 Open Access
 Total Downloads : 569
 Authors : B. Sakthi Sathya, S. Murugesan
 Paper ID : IJERTV2IS3503
 Volume & Issue : Volume 02, Issue 03 (March 2013)
 Published (First Online): 23032013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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 rpsIclosed sets and rpsIopen 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
4
properties.
Keywords:rpsI T0 spaces, rpsI T1 spaces, rpsI T2 spaces, rpsI T 1
4
spaces,
rpsI Q1 spaces
2010 AMS Subject Classification : 54A05.

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
4
are introduced, characterized and studied their relationships with T1 space and
semiIT2

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

semiIopen [6] if A cl(int(A)).

semi pre Iopen [6] if A cl(int(cl(A))).

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

regular Iopen [5] A = int(cl(A)).

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

regular generalized Iclosed (rgIclosed) [2] if cl(A) U whenever A U and
U is regular Iopen.

regular pre semi Iclosed [2] (rpsIclosed) if spIcl(A) A whenever A U and
U is regular generalized Iopen.
The complement of rpsIclosed set is rpsIopen set. rpsIcl(A) is the smallest
rpsIclosed set containing A.
Theorem 2.3. [2]

Every semiIclosed set is rpsIclosed.

Every Iclosed set is rpsIclosed.
Definition 2.4. A function f : (X, , I) (Y, ) is called

rpsIirresolute [3] if f 1(A) is rpsIclosed in X, for every rpsIclosed subset A
of Y .

rpsIopen map [3] if for every rpsIopen subset F of X, then the set f (F ) is
rpsIopen in Y .
Theorem 2.5. If f : (X, , I) (Y, ) is rpsIirresolute, then the inverse image of every rpsIopen set in Y is also a rpsIopen set in X.
Definition 2.6. An ideal topological space (X, , I) is said to be regular pre semi IT0 (briefly rpsI T0 space) if for each pair of distinct points x, y of X, there exists an rpsIopen 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 rpsIopen set G such that x G but y / G. Then there exists an rpsIopen 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

is a rpsI T0 space because {b, c, d} is rpsIopen set with a
not a.
b containing b but
Theorem 2.9. Given a map f : (X, , I) (Y, ) is bijective and rpsIopen, 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 rpsIopen set G in X such that x G but y / G.
Also, f is rpsIopen, f (G) is rpsIopen in Y containing f (x) = u but not containing
f (y) = v. Thus there exists a rpsIopen 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

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 rpsIopen set H in Y such that u H but v / H. Since f is rpsIirresolute, f 1(H) is rpsIopen in X containing f (x) = u but not containing f (y) = v. Thus there exists a rpsIopen 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 rpsIopen 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 rpsIopen 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 rpsIopen
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 rpsIopen sets in Y . Further x U, x Y implies
x U Y also y V, y Y implies y V Y . Thus there exists rpsIopen 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 rpsIirresolute, 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 rpsIopen 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 rpsIirresolute, we have X is rpsI T1 space.
Theorem 2.15. If {x} is rpsIclosed 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 rpsIclosed.
Then {x}c and {y}c are rpsIopen 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 Iopen 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 Iopen sets U, V in X such that x U and y V ,
x /
V, y /

. We know that every Iopen set is rpsIopen.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 rpsIopen 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 rpsIopen 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 rpsIopen sets is nonempty.
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 rpsIopen sets G, H in X such that a G
and b H, G H = . By definition of subspace G Y and H Y are rpsIopen
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 rpsIopen 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, rpsIopen and X
is rpsI T2 space. Then Y is rpsI T2 space.
Proof. Given f is a bijection and rpsIopen. 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 rpsIopen sets U and V of X such that x U and y V . Also
f is rpsIopen, we have f (U ) and f (V ) are rpsIopen sets containing f (x) = a and
f (y) = b. Thus there exists disjoint rpsIopen 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 rpsIclosed 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 rpsIopen 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
rpsIopen 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 rpsIopen sets G and H in X such that x G and y H. Therefore Hc is rpsIclosed 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 rpsIopen 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 rpsIclosed and
so {x} = rpsIcl({x}),{y} = rpsIcl({y}). Since X is rpsI T2 space, there exists
disjoint rpsIopen 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
4
space if for
any rpsIclosed 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 rpsIopen set, there exists a rpsIopen set H X such that x H H G. Then X is
rpsI T 1
4
space.
Proof. Let F X be rpsIclosed set with x F c. Since F c is rpsIopen by our assumption choose a rpsIopen set H with x H H X F . Let K = X H and so W is rpsIopen. Further, F X H = W and H K = . Hence X is
rpsI T 1
4
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 semiIT2[4], if for each pair of distinct points x and y in X, there exists two semiIopen sets U and V in X such that x U and y V , U V = .
Remark 2.31. Every semi IT2 space is a rpsI T2 space but the converse is not true because, every semi Iopen set is rpsIopen and rpsIopen set need not be semi Iopen [2].
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