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International Journal of Fuzzy Logic and Intelligent Systems 2020; 20(2): 129-137

Published online June 25, 2020

https://doi.org/10.5391/IJFIS.2020.20.2.129

© The Korean Institute of Intelligent Systems

New Approach to Intuitionistic Fuzzy Rough Sets

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 15, 2019; Revised: May 2, 2020; Accepted: May 5, 2020

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted noncommercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

The properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. In order to overcome this unnaturalness, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

Keywords: Intuitionistic fuzzy topology, Intuitionistic fuzzy rough sets

The notion of fuzzy sets was first introduced by Zadeh [1]. After that, many studies attempted to generalize the fuzzy set by using various approaches. Pawlak [2] introduced the concept of rough sets, Nanda and Majumda [3] and Coker [4] proposed the idea of fuzzy rough sets. Atanassov [5] introduced the idea of intuitionistic fuzzy sets. All these concepts provide useful means of expressing vagueness in real environments.

Combining the concepts of fuzzy rough sets and intuitionistic fuzzy sets, Samanta and Mondal [6] proposed the idea of intuitionistic fuzzy rough sets. By introducing further generalization, the authors [7, 8] also conducted a study on the intuitionistic fuzzy bitopology and intuitionistic smooth bitopology. Moreover, the categorical properties of the intuitionsitc fuzzy topological spaces were studied by the same research group [911].

Many attempts at combining fuzziness and roughness have been made. In [12], the measure of fuzziness in rough sets is provided and studied. In [13] a general framework for the study of fuzzy rough sets is presented, in which both constructive and axiomatic approaches are made. Lower and upper approximations of intuitionistic fuzzy sets with respect to an intuitionistic fuzzy approximation space are first defined by Zhou et al. [14, 15]. Several important properties of intuitionistic fuzzy approximation operators are examined by many researchers including us [1619].

However, the properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. This is because of the unnaturalness of the definition of fuzzy rough sets. For example, the double complement of a fuzzy rough set is different from itself. The property that the double complement of a set becomes the set itself is one of the essential properties of Boolean algebra. Hence this flaw is critical in expanding the related theory. In order to overcome this unnaturalness, we need a new approach to intuitionistic fuzzy rough sets.

In this paper, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. This new approach enables us to manipulate fuzzy rough sets more simply and easily. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

In [3], the definition of fuzzy rough sets has been introduced. The paper said:

“we shall consider ( , ) to be a rough universe where is a nonempty set and is a Boolean subalgebra of the Boolean algebra of all subsets of . Also consider a rough set with XLXU. A fuzzy rough set in X is an object of the form

A=(AL,AU),

where AL and AU are characterized by a pair of maps AL : XLL and AU : XUL with AL(x) ≤ AU(x), for all xXL, where (L,≤) is a fuzzy lattice.”

Furthermore, the complement Ā of a fuzzy rough set A = (AL, AU) is defined by (Ā)L(x) = (AU>L)′(x), ∀xXL and (Ā)U(x) = (AL<U)′(x), ∀xXU,

where AU>L(x) = AU(x), ∀xXL and

AL<U(x)={AL(x),if xXL,{AL(x)xXL},if xXU-XL.

Unfortunately, the double complement of a fuzzy rough set A is different from A, because XL and XU are different, i.e., XLXU, The property that double complement of a set becomes the set itself is one of the essential properties of Boolean algebra. Hence this flaw is critical in expanding the related theory. Thus we are going to introduce the new definition of a fuzzy rough set by weakening the condition of the old definition. Then the properties we obtain in this paper are different from the results in the above paper.

Definition 2.1

Ler X be an underlying set and (L, ≤) a fuzzy lattice. A fuzzy rough set in X is an object of the form

A=(AL,AU),

where AL and AU are defined by a pair of maps AL : XL and AU : XL with AL(x) ≤ AU(x), for all xX. Furthermore 0 = (0L, 0U) the null fuzzy rough set and 1 = (1L, 1U) the whole fuzzy rough set in X,

Definition 2.2

For any two fuzzy rough sets A = (AL, AU) and B = (BL, BU) of X, we define the following:

  • AB iff AL(x) ≤ BL(x) and AU(x) ≤ BU(x) for all xX.

  • A = B iff AB and BA.

If {Ai | iJ} is a family of fuzzy rough sets in X, with Ai = ((Ai)L, (Ai)U), then we define the following:

  • (∪i Ai)L(x) = ∨ (Ai)L(x) and (∪i Ai)U(x) = ∨ (Ai)U(x) for all xX.

  • (∩i Ai)L(x) = ∧ (Ai)L(x) and (∩i Ai)U(x) = ∧ (Ai)U(x) for all xX.

Definition 2.3

The complement Ā = ((Ā)L, (Ā)U) of a fuzzy rough set A = (AL, AU) in X is defined by (Ā)L(x) = (AU(x))′ and (Ā)U(x) = (AL(x))′ for all xX.

For any fuzzy rough set A = (AL, AU) and for any xX,

(A¯)L(x)=(AU(x))(AL(x))=(A¯)U(x).

So, Ā is again a fuzzy rough set. Furthermore, for any fuzzy rough set A = (AL, AU), we have the double complement property, i.e., (A¯)¯=A.

Theorem 2.4

Let A be any fuzzy rough set in X, then we have the following properties:

  • 0A1.

  • (0_)¯=1,(1_)¯=0_.

Theorem 2.5

If A, B, C, D and Bi, iJ are fuzzy rough sets in X, then we have the following properties:

  • AB and CD implies

    ACBD and ACBD.

  • AB and BC implies AC.

  • ABA, BAB.

  • A ∪ (∩i Bi) = ∩i(ABi), A ∩ (∪i Bi) = ∪i(ABi).

  • AB implies Ā.

  • iBi¯=iBi¯,iBi¯=iBi¯

Definition 2.6

Let f : XY be a mapping and A = (AL, AU) a fuzzy rough set in X. The image f(A) = ((f(A))L, (f(A))U) of A under f is defined by, for any yY

(f(A))L(y)=f(AL)(y)={0_L,if f-1(y)=,{AL(x)xf-1(y)},otherwise,(f(A))U(y)=f(AU)(y)={0_U,if f-1(y)=,{AU(x)xf-1(y)},otherwise.

Let B = (BL, BU) be a fuzzy rough set in Y . Then the inverse image f−1(B) = ((f−1(B))L, (f−1(B))U) of B under f is defined by, for any yY

(f-1(B))L(x)=f-1(BL)(x)=BL(f(x)),(f-1(B))U(x)=f-1(BU)(x)=BU(f(x)).

The image of A under f is well-defined, because

f(AL)(y)={AL(x)xf-1(y)}{AU(x)xf-1(y)}=f(AU)(y).

Moreover, the inverse image of B under f is well-defined, because

f-1(BL)(x)=BL(f(x))BU(f(x))=f-1(BU)(x).

Theorem 2.7

If f : XY is a mapping, then for any fuzzy rough sets A and B in X, we have the following:

  • AB implies f(A) ⊆ f(B).

  • If f is surjective, then f(A¯)f(A)¯.

  • f(∪iJ Ai) = ∪iJ f(Ai).

  • f(∩iJ Ai) ⊆ ∩iJ f(Ai).

Proof

(1) The proof is clear.

(2) Let f be surjective.

(f(A¯))L(y)=f((A¯)L)(y)={(A¯)L(x)xf-1(y)}={(AU(x))xf-1(y)}{(AU(x))xf-1(y)}=({(AU(x))xf-1(y)})=(f(AU)(y))=((f(A))U(y))=(f(A)¯)L(y).

The proof of the upper part is similar.

(3)

(f(iJAi))L(y)=f((iJAi)L)(y)={(iJAi)L(x)xf-1(y)}={iJ(Ai)L(x)xf-1(y)}=iJ{xf-1(y)(Ai)L(x)}=iJ{f((Ai)L)(y)}=iJ(f(Ai))L(y).

The proof of the upper part is similar.

(4)

(f(iJAi))L(y)=f((iJAi)L)(y)={(iJAi)L(x)xf-1(y)}={iJ(Ai)L(x)xf-1(y)}iJ{xf-1(y)(Ai)L(x)}=iJ{f((Ai)L)(y)}=iJ(f(Ai))L(y).

The proof of the upper part is similar.

The surjectiveness is essential in (2) of the above theorem. It can be shown by the following example.

Example 2.8

Let X = {1, 2, 3}, Y = {a, b, c}. Let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Then f is not surjective. Consider a fuzzy rough set

A=({1/0.4,2/0.3},{1/0.4,2/0.4}).

Then

A¯=({1/0.6,2/0.6},{1/0.6,2/0.7}).

So

f(A¯)=({a/0.6,b/0,c/0.6},{a/0.6,b/0,c/0.7}).

But

f(A)=({a/0.4,b/0,c/0.3},{a/0.4,b/0,c/0.4}),

and so

f(A)¯=({a/0.6,b/1,c/0.6},{a/0.6,b/1,c/0.7}).

Hence f(A¯)f(A)¯.

Remark 2.9

If f : XY and g : YZ are mappings, then for any fuzzy rough set C in Z, the inverse image of C under gf is defined by (gf)−1(C) = f−1(g−1(C)) where gf is the composition of g and f.

Theorem 2.10

If f : XY is a mapping, then for all fuzzy rough sets B, C, Bi, iJ in Y , we have the following:

  • f-1(B¯)=f-1(B)¯.

  • BC implies f−1(B) ⊆ f−1(C).

  • f−1(∩i Bi) = ∩i f−1(Bi).

  • f−1(∪i Bi) = ∪i f−1(Bi).

Proof

(1) (f-1(B¯))L(x)=f-1((B¯)L)(x)=(B¯)L(f(x))=(BU(f(x)))=(f-1(BU)(x))=((f-1(B))U(x))=(f-1(B)¯)L(x). The proof of upper part is similar. Hence f-1(B¯)=f-1(B)¯.

(2) Since BC, BL(y) ≤ CL(y) for all yY . (f−1(B))L (x) = f−1(BL)(x) = BL(f(x)) ≤ CL(f(x)) = f−1(CL)(x) = (f−1(C))L(x). The proof of upper part is similar. Hence BC implies f−1(B) ⊆ f−1(C).

(3) (f−1(∩i Bi))L(x) = f−1((∩i Bi)L)(x) = (∩i Bi)L (f(x)) = ∧i(Bi)L(f(x)) = ∧i f−1((Bi)L)(x) = ∧i(f−1(Bi))L (x) = (∩i f−1(Bi))L(x). The proof of upper part is similar. Hence f−1(∩i Bi) = ∩i f−1(Bi).

(4) Similarly.

Theorem 2.11

If f : XY is a mapping, then for any fuzzy rough set A in X and B in Y , we have the following:

  • Bf(f−1(B)).

  • Af−1(f(A)).

Definition 3.1

If A = (AL, AU) and B = (BL, BU) are two fuzzy rough sets in X with BĀ, then the ordered pair (A, B) is called an intuitionistic fuzzy rough set (briefly IF rough set) in X. The condition BĀ is called the intuitionistic condition (briefly IC).

Remark 3.2

In [6], the IC is that BĀ and A. But by the definition of IF rough set in this paper, we get the followings. If BĀ, then BL(x) ≤ (Ā)L(x) = (AU(x))′. i.e. BL(x) ≤ (AU(x))′ and BU(x) ≤ (AL(x))′. So AU(x) ≤ (BL(x))′ and AL(x) ≤ (BU(x))′. Thus A. Moreover, the converse is clear. Therefore BĀ iff A.

Definition 3.3

0* = (0, 1) and 1* = (1, 0) are called the null IF rough set and the whole IF rough set in X, respectively. Clearly 0*¯=1* and 1*¯=0*.

We denote by IFRS(X) the collection of all IF rough sets in X. Usually we shall use letters X, Y, Z, … to denote sets, letters A, B, C, … to denote fuzzy rough sets, and letters P, Q, R, … to denote IF rough sets.

Definition 3.4

53.4 Let P = (A, B) = ((AL, AU), (BL, BU)) and Q = (C, D) = ((CL, CU), (DL, DU)) be two IF rough sets in X. We define the following:

  • PQ iff AC and BD.

  • P = Q iff PQ and PQ.

  • The complement P̄ of P = (A, B) in X, is defined by = (B, A).

  • For IF rough sets Pi = (Ai, Bi) in X, iJ, we define

iJPi=(iJAi,iJBi),iJPi=(iJAi,iJBi).

Remark 3.5

Let Pi = (Ai, Bi) be an IF rough set for all i ∈ Δ, then AiBi¯ for all i ∈ Δ. For any k ∈ Δ, we have

AkBk¯iΔBi¯=iΔBi¯.

Thus iΔAiiΔBi¯. Hence ∪i∈ΔPi = (∪i∈ΔAi, ∩i∈ΔBi) is also an IF rough set.

Similarly, for any k ∈ Δ, iAiAkBk¯. So iAiiBi¯=iBi¯. Hence ∩i Pi = (∩i Ai, ∪i Bi) is also an IF rough set.

Theorem 3.6

Let P = (A, B), Q = (C, D), R = (E, F) and Pi = (Ai, Bi), iJ be IF rough sets in X, then we have the following:

  • PP = P = PP.

  • PQ = QP, PQ = QP

  • (PQ) ∩ R = P ∩(QR), (PQ) ∪ R = P ∪(QR).

  • PQP, QPQ.

  • PQ and QR implies PR.

  • PiQ, ∀iJ implies ∪i PiQ.

  • QPi, ∀iJ implies Q ⊆ ∩i Pi.

  • Q ∪ (∩i Pi) = ∩i(QPi).

  • Q ∩ (∪i Pi) = ∪i(QPi).

  • (P¯)¯=P.

  • PQ iff .

  • iPi¯=iPi¯,iPi¯=iPi¯.

Proof

(4) PQ = (AC, BD) ⊆ (A, B) ⊆ (AC, BD) = PQ.

(12) iPi¯=(iAi,iBi¯=(iBi,iAi)=i(Bi,Ai)=iPi¯.

Definition 3.7

Let f : XY be a mapping. Let P = (A, B) = ((AL, AU), (BL, BU)) be an IF rough set in X. Then we define an image of P under f by

f(P)=(f(A),f(B¯)¯),

where f(A) = (f(AL), f(AU)) and f(B¯)¯=((f(B¯)¯)L,(f(B¯)¯)U) is defined by (f(B¯)¯)L=f((B¯)L)¯ and (f(B¯)¯)U=f((B¯)U)¯.

We already know that f(A)(y) = (f(AL)(y), f(AU)(y)) = (∨{AL(x) | xf−1(y)}, ∨{AU(x) | xf−1(y)}) for any fuzzy rough set A in X and a function f : XY .

Theorem 3.8

By the above definition, we have the following conclusion:

For any fuzzy rough set B in X and a function f : XY , We have f(B¯)¯(y)=({(BL(x))xf-1(y)},{(BU(x))xf-1(y)}), ∧{(BU(x) | xf−1(y)}).

Prooff((B¯)L)(y)={(B¯)L(x)xf-1(y)}={(BU(x))xf-1(y)}=({(BU(x))xf-1(y)}),f((B¯)U)(y)={(B¯)U(x)xf-1(y)}={(BL(x))xf-1(y)}=({(BL(x))xf-1(y)}).

So, f()(y) = ((∧{(BU(x))| xf−1(y)})′, (∧{(BL(x)) | xf−1(y)})′). Thus, f(B¯)¯(y)=({(BL(x))xf-1(y)},{(BU(x))xf-1(y)}).

Remark 3.9

We will prove that the image of any IF rough set P = (A, B) satisfies the IC.

(f(A))L(y)=f(AL)(y)={AL(x)xf-1(y)}{(B¯)L(x)xf-1(y)}={(BU(x))xf-1(y)}=({(BU(x))xf-1(y)})=((f(B¯)¯)U(y))=((f(B¯)¯)¯)L(y).

The proof of the upper part is similar. So, if P = (A, B) be an IF rough set in X, then f(P)=(f(A),f(B¯)¯) is also an IF rough set in Y .

Remark 3.10

By the above theorem, we have the following conclusion:

For any IF rough set P = (A, B) in X and any map f : XY , The image f(P) = (f(A), f(B)) of P under f is actually expressed as follows:

For any yY

f(P)(y)=({A(x)xf-1(y)},{B(x)xf-1(y)}),

where

{A(x)xf-1(y)}={0_,if         f-1(y)=,({AL(x)xf-1(y)},{AU(x)xf-1(y)}),if         f-1(y),{B(x)xf-1(y)}={1_,if         f-1(y)=,({BL(x)xf-1(y)},{BU(x)xf-1(y)}),if         f-1(y).

Definition 3.11

Let f : XY be a mapping and Q = (C, D) = ((CL, CU), (DL, DU)) an IF rough set in Y . Then we define an inverse image of Q under f by

f-1(Q)=(f-1(C),f-1(D)),

where f−1(C) = ((f−1(C))L, (f−1(C))U) = (f−1(CL), f−1(CU)) and f−1(D) = ((f−1(D))L, (f−1(D))U) = (f−1(DL), f−1(DU)).

Remark 3.12

f-1(CL)(x)=CL(f(x))(D¯)L(f(x))=f-1((D¯)L)(x)=(f-1(D)¯)L(x). The upper part is also hold. Hence, the inverse image of an IF rough set under f is also an IF rough set.

Theorem 3.13

Let f : XY be a mapping. For any IF rough set P and Q on X, we have the following:

  • PQ implies f(P) ⊆ f(Q).

  • If f is surjective, then f(P¯)f(P)¯.

Proof

(1) It is clear.

(2) Let P = (A, B) be an IF rough set. Then f(P¯)=f((B,A))=(f(B),f(A¯)¯). Since f(P)=f((A,B))=(f(A),f(B¯)¯),f(P)¯=(f(B¯)¯,f(A)). Consider (f(B))L(y)=f(BL)(y)={BL(x)xf-1(y)}{BL(x)xf-1(y)}=(f(B¯)¯)L(y). And the proof of the upper part is similar. Thus f(B)f(B¯)¯. Similarly, f(A)f(A¯)¯. Therefore f(P¯)=(f(B),f(A¯)¯)(f(B¯)¯,f(A))=f(P)¯.

Definition 3.14

If f : XY and g : YZ are mappings, then the inverse image of IF rough set W in Z under (gf) is defined by (gf) −1(W) = f−1(g−1(W)) where gf is the composition of g and f.

Theorem 3.15

Let f : XY be a mapping. For any IF rough set R, S and Ri, iJ in Y , we have the following:

  • f-1(R¯)=f-1(R)¯.

  • RS implies f−1(R) ⊆ f−1(S).

  • f−1(∪i Ri) = ∪i f−1(Ri).

  • f−1(∩i Ri) = ∩i f−1(Ri).

Proof

(1) Let R = (A, B), then = (B, A). f-1(R¯)=(f-1(B),f-1(A))=(f-1(A),f-1(B))¯=f-1(R)¯.

(3) Let Ri = (Ai, Bi) for all iJ, then ∪i Ri = ∪i(Ai, Bi) = (∪i Ai,i Bi). f−1(∪i Ri) = f(( ∪i Ai,i Bi)) = (f−1(∪i Ai), f−1(∩i Bi)) = (∪i f−1(Ai), ∩i f−1(Bi)) = ∪i (f−1(Ai), f−1(Bi))= ∪i f−1(Ri).

Theorem 3.16

Let f : XY be a mapping. For any IF rough set P in X and R in Y , we have the following:

  • Rf(f−1(R)).

  • Pf−1(f(P)).

Proof

(1) Let R = (C, D) be an IF rough set for some fuzzy rough sets C, D. Then f−1(R) = (f−1(C), f−1(D)). We have

f(f-1(R))(y)=({f-1(C)(x)},{f-1(D)(x)})for all xf-1(y)={(0_,1_),if         f-1(y)=,({C(f(x))},{D(f(x))})for all xf-1(y),if         f-1(y)(C(y),D(y))=R(y).

Hence Rf(f−1(R)).

(2) Let P = (A, B) be an IF rough set. Then

f-1(f(P))(x)=f(P)(f(x))=f((A,B))(f(x))=({A(t)tf-1(f(x))},{B(t)tf-1(f(x))})(A(x),B(x))=P(x).

Hence Pf−1(f(P)).

In general, the equality does not hold in the above theorem. It can be shown by the following two examples.

Example 3.17

Let X = {1, 2, 3}, Y = {a, b, c}. And let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Consider an IF rough set R = (C, D) = (({a/0.4, b/0.3}, {a/0.4, b/0.4}), ({a/0.2, b/0.4}, {a/0.3, b/0.4})). Then we have R(b) = ((0.3, 0.4), (0.4, 0.4)). But f(f−1(R))(b) = ((0, 1), (1, 0)). Thus Rf(f−1(R)).

Example 3.18

Let X = {1, 2, 3}, Y = {a, b, c}. And let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Consider an IF rough set P = (A, B) = (({1/0.4, 2/0.3}, {1/0.4, 2/0.4}), ({1/0.2, 2/0.4}, {1/0.3, 2/0.4})). Then we have P(3) = ((0, 1), (1, 0)). But f−1(f(P))(3) = ((0.3, 0.4), (0.4, 0.4)). Thus Pf−1(f(P)).

Theorem 3.19

Let f : XY be a mapping. For any IF rough sets P, Q in X, we have f(PQ) = f(P) ∪ f(Q).

Proof

Let P = (A, B), Q = (C, D) be two IF rough sets.

f(PQ)(y)=f((A,B)(C,D))(y)=f((AC,BD))(y)=({(AC)(x)},{(BD)(x)})for all xf-1(y)=({(A(x)C(x)},{(B(x)D(x)})for all xf-1(y)=(({A(x)xf-1(y)})({C(x)xf-1(y)}),({B(x)xf-1(y)})({D(x)xf-1(y)}))=({A(x)xf-1(y)},{B(x)xf-1(y)})({C(x)xf-1(y)},{D(x)xf-1(y)})=f(P)(y)f(Q)(y)=(f(P)f(Q))(y).

Corollary 3.20

If P1, P2, · · ·, Pn are IF rough sets in X, then

f(P1P2Pn)=f(P1)f(P2)f(Pn).

Theorem 3.21

Let f : XY be a mapping. For any IF rough sets P, Q in X, we have f(PQ) ⊆ f(P) ∩ f(Q).

Proof

Let P = (A, B), Q = (C, D) for some fuzzy rough sets A, B, C and D.

f(PQ)(y)=f((A,B)(C,D))(y)=f((AC,BD))(y)=({(AC)(x)},{(BD)(x)})for all xf-1(y)=({(A(x)C(x)},{(B(x)D(x)})for all xf-1(y)(({A(x)xf-1(y)})({C(x)xf-1(y)}),({B(x)xf-1(y)})({D(x)xf-1(y)}))=({A(x)xf-1(y)},{B(x)xf-1(y)})({C(x)xf-1(y)},{D(x)xf-1(y)})=f(P)(y)f(Q)(y)=(f(P)f(Q))(y).

Hence f(PQ) ⊆ f(P) ∩ f(Q).

Corollary 3.22

If f : XY is one-one, then clearly f(PQ) = f(P) ∩ f(Q).

But in general f(PQ) ≠ f(P) ∩ f(Q), which can be shown by the following example.

Example 3.23

Let X = {x, y} and Y = {a} and a mapping f : XY be defined by f(x) = f(y) = a. Let P = (({x/0.4, y/0.3}, {x/0.4, y/0.4}), ({x/0.2, y/0.4}, {x/0.3, y/0.4})), and Q=(({x/0.3, y/0.4}, {x/0.3, y/0.4}), ({x/0.3, y/0.2}, {x/0.3, y/0.3})). Clearly P, Q are IF rough sets in X. And PQ = (({x/0.3, y/0.3}, {x/0.3, y/0.4}), ({x/0.3, y/0.4}, {x/0.3, y/0.4})).

Thus

f(PQ)=(({a/0.3},{a/0.4}),({a/0.3},{a/0.3})),f(P)=(({a/0.4},{a/0.4}),({a/0.2},{a/0.3})),f(Q)=(({a/0.4},{a/0.4}),({a/0.2},{a/0.3})).

So f(P)∩f(Q) = (({a/0.4}, {a/0.4}), ({a/0.2}, {a/0.3})). Hence f(PQ) ≠ f(P) ∩ f(Q).

Theorem 3.24

Let f : XY be a mapping. For any IF rough set Pi, iJ, we have the following:

  • f(∪iJ Pi) = ∪iJ (Pi).

  • f(∩iJ Pi) ⊆ ∩iJ (Pi).

Proof

(1) Let Pi = (Ai, Bi) where Ai, Bi are fuzzy rough sets for all iJ. Then

f(iJPi)(y)=f((iJAi,iJBi))(y)=(f(iJAi)(y),f(iJBi¯)¯(y))=({iJAi(x)},{iJBi(x)})for all xf-1(y)=(iJ{Ai(x)},iJ{Bi(x)})for all xf-1(y)=iJ({Ai(x)},{Bi(x)})for all xf-1(y)=iJ(f(Ai)(y),f(Bi¯)¯(y))=iJ(f(Pi))(y).

(2) Let Pi = (Ai, Bi) where Ai, Bi are fuzzy rough sets for all iJ. Then

f(iJPi)(y)=f((iJAi,iJBi))(y)=(f(iJAi)(y),f(iJBi¯)¯(y))=({iJAi(x)},{iJBi(x)})for all xf-1(y)(iJ{Ai(x)},iJ{Bi(x)}for all xf-1(y)=iJ({Ai(x)},{Bi(x)})for all xf-1(y)=iJ(f(Ai)(y),f(Bi¯)¯(y))=iJ(f(Pi))(y).

The properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. This is because of the unnaturalness of the definition of fuzzy rough sets. Hence this flaw is critical in expanding the related theory. In order to overcome this unnaturalness, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. This new approach enables us to manipulate fuzzy rough sets more simply and easily. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

No potential conflict of interest relevant to this article was reported.

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Sang Min Yun received his Ph.D. degree from Chungbuk National University in 2015. 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 his 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 University from 2003 to 2004. His research interests include general topology and fuzzy topology. He is a member of KIIS and KMS.

E-mail: sjl@cbnu.ac.kr


Article

Original Article

International Journal of Fuzzy Logic and Intelligent Systems 2020; 20(2): 129-137

Published online June 25, 2020 https://doi.org/10.5391/IJFIS.2020.20.2.129

Copyright © The Korean Institute of Intelligent Systems.

New Approach to Intuitionistic Fuzzy Rough Sets

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 15, 2019; Revised: May 2, 2020; Accepted: May 5, 2020

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted noncommercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. In order to overcome this unnaturalness, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

Keywords: Intuitionistic fuzzy topology, Intuitionistic fuzzy rough sets

1. Introduction

The notion of fuzzy sets was first introduced by Zadeh [1]. After that, many studies attempted to generalize the fuzzy set by using various approaches. Pawlak [2] introduced the concept of rough sets, Nanda and Majumda [3] and Coker [4] proposed the idea of fuzzy rough sets. Atanassov [5] introduced the idea of intuitionistic fuzzy sets. All these concepts provide useful means of expressing vagueness in real environments.

Combining the concepts of fuzzy rough sets and intuitionistic fuzzy sets, Samanta and Mondal [6] proposed the idea of intuitionistic fuzzy rough sets. By introducing further generalization, the authors [7, 8] also conducted a study on the intuitionistic fuzzy bitopology and intuitionistic smooth bitopology. Moreover, the categorical properties of the intuitionsitc fuzzy topological spaces were studied by the same research group [911].

Many attempts at combining fuzziness and roughness have been made. In [12], the measure of fuzziness in rough sets is provided and studied. In [13] a general framework for the study of fuzzy rough sets is presented, in which both constructive and axiomatic approaches are made. Lower and upper approximations of intuitionistic fuzzy sets with respect to an intuitionistic fuzzy approximation space are first defined by Zhou et al. [14, 15]. Several important properties of intuitionistic fuzzy approximation operators are examined by many researchers including us [1619].

However, the properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. This is because of the unnaturalness of the definition of fuzzy rough sets. For example, the double complement of a fuzzy rough set is different from itself. The property that the double complement of a set becomes the set itself is one of the essential properties of Boolean algebra. Hence this flaw is critical in expanding the related theory. In order to overcome this unnaturalness, we need a new approach to intuitionistic fuzzy rough sets.

In this paper, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. This new approach enables us to manipulate fuzzy rough sets more simply and easily. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

2. Fuzzy Rough Sets

In [3], the definition of fuzzy rough sets has been introduced. The paper said:

“we shall consider ( , ) to be a rough universe where is a nonempty set and is a Boolean subalgebra of the Boolean algebra of all subsets of . Also consider a rough set with XLXU. A fuzzy rough set in X is an object of the form

A=(AL,AU),

where AL and AU are characterized by a pair of maps AL : XLL and AU : XUL with AL(x) ≤ AU(x), for all xXL, where (L,≤) is a fuzzy lattice.”

Furthermore, the complement Ā of a fuzzy rough set A = (AL, AU) is defined by (Ā)L(x) = (AU>L)′(x), ∀xXL and (Ā)U(x) = (AL<U)′(x), ∀xXU,

where AU>L(x) = AU(x), ∀xXL and

AL<U(x)={AL(x),if xXL,{AL(x)xXL},if xXU-XL.

Unfortunately, the double complement of a fuzzy rough set A is different from A, because XL and XU are different, i.e., XLXU, The property that double complement of a set becomes the set itself is one of the essential properties of Boolean algebra. Hence this flaw is critical in expanding the related theory. Thus we are going to introduce the new definition of a fuzzy rough set by weakening the condition of the old definition. Then the properties we obtain in this paper are different from the results in the above paper.

Definition 2.1

Ler X be an underlying set and (L, ≤) a fuzzy lattice. A fuzzy rough set in X is an object of the form

A=(AL,AU),

where AL and AU are defined by a pair of maps AL : XL and AU : XL with AL(x) ≤ AU(x), for all xX. Furthermore 0 = (0L, 0U) the null fuzzy rough set and 1 = (1L, 1U) the whole fuzzy rough set in X,

Definition 2.2

For any two fuzzy rough sets A = (AL, AU) and B = (BL, BU) of X, we define the following:

  • AB iff AL(x) ≤ BL(x) and AU(x) ≤ BU(x) for all xX.

  • A = B iff AB and BA.

If {Ai | iJ} is a family of fuzzy rough sets in X, with Ai = ((Ai)L, (Ai)U), then we define the following:

  • (∪i Ai)L(x) = ∨ (Ai)L(x) and (∪i Ai)U(x) = ∨ (Ai)U(x) for all xX.

  • (∩i Ai)L(x) = ∧ (Ai)L(x) and (∩i Ai)U(x) = ∧ (Ai)U(x) for all xX.

Definition 2.3

The complement Ā = ((Ā)L, (Ā)U) of a fuzzy rough set A = (AL, AU) in X is defined by (Ā)L(x) = (AU(x))′ and (Ā)U(x) = (AL(x))′ for all xX.

For any fuzzy rough set A = (AL, AU) and for any xX,

(A¯)L(x)=(AU(x))(AL(x))=(A¯)U(x).

So, Ā is again a fuzzy rough set. Furthermore, for any fuzzy rough set A = (AL, AU), we have the double complement property, i.e., (A¯)¯=A.

Theorem 2.4

Let A be any fuzzy rough set in X, then we have the following properties:

  • 0A1.

  • (0_)¯=1,(1_)¯=0_.

Theorem 2.5

If A, B, C, D and Bi, iJ are fuzzy rough sets in X, then we have the following properties:

  • AB and CD implies

    ACBD and ACBD.

  • AB and BC implies AC.

  • ABA, BAB.

  • A ∪ (∩i Bi) = ∩i(ABi), A ∩ (∪i Bi) = ∪i(ABi).

  • AB implies Ā.

  • iBi¯=iBi¯,iBi¯=iBi¯

Definition 2.6

Let f : XY be a mapping and A = (AL, AU) a fuzzy rough set in X. The image f(A) = ((f(A))L, (f(A))U) of A under f is defined by, for any yY

(f(A))L(y)=f(AL)(y)={0_L,if f-1(y)=,{AL(x)xf-1(y)},otherwise,(f(A))U(y)=f(AU)(y)={0_U,if f-1(y)=,{AU(x)xf-1(y)},otherwise.

Let B = (BL, BU) be a fuzzy rough set in Y . Then the inverse image f−1(B) = ((f−1(B))L, (f−1(B))U) of B under f is defined by, for any yY

(f-1(B))L(x)=f-1(BL)(x)=BL(f(x)),(f-1(B))U(x)=f-1(BU)(x)=BU(f(x)).

The image of A under f is well-defined, because

f(AL)(y)={AL(x)xf-1(y)}{AU(x)xf-1(y)}=f(AU)(y).

Moreover, the inverse image of B under f is well-defined, because

f-1(BL)(x)=BL(f(x))BU(f(x))=f-1(BU)(x).

Theorem 2.7

If f : XY is a mapping, then for any fuzzy rough sets A and B in X, we have the following:

  • AB implies f(A) ⊆ f(B).

  • If f is surjective, then f(A¯)f(A)¯.

  • f(∪iJ Ai) = ∪iJ f(Ai).

  • f(∩iJ Ai) ⊆ ∩iJ f(Ai).

Proof

(1) The proof is clear.

(2) Let f be surjective.

(f(A¯))L(y)=f((A¯)L)(y)={(A¯)L(x)xf-1(y)}={(AU(x))xf-1(y)}{(AU(x))xf-1(y)}=({(AU(x))xf-1(y)})=(f(AU)(y))=((f(A))U(y))=(f(A)¯)L(y).

The proof of the upper part is similar.

(3)

(f(iJAi))L(y)=f((iJAi)L)(y)={(iJAi)L(x)xf-1(y)}={iJ(Ai)L(x)xf-1(y)}=iJ{xf-1(y)(Ai)L(x)}=iJ{f((Ai)L)(y)}=iJ(f(Ai))L(y).

The proof of the upper part is similar.

(4)

(f(iJAi))L(y)=f((iJAi)L)(y)={(iJAi)L(x)xf-1(y)}={iJ(Ai)L(x)xf-1(y)}iJ{xf-1(y)(Ai)L(x)}=iJ{f((Ai)L)(y)}=iJ(f(Ai))L(y).

The proof of the upper part is similar.

The surjectiveness is essential in (2) of the above theorem. It can be shown by the following example.

Example 2.8

Let X = {1, 2, 3}, Y = {a, b, c}. Let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Then f is not surjective. Consider a fuzzy rough set

A=({1/0.4,2/0.3},{1/0.4,2/0.4}).

Then

A¯=({1/0.6,2/0.6},{1/0.6,2/0.7}).

So

f(A¯)=({a/0.6,b/0,c/0.6},{a/0.6,b/0,c/0.7}).

But

f(A)=({a/0.4,b/0,c/0.3},{a/0.4,b/0,c/0.4}),

and so

f(A)¯=({a/0.6,b/1,c/0.6},{a/0.6,b/1,c/0.7}).

Hence f(A¯)f(A)¯.

Remark 2.9

If f : XY and g : YZ are mappings, then for any fuzzy rough set C in Z, the inverse image of C under gf is defined by (gf)−1(C) = f−1(g−1(C)) where gf is the composition of g and f.

Theorem 2.10

If f : XY is a mapping, then for all fuzzy rough sets B, C, Bi, iJ in Y , we have the following:

  • f-1(B¯)=f-1(B)¯.

  • BC implies f−1(B) ⊆ f−1(C).

  • f−1(∩i Bi) = ∩i f−1(Bi).

  • f−1(∪i Bi) = ∪i f−1(Bi).

Proof

(1) (f-1(B¯))L(x)=f-1((B¯)L)(x)=(B¯)L(f(x))=(BU(f(x)))=(f-1(BU)(x))=((f-1(B))U(x))=(f-1(B)¯)L(x). The proof of upper part is similar. Hence f-1(B¯)=f-1(B)¯.

(2) Since BC, BL(y) ≤ CL(y) for all yY . (f−1(B))L (x) = f−1(BL)(x) = BL(f(x)) ≤ CL(f(x)) = f−1(CL)(x) = (f−1(C))L(x). The proof of upper part is similar. Hence BC implies f−1(B) ⊆ f−1(C).

(3) (f−1(∩i Bi))L(x) = f−1((∩i Bi)L)(x) = (∩i Bi)L (f(x)) = ∧i(Bi)L(f(x)) = ∧i f−1((Bi)L)(x) = ∧i(f−1(Bi))L (x) = (∩i f−1(Bi))L(x). The proof of upper part is similar. Hence f−1(∩i Bi) = ∩i f−1(Bi).

(4) Similarly.

Theorem 2.11

If f : XY is a mapping, then for any fuzzy rough set A in X and B in Y , we have the following:

  • Bf(f−1(B)).

  • Af−1(f(A)).

3. Intuitionistic Fuzzy Rough Sets

Definition 3.1

If A = (AL, AU) and B = (BL, BU) are two fuzzy rough sets in X with BĀ, then the ordered pair (A, B) is called an intuitionistic fuzzy rough set (briefly IF rough set) in X. The condition BĀ is called the intuitionistic condition (briefly IC).

Remark 3.2

In [6], the IC is that BĀ and A. But by the definition of IF rough set in this paper, we get the followings. If BĀ, then BL(x) ≤ (Ā)L(x) = (AU(x))′. i.e. BL(x) ≤ (AU(x))′ and BU(x) ≤ (AL(x))′. So AU(x) ≤ (BL(x))′ and AL(x) ≤ (BU(x))′. Thus A. Moreover, the converse is clear. Therefore BĀ iff A.

Definition 3.3

0* = (0, 1) and 1* = (1, 0) are called the null IF rough set and the whole IF rough set in X, respectively. Clearly 0*¯=1* and 1*¯=0*.

We denote by IFRS(X) the collection of all IF rough sets in X. Usually we shall use letters X, Y, Z, … to denote sets, letters A, B, C, … to denote fuzzy rough sets, and letters P, Q, R, … to denote IF rough sets.

Definition 3.4

53.4 Let P = (A, B) = ((AL, AU), (BL, BU)) and Q = (C, D) = ((CL, CU), (DL, DU)) be two IF rough sets in X. We define the following:

  • PQ iff AC and BD.

  • P = Q iff PQ and PQ.

  • The complement P̄ of P = (A, B) in X, is defined by = (B, A).

  • For IF rough sets Pi = (Ai, Bi) in X, iJ, we define

iJPi=(iJAi,iJBi),iJPi=(iJAi,iJBi).

Remark 3.5

Let Pi = (Ai, Bi) be an IF rough set for all i ∈ Δ, then AiBi¯ for all i ∈ Δ. For any k ∈ Δ, we have

AkBk¯iΔBi¯=iΔBi¯.

Thus iΔAiiΔBi¯. Hence ∪i∈ΔPi = (∪i∈ΔAi, ∩i∈ΔBi) is also an IF rough set.

Similarly, for any k ∈ Δ, iAiAkBk¯. So iAiiBi¯=iBi¯. Hence ∩i Pi = (∩i Ai, ∪i Bi) is also an IF rough set.

Theorem 3.6

Let P = (A, B), Q = (C, D), R = (E, F) and Pi = (Ai, Bi), iJ be IF rough sets in X, then we have the following:

  • PP = P = PP.

  • PQ = QP, PQ = QP

  • (PQ) ∩ R = P ∩(QR), (PQ) ∪ R = P ∪(QR).

  • PQP, QPQ.

  • PQ and QR implies PR.

  • PiQ, ∀iJ implies ∪i PiQ.

  • QPi, ∀iJ implies Q ⊆ ∩i Pi.

  • Q ∪ (∩i Pi) = ∩i(QPi).

  • Q ∩ (∪i Pi) = ∪i(QPi).

  • (P¯)¯=P.

  • PQ iff .

  • iPi¯=iPi¯,iPi¯=iPi¯.

Proof

(4) PQ = (AC, BD) ⊆ (A, B) ⊆ (AC, BD) = PQ.

(12) iPi¯=(iAi,iBi¯=(iBi,iAi)=i(Bi,Ai)=iPi¯.

Definition 3.7

Let f : XY be a mapping. Let P = (A, B) = ((AL, AU), (BL, BU)) be an IF rough set in X. Then we define an image of P under f by

f(P)=(f(A),f(B¯)¯),

where f(A) = (f(AL), f(AU)) and f(B¯)¯=((f(B¯)¯)L,(f(B¯)¯)U) is defined by (f(B¯)¯)L=f((B¯)L)¯ and (f(B¯)¯)U=f((B¯)U)¯.

We already know that f(A)(y) = (f(AL)(y), f(AU)(y)) = (∨{AL(x) | xf−1(y)}, ∨{AU(x) | xf−1(y)}) for any fuzzy rough set A in X and a function f : XY .

Theorem 3.8

By the above definition, we have the following conclusion:

For any fuzzy rough set B in X and a function f : XY , We have f(B¯)¯(y)=({(BL(x))xf-1(y)},{(BU(x))xf-1(y)}), ∧{(BU(x) | xf−1(y)}).

Prooff((B¯)L)(y)={(B¯)L(x)xf-1(y)}={(BU(x))xf-1(y)}=({(BU(x))xf-1(y)}),f((B¯)U)(y)={(B¯)U(x)xf-1(y)}={(BL(x))xf-1(y)}=({(BL(x))xf-1(y)}).

So, f()(y) = ((∧{(BU(x))| xf−1(y)})′, (∧{(BL(x)) | xf−1(y)})′). Thus, f(B¯)¯(y)=({(BL(x))xf-1(y)},{(BU(x))xf-1(y)}).

Remark 3.9

We will prove that the image of any IF rough set P = (A, B) satisfies the IC.

(f(A))L(y)=f(AL)(y)={AL(x)xf-1(y)}{(B¯)L(x)xf-1(y)}={(BU(x))xf-1(y)}=({(BU(x))xf-1(y)})=((f(B¯)¯)U(y))=((f(B¯)¯)¯)L(y).

The proof of the upper part is similar. So, if P = (A, B) be an IF rough set in X, then f(P)=(f(A),f(B¯)¯) is also an IF rough set in Y .

Remark 3.10

By the above theorem, we have the following conclusion:

For any IF rough set P = (A, B) in X and any map f : XY , The image f(P) = (f(A), f(B)) of P under f is actually expressed as follows:

For any yY

f(P)(y)=({A(x)xf-1(y)},{B(x)xf-1(y)}),

where

{A(x)xf-1(y)}={0_,if         f-1(y)=,({AL(x)xf-1(y)},{AU(x)xf-1(y)}),if         f-1(y),{B(x)xf-1(y)}={1_,if         f-1(y)=,({BL(x)xf-1(y)},{BU(x)xf-1(y)}),if         f-1(y).

Definition 3.11

Let f : XY be a mapping and Q = (C, D) = ((CL, CU), (DL, DU)) an IF rough set in Y . Then we define an inverse image of Q under f by

f-1(Q)=(f-1(C),f-1(D)),

where f−1(C) = ((f−1(C))L, (f−1(C))U) = (f−1(CL), f−1(CU)) and f−1(D) = ((f−1(D))L, (f−1(D))U) = (f−1(DL), f−1(DU)).

Remark 3.12

f-1(CL)(x)=CL(f(x))(D¯)L(f(x))=f-1((D¯)L)(x)=(f-1(D)¯)L(x). The upper part is also hold. Hence, the inverse image of an IF rough set under f is also an IF rough set.

Theorem 3.13

Let f : XY be a mapping. For any IF rough set P and Q on X, we have the following:

  • PQ implies f(P) ⊆ f(Q).

  • If f is surjective, then f(P¯)f(P)¯.

Proof

(1) It is clear.

(2) Let P = (A, B) be an IF rough set. Then f(P¯)=f((B,A))=(f(B),f(A¯)¯). Since f(P)=f((A,B))=(f(A),f(B¯)¯),f(P)¯=(f(B¯)¯,f(A)). Consider (f(B))L(y)=f(BL)(y)={BL(x)xf-1(y)}{BL(x)xf-1(y)}=(f(B¯)¯)L(y). And the proof of the upper part is similar. Thus f(B)f(B¯)¯. Similarly, f(A)f(A¯)¯. Therefore f(P¯)=(f(B),f(A¯)¯)(f(B¯)¯,f(A))=f(P)¯.

Definition 3.14

If f : XY and g : YZ are mappings, then the inverse image of IF rough set W in Z under (gf) is defined by (gf) −1(W) = f−1(g−1(W)) where gf is the composition of g and f.

Theorem 3.15

Let f : XY be a mapping. For any IF rough set R, S and Ri, iJ in Y , we have the following:

  • f-1(R¯)=f-1(R)¯.

  • RS implies f−1(R) ⊆ f−1(S).

  • f−1(∪i Ri) = ∪i f−1(Ri).

  • f−1(∩i Ri) = ∩i f−1(Ri).

Proof

(1) Let R = (A, B), then = (B, A). f-1(R¯)=(f-1(B),f-1(A))=(f-1(A),f-1(B))¯=f-1(R)¯.

(3) Let Ri = (Ai, Bi) for all iJ, then ∪i Ri = ∪i(Ai, Bi) = (∪i Ai,i Bi). f−1(∪i Ri) = f(( ∪i Ai,i Bi)) = (f−1(∪i Ai), f−1(∩i Bi)) = (∪i f−1(Ai), ∩i f−1(Bi)) = ∪i (f−1(Ai), f−1(Bi))= ∪i f−1(Ri).

Theorem 3.16

Let f : XY be a mapping. For any IF rough set P in X and R in Y , we have the following:

  • Rf(f−1(R)).

  • Pf−1(f(P)).

Proof

(1) Let R = (C, D) be an IF rough set for some fuzzy rough sets C, D. Then f−1(R) = (f−1(C), f−1(D)). We have

f(f-1(R))(y)=({f-1(C)(x)},{f-1(D)(x)})for all xf-1(y)={(0_,1_),if         f-1(y)=,({C(f(x))},{D(f(x))})for all xf-1(y),if         f-1(y)(C(y),D(y))=R(y).

Hence Rf(f−1(R)).

(2) Let P = (A, B) be an IF rough set. Then

f-1(f(P))(x)=f(P)(f(x))=f((A,B))(f(x))=({A(t)tf-1(f(x))},{B(t)tf-1(f(x))})(A(x),B(x))=P(x).

Hence Pf−1(f(P)).

In general, the equality does not hold in the above theorem. It can be shown by the following two examples.

Example 3.17

Let X = {1, 2, 3}, Y = {a, b, c}. And let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Consider an IF rough set R = (C, D) = (({a/0.4, b/0.3}, {a/0.4, b/0.4}), ({a/0.2, b/0.4}, {a/0.3, b/0.4})). Then we have R(b) = ((0.3, 0.4), (0.4, 0.4)). But f(f−1(R))(b) = ((0, 1), (1, 0)). Thus Rf(f−1(R)).

Example 3.18

Let X = {1, 2, 3}, Y = {a, b, c}. And let f : XY be a mapping with f(1) = a, f(2) = f(3) = c. Consider an IF rough set P = (A, B) = (({1/0.4, 2/0.3}, {1/0.4, 2/0.4}), ({1/0.2, 2/0.4}, {1/0.3, 2/0.4})). Then we have P(3) = ((0, 1), (1, 0)). But f−1(f(P))(3) = ((0.3, 0.4), (0.4, 0.4)). Thus Pf−1(f(P)).

Theorem 3.19

Let f : XY be a mapping. For any IF rough sets P, Q in X, we have f(PQ) = f(P) ∪ f(Q).

Proof

Let P = (A, B), Q = (C, D) be two IF rough sets.

f(PQ)(y)=f((A,B)(C,D))(y)=f((AC,BD))(y)=({(AC)(x)},{(BD)(x)})for all xf-1(y)=({(A(x)C(x)},{(B(x)D(x)})for all xf-1(y)=(({A(x)xf-1(y)})({C(x)xf-1(y)}),({B(x)xf-1(y)})({D(x)xf-1(y)}))=({A(x)xf-1(y)},{B(x)xf-1(y)})({C(x)xf-1(y)},{D(x)xf-1(y)})=f(P)(y)f(Q)(y)=(f(P)f(Q))(y).

Corollary 3.20

If P1, P2, · · ·, Pn are IF rough sets in X, then

f(P1P2Pn)=f(P1)f(P2)f(Pn).

Theorem 3.21

Let f : XY be a mapping. For any IF rough sets P, Q in X, we have f(PQ) ⊆ f(P) ∩ f(Q).

Proof

Let P = (A, B), Q = (C, D) for some fuzzy rough sets A, B, C and D.

f(PQ)(y)=f((A,B)(C,D))(y)=f((AC,BD))(y)=({(AC)(x)},{(BD)(x)})for all xf-1(y)=({(A(x)C(x)},{(B(x)D(x)})for all xf-1(y)(({A(x)xf-1(y)})({C(x)xf-1(y)}),({B(x)xf-1(y)})({D(x)xf-1(y)}))=({A(x)xf-1(y)},{B(x)xf-1(y)})({C(x)xf-1(y)},{D(x)xf-1(y)})=f(P)(y)f(Q)(y)=(f(P)f(Q))(y).

Hence f(PQ) ⊆ f(P) ∩ f(Q).

Corollary 3.22

If f : XY is one-one, then clearly f(PQ) = f(P) ∩ f(Q).

But in general f(PQ) ≠ f(P) ∩ f(Q), which can be shown by the following example.

Example 3.23

Let X = {x, y} and Y = {a} and a mapping f : XY be defined by f(x) = f(y) = a. Let P = (({x/0.4, y/0.3}, {x/0.4, y/0.4}), ({x/0.2, y/0.4}, {x/0.3, y/0.4})), and Q=(({x/0.3, y/0.4}, {x/0.3, y/0.4}), ({x/0.3, y/0.2}, {x/0.3, y/0.3})). Clearly P, Q are IF rough sets in X. And PQ = (({x/0.3, y/0.3}, {x/0.3, y/0.4}), ({x/0.3, y/0.4}, {x/0.3, y/0.4})).

Thus

f(PQ)=(({a/0.3},{a/0.4}),({a/0.3},{a/0.3})),f(P)=(({a/0.4},{a/0.4}),({a/0.2},{a/0.3})),f(Q)=(({a/0.4},{a/0.4}),({a/0.2},{a/0.3})).

So f(P)∩f(Q) = (({a/0.4}, {a/0.4}), ({a/0.2}, {a/0.3})). Hence f(PQ) ≠ f(P) ∩ f(Q).

Theorem 3.24

Let f : XY be a mapping. For any IF rough set Pi, iJ, we have the following:

  • f(∪iJ Pi) = ∪iJ (Pi).

  • f(∩iJ Pi) ⊆ ∩iJ (Pi).

Proof

(1) Let Pi = (Ai, Bi) where Ai, Bi are fuzzy rough sets for all iJ. Then

f(iJPi)(y)=f((iJAi,iJBi))(y)=(f(iJAi)(y),f(iJBi¯)¯(y))=({iJAi(x)},{iJBi(x)})for all xf-1(y)=(iJ{Ai(x)},iJ{Bi(x)})for all xf-1(y)=iJ({Ai(x)},{Bi(x)})for all xf-1(y)=iJ(f(Ai)(y),f(Bi¯)¯(y))=iJ(f(Pi))(y).

(2) Let Pi = (Ai, Bi) where Ai, Bi are fuzzy rough sets for all iJ. Then

f(iJPi)(y)=f((iJAi,iJBi))(y)=(f(iJAi)(y),f(iJBi¯)¯(y))=({iJAi(x)},{iJBi(x)})for all xf-1(y)(iJ{Ai(x)},iJ{Bi(x)}for all xf-1(y)=iJ({Ai(x)},{Bi(x)})for all xf-1(y)=iJ(f(Ai)(y),f(Bi¯)¯(y))=iJ(f(Pi))(y).

4. Conclusion

The properties of the intuitionistic fuzzy rough sets are very complicated and inadequate in the sense of the extension of intuitionistic properties. This is because of the unnaturalness of the definition of fuzzy rough sets. Hence this flaw is critical in expanding the related theory. In order to overcome this unnaturalness, we introduce a new definition of intuitionistic fuzzy rough sets and investigate important properties about the image and inverse image of an intuitionistic rough sets under a mapping. This new approach enables us to manipulate fuzzy rough sets more simply and easily. All the results obtained from this new definition are different from the results in other papers, and will be proven useful in expanding the related theory.

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