Download SEPARATION AXIOMS VIA

International Journal of Mathematics and Computing Applications
ISSN: 0976-6790
Vol. 3, Nos. 1-2, January-December 2011, pp. 35-47
© International Science Press
SEPARATION AXIOMS VIA g-OPEN SETS
G. Navalagi1 and Debadatta Roy Chaudhuri2
1. INTRODUCTION
In a significant contribution to the theory of generalised closed sets, several
generalisations of α-open sets were studied by Maki et al [3]. The concepts of
gα-closed and αg-closed sets were introduced and all notions were defined through
α-open sets. Accordingly corresponding separation axioms were set using gα-open
and αg-open sets. In 2004 Rajamani and Viswanathan [14] defined αgs-closed sets
and quite recently Navalagi, Rajamani and Viswanathan [11] defined separation
axioms using these sets. In 1987, Andrijevic [1] introduced a new class of sets called
γ-open sets (A subset A of a topological space X is called γ-open set iff A ∩ B ∈
PO(X) for every B ∈ PO(X)). Quite recently G.B. Navalagi, M. Thivagar and R.
Rajarajeswari [12] defined and studied some properties of separation axioms using
these sets. In a parallel way the study of γg-open sets *evokes the consideration of
new separation axioms. The aim of this paper is to introduce concepts of γg-seperation
axioms using γg-closed and γg-open sets.
2. PRELIMINARIES
Throughout this paper (X, τ) or, simply X denotes a topological space. For any subset
A ⊂ X, Int(A) and Cl(A) denotes the interior and closure of A respectively.
Definition 2.1: A subset A of (X, τ) is called a preopen set [5] if A ⊂ Int(Cl(A).
The complement of a preopen set of (X, τ) is called a preclosed set [5]. The family of
all preopen (preclosed) sets of (X, τ) is denoted by PO(X) (resp. PC(X)). The family
of all preopen sets of X containing a point x ∈ X is denoted by PO(x). Clearly every
open set is preopen set and a dense set in X is also preopen in X.
For a subset A of X, the union of all preopen sets X which are contained in A is
called the preinterior of A and is denote it by pInt(A). Since the union of preopen sets
is preopen, pInt(A) is preopen in X. The intersection of all preclosed sets of X
1
Department of Mathematics, KLE Society’s, G.H. College, Haveri-581110, Karnataka, India.
E-mail: [email protected]
2
K.J. Somiaya College of Arts and Commerce, Vidyanagri, Vidyavihar, Mumbai-400077.
E-mail: [email protected]
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containing a set A is called the preclosure of A and is denoted by pCl(A), this is the
smallest preclosed set in X containing A.
Definition 2.2: A subset A of X is said to be generalized closed [2] (= g-closed)
iff Cl(A) ⊂ U whenever A ⊂ U and U is open in X. The complement of a g-closed set
in X is called g-open set. If a subset A of X is g-open iff F ⊂ Int(A) whenever F ⊂ A
and F is closed in X.
Definition 2.3: A subset A of X is said to be generalized preclosed [13] (= gp-closed)
iff pCl(A) ⊂ U whenever A ⊂ U and U is open in X. The complement of a gp-closed
set in X is called gp-open set. It can be easily obtained from the definition that a
subset A of X is gp-open iff F ⊂ pInt(A) whenever F ⊂ A and F is closed in X. The
family of all gp-open (gp-closed) sets of (X, τ) is denoted by GPO(X) (resp. GPC(X)).
The family of all gp-open (gp-closed) sets of X containing a point x ∈ X is denoted
by GPO(x) (resp. GPC(x)).
Definition 2.4: A subset A of a topological space X is called γg-open [9] set iff A
∩ B ∈ GPO(X) for every B ∈ GPO(X). The family of all γg-open subsets of X is
denoted by γgO(X). The complement of a γg-open set is called γg-closed set. The
family of all γg-closed subsets of X is denoted by γgC(X).
The union of all γg-open sets which are contained in A is called the γg-interior of
A and is denote it by γgInt(A). If A ⊂ X is γg-open in X then A = γgInt(A). The intersection
of all γg-closed sets containing a set A is called the γg-closure of A and is denoted by
γgCl(A). If A ⊂ X is γg-closed in X then A = γgCl(A).
Definition 2.5: A function f:(X, τ) → (Y, σ) is called
1. γ-continuous if[12] f –1(U) is γ-open in X whenever U is open in Y.
2. γ-irresolute if[12] f –1(U) is γ-open in X whenever U is γ-open in Y.
Definition 2.6: [14] A topological space is called
1. gp*-T0 if for any distinct pair of points x and y of X, there exists a gp-open set
of X containing x but not y or a gp-open set containing y but not x.
2. gp*-T1 if for any distinct pair of points x and y of X, there exists a gp-open set
of X containing x but not y and a gp-open set containing y but not x.
3. gp*-T2 if for any distinct pair of points x and y of X, there exist disjoint
gp-open sets U and V containing x and y respectively.
4. gp*-regular if for each closed set F of X and each point x in X–F, there exist
disjoint gp-open sets U and V such that x is in U and F is contained in V.
5. gp*-normal if for each pair of disjoint closed sets A and B, there exist disjoint
gp-open sets U and V such that A ⊂ U and B ⊂ V.
Seperation Axioms via gp-open Sets
37
Definition 2.7: [12] A topological space is called
1. γ-T0 if for any distinct pair of points x and y of X, there exists a γ-open set of X
containing x but not y or a γ-open set containing y but not x.
2. γ-T1 if for any distinct pair of points x and y of X, there exists a γ-open set of X
containing x but not y and a γ-open set containing y but not x.
3. γ-T2 if for any distinct pair of points x and y of X, there exist disjoint γ-open
sets U and V containing x and y respectively.
Definition 2.8: A topological space (X, τ) is said to be a T1/2 space [4] if and only
if every g-closed set is a closed or equivalently every g-open set is open.
3.
g-T0
SPACE
Definition 3.1: A topological space (X, τ) is said to be a γg-T0 space if and only
if given any pair of distinct points x and y of (X, τ) there exist a γg-open set G
containing one of them but not the other.
Theorem 3.2: A topological space (X, τ) in which every singleton subset {x} of
X is gp-open, is γg-T0 space.
Proof: Obvious.
The converse of this theorem is generally not true as shown by the following
example:
Example 3.3: Consider the set X = {a, b, c, d} and τ = { Ø ,{a, b}, {a, b, c}, X}.
Then it can be easily verified that τ is a topology on X and that,
γO(X) = { Ø , {a}, {b}, {a, b}, {a, d}, {b, c}, {a, b, c}, {a, b, d}, X}.
γgO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {a, b, d}, X}.
It is easy to check that (X, τ) is a γg-T0 space. But {d} is not γg-open set.
Theorem 3.4: A topological space (X, τ) is γg-T0 if for every pair of distinct
points x, y ∈ X, one of the singleton subset {x} or {y} of X is γg-closed.
Proof: Let x and y be two distinct points of X. Without loss of generality let us
assume that {x} is γg-closed. Thus X–{x} is a γg-open set containing y but not x and
hence X is γg-T0 space.
The converse of this theorem is generally not true as shown by the following
example:
Example 3.5: In Example 3.3 the space (X, τ) is a γg-T0 space but for the pair of
points a, b ∈ X, neither {a} nor {b} is γg-closed.
Theorem 3.6: A topological space is γg-T0 if the intersection of all γg-open sets
containing an arbitrary point of X is singleton.
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Proof: Let x, y be any two distinct points of X. By hypothesis the intersection of
all γg-open sets containing x is {x}. Hence there exists a γg-open set containing x
which does not contain y. Therefore X is γg-T0.
Theorem 3.7: If a topological space X is γg-T0 then distinct points have distinct
γg-closures.
Proof: Let X be a γg-T0 and let x, y be distinct points of X. Thus there exists
γg-open set G such that x ∈ G but y ∉ G or there exists γg-open set H such that y ∈ H
but x ∉ H. Suppose that there exists γg-open set G such that x ∈ G but y ∉ G. This
implies that X – G is a γg-closed set containing y but not x. Hence γgCl({y}) ⊆ X – G
and hence x ∉ γgCl({y}). Thus x ∈ γgCl({x}) and x ∉γgCl({y}) which implies that
γgCl({x} ≠ γgCl({y}). Suppose that there exists γg-open set H such that y ∈ H but x ∉
H. Then similar argument shows y ∈ γgCl({y}) and y ∉ γgCl({x}) which implies that
γgCl({x}) ≠ γgCl({y}).
4.
g-T1
SPACE
Definition 4.1: A topological space (X, τ) is said to be a γg-T1 space if and only
if given any pair of distinct points x and y of (X, τ) there exist γg-open sets G and H
such that x ∈ G but y ∉ G and y ∈ H but x ∉ H.
Obviously γg-T1 space implies γg-T0 space, but the reverse implication does not
hold generally as shown by the following example:
Example 4.2: Consider the set X = {a, b, c, d} and τ = { Ø , {a, b}, {a, b, c}, X}.
Then it can be easily verified that τ is a topology on X and that,
γO(X) = { Ø , {a}, {b}, {a, b}, {a, d}, {b, c}, {a, b, c}, {a, b, d}, X}.
γgO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {a, b, d}, X}.
It is easy to check that (X, τ) is a γg-T0 space but not γg-T1 because for the points
b, d ∈ X, there does not exists any γg-open sets G and H such that b ∈ G but d ∉ G
and d ∈ H but b ∉H.
Remark 4.3: Since every γ-open set is a γg-open set therefore γ-T1 space implies
γg-T1 space, but the reverse implication does not necessarily hold as shown by the
following example:
Example 4.4: Let X = {a, b, c, d} with topology τ = { Ø , {b}, {d}, {b, d}, X}.
Then, γO(X) = { Ø , {b}, {d}, {b, d}, {a, b, d}, {b, c, d}, X},
γgO(X) = { Ø , {a}, {b}, {c}, {d}, {c, d}, {a, d}, {a, b}, {b, c}, {b, d}, {a, b, d},
{b, c, d}, X}
It is easy to see that (X, τ) is a γg-T1 space but not a γ-T1 space because for the
points a, d ∈ X, there does not exists any γ-open sets G and H such that a ∈ G but d
∉ G and d ∈ H but a ∉ H.
Seperation Axioms via gp-open Sets
39
Remark 4.5: Since every γg-open set is a gp-open set therefore γg-T1 space implies
gp*-T1 space, but the reverse implication does not necessarily hold as shown by the
following example:
Example 4.6: Consider the set X = {a, b, c, d} and τ = { Ø , {a, b}, {a, b, c}, X}.
Then
GPO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {a, b, c}, {a,
b, d}, {a, c, d}, {b, c, d}, X}
γgO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {a, b, d}, X}.
It is easy to observe that (X, τ) is a gp*-T1 space. But for the points b, d ∈X, there
does not exists any γg-open sets G and H such that b ∈ G but d ∈ G and d ∈ H but
b ∉ H.
Theorem 4.7: Let x be any arbitrary point of a topological space X. If X is γg-T1
then for every y ∈ X –{x} there exists a γg-open set Gy such that y ∈Gy ⊂ X –{x}.
Proof: Let the topological space X be γg-T1 and let y ∈ X –{x}. Then x ≠ y. As X
is γg-T1, Therefore there exists Gy ∈ γgO(X) such that y ∈ Gy but x ∉ Gy. This implies
that y ∈ Gy ⊆ X – {x}.
Theorem 4.8: A topological space is γg-T1 if and only if the intersection of all
γg-open sets containing an arbitrary point of X is singleton.
Proof: Let the space X be γg-T1. Let N be the intersection of all γg-open sets
containing an arbitrary point x of X. Let y ∈ X be any point of X different from x.
Since the space is γg-T1, there exists a γg-open set which contains x but not y.
Consequently y ∉ N. Since y is arbitrary, no point of X other than x can belong to N.
It follows that N = {x}.
Conversely, let x, y be any two distinct points of X. By hypothesis the intersection
of all γg-open sets containing x is {x}. Hence there exists a γg-open set containing x
which does not contain y. Similarly there must be a γg-open set containing y but not
x. Hence X is γg-T1.
Theorem 4.9: A topological space in which every singleton is not nowhere dense
is a γg-T1 space.
Proof: Let x be an arbitrary point of a topological space X. As every subset of X
is either nowhere dense or preopen, therefore {x} not nowhere dense implies that
{x} is preopen. We claim that X – {x} is gp-closed. For,
•
Case 1: (when X – {x} is open) then only open sets containing X – {x} is X
– {x} and X and therefore for X – {x} ⊆ X –{x} then pCl[X –{x}] = X –pInt
{x} = X –{x}. For X –{x} ⊆ X then obviously
pCl(X – {x}) ⊆ X.
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•
Case 2: (when X –{x} is not open) then the only open set containing X –
{x} is X and hence X – {x} is obviously gp-closed.
Thus X – {x} is gp-closed which implies that {x} is gp-open. Thus all singleton
subsets of X are also γg-open. Hence the space X is γg-T1 space.
Theorem 4.10: If a topological space X is γg-T1 then for distinct points x, y of X,
x ∉ γgCl({y}) and y ∉ γgCl({x}).
Proof: Let X be a γg-T1 and let x, y be distinct points of X. Thus there exists
γg-open set G such that x ∈ G but y ∉ G. This implies that X – G is a γg-closed set
containing y but not x. Hence γgCl({y}) ⊆ X – G and hence x ∉ γgCl({y}). Also, there
exists γg-open set H such that y ∈ H but x ∉ H. This implies that X – γg is a γg-closed
set containing x but not y. Hence γgCl({x}) ⊆ X – H and hence y ∉ γgCl({x}).
Corollary 4.12: If X is a γg-T1 topological space, then for any x ∈ X, γgCl{x} =
{x}.
Proof: Follows easily from Theorem 4.10.
5.
g-T2
SPACE
Definition 5.1: A topological space (X, τ) is said to be a γg-T2 space if and only
if given any pair of distinct points x and y of (X, τ) there exist disjoint γg-open sets G
and H such that x ∈ G and y ∈ H. It is obvious from the definition that γg-T2 implies
γg-T1.
Remark 5.2: Since every γ-open set is a γg-open set therefore γ-T2 space implies
γg-T2 space, but the reverse implication does not necessarily hold as shown by the
following example:
Example 5.3: Consider the set X = {a, b, c} equipped with the topology τ = { Ø ,
{b}, X}. Then,
γO(X) = { Ø , {b}, {a, b}, {b, c}, X}
γgO(X) = { Ø , {b}, {c}, {a}, {a, b}, {b, c}, X} and
Clearly the space is γg-T2. But for the points a, b ∈ X, there does not exist any
disjoint γ-open sets G and H such that a ∈ G and b ∈ H. Hence X is not γ-T2.
Remark 5.4: Since every γg-open set is a gp-open set therefore γg-T2 space implies
gp*-T2 space, but the reverse implication does not necessarily hold as shown by the
following example:
Example 5.5: Consider the set X = {a, b, c, d} and τ = { Ø , {a, b}, {a, b, c},
X}. Then
GPO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {a, d},{b, c}, {b, d}, {a, b, c}, {a,
b, d}, {a, c, d}, {b, c, d}, X}
Seperation Axioms via gp-open Sets
41
γgO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {a, b, d}, X}.
It is easy to observe that (X, τ) is a gp*-T2 space. But for the points a, d ∈X, there
does not exists any disjoint γg-open sets G and H such that a ∈ G and d ∈ H.
Theorem 5.6: If a space X is γg-T2 then for distinct points x, y ∈ X, there exists
a γg-open set U containing x such that y ∉ γgCl(U).
Proof: Let x ≠ y be distinct points of X. As X is γg-T2, there exists disjoint
γg-open sets U, V such that x ∈ U and y ∈ V which implies that y ∉ γgCl(U).
Corollary 5.7: If a space X is γg-T2 then for distinct points x, y ∈ X, there exists
a γg-open set U containing x such that y ∈γgInt(X – U).
Proof: As X – γgCl(U) = γgInt(X – U), thus y ∉ γgCl(U) implies that y ∈
γgInt(X – U).
Theorem 5.8: If a space X is γg-T2 then for each x ∈ X, ∩ {γgCl(U) | x ∈ U and
U ∈ γgO(X)} = {x}.
Proof: Let x ∈ X and y ∈ X be a point distinct from x. Then there exists a γg-open
set U containing x such that y ∉γgCl(U), which implies that y ∉ ∩ {γgCl(U) | x ∈ U
and U ∈ γg O(X)} and hence ∩ {γgCl(U) | x ∈ U and U ∈ γg O(X)} = {x}.
Theorem 5.9: If a topological space is γg-T2 then the intersection of all γg-open
sets containing an arbitrary point of X is singleton.
Proof: Let the space X be γg-T2. Let N be the intersection of all γg-open sets
containing an arbitrary point x of X. Let y ∈ X be any other point of X different from
x. As X is γg-T2, there exists disjoint γg-open sets U, V such that x ∈ U and y ∈ V.
Thus y ∉ U and consequently y ∉ N. As y is arbitrarily chosen so N = {x}.
6.
g-REGULAR
SPACE
Definition 6.1: A topological space (X, τ) is said to be a γg-regular space if and
only if for each closed set F and each x ∉ F, there exist disjoint γg-open sets G and H
such that x ∈ G and F ⊂ H. A space which is γg-T2 need not be γg-regular as shown
by this example.
Example 6.2: Consider the set X = {a, b, c} equipped with the topology τ = { Ø ,
{b}, X}. Then,
γO(X) = { Ø , {b}, {a, b}, {b, c}, X}
γgO(X) = { Ø , {b}, {c}, {a},{a, b}, {b, c}, X} and
Clearly the space is γg-T2 but for the closed set {a, c} and the point b ∈ X there
does not exist any disjoint γg-open sets G and H such that b ∈ G and {a, c} ⊂ H.
Hence X is not γg-regular.
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Note that γg-T2 and γg-regular are independent of each other.
Theorem 6.3: If a topological space is γg-regular then for each x ∈ X and for
each open set U containing x, there exists a γg-open set V containing x such that x ∈V
⊂ γgCl(V) ⊂ U.
Proof: Let X be γg-regular. Let x ∈ X and U be an open set containing x which
implies that X–U is a closed set such that x ∉ X – U. Therefore there exists disjoint
γg-open sets V and W such that x ∈ V and X– U ⊂ W, that is X – W ⊂ U. Since V ∩ W
= Ø , therefore γgCl(v) ∩ W = φ and so γgCl(V) ⊂ X – W ⊂ U. Hence x ∈ V ⊂ γgCl
(V) ⊂ U.
The converse of this theorem is not true in general as illustrated by the
following example.
Example 6.4: Refer to the space (X, τ) in Example 6.2. The only open set
containing the points a and c is X and hence the condition x ∈ V γgCl(V) ⊂ U is
obviously satisfied for a, c in X. Also it is easy to see that γgCl({b}) = {b} and
therefore for b belonging to {b} there exists γg-open set {b} such that b ∈ {b} ⊂
γgCl({b}) = {b} ⊂ {b}. But the space is not γg-regular.
Theorem 6.5: If X is γg-regular, then for any closed subset F of X, ∩ {γgCl(V) |
F ⊂ V and V ∈ γgO(X)} = F.
Proof: Let F be a closed subset of X and x ∉ F. Then there exist disjoint γg-open
sets U and V such that x ∈ U and F ⊂ V. This implies that U ∩ γgCl(V) = Ø and thus
x ∉ γgCl(V). Hence, x ∉ ∩ {γgCl(V) | F ⊂ V and V ∈ γgO(X)} and since x is arbitrary
therefore ∩{γgCl(V) | F ⊂ V and V ∈ γgO(X)} = F.
Theorem 6.6: If X is γg-regular, then for each non-empty subset A of X and each
open set U such that A ∩ U ≠ Ø , there exists V ∈γgO(X) such that A ∩ V ≠ Ø and
γgCl(V) ⊂ U.
Proof: Let A be a non-empty subset of X and U an open set of X such that A ∩ U
≠ Ø . Therefore there exists x0 ∈ X such that x0 ∈ A ∩ U. Thus U is an open set
containing x0. As X is γg-regular, by Theorem 6.3 there exists a γg-open set V such
that x0 ∈ V ⊂ γgCl(V) ⊂ U. Since x0 ∈ A, so A ∩ V ≠ Ø .
The converse of the above theorem is not true in general as illustrated by the
following example:
Example 6.7: Consider the space (X, τ) in Example 6.2. The space is clearly not
γg-regular. But for each non-empty subset A of X and each open set U such that A ∩
U ≠ Ø , there exists V ∈ γgO(X) such that A ∩ V ≠ Ø and γgCl(V) ⊂ U.
Corollary 6.8: If X is γg-regular, then for each non-empty subset A of X and each
closed set F such that A ∩ F = Ø , there exists V ∈γgO(X) such that A ∩ V ≠ Ø and
F ⊂ γgInt(X – V).
Proof: Immediate consequence of Theorem 6.6.
Seperation Axioms via gp-open Sets
7.
g-NORMAL
43
SPACE
Definition 7.1: A topological space (X, τ) is said to be a γg-normal space if and
only if for each pair of disjoint closed set F1 and F2, there exist disjoint γg-open sets
G and H such that F1 ∈ G and F2 ⊂ H.
Note that γg-normal and γg-regular are independent of each other.
Theorem 7.2: If a topological space is γg-normal then for each closed set A and
for each open set U containing A, there exists a γg-open set V such that A ⊂ V ⊂
γgCl(V) ⊂ U.
Proof: Let A be a closed set of X and U be an open set containing A. This implies
that X – U is a closed set disjoint from A. Therefore there exists disjoint γg-open sets
V and W such that A ⊂ V and X – U ⊂ W, that is X – W ⊂ U. Since V ∩ W = Ø ,
therefore γgCl(V) ∩ W = Ø and so γgCl(V) ⊂ X – W ⊂ U. Hence A ⊂ V ⊂ γgCl(V)
⊂ U.
Theorem 7.3: If a topological space is γ g-normal then for each pair of
disjoint closed set A and B of X, there exists a γg-open set U such that A ⊂ U and
γgCl(U) ∩ B = Ø .
Proof: Let A and B be disjoint closed sets of X. Then A is a closed set such that A
is contained in X – B. Therefore, by Theorem 7.2 there exists a γg-open set U such
that A ⊂ U ⊂ γgCl(U) ⊂ X – B. Hence A ⊂ U and γgCl(U) ∩ B = Ø .
Theorem 7.4: Every topological space X which is γg-regular is γg-normal if γgO(X)
is closed under arbitrary unions and intersections.
Proof: Let X be γg-regular and A and B be any pair of disjoint closed sets of X.
Let x ∈ A. Then x ∉ B and therefore, as X is γg-regular, for each x ∈ A there exists
disjoint γg-open sets Ux and Vx such that x ∈ Ux and B ⊂ Vx. Hence A ⊂∪ Ux and B
⊂∩ Vx. As γgO(X) is closed under arbitrary unions and intersections therefore ∪ Ux
and ∩ Vx are disjoint γg-open sets containing A and B respectively. Hence X is
γg-normal.
Theorem 7.5: If X is a γg-normal space then every pair of disjoint closed set and
a g-closed set is separated by disjoint γg-open sets.
Proof: Let F be a closed set and A a g-closed set disjoint from F. Then A ⊂ X – F.
As X – F is an open set containing A, therefore by definition of g-closed sets Cl(A) ⊂
X – F and thus Cl(A) ∩ F = Ø . Since Cl(A) and F are disjoint closed subsets of X,
and X is γg-normal, so there exists γg-open sets U and V such that A ⊂ Cl(A) ⊂ U and
F ⊂ V.
Definition 7.6: A function is f : X → Y is said to be γg-closed if for each closed
set F of X, f(F) is γg-closed in X.
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Theorem 7.7: A surjective function f : X → Y is γg-closed iff for each subset B of
Y and each open set U of X containing f–1(B), there exists a γg-open set V of Y such
that B ⊂ V and f –1(V) ⊂ U.
Proof: (if part) Suppose that f is γg-closed. Let B be any subset of Y and U an
open set of X such that f–1(B) ⊂ U. Set V = Y – f(X – U). Then V is γg-open set in Y and
B ⊂ V and f –1(V) ⊂ U.
(only if part) Let F be any closed set of X. Put B = Y – f (F). Then f –1(B) ⊂ X – F
and X – F is open in X. By hypothesis, there exists a γg-open set V of Y such that B =
Y – f(F) ⊂ V and f –1(V) ⊂ X – F. Therefore Y – f (F) = V or Y – V = f (F). Hence f(F)
is γg-closed in Y. This proves that f is a γg-closed function.
Theorem 7.8: If f : X → Y is a continuous, γg-closed surjective function on a
normal space X, then Y is γg-normal space.
Proof: Let A and B be disjoint closed subsets of Y. As f is continuous f –1(A) and
f –1(B) are disjoint closed sets of X. Now X is normal implies that there exists disjoint
open sets U and V such that f –1(A) ⊂ U and f –1(B) ⊂ V. By Theorem, there exists
γg-open sets G and H of Y such that A ⊂ G and f –1(G) ⊂ U, B ⊂ H and f –1(H) ⊂ V.
Thus f –1(G) ∩ f –1(H) = Ø and hence G ∩ H = Ø . It follows from definition that Y
is γg-normal space.
8.
g-T1/2
SPACE
Definition 8.1: A topological space (X, τ) is said to be a γg-T1/2 space if and only
if every gp-closed set is γg-closed or equivalently every gp-open set is γg-open.
Example 8.2: Let X = {a, b, c, d} with topology τ = { Ø , {b}, {d}, {b, d}, X}.
Then GPC (X) = γgC (X) = { Ø , {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c}, {a, b}, {b,
c}, {c, d}, {a, d}, {a, c}, {c}, {a}, X}. Thus this topological space is a γg-T1/2 space.
A γ g-T 1/2 space is not necessarily a T 1/2 space. This is supported by the
following example:
Example 8.3: Let X ={a, b, c} with topology τ = { Ø , {b}, X}. Then, g-closed
sets = { Ø , {b, c}, {a, c}, {a, b}, {a}, {c}, X}
GPC(X) = γgC(X) = { Ø , {a, c}, {a, b}, {b, c}, {c}, {a}, X}.
This is a γg-T1/2 space which is not a T1/2.
Theorem 8.4: Every γg-T1/2 space is a γg-T0 space.
Proof: Let x and y be two distinct points of a γg-T1/2 space X. Now (by Theorem
4.8 of [7]) for each x ∈ X, X –{x} is either gp-closed or open. As X is γg-T1/2 space,
therefore if X – {x} is gp-closed implies that X – {x} is γg-closed and thus {x} is a
γg-open set containing x but not y. If however X – {x} is open and therefore γg-open,
Seperation Axioms via gp-open Sets
45
so X – {x} is a γg-open set containing y but not x. Thus in either case the space X is
γg-T0 space.
Every γg-T0 space is not necessarily a γg-T1/2 space as shown by the following
example.
Example 8.5: Consider the set X = {a, b, c, d} and τ = { Ø , {a, b}, {a, b, c}, X}.
Then it can be easily verified that τ is a topology on X and that,
GPC(X)={ Ø , {b, c, d}, {a, c, d}, {a, b, d}, {c, d}, {b, d}, {b, c}, {a, d}, {a, c},
{d}, {c}, {b}, {a}, X}.
γgO(X) = { Ø , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {a, b, d}, X}
γgC(X) = { Ø , {b, c, d}, {a, c, d}, {a, b, d}, {c, d}, {b, d}, {a, d}, {d}, {c}, X}.
It is easy to check that (X, τ) is a γg-T0 space. But {a} is a gp-closed set which is
not γg-closed and hence X is not γg-T1/2 space.
Theorem 8.6: If a topological space X is a γg-T1/2 space then for every subset A
of X, gpCl(A) = γgCl(A).
Proof: Let X be a γg-T1/2 space. Thus every gp-closed set is γg-closed. Therefore,
by definition of gp-closure and γg-closure we have gpCl(A) = γgCl(A).
Definition 8.7: [13] For a topological space X pτ* = {W ⊂ X|gpCl(X – W) =
X – W}.
Definition 8.8: For a topological space X γgτ* = {W ⊂ X|γgCl(X – W) = X – W}.
Theorem 8.9: If a topological space X is a γg-T1/2 space then γgτ* = pτ*.
Proof: From the previous theorem as gpCl(A) = γgCl(A) for a γg-T1/2 space,
therefore γgτ* = pτ*.
Remark 8.10: GPO(X) is not closed under finite intersections generally. This
fact is supported by the following example.
Example 8.11: Consider the set X = {a, b, c, d} and τ = { Ø ,{a, b}, {a, b, c}, X}.
It has been verified that τ is a topology on X and,
GPO(X)={ Ø , {a}, {b}, {c}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {a, b, c}, {a,
b, d}, {a, c, d}, {b, c, d}, X}.
Now {b, d} ∩ {a.c.d} = {d} ∉ GPO(X). Hence GPO(X) is not closed under
intersections.
However for a γg-T1/2 space GPO(X) is closed under finite intersections as shown
by the following theorem.
Theorem 8.12: If a topological space X is a γg-T1/2 space then GPO(X) is closed
under finite intersections.
46
International Journal of Mathematics and Computing Applications
Proof: Since for a γg-T1/2 space, every gp-open set is γg-open, therefore if A and
B are two gp-open sets then they are γg-open and as γg-open sets are closed under
finite intersections, the theorem follows.
Definition 8.13: A function f : x → y is called γg continuous if f –1(U) is γg-open
in X whenever U is open in Y.
Theorem 8.14: Let X be a γg-T1/2 space. If f : X → Y is a gp-continuous function
then f is γg-continuous.
Proof: Since for a γg-T1/2 space, every gp-open set is γg-open. Thus if U is an
open set in Y then, f being gp-continuous, f –1(U) is a gp-open set in X which implies
that f –1(U) is γg-open in X.
ACKNOWLEDGEMENT
The second author acknowledges the University of Mumbai for its financial support
for completing this paper under Minor Research Project (APD/237/119 of 2010).
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