The lattice and matroid representations of definable sets in generalized rough sets based on relations

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The lattice and matroid representations of definable sets in generalized rough sets based on relations Zhaohao Wang, Qinrong Feng, Hong Wang PII: DOI: Reference:

S0020-0255(19)30141-0 https://doi.org/10.1016/j.ins.2019.02.034 INS 14304

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Please cite this article as: Zhaohao Wang, Qinrong Feng, Hong Wang, The lattice and matroid representations of definable sets in generalized rough sets based on relations, Information Sciences (2019), doi: https://doi.org/10.1016/j.ins.2019.02.034

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The lattice and matroid representations of definable sets in generalized rough sets based on relations Zhaohao Wang∗, Qinrong Feng, Hong Wang

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School of Mathematics and Computer Science, Shanxi Normal University, Shanxi, Linfen, 041000, P.R. China

Abstract

The definable set is a core concept in rough set theory. It plays an important role in the characterizations of rough sets. In this paper, we study the lattice and matroid representations of definable sets in generalized rough sets based on relations. First, we propose the lower definable lattice, consisting of all lower definable sets with set inclusion order.

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Then we give some conditions under which the lower definable lattice is distributive (or geometric, or Boolean). Furthermore, we discuss the relationship between distributive lattices and the lower definable lattices in generalized approximation spaces based on reflexive and transitive relations. On the one hand, we show that the lower definable lattice in a generalized approximation space based on reflexive and transitive relation is distributive. On the other hand, we obtain the result that a distributive lattice can induce a lower definable lattice in a generalized approximation

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space based on a reflexive and transitive relation. Finally, we investigate the combination of generalized rough sets and matroids in terms of the lower definable lattice. We show that if a lower definable lattice is a lattice of closed sets of a matroid, then it must be a open-closed set lattice of a matroid. In addition, we prove that some lower definable

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lattices are not the lattices of closed sets of matroids. These results of this paper will benefit to our understanding of the relationship between matroids and generalized rough sets based on relations.

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1. Introduction

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Keywords: Generalized rough set, Definable set, Geometric lattice, Boolean lattice, Matroid

Rough set theory, which was first formulated by Z. Pawlak [17] in the early 1980s, has been successfully applied

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to many fields, such as, in artificial intelligence, computer science, decision theory, expert systems, knowledge representation, pattern recognition, etc. In order to deal with complex practical problems, Pawlak rough set model is extended, and various generalized rough set models have been established, for example, generalized rough set based on relation [19, 21, 32, 33, 39, 40], covering rough set model [5, 10, 20, 26, 35, 37, 41–44], rough set models in multigranulation spaces [13, 14, 22, 30, 36, 38, 46], etc. Moreover, many mathematical structures have been used to ∗ Corresponding

author Email address: [email protected] (Zhaohao Wang)

Preprint submitted to Elsevier

February 12, 2019

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study rough set theory, such as, topological structures [9, 19, 41, 47], lattice structures [3, 15, 18, 31] and matroidal structures [8, 11, 12, 23, 24, 26–29]. Matroid is a mathematical abstract structure for graph theory and linear algebra. It has been used in many fields such as the combinatorial optimization and integer programming. In theory, matroids can be defined in many different ways, which provides a well-established platform for combining them with other theories [45]. Furthermore, matroids appear in various combinatorial and algebraic contexts and have been proved to be essentially important in many fields

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[25]. Thus the researches on rough set theory via matroid become an interesting and natural topic in rough set theory. In the last decade, the studies of combining rough sets with matroids made progress both in theory and in application. Deng [6] used the rank function of matroids to study rough sets. Li and Liu [12] characterized the Pawlak rough set model through matroidal approaches. Zhu and Wang [28] proposed the concept of the upper approximation number function and studied matroids through this concept. Wang et al. [29] explored the attribute reduction by matroid

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approaches. Zhu and wang [45] proposed a new algebraic structure, known as rough matroids. Li et al. [11] studied the connections of rough sets and matriods from a lattice-theoretic viewpoint. These results established a sound foundation for combining matroids with rough sets. However, they do not refer to lower definable set. In fact, we can connect matroids with generalized rough sets by means of the lower definable sets. In addition, most of the researches are focused on the relationship between Pawlak rough sets and matroids. In this paper, we do some researches on the combination of generalized rough sets based on relations and matroids through definable sets from a viewpoint of

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lattice theory.

The rest of this paper is organized as follows. In Section 2, we present some fundamental definitions and prop-

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erties of rough sets, lattices and matroids. In Section 3, we propose a representation of the family of all lower (or upper) definable sets in generalized approximation spaces in terms of the approximation operation powers. Furthermore, we provide some conditions under which two different approximation spaces generate the same the family of

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lower (or upper) definable sets. In Section 4, we define the lower definable lattice, and give a lattice representation of generalized rough set model. In Section 5, we give some conditions under which the lower definable lattice is

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geometric (or Boolean), and prove that if a lower definable lattice is geometric then it must be Boolean. Then we compare generalized rough sets based on relation and matriods in terms of geometric lattices, and show that if the lower definable lattice is the lattice of closed sets of a matroid, then the closure of the matroid can be represented by

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the n power of the upper approximation operator. A concluding remark is given in the last section. 2. Basic concepts and properties In this section, we introduce some basic concepts and results on generalized rough sets, lattices and matroids, which will be used in this paper. In this paper, we always assume that U is a finite nonempty set of objects, and P(U) is the powerset of U, i.e., the

collection of all subsets of U. For X ∈ P(U), we write ∼ X for the complement of X with respect to U. 2

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2.1. Generalized Rough sets based on relation In this subsection, we shall briefly review basic concepts and results of the relation based rough sets. For more details, we refer to [19, 32–34, 39, 40]. R is a binary relation on U, that is, a subset of U 2 =U × U. R is reflexive if for each x ∈ U, (x, x) ∈ R; R is

symmetric if for all x, y ∈ U, (x, y) ∈ R implies (y, x) ∈ R; R is transitive if for all x, y, z ∈ U, (x, y) ∈ R and (y, z) ∈ R R s (x) = {y ∈ U : (x, y) ∈ R},

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imply (x, z) ∈ R. R is called an equivalence relation if R is reflexive, symmetric and transitive. For x ∈ U, the set

is called the successor neighborhood of x. The set-valued operator from U to the power set P(U) R s : x → R s (x)

is called the successor neighborhood operator. Obviously, R and R s determine each other.

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Assumed that R is a binary relation on U, we refer to the ordered pair (U, R) as a generalized approximation space based on relation.

Definition 1. Let (U, R) be a generalized approximation space based on relation. The lower approximation apr and R

the upper approximation aprR of X based on the binary relation R are defined by apr (X) = {x ∈ U : R s (x) ⊆ X}, R

(1)

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aprR (X) = {x ∈ U : R s (x) ∩ X , ∅}.

For X ∈ P(U), if apr (X) = X, then X is called a lower definable set of (U, R); if aprR (X) = X, then X is called an R

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upper definable set of (U, R); if apr (X) = aprR (X) = X, then X is called a definable set of (U, R). R

Remark 1. Let (U, R) be a generalized approximation space based on relation. The family of all lower definable sets in (U, R) is denoted by D(apr ), and the family of all upper definable sets in (U, R) is denoted by D(aprR ). Therefore,

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R

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D(apr ) = X ∈ P(U) : apr (X) = X R

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and

n o D(aprR ) = X ∈ P(U) : aprR (X) = X .

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Clearly, D(apr ) ∩ D(aprR ) is the family of all definable sets in (U, R). According to Proposition 1, U ∈ D(apr ) R

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and ∅ ∈ D(aprR ), that is, D(apr ) and D(aprR ) are nonempty sets, but D(apr ) ∩ D(aprR ) may be empty set in some R

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cases.

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Proposition 1. For any binary relation R on U, its lower and upper approximation operators satisfy the following properties: for all X, Y ∈ P(U),

(L1) apr (X) =∼ aprR (∼ X),

(U1) aprR (X) =∼ apr (∼ X);

(L2) apr (U) = U,

(U2) aprR (∅) = ∅;

R

R

R

(L3) apr (X ∩ Y) = apr (X) ∩ apr (Y), R

R

R

(L4) X ⊆ Y ⇒ apr (X) ⊆ apr (Y), R

R

(L5) apr (X ∪ Y) ⊇ apr (X) ∪ apr (Y), R

R

R

(U3) aprR (X ∪ Y) = aprR (X) ∪ aprR (Y);

(U4) X ⊆ Y ⇒ aprR (X) ⊆ aprR (Y);

(U5) aprR (X ∩ Y) ⊆ aprR (X) ∩ aprR (Y). 3

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Proposition 2. For any binary relation R on U. Then (1) R is reflexive ⇔ apr (X) ⊆ X, ∀X ∈ P(U) ⇔ X ⊆ aprR (X), ∀X ∈ P(U). R     (2) R is transitive ⇔ apr (X) ⊆ apr apr (X) , ∀X ∈ P(U) ⇔ aprR aprR (X) ⊆ aprR (X), ∀X ∈ P(U) R

R

R

2.2. Lattices In this subsection, some basic concepts and results on lattices are introduced [4, 7].

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If a binary relation on a set P is reflexive, antisymmetric and transitive, then we say the binary relation a partial ordering relation on P. A poset (P, ≤) consists of a nonempty set P and a partial ordering relation ≤ on P. Most of the time, we shall say that P is a poset. Let S ⊆ P, if there is a least element in the set {x ∈ P : y ≤ x, ∀y ∈ S }, it is

called the supremum of S and denoted by ∨S ; dually, ∧S can be defined. Particularly, we write x ∨ y for ∨{x, y} and x ∧ y for ∧{x, y}, the join and meet of x and y.

We need to be able to recognize when two posets, P and Q, are ‘essentially the same’. We say that P and Q

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are order-isomorphic, and write P  Q, if there exists a map ϕ from P onto Q such that x ≤ y in P if and only if ϕ(x) ≤ ϕ(y) in Q. Then ϕ is called an order-isomorphism.

Definition 2. (The dual of a poset) Let (P, ≤) be a poset. We can form a new poset Pδ (the dual) by defining x ≤ y to hold in Pδ if and only if y ≤ x holds in P.

Definition 3. (Lattice) Let (L, ≤) be a poset.

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In above definition, if P is finite, we obtain a diagram for Pδ simply by ‘turning upside down’ a diagram for P.

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(i) If x ∨ y and x ∧ y exist for all x, y ∈ L, then L is called a lattice;

(ii) If ∨S and ∧S exist for all S ⊆ L, then L is called a complete lattice.

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It is easy to check that the lattice L is complete when L is a finite set. Definition 4. Let U be a set and ∅ , L ⊆ P(U). If ∩i∈I Xi ∈ L for every nonempty family of {Xi : i ∈ I} ⊆ L, where I

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is an index set, then L is referred to as a closure system.

Proposition 3. Let L be a family of subsets of U, ordered by inclusion. If L is a closure system, and U ∈ L, then L

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is a complete lattice in which

∧i∈I Xi = ∩i∈I Xi and ∨i∈I Xi = ∩{Y ∈ P(U) : ∪i∈I Xi ⊆ Y}.

Remark 2. If U is a finite set, then the condition, ∩i∈I Xi ∈ L for {Xi : i ∈ I} ⊆ L, in Definition 4 is equivalent to the condition: ∀X, Y ∈ L, X ∩ Y ∈ L.

We now introduce some special classes of lattices. Let L be a poset and x, y ∈ L. If x ≤ y but x , y, we write

x < y. If x < y but there is no element z of L such that x < z < y, then we say that y covers x and write x ≺ y. X ⊆ L 4

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q

q

a c

b

a

c

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b

p

p (a) The lattice N5

(b) The lattice M3 Figure 1: The lattices M3 and N5

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is a sublattice of the lattice L if x ∨ y, x ∧ y ∈ X (∀x, y ∈ X). Note that we consider only finite lattices in this paper,

thus for a lattice L, ∨L(denoted by 1) and ∧L(denoted by 0) exist. An element x is called an atom of L if 0 ≺ x. A(L)

is the set of all atoms of L. A lattice L is atomic if ∀x ∈ L, x = ∨{y ∈ A(L) : y ≤ x}. For x, y ∈ L, we say x is a

complement of y if x ∨ y = 1 and x ∧ y = 0. Let L be a lattice and z ∈ L. L is said to be semimodular if it satisfies the Upper Covering Condition, that is, for all x, y ∈ L,

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L is said to be modular if it satisfies

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x ≺ y ⇒ x ∨ z ≺ y ∨ z or x ∨ z = y ∨ z.

x ≥ z ⇒ x ∧ (y ∨ z) = (x ∧ y) ∨ z,

∀x, y, z ∈ L.

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L is said to be distributive if it satisfies the distributive law, x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z).

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A finite semimodular atomic lattice is a geometric lattice. If every element of a distributive lattice L with 0 and 1 has a complement, then L is a Boolean lattice.

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In [4], Remark 4.5 (3) gives the following useful results. Proposition 4. Let L be a lattice. Then (1) L is modular if and only if Lδ is, and (2) L is distributive if and only if Lδ is.

It is well known that the lattice N5 is not semimodular, and the lattices N5 and M3 are not distributive (See Figure 1). The following conclusion is the famous M3 − N5 Theorem. Proposition 5. Let L be a lattice. Then the following statements hold. (i) L is modular if and only if it dose not contain a sublattice isomorphic to N5 . 5

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(ii) A lattice L is distributive if and only if L does not contain a sublattice isomorphic to N5 or a sublattice isomorphic to M3 . Remark 3. In [7], Corollary 376 shows that if L is a lattice of finite length, then L is modular if and only if L satisfies the Upper and the Lower Covering Conditions. Clearly, finite lattice is a lattice of finite length. Thus, according to the definition of semimodular lattices, a modular lattice is semimodular when the lattice is finite.

intersections.

In [11], the characterization of a Boolean lattice is given as follows:

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Definition 5. Let U be a finite nonempty set. L ⊆ P(U) is called a lattice of sets if it is closed under unions and

Proposition 6. A lattice is a Boolean lattice if and only if it is a geometric lattice and a distributive lattice.

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2.3. Matroids

In this subsection, some basic concepts and results on matroids are introduced, and they are from [16]. A matroid M is a pair (U, I) where U is a finite nonempty set and I ⊆ P(U) satisfies the following axioms: (I1) ∅ ∈ I;

(I2) If I ∈ I and I 0 ⊆ I, then I 0 ∈ I.

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(I3) (Extension axiom) If I1 , I2 ∈ I and |I1 | < |I2 |, then there is e ∈ I2 − I1 such that I ∪ {e} ∈ I.

If M is the matroid (U, I), then M is called a matroid on U. The members of I are the independent sets of M.

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We shall often write I(M) for I. A subset of U that is not in I is called dependent. The rank function of a matroid is

a function r : P(U) → N defined by r(X) = max{|Y| : Y ⊆ X, Y ∈ I} for X ⊆ U. A point x ∈ U is said to be dependent

on a subset X of U (denoted by x ∼ X) if and only if r(X) = r(X ∪ {x}). For each X ⊆ U, let cl M (X) = {x ∈ U : x ∼ X},

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called the closure of X in (U, I). When there is no confusion, we use the symbol cl(X) for short. X is called a flat or

a closed set if cl(X) = X. The set of all closed sets is denoted by CL(M).

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Proposition 7. The closure operator of a matroid on U has the following properties: (cl1) X ⊆ cl M (X), ∀X ⊆ U;

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(cl2) X ⊆ Y =⇒ cl M (X) ⊆ cl M (Y), ∀X, Y ⊆ U;   (cl3) cl M cl M (X) = cl M (X), ∀X ⊆ U;

(cl4) y ∈ cl M (X ∪ {x}) − cl M (X) =⇒ x ∈ cl M (X ∪ {y}), ∀X ⊆ U and x ∈ U. In [16], Lemma 1.7.3 shows that CL(M) is a lattice and, for all X, Y ∈ CL(M), X ∧Y = X ∩Y and X ∨Y = cl(X ∪Y).

This lattice is referred to as the lattice of closed sets of a matroid. In fact, CL(M) is a rather special type of lattice. Proposition 8. [16] A lattice is geometric if and only if it is the lattice of closed sets of a matroid.

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3. The structures of definable sets in generalized approximation spaces based on relations In this section, we give a characterization of lower (or upper) definable sets in generalized approximation spaces, and provide the conditions under which two different relations can generate the same lower (or upper) definable sets. For simplicity, throughout this paper, let (U, R) be a fixed generalized approximation space based on relation.

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3.1. The representations of lower (or upper) definable sets in generalized approximation spaces based on relations In this subsection, we provide a representation of lower (or upper) definable sets in generalized approximation spaces in terms of powers of approximation operations.

For computing the families D(apr ) and D(aprR ) (See Remark 1), we introduce the following notations. R

   k  apr0R (X) = X, apr1R (X) = aprR (X), apr2R (X) = aprR aprR (X) , aprk+1 R (X) = apr R apr R (X) ;     apr0 (X) = X, apr1 (X) = aprR (X), apr2 (X) = apr apr (X) , aprk+1 (X) = apr aprk (X) ; R

R

R

R

R

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R

(2)

R

R

where X ⊆ U and k ∈ N, and N is the set of all natural numbers.

Example 1. Let U = {1, 2, 3, 4, 5, 6}. Let R s (1) = {1} = R s (5), R s (2) = {3}, R s (3) = {2}, R s (4) = {1, 4} and

R s (6) = {4, 5}.

    apr2R ({1}) = aprR {1, 4, 5} = {1, 4, 5, 6}, apr3R ({1}) = aprR {1, 4, 5, 6} = {1, 4, 5, 6};     aprR ({2}) = {3}, apr2R ({2}) = aprR {3} = {2}, apr3R ({2}) = aprR {2} = {3}, · · · ;   apr ({1, 4, 6}) = {1, 4, 5}, apr2 ({1, 4, 6}) = apr {1, 4, 5} = {1, 4, 5, 6}, apr3 ({1, 4, 6}) = {1, 4, 5, 6}. R

R

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R

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aprR ({1}) = {1, 4, 5},

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In Example 1, we see that apr3R ({1}) ∈ D(aprR ) and apr2 ({1, 4, 6}) ∈ D(apr ), but ∀k ∈ N, aprkR ({2}) < D(aprR ). R

R

Then under what conditions is there k ∈ N such that aprkR (X) ∈ D(aprR ) and aprk (X) ∈ D(apr )? Next, we discuss R

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this issue.

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Combining Propositions 1 and 2 and the above formula (2), it is easy to prove the following results in terms of

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mathematical induction.

Proposition 9. Given (U, R) and k ∈ N, the following properties hold. (1) aprk (U) = U, (2)

aprkR (∅) = ∅;

=∼ aprk (∼ X), aprk (X) =∼ aprkR (∼ X), ∀X ⊆ U;

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R aprkR (X)

R

R

(3) aprk (X ∩ Y) = aprk (X) ∩ aprk (Y), aprkR (X ∪ Y) = aprkR (X) ∪ aprkR (Y); R

R

R

(4) X ⊆ Y =⇒ aprkR (X) ⊆ aprkR (Y) and aprk (X) ⊆ aprk (Y), ∀X, Y ⊆ U; R

R

(5) If R is reflexive, then X ⊆ aprkR (X) and aprk (X) ⊆ X, ∀X ⊆ U. R

Lemma 1. Given (U, R), for X ⊆ U, if aprR (X) = X, then ∀k ∈ N, aprkR (X) = X.

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Proof. If k = 1, then the result is obvious. If k ≥ 2, then by the formula (2) and the condition aprR (X) = X, we have apr2R (X) = aprR (X) = X ⇒ apr3R (X) = aprR (X) = X ⇒ · · · ⇒ aprkR (X) = aprR (X) = X, that is, aprkR (X) = X. This completes the proof of the lemma. n o Proposition 10. Given (U, R) and k ∈ N, k ≥ 1, we have that D(aprR ) ⊆ aprkR (X) : X ∈ P(U) and D(apr ) ⊆ R n o aprk (X) : X ∈ P(U) .

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R

n o Proof. We first prove D(aprR ) ⊆ aprkR (X) : X ∈ P(U) . ∀Z ∈ D(aprR ), by Remark 1, aprR (Z) = Z. It follows from n o n o Lemma 1 that Z = aprkR (Z) ∈ aprkR (X) : X ∈ P(U) . Consequently, D(aprR ) ⊆ aprkR (X) : X ∈ P(U) . By duality, n o we can prove D(apr ) ⊆ aprk (X) : X ∈ P(U) . This completes the proof. R

R

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n The converse of Proposition 10 is not true. In Example 1, ∀k ∈ N, aprkR ({2}) < D(aprR ), that is to say, aprkR (X) : o X ∈ P(U) * D(aprR ). However, if R is reflexive (or transitive) and k = |U|, then this converse holds. In the following section, we denote the cardinal number of a set U by |U|.

Lemma 2. Given (U, R) and |U| = n, the following assertions hold.

n n+1 (1) If R is reflexive, then ∀X ∈ P(U), aprn+1 (X) = aprn (X). R (X) = apr R (X) and apr R

(2) If R is transitive, then ∀X ∈ P(U),

aprn+1 R (X)

=

aprnR (X)

and

R

aprn+1 (X) R

= aprn (X). R

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n n+1 (X) = aprn (X). By Proposition 9 (2), we Proof. (1) We shall prove that ∀X ∈ P(U), aprn+1 R (X) = apr R (X) and apr R

R

n need to prove only that ∀X ∈ P(U), aprn+1 R (X) = apr R (X). Since R is reflexive, it follows from Proposition 2 (1) that

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X ⊆ aprR (X). Combining Proposition 1 (U4) and the formula (2), we have

X ⊆ aprR (X) ⊆ apr2R (X) ⊆ · · · ⊆ aprRn−1 (X) ⊆ aprnR (X) ⊆ · · · .

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k Since U is finite, there exists k ∈ N such that aprk+1 R (X) = apr R (X). We choose the least number k such that

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k aprk+1 R (X) = apr R (X). Next, we prove that k ≤ n. By our choice of k, we have

k k+1 X ⊂ aprR (X) ⊂ apr2R (X) ⊂ · · · ⊂ aprk−1 R (X) ⊂ apr R (X) = apr R (X).

Thus we have aprkR (X) ≥ k. Therefore, k ≤ aprkR (X) ≤ |U| = n. We have proved that there exists k ∈ N and

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k k+2 k ≤ n such that aprk+1 R (X) = apr R (X). In addition, according to the formula (2), it is easy to verify that apr R (X) =

k+3 k+2 n+1 n aprk+1 R (X), apr R (X) = apr R (X) and so on. Hence, by k ≤ n, we conclude that apr R (X) = apr R (X).

n (2) By duality, we need to prove only that ∀X ∈ P(U), aprn+1 R (X) = apr R (X). Since R is transitive, it follows from   Proposition 2 (2) that aprR aprR (X) ⊆ aprR (X), that is, apr2R (X) ⊆ aprR (X). Combining Proposition 1 (U4) and the

formula (2), we have

3 2 · · · ⊆ aprnR (X) ⊆ aprn−1 R (X) ⊆ · · · ⊆ apr R (X) ⊆ apr R (X) ⊆ apr R (X).

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k Since U is finite, there exists k ∈ N such that aprk+1 R (X) = apr R (X). We choose the least number k such that k aprk+1 R (X) = apr R (X). Next, we prove that k ≤ n. By our choice of k, we have

k 3 2 aprk+1 R (X) = apr R (X) ⊂ apr R (X) ⊂ · · · ⊂ apr R (X) ⊂ apr R (X) ⊂ apr R (X).

(3)

If aprR (X) = U, we can get that apr2R (X) = aprR (X) = U. Now, take k = 1 ≤ |U| = n. If aprR (X) , U, then we have apr (X) < |U| = n. In addition, by the formula (3), k − 1 ≤ apr (X) . Therefore, k − 1 ≤ apr (X) < |U| = n, that R

is, k ≤ n. We have proved that there exists k ∈ N and k ≤ n such that

R

aprk+1 R (X)

aprkR (X).

In addition, according to

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R

=

k+1 k+3 k+2 the formula (2), it is easy to verify that aprk+2 R (X) = apr R (X), apr R (X) = apr R (X) and so on. Hence, by k ≤ n, n we conclude that aprn+1 R (X) = apr R (X). This completes the proof.

By the formula (2), we can easily obtain the following corollary.

n Corollary 1. Given (U, R) and |U| = n, if R is reflexive (or transitive), then ∀m ≥ n and X ⊆ U, aprm R (X) = apr R (X) R

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and aprm (X) = aprn (X). R

Theorem 1. Given (U, R) and |U| = n, the following assertions hold. n o n o (1) If R is reflexive, then D(aprR ) = aprnR (X) : X ∈ P(U) and D(apr ) = aprn (X) : X ∈ P(U) . R o n o n R (2) If R is transitive, then D(aprR ) = aprnR (X) : X ∈ P(U) and D(apr ) = aprn (X) : X ∈ P(U) . R

R

M

n o Proof. (1) By duality, we only need to prove that D(aprR ) = aprnR (X) : X ∈ P(U) . For X ∈ P(U), write Z =

ED

aprnR (X). By Lemma 2 (1) and the formula (2), we have that aprR (Z) = Z. Thus aprnR (X) = Z ∈ D(aprR ). It follows n o n that aprnR (X) : X ∈ P(U) ⊆ D(aprR ). On the other hand, by Proposition 10, we have D(aprR ) ⊆ aprnR (X) : X ∈ o n o P(U) . In summary, D(aprR ) = aprnR (X) : X ∈ P(U) . (2) Similar to the proof of (1), it is straightforward from Lemma 2 (2), the formula (2) and Proposition 10.

PT

The above result shows that every lower (or upper) definable set can be represented by aprnR (or aprn ) when R R

is reflexive (or transitive). However, if R is symmetric, the above result is not necessarily true. See the following

CE

example.

AC

Example 2. Let U = {1, 2}. R is defined by R s (1) = {2} and R s (2) = {1}. Clearly, R is symmetric. It is easy to check n o that aprnR ({1}) < D(aprR ), that is to say, here, aprnR (X) : X ∈ P(U) * D(aprR ). Remark 4. In fact, we can easily prove that if R is reflexive, then aprnR (X) is the least upper definable set containing X, while aprn (X) is the greatest lower definable set contained in X. R

3.2. The conditions under which two different approximation spaces generate the same family of lower (or upper) definable sets We have given the characterizations of lower (or upper) definable sets. However, there is another question that need be considered. In fact, two different approximation spaces may have the same lower (or upper) definable sets 9

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as shown by Example 3. In this subsection, we give the conditions under which two different approximation spaces induce the same family of lower (or upper) definable sets. Example 3. Let U = {1, 2, 3}. The relation R1 is defined by (R1 ) s (1) = {1, 2}, (R1 ) s (2) = {2, 3} and (R1 ) s (3) = {1, 3}.

The relation R2 is defined by (R2 ) s (1) = (R2 ) s (2) = (R2 ) s (3) = {1, 2, 3}. The relation R3 is defined by (R3 ) s (1) = {2, 3},  (R3 ) s (2) = {1, 3} and (R3 ) s (3) = {1, 2}. It is easy to check that D(apr ) = D(apr ) = D(apr ) = ∅, U , but the R2

R3

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R1

three relations are not equal each other. In addition, R3 is not even reflexive.

Remark 5. Here, we introduce a notation for the family of subsets on U. Let A ⊆ P(U). We denote the family  ∼ X : X ∈ A by ∼ A, that is  ∼A= ∼X:X∈A . (4) By duality of apr and aprR , it is easy to see that

AN US

R

D(apr ) =∼ D(aprR ) and D(aprR ) =∼ D(apr ). R

(5)

R

Thus it is easy to verify that the map ϕ : X 7→∼ X from the dual D(apr )δ of D(apr ) to D(aprR ) is an orderR

R

isomorphism. Consequently, D(apr )  D(aprR ). Similarly, D(aprR )  D(apr ). δ

δ

R

R

Proposition 11. Let R1 and R2 be reflexive (or transitive) relations on U and |U| = n. If ∀x ∈ U, aprnR1 ({x}) = R1

R2

M

aprnR2 ({x}), then D(apr ) = D(apr ) and D(aprR1 ) = D(aprR2 ).

Proof. By Remark 5, we know that D(apr ) = D(apr ) ⇔ D(aprR1 ) = D(aprR2 ). Thus we only need to prove R1

R2

ED

D(aprR1 ) = D(aprR2 ). ∀X ∈ D(aprR1 ), then aprR1 (X) = X. By Lemma 1, we have aprnR1 (X) = X. According to Proposition 9(3) and the conditions ∀x ∈ U, aprnR1 ({x}) = aprnR2 ({x}), we conclude

PT

X = aprnR1 (X) =

[

aprnR1 ({x}) =

x∈X

[

aprnR2 ({x}) = aprnR2 (X),

x∈X

that is, aprnR2 (X) = X. According to Theorem 1(1), this implies X ∈ D(aprR2 ). Thus D(aprR1 ) ⊆ D(aprR2 ). Similarly,

CE

we can prove that D(aprR1 ) ⊇ D(aprR2 ). In summary, D(aprR1 ) = D(aprR2 ).

Similarly, by Theorem 1(2), we can prove the result holds when R1 and R2 are transitive. This completes the

AC

proof.

In Example 3, |U| = 3. It is easy to compute that ∀x ∈ U, apr3R1 ({x}) = apr3R2 ({x}) = apr3R3 ({x}), although R1 , R2

and R3 are different from each other. Thus by Proposition 11, we conclude that they induce the same lower definable sets. This coincides with the observation of Example 3. In the following example, we illustrates that the converse of the above result is not true. Example 4. Let U = {1, 2, 3}. Now, |U| = 3. We define R1 by (R1 ) s (1) = {2, 3} and (R1 ) s (2) = (R1 ) s (3) = {1}. We  define R2 by (R2 ) s (2) = {1, 3} and (R2 ) s (1) = (R1 ) s (3) = {2}. It is easy to check that D(apr ) = D(apr ) = ∅, U . R1

10

R2

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However, apr3R1 ({1}) = {2, 3} , {2} = apr3R2 ({1}), apr3R1 ({2}) = {1} , {1, 3} = apr3R2 ({2}) and apr3R1 ({3}) = {1} , {2} =

apr3R2 ({3}). This example indicates that it is difficult to give the necessary and sufficient conditions under which two arbitrary relations can induce the same lower definable sets. But if the two relations are reflexive, we can provide this necessary and sufficient condition. Theorem 2. Let R1 and R2 be two reflexive relations on U and |U| = n. Then the following conditions are equivalent. (1) D(apr ) = D(apr ); R2

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R1

(2) D(aprR1 ) = D(aprR2 );

(3) ∀x ∈ U, aprnR1 ({x}) = aprnR2 ({x}). Proof. By Remark 5, (1)⇔(2).

(2)⇒(3) ∀x ∈ U, we shall show that aprnR1 ({x}) = aprnR2 ({x}). Write X = aprnR2 ({x}). Since R2 is reflexive, it

follows from Proposition 9 (5) that {x} ⊆ X. According to Theorem 1 (1), we have X ∈ D(aprR2 ). Using the condition

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D(aprR1 ) = D(aprR2 ), we can obtain X ∈ D(aprR1 ) and {x} ⊆ X. This implies that aprR1 (X) = X and {x} ⊆ X. Thus,

by Proposition 1 (U4), we have

{x} ⊆ X ⇒ aprR1 ({x}) ⊆ aprR1 (X) = X ⇒ apr2R1 ({x}) ⊆ aprR1 (X) = X ⇒ · · · ⇒ aprnR1 ({x}) ⊆ aprR1 (X) = X.

In summary, aprnR1 ({x}) = aprnR2 ({x}).

M

By X = aprnR2 ({x}), we have proved that aprnR1 ({x}) ⊆ aprnR2 ({x}). Similarly, we can prove that aprnR1 ({x}) ⊇ aprnR2 ({x}). (3)⇒(2) It is straightforward from Proposition 11.

ED

Especially, if R1 and R1 are two equivalence relations, then aprnR1 ({x}) = [x]R1 and aprnR2 ({x}) = [x]R2 . So the conditions ∀x ∈ U, aprnR1 ({x}) = aprnR2 ({x}) means that the corresponding equivalence classes of the two equivalence

relations are the same. As a result, when R1 and R1 are two equivalence relations, D(apr ) = D(apr ) if and only if R1

R2

PT

R1 = R2 . This is clear.

Note that Theorem 2 is not necessarily true when R is transitive (or symmetric). Example 4 illustrates that if two

CE

relations are symmetric, it is not true. The following example shows that Theorem 2 is not true when we replace the condition R being reflexive with R being transitive.

AC

Example 5. Let U = {1, 2, 3} and R1 and R2 be two relations on U, where R1 is defined by (R1 ) s (1) = (R1 ) s (2) =

(R1 ) s (3) = {1, 2}, and R2 is defined by (R2 ) s (1) = (R2 ) s (2) = (R2 ) s (3) = {1, 3}. Clearly, R1 and R2 are transitive, and

D(aprR1 ) = D(aprR2 ). However, there exists 2 ∈ U such that apr3R1 ({2}) = {1, 2, 3} , ∅ = apr3R2 ({2}), where 3 = |U|.

Remark 6. The results of this section indicate that the powers of approximation operators are a key factor to study the lower (or upper) definable sets in generalized approximation spaces.

11

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4. The lattice representations of definable sets in generalized approximation spaces based on relations In this section, we formulate the lower (or upper) definable lattices for generalized rough sets based on relations and study the relationship between the lattices of sets (or distributive lattices) and the lower definable lattices in generalized approximation spaces based on relations. In fact, (D(apr ), ⊆) is a lattice structure. We will show this point. According to (L2) and (L3) of Proposition 1, R

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Definition 4, Remark 2, and Proposition 3, we can get the following result.

Theorem 3. Given (U, R), D(apr ) is a closure system and U ∈ D(apr ). Therefore (D(apr ), ⊆) is a complete R

R

lattice such that

R

n o X ∧ Y = X ∩ Y and X ∨ Y = ∩ Z ∈ D(apr ) : X ∪ Y ⊆ Z , ∀X, Y ∈ D(apr ). R

R

(6)

R

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According to Theorem 3 and the duality of apr and aprR , we can obtain the following result. Proposition 12. Given (U, R), (D(aprR ), ⊆) is a complete lattice such that

X ∧ Y = ∪{Z ∈ D(aprR ) : Z ⊆ X ∩ Y} and X ∨ Y = X ∪ Y, ∀X, Y ∈ D(aprR ).

(7)

According to Theorem 3 and Proposition 12, we can give the following definition. R

M

Definition 6. Given (U, R), we call (D(apr ), ⊆) a lower definable lattice based on relation, and more specifically

the lower definable lattice of (U, R). Similarly, we say (D(aprR ), ⊆) an upper definable lattice of (U, R).

ED

Remark 7. In this paper, for simplicity, when we say “D(apr ) is a lattice”, it means that D(apr ), equipped with R

R

the set-theoretic inclusion order, forms a lattice, that is, (D(apr ), ⊆) is a lattice. However, we may replace D(apr ) R

R

R

lattice (D(aprR ), ⊆).

PT

by (D(apr ), ⊆) if we want to emphasize the order relation. Analogously, we can also give the same statement for the

CE

Remark 8. Theorem 3 shows that a generalized approximation space can induce a complete lattice. However, the converse is not necessarily true. That is to say, a complete lattice may not induce a generalized approximation space. This is because some lattice structures can not become the lower definable lattices (See Remark 14). However, if

AC

the lower definable lattice is distributive, the case is different. Distributive lattices play a many-faceted role in the

development of lattice theory [7]. So, in the following section, we will discuss under what conditions the lower definable lattice is distributive. According to Remark 5 and Proposition 4, we can obtain the following result. Proposition 13. Given (U, R), the following assertions hold. (1) The lattice D(apr ) is modular if and only if the lattice D(aprR ) is, and R

(2) The lattice D(apr ) is distributive if and only if the lattice D(aprR ) is. R

12

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Remark 9. L ⊆ P(U) is a lattice of sets (See Definition 5). Since P(U) is distributive, it is easy to check that L is also distributive. That is to say, a lattice of sets is distributive. In fact, conversely, every distributive lattice is isomorphic to a lattice of sets [4]. Next, we provide conditions under which the lattice D(apr ) is a lattice of sets. R

Theorem 4. Given (U, R), if R is reflexive, the lattices D(apr ) and D(aprR ) are lattices of sets on U, therefore the R

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lattices (D(apr ), ⊆) and (D(aprR ), ⊆) are distributive R

Proof. By duality, we need to prove only that D(apr ) is a lattice of sets. For all X, Y ∈ D(apr ), by Theorem 3, R

R

it is clear that X ∩ Y ∈ D(apr ). Next, we shall show that X ∪ Y ∈ D(apr ). By (L5) of Proposition 1, we have R

R

X ∪ Y = apr (X) ∪ apr (Y) ⊆ apr (X ∪ Y). On the other hand, since R is reflexive, it follows from Proposition 2 R

R

R

(1) that apr (X ∪ Y) ⊆ X ∪ Y. Hence apr (X ∪ Y) = X ∪ Y and thus X ∪ Y ∈ D(apr ). According to Definition 5, R

R

R

D(apr ) is a lattice of sets on U. It follows from Remark 9 that D(apr ) and D(aprR ) are distributive. This completes R

AN US

R

the proof.

Conversely, D(apr ) and D(aprR ) are distributive lattices, but R is not necessarily reflexive. The following R

example illustrates this point.

M

Example 6. Let U = {1, 2, 3}. Let R s (1) = {2, 3}, R s (2) = {1, 3} and R s (3) = {1, 2}. It is easy to check that D(apr ) = R n o D(aprR ) = ∅, U . Clearly, D(apr ) and D(aprR ) are distributive lattices, but R is not reflexive. R

R

example.

ED

On the other hand, D(apr ) (or D(aprR )) may not be distributive when R is not reflexive. See the following

PT

Example 7. Let U = {1, 2, 3, 4, 5}. Let R s (1) = {1}, R s (2) = {3}, R s (3) = {2}, R s (4) = {1, 4} and R s (5) = {1, 2}. It n o is easy to check that D(apr ) = ∅, {1}, {1, 4}, {2, 3}, U . Clearly, R is not reflexive. See (a) of Figure 3, D(apr ) R

R

is isomorphic to N5 . Therefore D(apr ) is not a modular lattice by Proposition 5. Certainly, D(apr ) is also not a R

R

CE

distributive lattice.

In addition, D(apr ) is not necessarily equal to D(aprR ) when D(apr ) and D(aprR ) are distributive. We give R

R

AC

the following example.

Example 8. Let U = {1, 2, 3, 4} and let R s (1) = {1, 2}, R s (2) = {2}, R s (3) = {2, 3} and R s (4) = {4}. Clearly, R is a

reflexive relation. According to Remark 1, we can get that

n o D(apr ) = ∅, {2}, {4}, {1, 2}, {2, 3}, {2, 4}, {1, 2, 3}, {1, 2, 4}, {2, 3, 4}, U , R n o D(aprR ) = ∅, {1}, {3}, {4}, {1, 3}, {1, 4}, {3, 4}, {1, 2, 3}, {1, 3, 4}, U . 13

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{1,2,3,4}

{1,2,3,4}

{1,2,3}

{1,3,4}

{2,3}

{2,4}

{1,3}

{1,4}

{3,4}

{2}

{4}

{1}

{3}

{4}

{1,2,4}

{1,2}

(a) The lattice D(apr )

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{2,3,4}

{1,2,3}

(b) The lattice D(aprR )

R

Figure 2: The Three lattices in Example 8

R

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According to Theorem 4, (D(apr ), ⊆) and (D(aprR ), ⊆) are all distributive lattices (See Figure 2), but D(apr ) , R

D(aprR ). It is worth noting that {2} ∈ D(apr ), but ∼ {2} = {1, 3, 4} < D(apr ). So D(apr ) is not a Boolean R

R

R

lattice from the definition of Boolean lattices. In the following, we shall show that D(apr ) = D(aprR ) if and only if D(apr ) is a Boolean lattice (See Proposition 15). R

R

According to Theorem 4, we know that if R is a reflexive relation on U, then the lattice D(apr ) is distributive. R

M

Conversely, assume that L is a distributive lattice, then we can obtain a reflexive relation such that L is the lower

definable lattice in the approximation space based on this relation. Since every distributive lattice is isomorphic to a

lattice of sets [4], we may assume that L is a lattice of sets on U. In addition, it is clear that if R is a reflexive relation

ED

on U, then apr (∅) = ∅ and apr (U) = U, that is to say, ∅, U ∈ D(apr ). Therefore we should choose L to satisfy R

R

R

∅, U ∈ L. Then a way to define a binary relation LRL on U is as follows:

(8)

PT

∀x ∈ U, (LRL ) s (x) = ∩{X ∈ L : x ∈ X}.

Clearly, LRL is a binary relation on U.

AC

CE

n o Example 9. Let U = {1, 2, 3} and L = ∅, {2}, {1, 2}, {2, 3}, U ⊆ P(U). See (b) of Figure 3, L is a lattice of sets on U. n o Using the formula (8), we can get the binary relation LRL on U, where (LRL ) s (1) = ∩ {1, 2}, U = {1, 2}, (LRL ) s (2) = n o n o ∩ {2}, {1, 2}, {2, 3}, U = {2} and (LRL ) s (3) = ∩ {2, 3}, U = {2, 3}. It is easy to check that D(apr ) = L. In fact, for any lattice of sets, this equation is also correct. So we give the following results.

LRL

Proposition 14. Let L be a lattice of sets on U and ∅, U ∈ L. Then the following properties hold: (1) LRL is reflexive and transitive;

(2) ∀x ∈ U, (LRL ) s (x) ∈ L.  Proof. (1) ∀x ∈ U, by U ∈ L, we conclude that U ∈ X ∈ L : x ∈ X , ∅. Since U is a finite set, it follows that  x ∈ (LRL ) s (x) = ∩ X ∈ L : x ∈ X . This implies that LRL is reflexive. Next, we shall prove that LRL is transitive. 14

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{1,2,3}

{1,2,3,4,5}

{1,4}

{1,2}

{2,3}

{2,3} {1}

(a) The lattice D(apr ) of Example 7

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{2}

(b) The set lattice L of Example 9

R

Figure 3: The lattices D(apr ) and L R

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Applying The formula (8), ∀y ∈ (LRL ) s (x), clearly, {X ∈ L : x ∈ X} ⊆ {X ∈ L : y ∈ X}. Thus (LRL ) s (x) = ∩{X ∈ L : x ∈ X} ⊇ ∩{X ∈ L : y ∈ X} = (LRL ) s (y). That is to say, LRL is transitive. This completes the proof.

(2) By (1), (LRL ) s (x) , ∅. Therefore we have that {X ∈ L : x ∈ X} , ∅ from the formula (8). Since L is a lattice of

M

sets on U, it follows from Definition 5 that it is closed under intersections. So (LRL ) s (x) = ∩{X ∈ L : x ∈ X} ∈ L. Theorem 5. Let L be a lattice of sets on U and ∅, U ∈ L. Then the following statements hold.

ED

(i) There exists a reflexive and transitive relation LRL given by the formula (8) such that D(apr

(ii) There exists a reflexive relation LR∼L such that D(apr LR∼L ) = L.

LRL

) = L.

PT

Proof. (i) The reflexivity and transitivity of LRL hold by Proposition 14. Next, we shall prove that D(apr ∀X ∈ L, by the definition of LRL and Definition 1, we can get that apr Remark 1. Hence L ⊆ D(apr

LRL

). On the other hand, ∀X ∈ D(apr

CE

our condition ∅ ∈ L, we have X ∈ L. If X , ∅, then apr

LRL

LRL

LRL

(X) = X and thus X ∈ D(apr

), then apr

LRL

LRL LRL

) = L. ) from

(X) = X. If X = ∅, then by

(X) = X implies ∀x ∈ X, (LRL ) s (x) ⊆ X and thus

∪ x∈X (LRL ) s (x) ⊆ X. By the reflexivity of LRL , we have X ⊆ ∪ x∈X (LRL ) s (x). Thus, ∪ x∈X (LRL ) s (x) = X. Since

AC

L is a lattice of sets on U, it follows from Definition 5 that it is closed under unions. By (2) of Proposition 14, X = ∪ x∈X (LRL ) s (x) ∈ L. We have proved that ∀X ∈ D(apr

D(apr

LRL

) = L.

LRL

), X ∈ L. Therefore, D(apr

LRL

) ⊆ L. In summary,

(ii) Since L is closed under unions and intersections, it follows that ∼ L is closed under unions and intersections,

which implies that ∼ L is a lattice of sets. By (i), there exists a reflexive relation LR∼L such that D(apr

Thus, we conclude that L =∼ D(apr

∼LRL

) = D(apr∼LRL ).

∼LRL

) =∼ L.

Theorem 5 also shows that for every lattice of sets, there is a reflexive relation such that its lower definable lattice equals the lattice of sets. Theorem 4 indicates that a reflexive relation R can induce a lower definable lattice 15

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D(apr ) that is the lattice of sets. However, this correspondence is not one-to-one. The reason is that two different R

approximation spaces based reflexive relations may have the same lower definable lattices as shown by Example 3. In fact, there exists a one-to-one correspondence between the lattices of sets and the approximation spaces based on reflexive and transitive relations. Next, we will prove this point. Lemma 3. Let R be a reflexive and transitive relation on U. Then ∀x ∈ U, R s (x) ∈ D(apr ). R

R

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  Proof. Since R is transitive, it follows that ∀y ∈ R s (x), R s (y) ⊆ R s (x). This implies that R s (x) ⊆ apr R s (x) . R     By the reflexivity of R and Proposition 2(1), we have apr R s (x) ⊆ R s (x). Thus apr R s (x) = R s (x), that is, R

R s (x) ∈ D(apr ). R

Theorem 6. Let R be a reflexive and transitive relation on U. Then D(apr ) is a lattice of sets and ∅, U ∈ D(apr ), R

and LRD(apr ) = R. R

R

R

R

x ∈ LRD(apr

R




) s (x).

R




) s (x)

= apr

R



s

LRD(apr

R




) s (x)

 .

On the other hand, since R is reflexive and transitive, it follows from Lemma 3

M

Therefore, R s (x) ⊆ LRD(apr

R

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Proof. Since R is reflexive, it follows from Theorem 4 that D(apr ) is a lattice of sets. In addition, clearly, ∅, U ∈ R
D(apr ). Next, we shall show that LRD(apr ) = R. We need to prove only that ∀x ∈ U, LRD(apr ) s (x) = R s (x). By R R R  
Proposition 14, we have LRD(apr ) s (x) ∈ D(apr ) and x ∈ LRD(apr ) (x). This implies

that R s (x) ∈ D(apr ) and x ∈ R s (x). Thus by the formula (8), we conclude R

In summary, LRD(apr

R




) s (x)

R

= ∩{X ∈ D(apr ) : x ∈ X} ⊆ R s (x). R

= R s (x). We have proved that ∀x ∈ U, LRD(apr

PT

This completes the proof.




) s (x)

ED

LRD(apr

R




) s (x)

= R s (x). Therefore, LRD(apr ) = R. R

Remark 10. Theorems 5 and 6 show that there exists a one-to-one correspondence between the lattices of sets and

CE

the lower definable lattices in approximation spaces based on reflexive and transitive relations. By Remark 9, that is to say, there exists a one-to-one correspondence between the distributive lattices and the lower definable lattices in

AC

approximation spaces based on reflexive and transitive relations. 5. The matroid representation of definable sets in generalized approximation spaces based on relations In this section, we give the conditions under which a lower definable lattice is geometric (or Boolean) and show

that if a lower definable lattice is the lattice of closed sets of a matroid, then it must be the open-closed set lattice of a matroid. This indicates that not every lattice of closed sets of a matroid can be seen as a lower definable lattice in a generalized approximation space based on relation. 16

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Since the family of all closed sets of a matroid is a geometric lattice, we need to discuss under what conditions the lower definable lattices are geometric (or Boolean) in order to study the connection between generalized rough sets and matroids. In Subsection 2.2, the definition of Boolean lattices is given that “L is said a Boolean lattice if every element of a distributive lattice L with 0 and 1 has a complement”. Clearly, if L is finite, then 1, 0 ∈ L. Thus,

in order to prove that L is a Boolean lattice, we need to prove only that every element of a distributive lattice L has a

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complement. Proposition 15. Given (U, R), the lattice (D(apr ), ⊆) is Boolean if and only if D(apr ) = D(aprR ). R

R

Proof. We first prove the necessity. Since (D(apr ), ⊆) is a Boolean lattice, it follows from the definition of Boolean R

lattices that D(apr ) =∼ D(apr ). In addition, by Remark 5, D(aprR ) =∼ D(apr ). Therefore, D(apr ) = D(aprR ). R

R

R

R

Conversely, using Remark 5, we can get that D(aprR ) =∼ D(apr ). By the condition D(apr ) = D(aprR ). Thus R

R

D(apr ) =∼ D(apr ), which shows that every element of D(apr ) has a complement. Therefore, according to the R

R

R

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definition of Boolean lattice, we need to prove only that (D(apr ), ⊆) is a distributive lattice. Moreover, by Remark R

9, we need to prove only that D(apr ) is a lattice of sets. Next, we show this point. R

(i) For all X, Y ∈ D(apr ), it follows from Theorem 3 that X ∩ Y ∈ D(apr ). R

R

(ii) For all X, Y ∈ D(apr ), clearly, ∼ X, ∼ Y ∈ D(apr ). By (i), ∼ X∩ ∼ Y ∈ D(apr ). Since every element of R

R

R

D(apr ) has a complement, it follows that X ∪ Y =∼ (∼ X∩ ∼ Y) ∈ D(apr ). R

R

By (i) and (ii), D(apr ) is a lattice of sets. In summary, (D(apr ), ⊆) is a Boolean lattice. R

M

R

This completes the proof.

In fact, D(apr ) is not necessarily equal to D(aprR ) when D(apr ) is a distributive lattice (See Example 8). R

ED

R

A obvious corollary is obtained as follows.

Corollary 2. Given (U, R), the lattice (D(aprR ), ⊆) is Boolean if and only if D(apr ) = D(aprR ).

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R

Obviously, if R is an equivalence relation then D(apr ) = D(aprR ), that is, D(apr ) = D(aprR ) when (U, R) is R

R

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the Pawlak approximation space.

Corollary 3. Let (U, R) be the Pawlak approximation space. Then the lattice D(aprR ) is Boolean. In fact, R is not necessarily an equivalence relation when D(apr ) is a Boolean lattice. In Example 6, D(apr ) = R

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D(aprR ) and thus D(apr ) is a Boolean lattice. However, R is not an equivalence relation. R

Remark 11. In general, a Boolean lattice is a geometric lattice, but not vice versa [4, 11]. Lemma 4. Given (U, R) and |U| = n, for any A, B ∈ D(apr ), the following assertions hold. R

(1) W ⊆∼ (A − B) and W ∈ D(apr ) ⇒ W ⊆ aprk (∼ (A − B)), ∀k ∈ N; R

R

(2) aprk+1 (∼ (A − B)) ⊆ aprk (∼ (A − B)), ∀k ∈ N and k , 0; R

R

(3) ∀a ∈ A − B, ∀k ∈ N, a < aprk (∼ (A − B)); R

(4)

aprn (∼ R

(A − B)) ∈ D(apr ). R

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Proof. (1) Let W ⊆∼ (A − B) and W ∈ D(apr ). We use mathematical induction to prove W ⊆ aprk (∼ (A − B)). By R

R

W ∈ D(apr ), we have W = apr (W). It follows from W ⊆∼ (A − B) and Proposition 1 (L4) that W = apr (W) ⊆ R

R

R

apr (∼ (A − B)). We have proved that the result holds when k = 1. Suppose that the result holds when k = i, that is, R

W ⊆ apri (∼ (A − B)).

(9)

R

Next, we show that W ⊆ apri+1 (∼ (A − B)). By Proposition 1 (L4) and the formula (9), we can conclude that

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R

W = apr (W) ⊆ apr (apri (∼ (A − B))) = apri+1 (∼ (A − B)). R

R

R

R

This completes the proof of (1) by mathematical induction.

(2) We first prove that apr2 (∼ (A − B)) ⊆ apr (∼ (A − B)). For all x ∈ apr2 (∼ (A − B)), we have that R

R

R

R s (x) ⊆ apr (∼ (A − B)). Suppose that R s (x) *∼ (A − B), then R s (x) ∩ (A − B) , ∅. Choose t ∈ R s (x) ∩ (A − B), then R

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t ∈ R s (x) and t ∈ (A − B). Clearly,

t ∈ R s (x) and R s (x) ⊆ apr (∼ (A − B)) ⇒ t ∈ apr (∼ (A − B)) R

R

⇒ R s (t) ⊆∼ (A − B) ⇒ R s (t) ∩ (A − B) = ∅ ⇒ R s (t) ∩ A ∩ (∼ B) = ∅; t ∈ (A − B) ⇒ t ∈ A = apr (A) (Since A ∈ D (apr )) ⇒ R s (t) ⊆ A. R

R

Thus R s (t) ∩ (∼ B) = ∅ and so R s (t) ⊆ B. This implies t ∈ apr (B) = B, which contradicts with our choice t ∈ A − B. R

Consequently, R s (x) ⊆∼ (A − B), which implies x ∈ apr (∼ (A − B)). Thus apr2 (∼ (A − B)) ⊆ apr (∼ (A − B)). We R

M

R

R

have proved that the result holds when k = 1. Suppose that the result holds when k = i, that is,

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apri+1 (∼ (A − B)) ⊆ apri (∼ (A − B)). R

(10)

R

Next, we shall prove that apri+2 (∼ (A − B)) ⊆ apri+1 (∼ (A − B)). By Proposition 1 (L4) and the formula (10), we R

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have that

R

apri+2 (∼ (A − B)) = apr (apri+1 (∼ (A − B))) ⊆ apr (apri (∼ (A − B))) = apri+1 (∼ (A − B)). R

R

R

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R

R

R

This completes the proof of (2) by mathematical induction. (3) Take a ∈ A − B, then a ∈ A but a < B. We first prove that a < apr (∼ (A − B)). By A ∈ D(apr ), we R

R

have A = apr (A). Thus R s (a) ⊆ A. Suppose that a ∈ apr (∼ (A − B)), then R s (a) ⊆∼ (A − B). This implies

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R

R

R s (a) ∩ (A − B) = R s (a) ∩ A ∩ (∼ B) = ∅. Using R s (a) ⊆ A, we can get that R s (a) ∩ (∼ B) = ∅, and thus R s (a) ⊆ B. Therefore, a ∈ apr (B) = B, which contradicts with our choice a ∈ A − B. Consequently, a < apr (∼ (A − B)). In R

R

addition, applying (2), we can get that

aprk (∼ (A − B)) ⊆ aprk−1 (∼ (A − B)) ⊆ · · · ⊆ apr2 (∼ (A − B)) ⊆ apr (∼ (A − B)). R

R

R

R

Therefore a < aprk (∼ (A − B)). R

(4) Applying (2), similar to the proof of Theorem 1, we can get that aprn (∼ (A − B)) ∈ D(apr ). R

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B

B

Y Z

Y

X

Z

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X A

A (a) The sublattice S

(b) The sublattice N

Figure 4: The sublattices S and N of D(apr ) R

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Proposition 16. Given (U, R), the following assertions hold.

(1) The lattice D(apr ) (or D(aprR )) is modular if and only if the lattice D(apr ) (or D(aprR )) is distributive; R

R

(2) If the lattice D(apr ) is atomic, then the lattice D(apr ) is modular; R

R

(3) If the lattice D(apr ) is atomic, then the lattice D(apr ) is distributive. R

R

Proof. (1) According to Proposition 13, we need to prove only that D(apr ) is modular ⇔ D(apr ) is distributive. R

R

M

The sufficiency is obvious by the definition of modular lattices.

Conversely, by Proposition 5, there does not exist a sublattice of D(apr ) such that it is isomorphic to N5 . Again R

applying Proposition 5, it will be enough to prove that D(apr ) does not contain a sublattice isomorphic to M3 .

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R

Suppose that there exists a sublattice S of D(apr ) isomorphic to M3 , without loss of generality, we may let R

S = {A, X, Y, Z, B} ⊆ D(apr ) with A ⊂ X, A ⊂ Y, A ⊂ Z, X ⊂ B, Y ⊂ B and Z ⊂ B. (As shown by Figure 4(a))

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R

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By Figure 4 (a), X ∨ Y = B and X ∨ Z = B. According to the formula (6), this implies (11)

X ∪ Y ∪ Z ⊆ B.

Clearly, Z, A ∈ D(apr ) and X ∪ Y ⊆∼ (Z − A). Let |U| = n. Applying (1) and (4) of Lemma 4, we can obtain that R

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X ∪ Y ⊆ aprn (∼ (Z − A)) and aprn (∼ (Z − A)) ∈ D(apr ). Thus, by the formula (6), we conclude that R

R

R

o n B = X ∨ Y = ∩ Z ∈ D(apr ) : X ∪ Y ⊆ Z ⊆ aprn (∼ (Z − A)). R

R

Thus B ⊆ aprn (∼ (Z − A)). By the formula (11), it is clear that X ∪ Y ∪ Z ⊆ aprn (∼ (Z − A)). By Figure 4(a), we R

R

know that Z − A , ∅. Take z ∈ Z − A, according to (3) of Lemma 4, we have that z < aprn (∼ (Z − A)), which is a R

contradiction to X ∪ Y ∪ Z ⊆

aprn (∼ R

This completes the proof of (1).

(Z − A)). Therefore, D(apr ) dose not contain a sublattice isomorphic to M3 . R

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(2) By Proposition 5, we need to prove only that there does not exist a sublattice of D(apr ) such that it is R

isomorphic to N5 . Suppose that there exists a sublattice N of D(apr ) isomorphic to N5 , without loss of generality, R

we may let

N = {A, X, Y, Z, B} ⊆ D(apr ) with A ⊂ X ⊂ Y ⊂ B, A ⊂ Z ⊂ B. (See Figure 4(b)) R

By Figure 4(b), X ∨ Z = B and Y ∨ Z = B. According to the formula (6), this implies

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X ∪ Y ∪ Z ⊆ B.

(12)

n o Since D(apr ) is atomic, it follows from the definition of atomic lattices that Y = ∪ C ∈ A(D(apr )) : C ⊆ Y . R

R

(*) We shall show that ∃W ∈ A(D(apr )) and W ⊆ Y but W ∩ X = ∅. Suppose that ∀W ∈ A(D(apr )) and W ⊆ Y R

R

such that W ∩ X , ∅. Since W is an atom, W ∩ X , ∅ implies W ⊆ X. Thus if ∀W ∈ A(D(apr )) and W ⊆ Y such that R n o n o W ∩ X , ∅, then ∪ C ∈ A(D(apr )) : C ⊆ Y ⊆ X ⊂ Y, which contradicts with Y = ∪ C ∈ A(D(apr )) : C ⊆ Y . R

R

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R

Thus ∃W ∈ A(D(apr )) and W ⊆ Y but W ∩ X = ∅.

(**) We shall prove that W ∩ Z = ∅, where W satisfies the conditions W ∈ A(D(apr )), W ⊆ Y and W ∩ X = ∅. R

Suppose that W ∩ Z , ∅, then W ⊆ Z by W being atomic. Thus W ⊆ Z ∩ Y = A ⊆ X (See Figur 4(b)), which contradicts to W ∩ X = ∅. Thus W ∩ Z = ∅.

By (*), we can choose W ∈ A(D(apr )) and W ⊆ Y but W ∩ X = ∅. Clearly, W, X ∈ D(apr ). By (**), we know R

R

X, Z ⊆∼ W. Since W ∩ X = ∅, it follows that W = W − X. Thus, according to (1)and (4) of Lemma 4, we can get that R

R

M

X ∪ Z ⊆ aprn (∼ (W − X)) = aprn (∼ W) ∈ D(apr ). Thus, by the formula (6), we conclude that R

n o B = X ∨ Z = ∩ Z ∈ D(apr ) : X ∪ Z ⊆ Z ⊆ aprn (∼ W).

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R

R

Thus B ⊆ aprn (∼ W). By the formula (12), it is clear that X ∪ Y ∪ Z ⊆ aprn (∼ W). Since W is an atom, it follows R

R

that W , ∅. Take z ∈ W, according to (3) of Lemma 4, we have that z < aprn (∼ W), which is a contradiction to R

R

proof of (2).

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X ∪ Y ∪ Z ⊆ aprn (∼ W). Therefore, D(apr ) dose not contain a sublattice isomorphic to N5 . This completes the R

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(3) It is straightforward from (1) and (2). Remark 12. The above theorem demonstrates that there does not exist a lower definable lattice such that it is modular

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but not distributive. That is to say, M3 can not become a sublattice of a lower definable lattice, especially, M3 can not

become a lower definable lattice. Conversely, a lower definable lattice can not be isomorphic to M3 . This indicates

that lattice structures are a more general framework than the lower definable lattices. Corollary 4. Given (U, R), the lattice D(apr ) is geometric if and only if the lattice D(apr ) is Boolean. R

R

Proof. According to Remark 11, the sufficiency is obvious. Conversely, since D(apr ) is geometric, it follows that D(apr ) is atomic. By Proposition 16(3), D(apr ) is R

R

distributive. By Proposition 6, we conclude that D(apr ) is a Boolean lattice. R

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Remark 13. In [11], the authors show that a lattice is a Boolean lattice if and only if it is the open-closed set lattice of a matroid. Theorem 7. Given (U, R), the following conditions are equivalent: (1) D(apr ) is the lattice of closed sets of a matroid; R

(2) D(apr ) is a geometric lattice; R

(3) D(apr ) is a Boolean lattice;

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R

(4) D(apr ) = D(aprR ) R

(5) D(aprR ) is a Boolean lattice;

(6) D(apr ) is the open-closed set lattice of a matroid. R

Proof. By Proposition 8, (1) ⇔ (2).

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By Corollary 4, (2) ⇔ (3).

By Proposition 15, (3) ⇔ (4). By Corollary 2, (4) ⇔ (5).

By the above results, we can get that (5) ⇔ (3). According to Remark 13, we conclude that (3) ⇔ (6). Conse-

quently, (5) ⇔ (6). This completes the proof of this theorem.

In Example 8, R is reflexive, but D(apr ) , D(aprR ). Thus, now, D(apr ) can not be seen as the lattice of closed R

R

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M

sets of a matroid according to Theorem 7. On the other hand, we can also check that D(apr ) is not the lattice of closed R o   n sets of a matroid in terms of the famous conclusion given by Proposition 8. In Example 8, A D(apr ) = {2}, {4} . R n o Clearly, there exists {2, 3} ∈ D(apr ) but {2, 3} , ∪ X ∈ A(D(apr )) : X ⊆ {2, 3} . This shows that D(apr ) is not R

R

R

atomic and thus it is also not geometric. By Proposition 8, we have that D(apr ) can not be taken as the lattice of R

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closed sets of a matroid. This observation coincides that obtained by Theorem 7.

Remark 14. (2) and (3) of Theorem 7 show that the lower definable lattice D(apr ) can not be geometric but not R

Boolean. On the other hand, Theorem 4 and Example 8 illustrate that D(apr ) may be distributive but not Boolean

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R

(or geometric). These conclusions demonstrate that the generalized approximation space based on relation is not a

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more general framework than matroid from a lattice-theoretic viewpoint, and vice versa (See Figure 5). In the rest of this section, we give the representation of closure of the matroid induced by the lower definable

lattices.

Proposition 17. Given (U, R) and |U| = n, if R is reflexive and D(apr ) is the lattice of closed sets of a matroid M, R

then aprnR = cl M , that is to say, aprnR is the closure of the matroid M.

Proof. For all X ⊆ U, we shall show that aprnR (X) = cl M (X). Since R is reflexive, it follows from Theorem 1 that aprnR (X) ∈ D(apr ). Thus aprnR (X) is a closed set of M. Clearly, cl M (X) is also a closed set of M. As a result, R

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Finite lattices

Lower

Lower definable

definable lattices based on relation



lattices

lattices D apr R

based on

when



reflexive

D  apr R   D  apr R 

relation

Open  closed lattices

( Distributive

of matroids ( Boolean lattices )

lattices )

Lattices of closed sets of matroids (Geometric lattices )

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Lower definable

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Figure 5: The relationships among different lattices

aprnR (X) ∩ cl M (X) is a closed set of M, and thus aprnR (X) ∩ cl M (X) ∈ D(apr ). Again by Theorem 1, there exists R

Z ⊆ U such that

aprnR (X) ∩ cl M (X) = aprnR (Z).

(13)

X ⊆ aprnR (X) ∩ cl M (X).

(14)

M

According to Propositions 7 and 9, we have

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Thus, by the formula (13), we have X ⊆ aprnR (Z). It follows from Proposition 9 (4) and Corollary 1 that   n n aprnR (X) ⊆ aprnR aprnR (Z) = apr2n R (Z) = apr R (Z) = apr R (X) ∩ cl M (X),

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that is, aprnR (X) ⊆ aprnR (X) ∩ cl M (X). Therefore aprnR (X) ⊆ cl M (X). On the other hand, by the formula (14) and   Proposition 7 (cl2), we have cl M (X) ⊆ cl M aprnR (X) ∩ cl M (X) . In addition, since aprnR (X) ∩ cl M (X) is a closed set

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of M, it follows that cl M (aprnR (X) ∩ cl M (X)) = aprnR (X) ∩ cl M (X). Therefore, cl M (X) ⊆ aprnR (X) ∩ cl M (X) and thus

cl M (X) ⊆ aprnR (X). In summary, cl M (X) = aprnR (X). We have proved that ∀X ⊆ U, aprnR (X) = cl M (X). Consequently, aprnR = cl M . This completes the proof.

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The above result shows that aprnR must be the closure of the matroid M given by Proposition 17. However, aprR

is not necessarily the closure of the matroid M. The following example illustrates this point. Example 10. Let U = {1, 2, 3, 4} and let R s (1) = {1, 2}, R s (2) = {2, 3}, R s (3) = {3, 4} and R s (4) = {1, 4}. It is easy to n o check that D(apr ) = D(aprR ) = ∅, U . Applying Theorem 7, we know that D(apr ) is the lattice of closed sets of R

R

a matroid M. Take {1} ⊆ U, aprR ({1}) = {1, 4} < D(apr ). Thus aprR can not be the closure of the matroid M. In R

fact, we can easily check that apr4R is the closure of the matroid M. 22

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Note that Proposition 17 indicates that if R is reflexive, then we can use the n powers of aprR to represent the closure of the matroid induced by the lower definable lattice. However, if we replace the condition R being reflexive in Proposition 17 with the condition R being transitive (or symmetric), then Proposition 17 is not necessarily true. Example 2 illustrates that aprnR (X) may not be a lower definable set (a closed set) when R is symmetric. Thus aprnR can not become the closure of a matroid induced by the lower definable lattices. While Example 5 illustrates that X ⊆ aprn (X) may not true when R is transitive, so aprn can not become the closure of a matroid. R

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6. Conclusions

The family of all lower definable sets in a approximation space is proved to be a lattice with set inclusion order. It is referred to as the lower definable lattice of a approximation space. This paper gives the conditions under which the lower definable lattice is distributive (or geometric, or Boolean). In view of the family of closed sets of a matroid

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being a geometric lattice, we establish the relationship between matroids and generalized approximation spaces based on relations by means of the lower definable lattices. There are still many interesting issues worth studying: (1) In this paper, we give the necessary and sufficient conditions under which two reflexive relations can induce the same lower definable sets. However, for two arbitrary relations, under what the necessary and sufficient conditions they can induce the same lower definable sets? (See Example 4)

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(2) Similar issues should be discussed about other generalized approximation spaces, such as, covering approximation spaces, multigranulation spaces and so on.

(3) The granular partition lattice of information tables is an important order structure in granular computing. It

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provides a complete hierarchical classification of the knowledge obtained from all possible choices of attribute subsets [1, 2]. So we may establish the relationship between rough sets and matroid by means of the granular partition lattice

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structure from a lattice-theoretic viewpoint.

Acknowledgments The authors thank all of the editors and reviewers for their constructive comments as well

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as helpful suggestions, which have substantially improved this paper. This work is supported by The Foundation of

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Shanxi Normal University (Grant No. 872022).

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