# Remarks on Pre-I-Regular Pre-I-Open Sets

Authors: R. Sajuntha
Journal: International Scholarly Research Notices
 (2014)

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## Abstract

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


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## Body

## 1. Introduction

The subject of ideals in topological space has been studied by Kuratowski [1] and Vaidyanathaswamy [2]. An ideal I on a topological space (X,τ) is a nonempty collection of subsets of X which satisfies (i) A∈I and B⊂A implies B∈I and (ii) A∈I and B∈I implies A∪B∈I. Given a topological space (X,τ) with an ideal I on X and if ℘(X) is the set of all subsets of X, a set operator (·)⋆:℘(X)→℘(X), called a local function [1] of A with respect to τ and I, is defined as follows: for A⊂X, A⋆(τ,I)={x∈X∣U∩A∉I}, for every {U∈τ(x)} where τ(x)={U∈τ∣x∈U}. A Kuratowski closure operator cl⋆() for a topology τ⋆(X,τ), called the ⋆-topology, finer than τ is defined by cl⋆(A)=A∪A⋆(I,τ) [3]; when there is no chance for confusion, we will simply write A⋆ for A⋆(I,τ) and τ⋆ for τ⋆(I,τ). If I is an ideal on X, then (X,τ,I) is called an ideal space. A subset A of an ideal space (X,τ,I) is said to be I-open [4] if A⊂int⁡(A⋆). A subset A of an ideal space (X,τ,I) is said to be pre-I-open [5] if A⊂int⁡(cl⋆(A)). The complement of pre-I-open set is called pre-I-closed. The family of all pre-I-open sets in (X,τ,I) is denoted by PIO(X,τ) or simply PIO(X). Clearly τ⊂PIO(X). The largest pre-I-open set contained in A, denoted by pIint⁡(A), is called the pre-I-interior of A. The smallest pre-I-closed set containing A, denoted by pIcl(A), is called the pre-I-closure of A. A subset A of an ideal space (X,τ,I) is said to be α-I-open [6] if A⊂int⁡(cl⋆(int⁡(A))). The family of all α-I-open sets is a topology finer than τ. We will denote the α-I-interior subset of A of X by αIint⁡(A). A subset A of an ideal space (X,τ,I) is τ⋆-dense [7] if cl⋆(A)=X. A space (X,τ,I) is I-submaximal [8] if every τ⋆-dense subset of X is open.The following lemmas will be useful in the sequel.Lemma 1.
Let(X,τ,I) be an ideal space and let A be a subset of X. Then pIint⁡(A)=A∩int⁡(cl⋆(A)) [9].


## 2. Pre-I-Regular Pre-I-Open Sets

A pre-I-open set A of a space (X,τ,I) is said to bepre-I-regular pre-I-open if A=pIint⁡(pIcl(A)). The complement of a pre-I-regular pre-I-open set is calledpre-I-regular pre-I-closed set, equivalently A=pIcl(pIint⁡(A)). A subset A of a space (X,τ,I) with an ideal I is said to be pre-I-regular if it is pre-I-open and pre-I-closed. Clearly, X and ∅ are pre-I-regular pre-I-open.Remark 2.
Also every pre-I-regular set is pre-I-regular pre-I-open.Proof.
AssumeA is pre-I-regular that implies A is pre-I-open and pre-I-closed. A is pre-I-open which implies A=pIint⁡(A)=pIint⁡(pIcl(A)). Hence A is pre-I-regular pre-I-open.But the converse is not true as shown by Example3.Example 3.
Consider the ideal space(X,τ,I) where X={a,b,c}, τ={∅,{a},{b},{a,b},X}, and I={∅,{a}}. Here {a} is pre-I-regular pre-I-open but not pre-I-closed.Moreover, the intersection of two pre-I-regular pre-I-open sets is not pre-I-regular pre-I-open in general as Example 4 shows.Example 4.
Consider the ideal space(X,τ,I) where X={a,b,c}, τ={∅,{b,c},X}, and I={∅,{a}}. Here PIO(X)={∅,{b},{c},{a,b},{a,c},{b,c},X}. Thus, A={a,b}, B={a,c} are both pre-I-regular pre-I-open. But A∩B is not, since it is not even pre-I-open.The notions of pre-I-regular pre-I-open and I-open sets are independent of each other. Consider the space (X,τ) with an ideal I as in Example 4. Here {b,c} is I-open but not pre-I-regular pre-I-open. Consider the space X={a,b,c} and τ={∅,{a},{b},{a,b},X}, I={∅,{a}}. Here {a} is pre-I-regular pre-I-open but not I-open. Observe that every pre-I-regular pre-I-open set is pre-I-open but the converse need not be true. Here {b,c} is pre-I-open but not pre-I-regular pre-I-open.Theorem 5.
Let(X,τ,I) be an ideal space and let A, B be any subsets of X. Then the following hold.(a)
IfA⊂B, then pIint⁡(pIcl(A))⊂pIint⁡(pIcl(B)).(b)
IfA∈PIO(X), then A⊂pIint⁡(pIcl(A)).(c)
For everyA∈PIO(X), pIint⁡(pIcl(pIint⁡(pIcl(A))))=pIint⁡(pIcl(A)).(d)
IfA and B are disjoint pre-I-open sets, then pIint⁡(pIcl(A)) and pIint⁡(pIcl(B)) are disjoint.(e)
IfA is a pre-I-regular pre-I-open, then pIcl(X-A) is pre-I-regular pre-I-closed.(f)
IfA is pre-I-regular pre-I-open, then pIint⁡(A) is pre-I-regular pre-I-open.Proof.
(a)
SupposeA⊂B⇒pIcl(A)⊂pIcl(B). Therefore pIint⁡(pIcl(A))⊂pIint⁡(pIcl(B)).
(b)
Suppose thatA∈PIO(X,τ)⇒A=pIint⁡(A)⊂pIint⁡(pIcl(A)).
(c)
It is obvious thatpIint⁡(pIcl(A))∈PIO(X,τ), so by (b) we have pIint⁡(pIcl(A))⊂pIint⁡(pIcl(pIint⁡(pIcl(A)))). On the other hand pIint⁡(pIcl(A))⊂pIcl(A) which implies pIcl(pIint⁡(pIcl(A)))⊂pIcl(pIcl(A))=pIcl(A). Therefore, pIint⁡(pIcl(pIint⁡(pIcl(A))))⊂pIint⁡(pIcl(A)). Hence pIint⁡(pIcl(pIint⁡(pIcl(A))))=pIint⁡(pIcl(A)).
(d)
SinceA and B are disjoint pre-I-open sets, we have A∩B=∅ which implies A∩pIcl(B)=∅⇒A∩pIint⁡(pIcl(B))=∅. Since pIint⁡(pIcl(B)) is pre-I-open, pIcl(A)∩pIint⁡(pIcl(B))=∅. HencepIint⁡(pIcl(A))∩pIint⁡(pIcl(B))=∅.
(e)
Given thatA is pre-I-regular pre-I-open, A=pIint⁡(pIcl(A)) implies X-A=X-pIint⁡(pIcl(A))=pIcl(pIint⁡(X-A)). Therefore, pIcl(X-A)=pIcl(pIint⁡(pIcl(X − A))). Hence pIcl(X-A) is pre-I-regular pre-I-closed.
(f)
By (e) ifA is pre-I-regular pre-I-open, then pIcl(X-A) is pre-I-regular pre-I-closed. Hence X-(pIcl(X-A)) is pre-I-regular pre-I-open that implies pIint⁡(A) is pre-I-regular pre-I-open.Lemma 6.
For an ideal topological space(X,τ,I) the following are equivalent. (a)
Every pre-I-open set is open.(b)
Everyτ⋆-dense set is open.Proof.
(a)⇒ (b): let A be a τ⋆-dense set that implies cl⋆(A)=X which implies int⁡(cl⋆(A))=X, so that A⊂int⁡(cl⋆(A))=X. By (a) every pre-I-open set is open and hence A is open.
(b)⇒ (a): let B be a pre-I-open subset of X, so that B⊂int⁡(cl⋆(B))=G, say. Then cl⋆(B)=cl⋆(G), so that cl⋆((X-G)∪B)=cl⋆(X-G)∪cl⋆(B)=(X-G)∪cl⋆(B)=X, and thus (X-G)∪B is τ⋆-dense in X. Thus (X-G)∪B is open. Now B=((X-G)∪B)∩G is the intersection of two open sets, so that B is open.Lemma 7.
If a space(X,τ) with an ideal I is I-submaximal, then any finite intersection of pre-I-open set is pre-I-open.Proof.
From Lemma6, every pre-I-open set is open and hence a finite intersection of pre-I-open set is pre-I-open.Theorem 8.
If a space(X,τ) with an ideal I is I-submaximal, then any finite intersection of pre-I-regular p-I-open set is pre-I-regular pre-I-open.Proof.
Let{Oi∣i=1,2,…,n} be a finite family of pre-I-regular pre-I-open sets. Since the space X is I-submaximal, then by Lemma 6, ∩{Oi∣i=1,2,…,n} is pre-I-open. Therefore, ∩{Oi∣i=1,2,…,n}⊂pIint⁡(pIcl(∩Oi)). Also, for each i=1,2,…,n,∩Oi⊂Oi which implies pIint⁡(pIcl(∩Oi))⊂pIint⁡(pIcl(Oi)). Also, each Oi is pre-I-regular pre-I-open that implies Oi=pIint⁡(pIcl(Oi)) which implies pIint⁡(pIcl(∩Oi))⊂∩Oi and so ∩Oi=pIint⁡(pIcl(∩Oi)). Hence ∩Oi is pre-I-regular pre-I-open.It should be noted that an arbitrary union of pre-I-regular pre-I-open set is pre-I-regular pre-I-open. But the intersection of two pre-I-regular pre-I-closed sets fails to be pre-I-regular pre-I-closed as shown by Example 9.Example 9.
Consider the ideal space(X,τ,I) as in Example 3. Clearly, {a,c}, {a,b} are pre-I-regular pre-I-closed but their intersection is not pre-I-regular pre-I-closed.Theorem 10.
The following hold for a subsetA of a space (X,τ,I).(a)
IfA is pre-I-closed, then pIint⁡(A) is pre-I-regular pre-I-open.(b)
IfA=pIint⁡(A), then pIcl(A) is pre-I-regular pre-I-closed.(c)
IfA and B are pre-I-regular pre-I-closed sets, then A⊂B if and only if pIint⁡(A)⊂pIint⁡(B).(d)
IfA and B are pre-I-regular pre-I-open sets, then A⊂B if and only if pIcl(A)⊂pIcl(B).Proof.
(a) SinceA is pre-I-closed, A=pIcl(A).
Now,pIint⁡(pIcl(pIint⁡(A)))=pIint⁡(pIint⁡(A))=pIint⁡(A). Hence pIint⁡(A) is pre-I-regular pre-I-open.
(b) NowpIcl(pIint⁡(pIcl(A)))=pIcl(pIcl(A))=pIcl(A). Hence pIcl(A) is pre-I-regular pre-I-closed.
(c) Given thatA and B are pre-I-regular pre-I-closed sets, therefore, A=pIcl(pIint⁡(A)) and B=pIcl(pIint⁡(B)). Clearly, pIint⁡(A)⊂pIint⁡(B) if A⊂B.
Conversely,pIint⁡(A)⊂pIint⁡(B). Now A=pIcl(pIint⁡(A))⊂pIcl(pIint⁡(B))⊂B. Hence A⊂B.
(d) Given thatA and B are pre-I-regular pre-I-open, therefore, A=pIint⁡(pIcl(A)) and B=pIint⁡(pIcl(B)). Suppose A⊂B, pIcl(A)=pIcl(pIint⁡(pIcl(A)))⊂pIcl(pIint⁡(pIcl(B)))⊂pIcl(B). Therefore, pIcl(A)⊂pIcl(B).
Conversely,pIcl(A)⊂pIcl(B). Now A=pIint⁡(pIcl(A))⊂pIint⁡(pIcl(B))⊂B.A subsetA of an ideal topological space (X,τ,I) is said to be I-rare if it has no interior points in τ⋆.Theorem 11.
Let(X,τ,I) be an ideal space. Then the following hold.(a)
The empty set is the only subset which is nowhere dense and pre-I-regular pre-I-open.(b)
IfA is pre-I-regular pre-I-closed, then every I-rare set is pre-I-open.Proof.
(a) SupposeA is nowhere dense and A is pre-I-regular pre-I-open. Then A=pIint⁡(pIcl(A))=pIcl(A)∩int⁡(cl⋆(pIcl(A))), by Lemma 1. Therefore, A⊂pIcl(A)∩int⁡(cl⋆(cl(A)))⊂pIcl(A)∩int⁡(cl(A))=pIcl(A)∩∅=∅.
(b) SupposeA is pre-I-regular pre-I-closed. Then A=pIcl(pIint⁡(A))=pIint⁡(A)∪cl(int⁡⋆(pIint⁡(A)))⊂pIint⁡(A)∪cl(int⋆(A))=pIint⁡(A)∪∅=pIint⁡(A). Therefore, A=pIint⁡(A). Hence A is pre-I-open.An ideal space(X,τ,I) is called extremally pre-I-disconnected if the pre-I-closure of every pre-I-open set is pre-I-open.Theorem 12.
For a topological space(X,τ,I) the following are equivalent.(a)
(X,τ,I) is extremally pre-I-disconnected.(b)
Every pre-I-regular pre-I-open subset is pre-I-regular.Proof.
(a)⇒ (b): assume (X,τ,I) is extremally pre-I-disconnected. Suppose A is pre-I-regular pre-I-open. Then A is pre-I-open and so pIcl(A) is a pre-I-open set. Hence A=pIint⁡(pIcl(A))=pIcl(A). Hence A is pre-I-closed which implies A is pre-I-regular.
(b)⇒ (a): suppose A is pre-I-open. Then pIcl(A) is pre-I-regular pre-I-closed which implies X-pIcl(A) is pre-I-regular pre-I-open. Hence X-pIcl(A) is pre-I-regular. Therefore, X-pIcl(A) is pre-I-closed and so pIcl(A) is pre-I-open. Hence (X,τ,I) is extremally pre-I-disconnected.Theorem 13.
Let(X,τ,I) be an extremally pre-I-disconnected space and A⊂X. Then the following are equivalent:(a)
A is pre-I-regular,(b)
A=pIcl(pIint⁡(A)),(c)
X-A is pre-I-regular pre-I-open,(d)
A is pre-I-regular pre-I-open.Proof.
(a)⇒ (b): suppose A is pre-I-regular. Then A is pre-I-open and pre-I-closed and so A=pIint⁡(A) and A=pIcl(A). Hence A=pIcl(pIint⁡(A)).
(b)⇒ (c): let A=pIcl(pIint⁡(A)). Then X-A=X-pIcl(pIint⁡(A))=pIint⁡(pIcl(A)) so X-A is pre-I-regular pre-I-open.
(c)⇒ (d) is clear.
(d)⇒ (a) follows from Theorem 12.An ideal space(X,τ,I) is calledlocally pre-I-indiscrete if every pre-I-open subset of X is pre-I-closed (or) if every pre-I-closed subset of X is pre-I-open.Theorem 14.
Let(X,τ,I) be an ideal space. Then the following are equivalent.(a)
(X,τ,I) is locally pre-I-indiscrete.(b)
Every pre-I-open subset is pre-I-regular.(c)
Every pre-I-open subset is pre-I-regular pre-I-open.(d)
pIint⁡(pIcl({x}))≠∅, for every x∈X.(e)
The empty set is the only nowhere dense subset ofX.Proof.
(a)⇒ (b): assume that (X,τ,I) is locally pre-I-indiscrete. Let A be a pre-I-open subset of X. By hypothesis, A is pre-I-closed. Hence A is pre-I-regular.
(b)⇒ (c): if A is pre-I-open, then A=pIint⁡(A). Also by hypothesis, A is pre-I-closed. Therefore, A=pIint⁡(pIcl(A)). Hence A is pre-I-regular pre-I-open.
(c)⇒ (d): since {x} is preopen, {x} is pre-I-open. By (c), {x} is a pre-I-regular pre-I-open set. Therefore, {x}=pIint⁡(pIcl({x})).
(d)⇒ (e): by Theorem 11, in any space, the empty set is the only subset which is nowhere dense and pre-I-regular pre-I-open.
(e)⇒ (a): suppose that A is a pre-I-closed set. Now int⁡(cl(A-pIint⁡(A)))=int⁡(cl(A-(A∩int⁡(cl⋆(A)))))=int⁡(cl(A∩(X-A)∪(int⁡(cl⋆(X-A)))))=int⁡(cl(A∩(X-A)∪(A-int⁡(cl⋆(A)))))=int⁡(cl(A-int⁡(cl⋆(A))))⊂int⁡(cl(A)-int⁡(cl⋆(A)))=int⁡(cl(A)-cl(int⁡(cl⋆(A))))⊂int⁡(cl(A) − (int⁡(cl⋆(A))))=∅. Therefore A-pIint⁡(A) is nowhere dense which implies A=pIint⁡(A), and so A is pre-I-open. Hence is A locally pre-I-indiscrete.An ideal space(X,τ,I) is said to be pIR-door if every subset of X is either pre-I-regular pre-I-open or pre-I-regular pre-I-closed.Theorem 15.
Let(X,τ,I) be a pIR-door space; then every pre-I-open set in the space is pre-I-regular pre-I-open.Proof.
LetA be a pre-I-open subset of X. Since X is pIR-door, A is pre-I-regular pre-I-closed and so A=pIint⁡(pIcl(A)) which implies that pIint⁡A=pIint⁡(pIint⁡(pIcl(A))). Since A is pre-I-open, A=pIint⁡(pIcl(A)). Hence A is pre-I-regular pre-I-open.Theorem 16.
Let(X,τ,I) be an ideal space. A subset A of X is both α-I-open and α-I-closed; then A is a pre-I-regular pre-I-open set.Proof.
LetA be an α-I-open and α-I-closed set. Then A is a pre-I-open and pre-I-closed set and hence A is a pre-I-regular pre-I-open set.Theorem 17.
Let(X,τ,I) be an ideal space. A subset A of X is α-I-open and pre-I-regular pre-I-open; then A=int⁡(cl⋆(int⁡(A))).Proof.
SupposeA is α-I-open. Then A⊂int⁡(cl⋆(int⁡(A))). And A is pre-I-regular pre-I-open which implies A=pIint⁡(pIcl(A))⊃pIint⁡(cl⋆(int⁡(pIcl(A))))⊃pIint⁡(cl⋆(int⁡(A)))⊃int⁡(cl⋆(int⁡(A))). Therefore A=int⁡(cl⋆(A)).



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