a saturated IF closure operator on X. Then there exists an IF preorder R on X such that k = R? iff

(i) k(?jJAj) = ?jJk(Aj) for any Aj ∈ IF(X), and

(ii) k((a, b) ∧ A) = (a, b) ∧ k(A) for any A ∈ IF(X) and (a, b) ∈ I ? I.

Proof

Suppose that k satisfies (i) and (ii). By using k, we define an IF relation R on X as

R(x,y)=k(χ{y})(x),?????????x,yX.

For each A ∈ IF(X), if x = ≠ y, we have

(χ{y}A(y))(x)=(χ{y})(x)A(y)=(0.1)A(y)=(0,1).

So (?yX(χ{y}A(y)) (x) = (χ{x}A(x))(x) = A(x), hence A = ?yX(χ{y}A(y)).

For every xX, we have

R?(A)(x)=?yX(R(x,y)A(y))=?yX(k(χ(y))(x)A(y))=?yX(k(χ{y})A(y))(x)=?yXk((χ{y})A(y))(x)=k(?yX(χ{y})A(y))(x)=k(A)(x),

which implies k(A) = R?(A).

Conversely suppose the assumptions. Since k is a Kuratowski IF closure operator, = {AC ∈ IF(X) | k(A) = A} is an IF topology on X, and it satisfies (i). By Lemma 4.9, there exists an IF preorder R on X with respect to the family . Since k = R?, k satisfies (ii) by Proposition 2.4.

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Sang Min Yun is pursuing a doctorate at Chungbuk National University. His research interests include general topology and fuzzy topology. He is a member of KIIS and KMS.

E-mail: jivesm@naver.com

Seok Jong Lee received the M.S. and Ph.D. degrees from Yonsei University in 1986 and 1990, respectively. He is a professor at the Department of Mathematics, Chungbuk National University since 1989. He was a visiting scholar in Carleton University from 1995 to 1996, and Wayne State Univerisity from 2003 to 2004. His research interests include general topology and fuzzy topology. He is a member of KIIS, KMS, and CMS. He served as a general chair of the 12th International Symposium on Advanced Intelligent Systems(ISIS 2011).

E-mail: sjl@cbnu.ac.kr

### Article

#### Original Article

Int. J. Fuzzy Log. Intell. Syst. 2015; 15(1): 79-86

Published online March 31, 2015 https://doi.org/10.5391/IJFIS.2015.15.1.79

## Intuitionistic Fuzzy Topology and Intuitionistic Fuzzy Preorder

Sang Min Yun, and Seok Jong Lee

Department of Mathematics, Chungbuk National University, Cheongju, Korea

Correspondence to:Seok Jong Lee (sjl@cbnu.ac.kr)

Received: December 24, 2014; Revised: January 20, 2015; Accepted: March 14, 2015

### Abstract

This paper is devoted to finding relationship between intuitionistic fuzzy preorders and intuitionistic fuzzy topologies. For any intuitionistic fuzzy preordered space, an intuitionistic fuzzy topology will be constructed. Conversely, for any intuitionistic fuzzy topological space, we obtain an intuitionistic fuzzy preorder on the set. Moreover, we will show that the family of all intuitionistic fuzzy preorders on an underlying set has a very close link to the family of all intuitionistic fuzzy topologies on the set satisfying some extra condition.

Keywords: Intuitionistic fuzzy topology, Intuitionistic fuzzy preorder

### 1. Introduction

The theory of rough sets was introduced by Pawlak [1]. It is an extension of set theory for the research of intelligent systems characterized by insufficient and incomplete information. The relations between rough sets and topological spaces have been studied in some papers [2?4]. It is proved that the pair of upper and lower approximation operators is a pair of closure and interior of a topological space under a crisp reflexive and transitive relation.

The concept of fuzzy rough sets was proposed by replacing crisp binary relations with fuzzy relations in [5]. Furthermore in [6], the axiomatic approach for fuzzy rough sets were provided. In [7], the authors presented a general framework for the research of fuzzy rough sets in which both constructive and axiomatic approaches are used.

In [8], the authors showed that there is one to one correspondence between the family of fuzzy preorders on a nonempty set and the family of fuzzy topologies on this set satisfying certain extra conditions, and hence they are essentially equivalent.

This paper is devoted to finding relationship between intuitionistic fuzzy preorders and intuitionistic fuzzy topologies. For any intuitionistic fuzzy preordered space, an intuitionistic fuzzy topology will be constructed. Conversely, for any intuitionistic fuzzy topological space, we obtain an intuitionistic fuzzy preorder on the set. Moreover, we will show that the family of all intuitionistic fuzzy preorders on an underlying set has a very close link to the family of all intuitionistic fuzzy topologies on the set satisfying some extra condition.

### 2. Preliminaries

Let X be a nonempty set. An intuitionistic fuzzy set A is an ordered pair

$A=(μA,νA).$

where the functions μA : XI and νA : XI denote the degree of membership and the degree of nonmembership, respectively and μA + νA ≤ 1 (See [9]). Obviously every fuzzy set μ in X is an intuitionistic fuzzy set of the form (μ, 1? ? μ).

Throughout this paper, I ? I denotes the family of all intuitionistic fuzzy numbers (a, b) such that a, b ∈ [0, 1] and a + b ≤ 1, with the order relation defined by

$(a,b)≤(c,d)?iff?a≤c?and?b≥d.$

And IF(X) denotes the family of all intuitionistic fuzzy sets in X, and ‘IF’ stands for ‘intuitionistic fuzzy.’

Any IF set A = (μA, νA) on X can be naturally written as a function A : XI ? I defined by A(x) = (μA(x), νA(x)) for any xX.

For any IF set A = (μA, νA) of X, the value

$πA(x)=1-μA(x)-νA(x)$

is called an indeterminancy degree (or hesitancy degree) of x to A (See [9]). Szmidt and Kacprzyk [10] call πA(x) an intuitionistic index of x in A. Obviously

$0≤πA(x)≤1,?????????∀x∈X.$

Note πA(x) = 0 iff νA(x) = 1 ? μA(x). Hence any fuzzy set μA can be regarded as an IF set (μA, νA) with πA = 0.

### Definition 2.1

([11]). An IF set R on X × X is called an IF relation on X. Moreover, R is called

• (i) reflexive if R(x, x) = (1, 0) for all xX,

• (ii) symmetric if R(x, y) = R(y, x) for all x, yX,

• (iii) transitive if R(x, y) ? R(y, z) ≤ R(x, z) for all x, y, zX.

A reflexive and transitive IF relation is called an IF preorder. A symmetric IF preorder is called an IF equivalence. An IF preorder on X is called an IF partial order if for any x, yX, R(x, y) = R(y, x) = (1, 0) implies that x = y. In this case, (X, R) is called an IF partially ordered space. An IF preorder R is called an IF equality if R is both an IF equivalence and an IF partial order.

### Remark 2.2

R? 1 is called the inverse of R if R? 1(x, y) = R(y, x) for any x, yX. If R is an IF preorder, so is R? 1. RC is called the complement of R if RC(x, y) = (νR(x,y), μR(x,y)) where R(x, y) = (μR(x,y), νR(x,y)). It is obvious that R?1 ≠ = RC.

### Definition 2.3

([12]). Let R be an IF relation on X. Then the two functions R?, R : IF(X) → IF(X), defined by

$R?(A)(x)=?y∈X(R(x,y)∧A(y)),R_(A(x))=?y∈X(RC(x,y)∨A(y)),$

are called the upper approximation operator and the lower approximation operator on X, respectively. Moreover, (X, R) is called an IF approximation space.

For any IF number (a, b) ∈ I ? I, (?, b?) is an IF set which has the membership value a constant “a” and the nonmember-ship value a constant “b” for all xX.

### Proposition 2.4

([12, 13]). Let (X, R) be an IF approximation space. Let A, B ∈ IF(X), {Aj| jJ} ⊆ IF(X) and (a, b) ∈ I ? I. Then we have

• R((1?, 0?)) = (1?, 0?),

R?((0?, 1?)) = (0?, 1?),

• ABR(A) ⊆ R(B),

R?(A) ⊆ R?(B),

• R(AC) = (R?(A))C,

R?(AC) = (R(A))C,

• R(AB) = R(A) ∩ R(B),

R?(AB) = R?(A) ∪ R?(B),

• R(?jJAj) = ?jJ (R(Aj)),

R?(?jJAj) = ?jJ (R?(Aj)),

• R((?,b?) ∪ A) = (?,b?) ∪ R(A),

R?((?,b?) ∩ A) = (?,b?) ∩ R?(A).

### Theorem 2.5

([12, 13]). Let (X, R) be an IF approximation space. Then

• R is reflexive $⇔∀A∈IF(X),R_(A)⊆A⇔∀A∈IF(X),A⊆R?(A).$

• R is transitive $⇔∀A∈IF(X),R_(A)⊆R_(R_(A))⇔∀A∈IF(X),R?(R?(A))⊆R?(A).$

### Theorem 2.6

Let (X, R) be a reflexive IF approximation space. If R(Aj) = Aj for each jJ, then R(?jJAj) = ?jJAj.

Proof

By the reflexivity of R and Theorem 2.5, R(?jJAj) ⊆ ?jJAj. By Proposition 2.4,

$R_(?j∈JAj)⊇?j∈JR_(Aj)=?j∈JAj.$

Thus R(?jJAj) = ?jJAj.

### Example 2.7

Let X = {x1, x2}. Let R = {?(x1, x1), 1, 0?, ?(x1, x2), 0.2, 0.5?, ?(x2, x1), 0.4, 0.3?, ?(x2, x2), 1, 0?}. Then (X, R) is an IF approximation space. Let A = {?x1, 0.6, 0.3?, ?x2, 0.5, 0.4?} be an IF set on X, then

$?y∈X(μR(x1,y)∧μA(y))=(1∧0.6)∨(0.2∧0.5)=0.6,νR?(A)(x1)=?y∈X(νR(x1,y)∨νA(y))=(0∨0.2)∧(0.5∨0.4)=0.2.$

Similarly, we obtain

$μR?(A)(x2)=0.5,νR?(A)(x2)=0.3.$

Hence,

$R?(A)={?x1,0.6,0.2?,?x2,0.5,0.3?}.$

Similarly, we have

$R_(A)={?x1,0.5,0.3?,?x2,0.5,0.4?}.$

### Proposition 2.8

([12]). For an IF relation R on X and A ∈ IF(X), the pair R? and R are “dual”, i.e.,

$R?(A)=(R_(AC))C,R_(A)=(R?(AC))C,$

where AC is the complement of A.

### Definition 2.9

([13, 14]). An IF topology on X in the sense of Lowen [15] is a family of IF sets in X that is closed under arbitrary suprema and finite infima, and contains all constant IF sets. The IF sets in are called open, and their complements, closed.

### Definition 2.10

([8]). A Kuratowski IF closure operator on X is a function k : IF(X) → IF(X) satisfying for (a, b) ∈ I ? I, A, B ∈ IF(X),

• (i) k((?,b?)) = (?,b?),

• (ii) Ak(A),

• (iii) k(AB) = k(A) ∨ k(B),

• (iv) k(k(A)) = k(A).

A Kuratowski IF closure operator k on X is called saturated if for all Aj ∈ IF(X), jJ,

$k(?{Aj?j∈J})=?{k(Aj)?j∈J}.$

Furthermore, an IF topology is called saturated if it has a saturated IF closure operator.

### Remark 2.11

([13]). Every Kuratowski IF closure operator k on X gives rise to an IF topology on X in which an IF set B is closed iff k(B) = B.

### 3. Intuitionistic Fuzzy Implication Operator

Generally, for (a1, a2), (b1, b2) ∈ I ? I, the implication operator (or residual implicator) [16, 17] is defined as follows;

$(a1,a2)→(b1,b2)=sup{(d1,d2)∈I?I?(a1,a2)∧(d1,d2)≤(b1,b2)}.$

If given IF numbers (a1, a2), (b1, b2) are comparable, then the IF implication operator is clearly given by

$(a1,a2)→(b1,b2)={(1,0)if(a1,a2)≤(b1,b2),(b1,b2)if(a1,a2)>(b1,b2),$

for all (a1, a2), (b1, b2) ∈ I ? I.

But it is not always able to compare given IF numbers. Nevertheless, in many papers the IF implication operator is studied where given IF numbers are comparable. In this paper, we consider the IF implication operator to the extend that the given IF numbers are not comparable with some restrictions.

### Remark 3.1

The IF residual implicator is clearly given by

$(a1,a2)→(b1,b2)={(1,0)if(a1,a2)≤(b1,b2),(b1,b2)if(a1,a2)>(b1,b2),(1-b2,b2)ifa1b1,a2≥b2,$

for all (a1, a2), (b1, b2) ∈ I ? I.

### Definition 3.2

(1) For (a1, a2) ∈ I ? I, A ∈ IF(X) and xX, the map ((a1, a2) ∧ A) : I ? I × IF(X) → IF(X) is defined by $((a1,a2)∧A)(x)=(a1,a2)∧A(x).$

(2) For (a1, a2) ∈ I ? I, A ∈ IF(X) and xX, the map ((a1, a2) → A) : I ? I × IF(X) → IF(X) is defined by $((a1,a2)→A)(x)=(a1,a2)→A(x).$

(3) For A, B ∈ IF(X), the map (AB) : IF(X)×IF(X) → IF(X) is defined by $(A→B)(x)=A(x)→B(x),?????????∀x∈X.$

### Remark 3.3

(1) We denote by [(a1, a2), (1, 0)] the rectangular plane which represents [a1, 1]× [0, a2]. For a set AX, an IF set χA : XI ? I is a map defined by

$χA(x)={(1,0)ifx∈A,(0,1)ifx?A.$

From the above, if (a1, a2) ∈ I ? I and A ∈ IF(X) are comparable, then we have that

$(a1,a2)→A=χA-1[(a1,a2),(1,0)]∨A.$

(2) If (a1, a2), (b1, b2), (c1, c2) ∈ I ? I are all comparable, then

$(a1,a2)∧(b1,b2)≤(c1,c2)iff?(b1,b2)≤(a1,a2)→(c1,c2).$

Clearly the following holds;

$((a1,a2)→(b1,b2))∧((b1,b2)→(c1,c2))≤((a1,a2)→(c1,c2)).$

### Theorem 3.4

The implication operator ”→” is an IF preorder on I ? I.

Proof

By the above property and the fact

$(a1,a2)→(a1,a2)=(1,0),$

it follows.

### 4. Intuitionistic Fuzzy Preorder and Intuitionistic Fuzzy Topology

Let (X, ≤) be a preordered space and A ⊆ X. Let ↑ A = {yX | yx, for some xA}. If ↑ A = A, then A is called an upper set. Dually if B =↓ B = {yX | yx, for some xB}, then B is called a lower set. The family of all the upper sets of X is clearly a topology on X, which is called the Alexandrov topology (See [18]) on X, and denoted Γ(≤). We write simply Γ(X) for the topological space (X, Γ(≤)).

On the other hand, for a topological space (X, ) and x, yX, let xy if xU implies yU for any open set U of X, or equivalently, $x∈{y}?$. Then ≤ is a preorder on X, called the specialization order (See [18]) on X. Denote this preorder by Ω( ). We also write simply Ω(X) for (X, Ω( )).

A function f : (X, ≤1) → (Y, ≤2) between two preordered sets is called order-preserving if x1y implies f(x) ≤2f(y).

From now on we are going to enlarge the above ideas to the IF theories in a natural way.

### Definition 4.1

Let (X, R) be an IF preordered space. Then A ∈ IF(X) is called an IF upper set in (X, R) if

$A(x)∧R(x,y)≤A(y),???∀x,y∈X.$

Dually, A is called an IF lower set if A(y) ∧ R(x, y) ≤ A(x) for all x, yX.

Let R be an IF preorder on X. For x, yX, the real number R(x, y) can be interpreted as the degree to which x is less than or equal to y. The condition A(x) ∧ R(x, y) ≤ A(y) can be interpreted as the statement that if x is in A and xy then y is in A. Particularly, if R is an IF equivalence relation, then an IF set A is an upper set in (X, R) if and only if it is a lower set in (X, R).

The classical preorder relation xy can be naturally extended to R(x, y) = (1, 0) in IF preorder relation. Since (1, 0) = R(x, y) ≤ A(x) → A(y), A(x) ≤ A(y) for any IF upper set A. That is, xy means A(x) ≤ A(y). Obviously, the notion of IF upper sets and IF lower sets agrees with that of upper sets and lower sets in classical preordered space.

### Definition 4.2

A function f : (X, R1) → (Y, R2) between IF preordered spaces is called order-preserving if

$R1(x,y)≤R2(f(x),f(y)),???∀x,y∈X.$

### Definition 4.3

Let X be a totally ordered set. An IF set A on X is called

• (i) increasing if for all x, yX with x < y, A(x) ≤ A(y),

• (ii) decreasing if for all x, yX with x < y, A(x) ≥ A(y),

• (iii) monotone if A is increasing or decreasing.

### Definition 4.4

Let A = (μA, νA) be an IF set. A is called simple if μA or νA is a constant function.

### Remark 4.5

(1) If indeterminancy degree πA is a constant function t, then for any two different elements x, yX, πA(x) = πA(y) = t. So, if μA(x) ≤ μA(y), then νA(x) ≥ νA(y). Hence A(x) ≤ A(y). Therefore A(x) and A(y) are comparable for any x, yX provided that the hesitancy degree of an IF set A is a constant function.

(2) Suppose that the universal set X is a totally ordered set. If A is monotone or simple, then A(x) and A(y) are comparable for any different element x, yX.

From now on, in order to avoid getting imprecise value in acting with the implication operator, we will consider only the IF set A : XI ? I such that A(x) and A(y) are comparable for any x, yX. By Remark 4.5, any function which is monotone or simple or of constant hesitancy degree is an example of such function.

### Lemma 4.6

For a given IF preordered space (X, R), an IF set B : XI ? I is an IF upper set of (X, R) if and only if B : (X, R) → (I ? I, →) is an order-preserving function.

Proof

For any B ∈ IF(X), we have the following relations.

B is an IF upper set in (X, R)

$⇔B(x)∧R(x,y)≤B(y)?for?all?x,y,∈X⇔R(x,y)≤B(x)→B(y)?for?all?x,y∈X⇔The?map?B:(X,R)→(I?I,→)?defined?by?x?B(x)=(μB(x),νB(x))?preserves?order.$

### Theorem 4.7

If (X, R) is an IF preordered space, then the family of all the upper sets in X satisfies the following conditions, and hence it is an IF topology on X.

For any IF sets Aj, A;

• (i) ∀(a, b) ∈ I ? I, (?,b?) ∈ ,

• (ii) ?jJAj,

• (iii) ?jJAj,

• (iv) ∀(a, b) ∈ I ? I, ((a, b) ∧ A) ∈ ,

• (v) Suppose that for any A and x, yX and (a, b) ∈ I ? I, A(x) and A(y) and (a, b) are all comparable. Then ((a, b) → A) ∈ .

Proof

(i) Let (a, b) ∈ I ? I, then (?, b?)(x) ∧ R(x, y) = (a, b) ∧ R(x, y) ≤ (a, b) = (?, b?) (y). So (?, b?) ∈ .

(ii) (?Aj)(x)∧R(x, y) = ?(Aj(x)∧R(x, y)) ≤ ?(Aj(y)) = (?Aj)(y).

(iii) (?Aj)(x)∧R(x, y) = ?(Aj(x)∧R(x, y)) ≤ ?(Aj(y)) = (?Aj)(y).

(iv) If A is an upper set, then A(x) ∧ R(x, y) ≤ A(y) for any x, yX. So, we obtain that

$((a,b)∧A(x))∧R(x,y)≤((a,b)∧A(y)),$

for all (a, b) ∈ I ? I. This means that ((a, b) ∧ A) ∈ .

(v) By Remark 3.3, ((a, b) → A(x)) ∧ (A(x) → A(y)) ≤ ((a, b) → A(y)). Thus ((a, b) → A(x)) → ((a, b) → A(y)) ≥ (A(x) → A(y)) ≥ R(x, y). Hence ((a, b) → A(x)) ∧ R(x, y) ≤ ((a, b) → A(y)).

### Definition 4.8

For an IF preordered space (X, R), let

$TR?={A∈IF(X)?A=R?(A)},TR_={A∈IF(X)?A=R_(A)}.$

If is a family of IF sets in X which satisfies conditions (i)?(v) of Theorem 4.7, then there exists an IF preorder R on X such that consists of all the upper sets of the IF preordered space (X, R). It will be shown in the following theorem.

### Lemma 4.9

Let Λ be a subfamily of IF(X) such that for any A ∈ Λ and for any x, yX, A(x) and A(y) are comparable. Let R(x, y) = ?B∈Λ(B(x) → B(y)). Then R is an IF preorder.

Proof

(i) R(x, x) = ?B∈Λ (B(x) → B(x)) = (1, 0).

(ii) R(x, y)∧R(y, z) = (?B∈Λ(B(x) → B(y)))∧(?D∈Λ(D(y) → D(z))) ≤ ?B∈Λ ((B(x) → B(y)) ∧ (B(y) → B(z))) ≤ ?B∈Λ(B(x) → B(z)) = R(x, z).

### Theorem 4.10

Let Λ be a subfamily of IF(X) satisfying (i)?(v) of Theorem 4.7 such that for any A ∈ Λ and for any x, yX, A(x) and A(y) are comparable. Then there exists an IF preorder R such that Λ is the family of all upper sets in X with respect to R.

Proof

Suppose that Λ ⊆ IF(X) satisfy the conditions (i)?(v) of Theorem 4.7. Define R(x, y) =?B∈Λ(B(x) → B(y)). By the above lemma, R is an IF preorder on X. Let B ∈ Λ. Since B(x) and B(y) are comparable for any x, yX, we have R(x, y) ≤ B(x) → B(y). Hence B(x) ∧ R(x, y) ≤ B(y), i.e. B is an IF upper set. Thus Λ ⊆ .

What remains is to show that ⊆ Λ. Take D. For a given xX, define mx : XI ? I by mx(z) = D(x) ∧ R(x, z) for all zX. Then mx(x) = D(x) and mx(z) ≤ D(z) for all zX. Thus D = ?xXmx.

For each B ∈ Λ and previously given x, define gB : XI ? I by gB(z) = B(x) → B(z) for all zX. By (v), gB ∈ Λ. By (iii), we obtain that ?B∈ΛgB ∈ Λ. Since

$mx(z)=D(x)∧R(x,z)=D(x)∧(?B∈Λ(B(x)→B(z))=D(x)∧(?B∈ΛgB(z))),$

we have mx ∈ Λ. Note that D = ?xXmx. Therefore D ∈ Λ by (ii).

### Example 4.11

Let X = [0, 1] be the universal set. Let Λ be the family of all constant IF sets on X and the IF set A = (μA, νA), where μA(x) = 1 ? x, νA(x) = x. Take Γ by arbitrary suprema and arbitrary infima with members of Λ. Then clearly Γ ⊆ IF(X) and it satisfies (i)?(v) of Theorem 4.7. We can define the order R on X by

$R(x,y)=?B∈T(B(x)→B(y))?for?all?x,y∈X.$

Then clearly R is reflexive. Take x, yX such that xy, then R(x, y) = A(y) and R(y, x) = (1, 0). So we know that R is transitive. Therefore R is a preorder on X. Consider A(x) ∧ R(x, y) when xy. Then R(x, y) = A(y), and hence A(x) ∧ R(x, y) = A(x) ∧ A(y) ≤ A(y). Consequently we know that Γ is the family of all upper sets in X with respect to R.

The following result relates lower sets and upper sets in an IF preordered space (X, R) with the upper approximation operator R? and $R-1?$, respectively.

### Proposition 4.12

Let (X, R) be an IF preordered space and A ∈ IF(X). Then A is a lower set iff A = R?(A).

Proof

A is a lower set

$iff?A(y)∧R(x,y)≤A(x),∀x,y∈Xiff??{A(y)∧R(x,y)?y∈X}≤A(x),∀x∈Xiff?R?(A)(x)≤A(x),∀x∈Xiff?R?(A)≤Aiff?R?(A)=A.$

### Proposition 4.13

Let (X, R) be an IF preordered space and A ∈ IF(X). Then A is an upper set iff $A=R-1?(A)$.

Proof

A is an upper set

$iff?A(x)∧R(x,y)≤A(y),∀x,y∈Xiff??{A(x)∧R(x,y)?x∈X}≤A(y),∀y∈Xiff??{A(x)∧R-1(y,x)?x∈X}≤A(y),∀y∈Xiff?R-1?(A)(y)≤A(y),∀y∈Xiff?R-1?(A)≤Aiff?R-1?(A)=A.$

### Remark 4.14

Let be the family of all upper sets of an IF preordered space (X, R). By Theorem 4.7, is an Alexandrov IF topology on X. Furthermore $T=TR-1?$ by the above proposition.

### Proposition 4.15

Let (X, R) be an IF preordered space. Then = {A ∈ IF(X) | A = R(A)} is an Alexandrov IF topology on X.

Proof

(1) Take (a, b) ∈ I ? I, then we know that R((?,b?))(x) = ?yX((?,b?)(y) ∨ RC(x, y)) ≥ ?yX(?,b?)(y) = (a, b) = (?,b?)(x). So R((?,b?)) ≥ (?,b?). Clearly R((?,b?) ≤ (?,b?). Hence (?,b?) = R((?,b?)). Therefore (?,b?) ∈ for any (a, b) ∈ I ? I.

(2) Take Aj. Then Aj = R(Aj) for each jJ. So (?jJAj) (x) = (?jJR(Aj)) (x) = ?jJR(Aj) (x) = ?J (?yX(Aj(y) ∨ RC(x, y))) = ?yX ((?jJAj)(y)∨ RC(x, y) = R(?jJAj)(x). Thus ?jJAj.

(3) Take Aj. Then Aj = R(Aj) for each jJ. So (?jJAj) (x) = (?jJR(Aj)) (x) = ?jJR(Aj) (x) = ?jJ (?yX (Aj(y) ∨ RC(x, y))) = ?yX ((?jJAj)(y) ∨ RC(x, y)) = R(?jJAj)(x). Thus ?jJAj.

### Remark 4.16

The IF topology is dual to the IF topology . It follows from the fact that for A ∈ IF(X), AA = R(A) ⇔ A = (R?(AC))CAC = R?(AC) ⇔ AC. In addition, the IF topologies and are Alexandrov IF topologies.

### Proposition 4.17

Let (X, R) be an IF preordered space and A ∈ IF(X). Then

(i) R is the IF interior operator for the IF topology ,

(ii) R is the IF closure operator for the IF topology .

Proof

(i) We will show that int(A) = R(A) for any A ∈ IF(X). Since int(A) = ?{B | B, BA}, int(A) = ?{R(B) | R(B) ≤ A}. Since ?{R(B) | R(B) ≤ A} ≤ A, R(?{R(B) | R(B) ≤ A}) ≤ R(A). By Theorem 2.6, R(?{R(B) | R(B) ≤ A}) = ?{R(B) | R(B) ≤ A}. So ?{R(B) | R(B) ≤ A} ≤ R(A). On the other hand, by R(A) ≤ A, we obtain R(A) ≤ ?{R(B) | R(B) ≤ A}. Hence ?{R(B) | R(B) ≤ A} = R(A). Therefore int(A) = R(A).

(ii) We will show that cl(A) = R? (A) for any A ∈ IF(X). Since cl(A) = ?{BC | B, BCA}, cl(A) = ?{R(B)C | R(B)CA}. By the duality, R?(A) = (R(AC))C = (?{R(B) | R(B) ≤ AC})C = ?{R(B)C | R(B)CA}. Hence cl(A) = R?(A).

### Proposition 4.18

Let k be a saturated IF closure operator on X. Then there exists an IF preorder R on X such that k = R? iff

(i) k(?jJAj) = ?jJk(Aj) for any Aj ∈ IF(X), and

(ii) k((a, b) ∧ A) = (a, b) ∧ k(A) for any A ∈ IF(X) and (a, b) ∈ I ? I.

Proof

Suppose that k satisfies (i) and (ii). By using k, we define an IF relation R on X as

$R(x,y)=k(χ{y})(x),?????????x,y∈X.$

For each A ∈ IF(X), if x = ≠ y, we have

$(χ{y}∧A(y))(x)=(χ{y})(x)∧A(y)=(0.1)∧A(y)=(0,1).$

So (?yX(χ{y}A(y)) (x) = (χ{x}A(x))(x) = A(x), hence A = ?yX(χ{y}A(y)).

For every xX, we have

$R?(A)(x)=?y∈X(R(x,y)∧A(y))=?y∈X(k(χ(y))(x)∧A(y))=?y∈X(k(χ{y})∧A(y))(x)=?y∈Xk((χ{y})∧A(y))(x)=k(?y∈X(χ{y})∧A(y))(x)=k(A)(x),$

which implies k(A) = R?(A).

Conversely suppose the assumptions. Since k is a Kuratowski IF closure operator, = {AC ∈ IF(X) | k(A) = A} is an IF topology on X, and it satisfies (i). By Lemma 4.9, there exists an IF preorder R on X with respect to the family . Since k = R?, k satisfies (ii) by Proposition 2.4.

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