Mathematical Problems in Engineering

Mathematical Problems in Engineering / 2014 / Article

Research Article | Open Access

Volume 2014 |Article ID 502724 | 10 pages | https://doi.org/10.1155/2014/502724

Comparative Studies of Covering Rough Set Models

Academic Editor: Roque J. Saltarén
Received28 Apr 2014
Accepted05 Aug 2014
Published28 Aug 2014

Abstract

Many different proposals exist for the definition of lower and upper approximation operators in covering-based rough sets and so many different covering rough set models are built correspondingly. It is meaningful to explore the connection of these covering rough set models for their applications in practice. In this paper, we establish relationships between the most commonly used operators in covering rough set models. We investigate the conditions under which two types of upper approximation operations are identical.

1. Introduction

Rough set theory is a useful tool for dealing with the vagueness, granularity, and uncertainty in information systems and is also a data mining tool. It was proposed by Pawlak in the early 1980s [1, 2]. Since then we have witnessed a systematic, world-wide growth of interest in rough set theory [311] and its applications [1217]. Nowadays, it turns out that this approach is of fundamental importance to artificial intelligence and cognitive sciences, especially in the areas of data mining, machine learning, decision analysis, knowledge management, expert systems, and pattern recognition.

The classical rough set theory is based on equivalence relations, but in some situations, equivalence relations are not suitable for coping with granularity, and thus many practical data sets cannot be handled well. In light of this, equivalence relation has been generalized to similarity relation [18], tolerance relation [1921], and even arbitrary binary relation [2225] in some extensions of the classical rough sets. Another approach is the relaxation of the partition arising from equivalence relation to a covering. The covering of a universe is used to construct the lower and upper approximations of any subset of the universe [3, 6, 8, 11, 26].

In [27], Yang and Li presented a summary of seven pairs of approximation operators used by Bonikowski et al. [3], Pomykała [6], Tsang et al. [28], Zhu [29], Zhu and Wang [30], and Xu and Zhang [31]. In particular, Bonikowski et al. first extended Pawlak rough set theory from a partition to a covering in [3]; The second type of covering rough set model was presented by Pomykała in [6], while Tsang et al. [28] studied the third type; Xu and Wang gave the definition of the sixth type in [32]; Zhu defined the fourth and the fifth types of covering-based approximation in [29, 30]; The seventh type of approximation operations can be found in [31]. For the seven types of covering rough sets, their basic properties and axiomization problem were studied in [6, 2832], their redundancy problem were considered in [11, 27], and their topological properties were explored in [33]. Therefore, these seven pairs of approximation operators play an important role in the study of covering rough set models [34]. It is necessary to explore the relationships among the seven types of covering rough sets. Through the comparison of approximation operations, we can find the difference and relationship between them and give the classification of covering rough sets, which can help a decision maker to choose a suitable rough set model for data analysis. Zhu did some research on the relationships among four types of covering rough sets [30, 35]. However, Zhu considered only the case that the covering is finite. When we do not restrict the covering to the finite, that is, the covering can be infinite, the corresponding problems become complex. In this paper, we do not limit the covering to the finite and investigate the relationships among seven types of covering rough sets.

The remainder of the paper is organized as follows: Section 2 reviews some basic concepts about covering rough set model. In Section 3, we present some results about relationships between the first type of covering rough sets and the second, the third, the fourth, the fifth, the sixth, and the seventh types, and give the conditions under which two types of upper approximation operations are identical. Similarly, in Section 4, we investigate the relationships between the second type of covering rough sets and the third, the fourth, the fifth, the sixth, and the seventh types. In Section 5, we do some research on the relationships between the third type of covering rough sets and the fourth, the fifth, the sixth, and the seventh types. In Section 6, we explore the relationships between the fourth type of covering rough sets and the fifth, the sixth, and the seventh types. In Section 7, we present some results about relationships among the fifth, the sixth, and the seventh types of covering rough sets.

2. Preliminaries

In this section, we briefly introduce the basic ideas of rough sets and coverings.

Definition 1 (see [3], covering). Let be a universe of discourse, a family of subsets of . If no subsets in are empty and , is called a covering of .

It is clear that a partition of is a covering of ; therefore the concept of a covering is an extension of the concept of a partition. In the following discussion, can be an infinite covering; that is, we do not restrict the covering to the finite.

Definition 2. Let be a nonempty set, a covering of . The pair is called a covering approximation space.

Definition 3 (see [3], minimal description). Let be a covering approximation space and . The family of sets is called a minimal description of . When there is no confusion, we omit the subscript .

Definition 4 (see [3, 29], neighborhood). Let be a covering of ; is called a neighborhood of . Generally, we omit the subscript when there is no confusion.

By Definition 4, it is easy to see that ,  .

In covering rough set models, seven pairs of the most commonly used approximation operators are defined as follows [27, 34].

Definition 5 (covering lower and upper approximation operations). Let be a covering of . The operations, and , are defined as follows: for all ,
We call the first, the second, the third, the fourth, or the fifth covering lower approximation operations and the sixth and the seventh covering lower approximation operations with respect to the covering .
The operations are defined as follows: for all ,,,,,,,.
, , , , , , are called the first, the second, the third, the fourth, the fifth, the sixth, and the seventh covering upper approximation operations with respect to , respectively. We leave out at the subscript when there is no confusion.

Note 1. Since for all , , it is easy to check that when is a finite (or infinite) covering. Therefore is still held when we do not restrict the covering to the finite.

Definition 6 (see [35] unary covering). Let be a covering of a set . is called unary if for all , .

3. Relationships between and , , , , , and

In this section, we study the relationships between the first type of covering rough sets and others and present some conditions under which two types of upper approximation operations are identical.

When the covering is finite, Zhu and Wang presented the condition, “ is a partition,” under which [35]. In the following, we show that the result is held when is not restricted to the finite.

Theorem 7. Let be a covering approximation space. Then if and only if is a partition.

Proof. We only need to prove the necessity. Suppose that is not a partition. Then there exist such that and . Hence or . We may assume that . By Definition 5, we conclude that and , thus . This is a contradiction with the condition . Hence is a partition.

When the covering, , is finite, Zhu and Wang gave a condition under which [35]. His result is that “ if and only if is unary.” However, when the covering, , is infinite, the above result is not always true. In fact, Example 10 illustrates that “ is unary,” and the following example illustrates that “ is unary .”

Example 8. Let , , , , and . Clearly, is a covering of . It is easy to check that , , and , . Hence is unary. It is easy to see that and , which implies . Therefore, .

In the following, we give a condition under which when is a general covering.

Theorem 9. Let be a covering approximation space. Then if and only if for all , .

Proof. We first prove the necessity. Let . Then it follows by that . By the definitions of and , .
Conversely, let . By the definition of , there exists such that . By the condition , , we have that that is, . Hence We have proved that, , . Therefore, .

Applying the above result, we can give the following example which illustrates that “ is unary.”

Example 10. Let , , , and . Clearly, is a covering. It is clear that , . Hence , . It follows from Theorem 9 that . However, we can prove that ; that is, , and thus is not unary. Suppose that ; that is, ; then there exists . However, by the definition of , we have that and , which is a contradiction with . Hence .

When the covering, , is finite, Zhu and Wang obtained a condition under which [30]. The result is still true when is a general covering and the proof is similar.

Theorem 11 (see [30]). Let be a covering approximation space. Then if and only if satisfies the following condition: for all , if , then, and , .

There are no studies on the relationships and , , and . In the following, we discuss these problems. First, we establish a condition under which and are identical.

Theorem 12. Let be a covering approximation space. Then if and only if, for all , .

Proof. The sufficiency follows directly from the definitions of and .
Conversely, suppose that there exists such that . We will show that . Otherwise, ; then we can get that , a contradiction. Thus and so . It follows from Definition 5 that and . Thus , which contradicts the condition, . This completes the proof of the necessity.

It is worth noting that in Example 8, is unary; however, . That is, is unary , .

Lemma 13. Let be a covering approximation space and . If, , , , then .

Proof. (i) We will show that, , . Let . For any , clearly, , and thus and so . On the other hand, for all , there exists such that and . Suppose ; then it follows from the assumed condition that , which is a contradiction. Hence and so , which implies that . Thus, . In summary, .
(ii) By the definition of , there exists such that . By (i), we have that that is, .
(iii) We will show that , . By Definition 4, it is clear that . Clearly, ; thus by (i), . Therefore, .
(iv) By (iii), the proof of is similar to that of (ii).
Thus, by (ii) and (iv), .

Lemma 14. Let be a covering approximation space. If, , , then, , .

Proof. Let . Since, , , it follows that Therefore, .

In the following, we give a condition under which and are identical.

Theorem 15. Let be a covering approximation space. Then if and only if, , , and , .

Proof. We first prove the necessity.
(i) We will show that, , , . Suppose that, , , . Then by Definition 5, we have that and ; thus by , . Since , it follows that and so . Hence , which is a contradiction with the assumed condition . Thus, , , .
(ii) We will show that, , . Let . If , then and by Definition 3, we have that . Hence and . Therefore, by , we can get that that is, . If , then and so . In addition, . By , we conclude that and so .
By (i) and (ii), the necessity holds.
Conversely, let . By Lemma 13, we have that . By Lemma 14, we have that . Thus We have proved that, , . Therefore, .

Lemma 16. Let be a covering approximation space. If, , , then, , .

Proof. Let . Since, , , it follows that Therefore, .

In the following, we present a condition under which and are identical.

Theorem 17. Let be a covering approximation space. Then if and only if, , , and , .

Proof. We first prove the necessity.
(i) We will show that, , , . Suppose that, , , . Then by Definition 5, we have that and ; thus by , . Since , it follows that and so . Hence , which is a contradiction with the condition, . Thus, , , .
(ii) We will show that, , . Let . If , then and by Definition 3, we have that . Hence and . By , we have that . Thus . If , then and so . In addition, . By , we conclude that and so .
(iii) By (i) and (ii), the necessity holds.
Conversely, let . By Lemma 13, we have that . By Lemma 16, . Thus We have proved that, , . Therefore, .

4. Relationships between and , , , , and

In this section, we explore the relationships between the second type of covering rough sets and later five types, and establish the conditions under which two types of upper approximation operations are identical.

When the covering, , is finite, Zhu and Wang gave a condition under which and are identical in [35]. In the following part, we show that their conclusion is still held when the covering is a general covering.

Theorem 18. Let be a covering approximation space. Then if and only if, ,  , .

Proof. Let and ; then . By Definition 5, we conclude that and . By , and so . This finishes the proof of necessity.
Conversely, let . By Definitions 3 and 5, it is easy to see that . On the other hand, , then, , and . Taking , then and . Thus . By our condition, we have that . Hence . Thus . In summary, . We have proved that, , . Therefore, .

Now we present some conditions under which , , and .

Theorem 19. Let be a covering approximation space. Then the following conditions are equivalent.(1).(2).(3).(4) is a partition.

Proof. Suppose that is not a partition; then there exist such that and . Hence or . We may assume that . By Definition 5, we have that and ; thus . This is a contradiction with the condition, . Hence is a partition.
The proof is similar to that of “”.
Suppose that is not a partition; then there exist such that and . Thus or . Without loss of generality, we assume that . Taking and , by Definition 4, we have that . Since , it follows that ; that is, . Hence by the definition of , we have that . On the other hand, by and the definition of , we have that . Since , it follows that . Thus , which is a contradiction with the condition . This completes the proof.
It follows directly from Definition 5.

By Theorems 7 and 19, we can conclude that .

In the following, we give a condition under which and are identical.

Theorem 20. Let be a covering approximation space. Then if and only if, , , there exists such that .

Proof. We first prove the necessity. , , we will show that, , . Clearly, , hence by the definition of we have that and so . By , we can get that , and thus . According to the definition of , we conclude that there exists such that and . Since implies that , it follows that . This finishes the proof of the necessity.
Conversely, let . Then by the definitions of and , it is easy to check that . On the other hand, , by the definition of , there exists such that and . Taking , then . By the assumed condition, there exists such that . Clearly, , and thus by the definition of , we have that and so . Hence . In summary, . We have proved that, , . Therefore, .

5. Relationships between and , , , and

In this section, we study the relationships between the third type of covering rough sets and later four types and establish the conditions under which two types of upper approximation operations are identical.

Lemma 21. Let be a covering approximation space and . Then, , .

Proof. Let . Then and . If , then by and the definition of we have that , a contradiction. Therefore, .

Lemma 22. Let be a covering approximation space and . If , and imply that , then the following assertions hold.(1), .(2).

Proof. (1) Let . For all , if , then . If , then, by Definition 3, there exists such that and . Hence and , which implies that by the assumed condition. Thus and so . It follows that . On the other hand, , then, , . Thus, , . If , then . If , then by, , it follows from the assumed condition that . Hence . By , we have that . By , we have that and so . Therefore, . In summary, .
(2) , then, , . This implies that, , . Since , it follows that and so . Therefore, . On the other hand, , then there exists such that and . Taking , then and by Lemma 21, we have that . Suppose that , then by Definition 3, there exists such that and . Thus , by our condition, this implies that , a contradiction with . Hence and so by , . It follows that . Thus . In summary, .

When a covering is finite, Zhu and Wang presented a condition under which [30]. In the following, we show that the result is held when we do not restrict to the finite.

Theorem 23. Let be a covering approximation space. Then if and only if for , and , .

Proof. Suppose that, , , and , . Then . By the definitions of and , we have that and . On the other hand, by , we have that and . Thus and . Hence by and , we can get that and . This implies that and so . It follows by that ; that is , which is a contradiction with the condition . This finishes the proof of necessity.
Conversely, let . By the definition of , we have that, , . By the part (1) of Lemma 22, we know that, , . Thus, that is, . Further, by the part (2) of Lemma 22, we know that . Therefore, that is, . We have proved that, , . Therefore, .

Now we provide a condition under which and are identical.

Theorem 24. Let be a covering approximation space. Then if and only if, , .

Proof. The sufficiency follows directly from the definitions of and .
Conversely, suppose that there exists such that . Then by Definition 5, we have that and . Thus , which contradicts the condition . This finishes the proof of the necessity.

By Theorems 12 and 24, we can conclude that .

In the following, we give a condition of necessity under which and are identical.

Proposition 25. Let be a covering approximation space. If , then, , .

Proof. Let ; then by Definition 5, we have that and . Since , it follows that .

The following example shows that the converse of the above result is not true.

Example 26. Let , , , and . Clearly, is a covering. By Example 10, and , . It is easy to see that, , . However, by , it follows from Theorem 24 that .

However, when the covering is finite, the converse of Proposition 25 is true.

Theorem 27. Let be a covering approximation space. If is a finite covering, then the following conditions are equivalent.(1).(2), .(3), .

Proof. It comes from Proposition 25.
Suppose that, , . Since is a finite covering, it follows from Definition 3 that . Taking , we will show that, and , . Otherwise, and , , by Definitions 3 and 4, which implies that . This contradicts the assumption . Hence there exists and such that . and imply that and so , which is a contradiction with the condition (2). This completes the proof.
It comes from Theorem 24.

When the covering is finite, by Theorems 9, 12, and 27, we can conclude that .

In the following, we present a condition under which and are identical.

Theorem 28. Let be a covering approximation space. Then if and only if, , .

Proof. We first prove the necessity. , by Definition 5, we have that and . By the condition we have that ; thus .
Conversely, let . By Lemma 14, we have that . It follows from Definition 5 that . We have proved that, , . Therefore, .

By Theorems 15 and 28, we can conclude that .

In the following, we give a condition under which and are identical.

Theorem 29. Let be a covering approximation space. Then if and only if, , .

Proof. We first prove the necessity. , by Definition 5, we have that and . By the condition we have , and thus .
Conversely, let . By Lemma 16, we have that . It follows from Definition 5 that . We have proved that, , . Therefore, .

By Theorems 17 and 29, we can conclude that .

6. Relationships between and , , and

In this section, we explore the relationships between the fourth type of covering rough sets and later three types and establish some conditions under which two types of upper approximation operations are identical.

Lemma 30. Let be a covering approximation space and . If, , , or , then .

Proof. , there exists such that and . Taking , then and . By Lemma 21, we have that . By our condition, this implies that . Further, since , it follows that and so . Therefore, On the other hand, , then, , . By Definition 4, implies that there exists such that . Clearly, ; thus and so . Thus . Therefore, In summary, .

In the following, we present a condition under which and are identical.

Theorem 31. Let be a covering approximation space. Then if and only if, , , or .

Proof. We first prove the necessity, suppose that there exist and such that and ; then . By the definition of , implies that ; thus by Definition 5, we have that and (because ). Since , it follows that and so , which contradicts the condition . This completes the proof of the necessity.
Conversely, let . By Lemma 30, we have that . It follows from the definitions of and that . We have proved that, , . Therefore, .

Proposition 32. Let be a covering approximation space. If , then .

Proof. By Theorem 23, we only need to prove that, , , and , . Suppose that , then by and Theorem 31, it follows from and that and . Thus , a contradiction with . This completes the proof.

Lemma 33. Let be a covering approximation space, , and . If and , then , .

Proof. Suppose that, , , then by , we have that . Since , it follows that . Thus and so by the definition of , we get that . Hence . On the other hand, by , we have that , and by , we can conclude that . It follows from the definition of that . Thus,