# Regular pre semi I Separation axioms in Ideal Topological Spaces

DOI : 10.17577/IJERTV2IS3503

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#### Regular pre semi I Separation axioms in Ideal Topological Spaces

Regular pre semi I Separation axioms in Ideal Topological Spaces

B. Sakthi @ Sathya

Department of Mathematics G.Venkataswamy Naidu College Kovilpatti.

S. Murugesan

Department of Mathematics S.Ramasamy Naidu Memorial College Sattur.

Abstract

The authors introduced rpsI-closed sets and rpsI-open sets in ideal topo- logical spaces and established their relationships with some generalized sets in ideal topological spaces. The aim of this paper is to introduce rpsI T0,

rpsI T1, rpsI T2, rpsI T 1 , rpsI Q1 spaces and characterize their basic

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properties.

Keywords:rpsI T0 spaces, rpsI T1 spaces, rpsI T2 spaces, rpsI T 1

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spaces,

rpsI Q1 spaces

2010 AMS Subject Classification : 54A05.

1. Introduction

Separation axioms in topological spaces play a dominated role in analysis. Recently general topologists concentrate on separation axioms between T0 and T1. In this

paper the concepts of rpsI T0, rpsI T1, rpsI T2, rpsI T 1 , rpsI Q1 spaces

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are introduced, characterized and studied their relationships with T1 space and

semi-I-T2

2. Preliminaries

For a subset A of an ideal topological space(X, , I), cl(A) and int(A) denote the closure of A and interior of A respectively. X A denotes the complement of A in X. We recall the following definitions and results.

Definition 2.1. A subset A of an ideal topological space (X, , I) is called

1. semi-I-open [6] if A cl(int(A)).

2. semi pre I-open [6] if A cl(int(cl(A))).

3. I- open [6] if A int(cl(int(A))).

4. regular I-open [5] A = int(cl(A)).

Definition 2.2. A subset A of an ideal topological (X, , I) is called

1. regular generalized I-closed (rgI-closed) [2] if cl(A) U whenever A U and

U is regular I-open.

2. regular pre semi I-closed [2] (rpsI-closed) if spIcl(A) A whenever A U and

U is regular generalized I-open.

The complement of rpsI-closed set is rpsI-open set. rpsIcl(A) is the smallest

rpsI-closed set containing A.

Theorem 2.3. [2]

1. Every semi-I-closed set is rpsI-closed.

2. Every I-closed set is rpsI-closed.

Definition 2.4. A function f : (X, , I) (Y, ) is called

1. rpsI-irresolute [3] if f 1(A) is rpsI-closed in X, for every rpsI-closed subset A

of Y .

2. rpsI-open map [3] if for every rpsI-open subset F of X, then the set f (F ) is

rpsI-open in Y .

Theorem 2.5. If f : (X, , I) (Y, ) is rpsI-irresolute, then the inverse image of every rpsI-open set in Y is also a rpsI-open set in X.

Definition 2.6. An ideal topological space (X, , I) is said to be regular pre semi I-T0 (briefly rpsI T0 space) if for each pair of distinct points x, y of X, there exists an rpsI-open set containing one point but not the other.

Theorem 2.7. Every subspace of an rpsI T0 space is a rpsI T0 space.

Proof. Let X be a rpsI T0 space and Y be a subspace of X. Let x, y be two distinct points of Y . Since Y X, we have x and y are distinct points of X. Since X is rpsI T0 space, there exists an rpsI-open set G such that x G but y / G. Then there exists an rpsI-open set G Y in Y which contains x but does not contain

y. Hence Y is a rpsI T0 space.

Example 2.8. Consider the ideal topological space (X, , I), where X = {a, b, c, d}

with = {, X, {a}, {b}, {a, b}, {b, c}, {a, b, c}} and I = {, {a}}. In this ideal space

1. is a rpsI T0 space because {b, c, d} is rpsI-open set with a

not a.

b containing b but

Theorem 2.9. Given a map f : (X, , I) (Y, ) is bijective and rpsI-open, X is

rpsI T0 space, then Y is rspI T0 space.

Proof. Let Y be an ideal topological space and let u, v be two distinct points of Y . Since f is a bijection, we have , y X such that f (x) = u, f (y) = v. Since X is

rpsI T0 space, there exists rpsI-open set G in X such that x G but y / G.

Also, f is rpsI-open, f (G) is rpsI-open in Y containing f (x) = u but not containing

f (y) = v. Thus there exists a rpsI-open set f (G) in Y such that u f (G) but

v / f (G) and hence Y is a rpsI T0 space.

Theorem 2.10. Let f : (X, , I) (Y, ) be an rpsI- irresolute bijective map. If

2. is rpsI T0 space, then X is rpsI T0 space.

Proof. Assume that Y is a rpsI T0 space. Let u, v be two distinct points of Y . Since f is a bijection, we have x, y X such that f 1(u) = x; f 1(v) = y. since Y is a rpsI T0 space, there exists rpsI-open set H in Y such that u H but v / H. Since f is rpsI-irresolute, f 1(H) is rpsI-open in X containing f (x) = u but not containing f (y) = v. Thus there exists a rpsI-open set f 1(H) in X such that x f 1(H) but y / f 1(H) and hence X is a rpsI T0 space.

Definition 2.11. An ideal topological space (X, , I) is said to be rpsI T1 space if for each pair of distinct points x, y of X, there exists a pair of rpsI-open sets, one containing x but not y and the other containing y but not x.

Example 2.12. Consider the ideal topological space (X, , I), where X = {a, b, c, d} with = {, X, {a}, {b}, {a, b}, {b, c}, {a, b, c}} and I = {, {a}}. In this ideal topological space X is rpsI T1 space because {b, c, d} and {a, d} are rpsI-open sets with a 6= c, c {b, c, d} but a / {b, c, d} and a {a, d} but c / {a, d}.

Theorem 2.13. Every subspace of a rpsI T1 space is also a rpsI T1 space.

Proof. Let X be a rpsI T1 space and let Y be a subspace of X. Let x, y Y X

such that x 6= y. By hypothesis X is arpsI T1 space, hence there exists rpsI-open

set U, V in X such that x U and y V , x /

V and y /

U . By definition of

subspace, U Y and V Y are rpsI-open sets in Y . Further x U, x Y implies

x U Y also y V, y Y implies y V Y . Thus there exists rpsI-open sets U Y and V Y in Y such that x U Y, y V Y and x / V Y, y / U Y . Hence Y is a rpsI T1 space.

Theorem 2.14. Let f : (X, , I) (Y, ) be an rpsI-irresolute, injective map. If

Y is rpsI T1 space, then X is rpsI T1 space.

Proof. Assume that Y is a rpsI T1 space. Let x, y Y such that x 6= y. Then there exists a pair of rpsI-open sets U, V in Y such that f (x) U and f (y) V ,

f (x) /

V , f (y) /

U which implies x f 1(U ), y f 1(V ) and x /

f 1(V ) and

y / f 1(U ). Since f is rpsI-irresolute, we have X is rpsI T1 space.

Theorem 2.15. If {x} is rpsI-closed in X, for every x X. Then X is rpsI T1

space.

Proof. Let x, y be two distinct points of X such that {x} and {y} are rpsI-closed.

Then {x}c and {y}c are rpsI-open in X such that y {x}c but x / x {y}c but y / {y}c. Hence X is rpsI T1 space.

{x}c and

Definition 2.16. An ideal topological (X, , I) is said to be I T1 space if for each pair of distinct points x, y of X, there exists a pair of I-open sets one containing x but not y and the other containing y but not x.

Theorem 2.17. Every I T1 space is a rpsI T1 space.

Proof. Let (X, , I) be an I T1 space and x y of X. Since X is I T1

space, there exists a pair of I-open sets U, V in X such that x U and y V ,

x /

V, y /

1. . We know that every I-open set is rpsI-open.Therefore X is

rpsI T1 space.

Definition 2.18. A ideal topological space (X, , I) is said to be rpsI T2 space if for each pair of distinct points x, y of X, there exists disjoint rpsI-open sets U and V such that x U and y V .

Remark 2.19. Every rpsI T2 space is rpsI T1 space. The converse need not be true as seen from the following example.

Example 2.20. Consider the ideal topological space (X, , I), where X = {a, b, c, d} with = {, X, {a}, {b}, {a, b}, {b, c}, {a, b, c}} and I = {, {a}}. In this ideal space, X is rpsI T1 space but not rpsI T2 space. That is there exists a pair of rpsI-open sets U = {b, c, d} and V = {a, b} such that c U and a V but c / V and a / U . Here the intersection of rpsI-open sets is non-empty.

Theorem 2.21. Every subspace of a rpsI T2 space is also a rpsI T2 space.

Proof. Let X be a rpsI T2 space and let Y be a subspace of X. Let a, b Y X

with a b. By hypothesis there exists rpsI-open sets G, H in X such that a G

and b H, G H = . By definition of subspace G Y and H Y are rpsI-open

sets in Y . Further a G, a Y implies a G Y and b H, b Y implies b H Y . Since G H = , (Y G) (Y H) = Y (G H) = Y = . Therefore G Y and H Y are disjoint rpsI-open sets in Y such that a G Y and b H Y . Thus Y is rpsI T2 space.

Theorem 2.22. Given a map f : (X, , I) (Y, ) is bijection, rpsI-open and X

is rpsI T2 space. Then Y is rpsI T2 space.

Proof. Given f is a bijection and rpsI-open. Let X be a rpsI T2 space and Y be

any ideal space. Let a, b Y with a b. Since f is a bijection, there exists distinct

points x, y X such that f (x) = a and f (y) = b. Since X is a rpsI T2 space,

there exists disjoint rpsI-open sets U and V of X such that x U and y V . Also

f is rpsI-open, we have f (U ) and f (V ) are rpsI-open sets containing f (x) = a and

f (y) = b. Thus there exists disjoint rpsI-open sets f (U ) and f (V ) in Y such that

a f (U ) and b f (V ). Hence Y is rpsI T2 space.

Theorem 2.23. If {x} is rpsI-closed in X, for every x X, then X is rpsI T2

space.

Proof. Let x and y be two distinct points of X such that {x} and {y} are rpsI- closed set. Then {x}c and {y}c are rpsI-open in X such that x {y}c and y {x}c. Hence X is rpsI T2 space.

Theorem 2.24. If X is rpsI T2 space, then for y 6= x X, there exists an

rpsI-open set G such that x G and y / rpsIcl(G).

Proof. Let x, y X such that y 6= x. Since X is rpsI T2 space, there exists disjoint rpsI-open sets G and H in X such that x G and y H. Therefore Hc is rpsI-closed set such that rpsIcl(G) Hc. Since y H, we have y / Hc. Hence y / rpsIcl(G).

Definition 2.25. An ideal topological space (X, , I) is rpsI Q1 space, if for x, y X with rpsIcl({x}) 6= rpsIcl({y}), then there exists disjoint rpsI-open sets U and V such that rpsIcl({x}) U and rpsIcl({y}) V .

Theorem 2.26. If (X, , I) is rpsI T2 space, then it is rpsI Q1 space.

Proof. Let {x}, {y} be two distinct closed set in X such that for every x, y X

with rpsIcl({x}) rpsIcl({y}). We know that, every closed set is rpsI-closed and

so {x} = rpsIcl({x}),{y} = rpsIcl({y}). Since X is rpsI T2 space, there exists

disjoint rpsI-open sets U and V in X such that x U and y V . Therefore

rpsIcl({x}) U and rpsIcl({y}) V . Hence X is rpsI Q1 space.

Definition 2.27. An ideal topological space (X, , I) is called rpsI T 1

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space if for

any rpsI-closed set F X and any point x X F there exists disjoint open sets

U, V X and x U and F V .

Theorem 2.28. For any ideal space (X, , I), if x G X and G is a rpsI-open set, there exists a rpsI-open set H X such that x H H G. Then X is

rpsI T 1

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space.

Proof. Let F X be rpsI-closed set with x F c. Since F c is rpsI-open by our assumption choose a rpsI-open set H with x H H X F . Let K = X H and so W is rpsI-open. Further, F X H = W and H K = . Hence X is

rpsI T 1

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space.

Remark 2.29. We have the following diagram,

rpsI T2 = rpsI T1 = rpsI I0

Definition 2.30. An ideal space (X, , I) is said to be semi-I-T2[4], if for each pair of distinct points x and y in X, there exists two semi-I-open sets U and V in X such that x U and y V , U V = .

Remark 2.31. Every semi I-T2 space is a rpsI T2 space but the converse is not true because, every semi I-open set is rpsI-open and rpsI-open set need not be semi- I-open [2].

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