Remarks on Pre--Regular Pre--Open Sets

Abstract

We deal with the new class of pre--regular pre--open sets in which the notion of pre--open set is involved. We characterize these sets and study some of their fundamental properties. We also present other notions called extremally pre--disconnectedness, locally pre--indiscreetness, and pre--regular sets by utilizing the notion of pre--open and pre--closed sets by which we obtain some equivalence relation for pre--regular pre--open sets.

1. Introduction

The subject of ideals in topological space has been studied by Kuratowski [1] and Vaidyanathaswamy [2]. An ideal on a topological space is a nonempty collection of subsets of which satisfies (i) and implies and (ii) and implies . Given a topological space with an ideal on and if is the set of all subsets of , a set operator , called a local function [1] of with respect to and , is defined as follows: for , , for every where . A Kuratowski closure operator for a topology , called the -topology, finer than is defined by [3]; when there is no chance for confusion, we will simply write for and for . If is an ideal on , then is called an ideal space. A subset of an ideal space is said to be -open [4] if . A subset of an ideal space is said to be pre--open [5] if . The complement of pre--open set is called pre--closed. The family of all pre--open sets in is denoted by or simply . Clearly . The largest pre--open set contained in , denoted by , is called the pre--interior of . The smallest pre--closed set containing , denoted by , is called the pre--closure of . A subset of an ideal space is said to be --open [6] if . The family of all --open sets is a topology finer than . We will denote the --interior subset of of by . A subset of an ideal space is [7] if . A space is [8] if every subset of is open.

The following lemmas will be useful in the sequel.

Lemma 1. Let be an ideal space and let be a subset of . Then [9].

2. Pre--Regular Pre--Open Sets

A pre--open set of a space is said to be pre--regular pre--open if . The complement of a pre--regular pre--open set is called pre--regular pre--closed set, equivalently . A subset of a space with an ideal is said to be pre--regular if it is pre--open and pre--closed. Clearly, and are pre--regular pre--open.

Remark 2. Also every pre--regular set is pre--regular pre--open.

Proof. Assume is pre--regular that implies is pre--open and pre--closed. is pre--open which implies . Hence is pre--regular pre--open.

But the converse is not true as shown by Example 3.

Example 3. Consider the ideal space where , , and . Here is pre--regular pre--open but not pre--closed.

Moreover, the intersection of two pre--regular pre--open sets is not pre--regular pre--open in general as Example 4 shows.

Example 4. Consider the ideal space where , , and . Here . Thus, , are both pre--regular pre--open. But is not, since it is not even pre--open.

The notions of pre--regular pre--open and -open sets are independent of each other. Consider the space with an ideal as in Example 4. Here is -open but not pre--regular pre--open. Consider the space and , . Here is pre--regular pre--open but not -open. Observe that every pre--regular pre--open set is pre--open but the converse need not be true. Here is pre--open but not pre--regular pre--open.

Theorem 5. Let be an ideal space and let , be any subsets of . Then the following hold.(a)If , then .(b)If , then .(c)For every , .(d)If and are disjoint pre--open sets, then and are disjoint.(e)If is a pre--regular pre--open, then is pre--regular pre--closed.(f)If is pre--regular pre--open, then is pre--regular pre--open.

Proof. (a)Suppose . Therefore .(b)Suppose that .(c)It is obvious that , so by (b) we have . On the other hand which implies . Therefore, . Hence .(d)Since and are disjoint pre--open sets, we have which implies . Since is pre--open, . Hence.(e)Given that is pre--regular pre--open, implies . Therefore,  − . Hence is pre--regular pre--closed.(f)By (e) if is pre--regular pre--open, then is pre--regular pre--closed. Hence is pre--regular pre--open that implies is pre--regular pre--open.

Lemma 6. For an ideal topological space the following are equivalent. (a)Every pre--open set is open.(b)Every -dense set is open.

Proof. (a) (b): let be a -dense set that implies which implies , so that . By (a) every pre--open set is open and hence is open.
(b) (a): let be a pre--open subset of , so that , say. Then , so that , and thus is -dense in . Thus is open. Now is the intersection of two open sets, so that is open.

Lemma 7. If a space with an ideal is -submaximal, then any finite intersection of pre--open set is pre--open.

Proof. From Lemma 6, every pre--open set is open and hence a finite intersection of pre--open set is pre--open.

Theorem 8. If a space with an ideal is -submaximal, then any finite intersection of pre--regular --open set is pre--regular pre--open.

Proof. Let be a finite family of pre--regular pre--open sets. Since the space is -submaximal, then by Lemma 6, is pre--open. Therefore, . Also, for each which implies . Also, each is pre--regular pre--open that implies which implies and so . Hence is pre--regular pre--open.

It should be noted that an arbitrary union of pre--regular pre--open set is pre--regular pre--open. But the intersection of two pre--regular pre--closed sets fails to be pre--regular pre--closed as shown by Example 9.

Example 9. Consider the ideal space as in Example 3. Clearly, , are pre--regular pre--closed but their intersection is not pre--regular pre--closed.

Theorem 10. The following hold for a subset of a space .(a)If is pre--closed, then is pre--regular pre--open.(b)If , then is pre--regular pre--closed.(c)If and are pre--regular pre--closed sets, then if and only if .(d)If and are pre--regular pre--open sets, then if and only if .

Proof. (a) Since is pre--closed, .
Now, . Hence is pre--regular pre--open.
(b) Now . Hence is pre--regular pre--closed.
(c) Given that and are pre--regular pre--closed sets, therefore, and . Clearly, if .
Conversely, . Now . Hence .
(d) Given that and are pre--regular pre--open, therefore, and . Suppose , . Therefore, .
Conversely, . Now .

A subset of an ideal topological space is said to be -rare if it has no interior points in .

Theorem 11. Let be an ideal space. Then the following hold.(a)The empty set is the only subset which is nowhere dense and pre--regular pre--open.(b)If is pre--regular pre--closed, then every -rare set is pre--open.

Proof. (a) Suppose is nowhere dense and is pre--regular pre--open. Then , by Lemma 1. Therefore, .
(b) Suppose is pre--regular pre--closed. Then . Therefore, . Hence is pre--open.

An ideal space is called extremally pre--disconnected if the pre--closure of every pre--open set is pre--open.

Theorem 12. For a topological space the following are equivalent.(a) is extremally pre--disconnected.(b)Every pre--regular pre--open subset is pre--regular.

Proof. (a) (b): assume is extremally pre--disconnected. Suppose is pre--regular pre--open. Then is pre--open and so is a pre--open set. Hence . Hence is pre--closed which implies is pre--regular.
(b) (a): suppose is pre--open. Then is pre--regular pre--closed which implies is pre--regular pre--open. Hence is pre--regular. Therefore, is pre--closed and so is pre--open. Hence is extremally pre--disconnected.

Theorem 13. Let be an extremally pre--disconnected space and . Then the following are equivalent:(a) is pre--regular,(b),(c) is pre--regular pre--open,(d) is pre--regular pre--open.

Proof. (a) (b): suppose is pre--regular. Then is pre--open and pre--closed and so and . Hence .
(b) (c): let . Then so is pre--regular pre--open.
(c) (d) is clear.
(d) (a) follows from Theorem 12.

An ideal space is called locally pre--indiscrete if every pre--open subset of is pre--closed (or) if every pre--closed subset of is pre--open.

Theorem 14. Let be an ideal space. Then the following are equivalent.(a) is locally pre--indiscrete.(b)Every pre--open subset is pre--regular.(c)Every pre--open subset is pre--regular pre--open.(d), for every .(e)The empty set is the only nowhere dense subset of .

Proof. (a) (b): assume that is locally pre--indiscrete. Let be a pre--open subset of . By hypothesis, is pre--closed. Hence is pre--regular.
(b) (c): if is pre--open, then . Also by hypothesis, is pre--closed. Therefore, . Hence is pre--regular pre--open.
(c) (d): since is preopen, is pre--open. By (c), is a pre--regular pre--open set. Therefore, .
(d) (e): by Theorem 11, in any space, the empty set is the only subset which is nowhere dense and pre--regular pre--open.
(e) (a): suppose that is a pre--closed set. Now  −  . Therefore is nowhere dense which implies , and so is pre--open. Hence is locally pre--indiscrete.

An ideal space is said to be -door if every subset of is either pre--regular pre--open or pre--regular pre--closed.

Theorem 15. Let be a -door space; then every pre--open set in the space is pre--regular pre--open.

Proof. Let be a pre--open subset of . Since is -door, is pre--regular pre--closed and so which implies that . Since is pre--open, . Hence is pre--regular pre--open.

Theorem 16. Let be an ideal space. A subset of is both --open and --closed; then is a pre--regular pre--open set.

Proof. Let be an --open and --closed set. Then is a pre--open and pre--closed set and hence is a pre--regular pre--open set.

Theorem 17. Let be an ideal space. A subset of is --open and pre--regular pre--open; then .

Proof. Suppose is --open. Then . And is pre--regular pre--open which implies . Therefore .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.