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Imperial Journal of Interdisciplinary Research (IJIR)
Vol-3, Issue-1, 2017
ISSN: 2454-1362, http://www.onlinejournal.in
On gωα -Separation Axioms In Topological Spaces
S. S. Benchalli1, P. G. Patil2 & Pushpa M. Nalwad3
123
Department of Mathematics, Karnatak University, Dharwad-580 003, Karnataka, India.
Abstract: The aim of this paper is to study and
investigate the separation axioms: g - Ti spaces
Definition 2.1 A subset A of a space X is called
(i) a semiopen set [11] if A cl (int ( A)) .
(for i=1, 2, 3) and weaker forms of regular and
normal spaces by using the notions of g -open
and g -closed sets. Moreover, some of their
properties, which belong to them are studied.
(ii) an -open set [19] if A int (cl (int ( A))) .
(iii) a regular-open set [9] if A = int(cl(A)).
The complements of the above mentioned sets are
called their respective closed sets.
1. Introduction
Definition 2.2 [3] A subset A of X is g -closed if
It is observed from literature that there has been a
considerable work on different relatively weak forms
of separation axioms, like Ti (i = 0,1,2) spaces,
X is denoted by
regularity and normality axioms in particular, several
other neighbouring forms of them have also been
studied in many papers. For instance semiTi (i = 0,1,2) spaces which has been defined by
Maheshwari and Prasad [14, 15] which are weaker
then Ti (i = 0,1, 2) spaces. Maheshwari and Prasad
introduced the new class of spaces called s-regular
[14] and s-normal [15] spaces using semi-open sets.
It was further studied by Noiri and Popa [13], Dorsett
[8] and Arya [1]. Munshi [16] introduced g-regular
and g-normal spaces using g-closed sets of Levine
[12]. Later, El-Deeb et al. [5] and Paul et al. [18]
introduced p-regular and p-normal spaces. Later,
Benchalli et al. [2], Shik John [10] and P. G. Patil et
*
*
al. [17] studied the concept of g -pre regular, g pre normal, -normal, -regular and regular, -normal spaces in topological spaces.
Recently, Benchalli et al. [3] [4] introduced and
studied the properties of g - closed sets and
g -continuous functions. In this paper, the
notion of g -open sets in topological space is
used to define new separation axioms and obtained
some of their properties. Also, the relationship with
some other functions are discussed.
cl(A) U whenever A U and U is open in X.
The family of all g -closed subsets of the space
G C(X).
Definition 2.3 [3] The intersection of all g closed sets containing a set A is called g closure of A and is denoted by g -cl(A).
A set A is g -closed set if and only if g cl(A) = A.
Definition 2.4 [3] The union of all g -open sets
contained in A is called g -interior of A and is
denoted by g -int(A).
A set A is g -open if and only if g -int(A) =
A.
Definition 2.5 A topological space X is said to be a
(i) g-regular [13] (resp. -regular [6], -regular
[10], -regular [17]) if for each g-closed (resp.
closed, -closed, -closed) set F of X and each
point x F , there exist disjoint open (resp. open, open, open) sets U and V such that F U
and x V .
(ii) s-regular [14] (semi regular [7]) if for each closed
(resp. semiclosed) set F of X and each point x F ,
there exist disjoint semiopen sets U and V such that
F V and X U .
2. Preliminaries
Throughout this paper (X, ) and (Y, ) (or
simply X and Y ) always mean topological spaces
on which no separation axioms are assumed unless
explicitly stated. For a subset A of a space X the
closure and interior of A with respect to are
denoted by cl(A) and int(A) respectively.
Imperial Journal of Interdisciplinary Research (IJIR)
Definition 2.6 A topological space X is said to be a
(i) g-normal [13] (resp. -normal [6], -normal
[10], -normal [17] if for any pair of disjoint gclosed (resp. closed, -closed, -closed) sets A
and B, there exist disjoint open (resp. -open, open,
open) sets U and V such that A U and B V .
(ii) s-normal [15] (semi normal [8]) if for each pair
of disjoint closed (semiclosed) sets A and B in X
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Imperial Journal of Interdisciplinary Research (IJIR)
Vol-3, Issue-1, 2017
ISSN: 2454-1362, http://www.onlinejournal.in
there exist disjoint semiopen (open) sets U and V
such that A U and B V .
3. Separation Axioms
In this section, we introduce and study weak
separation axioms such as g T0 , g T1
y1 = f ( x1 ) f (G) and y2 = f ( x2 ) f (G) .
Since f being strongly g -open function,
f (G) is g -open in Y. Thus, there exists a
g -open set f (G) in Y such that y1 f (G)
and y2 f (G) . Therefore Y is g T0 space.
and g T2 spaces and obtain some of their
properties.
Definition 3.2 A topological space X is said to be
g T1 space if for each pair of disjoint points
x and y there exist g -open sets G and H such
Definition 3.1 A topological space X is said to be a
x space if for each pair of disjoint points x and y
of X there exists a g -open set containing one
point but not the other.
that
x G , y G and x H , y H .
Theorem 3.3 A topological space X is g T1 if
and only if singletons are g - closed sets.
Theorem 3.1 A topological space X is g T0 -
Proof. Let X be a g T1 space and
is g T0 -space, there exists a g -open set
consequently,
space if and only if g -closure of distinct points
are distinct.
Proof. Let x and y be distinct points of X. Since X
G such that x G and y G consequently, X G is a g -closed set containing y but not x ,
but g -cl( y ) is the intersection of all g closed set containing y . Hence y g -cl( y )
but x g -cl(y) as x X - G. Therefore,
g -cl( x ) g -cl( y ).
Conversely: Let g -cl( x ) g -cl( y ) for
x y . Then there exists at least one point z
z g -cl {x} but z g cl {y} . We claim x g -cl {y} , because if x
g -cl {y} then {x} g -cl {y}
implies g -cl {x} g -cl {y} . So z
g -cl {y} , which is contradiction. Hence x
g -cl {y} which implies x X - g cl {y} , which is g -open set containing x but
not y . Hence X is g T0 space.
X such that
f : X Y is a bijective strongly
g -open and X is g T0 space then Y is
also g T0 space.
Proof. Let y1 and y 2 be two distinct points of Y.
Since f is bijective there exist x1 and x2 of X such
that f ( x1 ) = y1 and f ( x2 ) = y2 . Since X is
g T0 there exists a g -open set G such
that
and
Therefore
x1 G
x2 G .
Theorem 3.2 If
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x X . Let
y X {x} . Then for x y there exists g open set U y such that y U y and x U y
y U y X {x} .
X {x} = {U y : y X {x}} ,
That
is
which
is
g -open set. Hence {x} is g -closed set.
Conversely. Suppose {x} is g -closed set for
every x X . Let x , y X with x y . Now
x y implies y X {x} . Hence X {x} is
g -open set containing y but not x . Similarly,
X {y} is g -open set containing x but not
y . Therefore X is g T1 space.
Theorem 3.4 The property being g T1 space
is preserved under bejection and strongly g open function.
Proof. Let f : X Y be bijective and strongly
g -open function. Let X be g T1 space
and y1 and y 2 be any two distinct points of Y.
Since f is bijective there exist distinct points x1 ,
x2 of X such that y1 = f ( x1 ) and y2 = f ( x2 ) .
Now X being a g T1 space, there exist g open sets G and H such that x1 G , x2 G and
x1 H , x2 H . Therefore y1 = f ( x1 ) f (G)
but
and
y2 = f ( x2 ) f (G)
y2 = f ( x2 ) f ( H ) and y1 = f ( x1 ) f ( H ) .
Now f being strongly g -open, f (G) and
f (H ) are g -open subset of Y such that
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Imperial Journal of Interdisciplinary Research (IJIR)
Vol-3, Issue-1, 2017
ISSN: 2454-1362, http://www.onlinejournal.in
y1 f (G) but y2 f (G) and y2 f ( H ) and
y1 f ( H ) . Hence Y is g T1 space.
f : X Y be bijective and
g -open function. If X is g T1 and Tg -
Theorem 3.5 Let
space then Y is g T1 space.
Proof. Let y1 and y 2 be two distinct points of Y.
f is bijective there exist disjoint points x1 ,
x2 of X such that y1 = f ( x1 ) and y2 = f ( x2 ) .
Now X being a g T1 space, there exist g open sets G and H such that x1 G , x2 G and
x1 H , x2 H . Therefore y1 = f ( x1 ) f (G)
but
and
y2 = f ( x2 ) f (G)
y2 = f ( x2 ) f ( H ) and y1 = f ( x1 ) f ( H ) .
Not X is Tg -space which implies G and H are
open sets in X and f is g -open function,
f (G) and f (H ) are g -open subsets of Y.
Thus, there exist g -open sets such that
y1 f (G) but y2 f (G) and y2 f ( H ) but
y1 f ( H ) . Therefore Y is g T1 space.
Since
f : X Y is g -continuous
injection and Y is T1 , then X is g T1 space.
Proof. Let f : X Y is g -continuous
injection and Y be T1 . For any two distinct points x1
and x2 of X there exist distinct points y1 , y 2 of Y
such that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is
T1 space there exist distinct open sets U and V in Y
such that y1 U , y2 U and y1 V , y2 V .
That
is
and
x1 f 1 (U ), x1 f 1 (V )
1
1
x2 f (V ), x2 f (U ) . Since f is g 1
1
continuous f (U ) and f (V ) are g -open
sets in X. Thus, for two distinct points x1 , x2 of X
1
1
there exist g -open sets f (U ) and f (V )
such
that
and
x1 f 1 (U ), x1 f 1 (V )
1
1
x2 f (V ), x2 f (U ) . Therefore X is
g T1 space.
Theorem 3.6 If
f : X Y is g -irresolute
injective function and Y is g T1 space, then X
is g T1 space.
Theorem 3.7 Let
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Proof. Let x1 , x2 be pair of distinct points of X.
f is injective there exists distinct points y1
and y 2 of Y such that y1 = f ( x1 ) and
y2 = f ( x2 ) . Since Y is g T1 space, there
exist g -open sets U and V in Y such that
y1 U , y2 U and y1 V , y2 V . That is
and
x1 f 1 (U ), x1 f 1 (V )
1
1
x2 f (V ), x2 f (V ) . Since f is g 1
1
irresolute f (U ) and f (V ) are g -open
sets in X. Thus for two distinct points x1 , x2 of X
1
1
there exist g -open sets f (U ) and f (V )
and
such
that
x1 f 1 (U ), x1 f 1 (V )
1
1
x2 f (V ), x2 f (U ) . Therefore X is
g T1 space.
Since
Definition 3.3 A topological space X is said to be a
g T2 space if for any pair of distinct points x
and y , there exist disjoint g -open sets G and H
such that x G and
yH .
f : X Y is g -continuous
injection and Y is T2 , then X is g T2 space.
Proof. Let f : X Y is g -continuous
injection and Y is T2 . For any two distinct points x1 ,
x2 of X there exist distinct points y1 , y2 of Y such
that y1 = f ( x1 ) and y2 = f ( x2 ) . Since Y is T2
Theorem 3.8 If
space there exist disjoint open sets U and V in Y such
that y1 U and y2 V . That is x1 f
and x2 f
1
1
(U )
(V ) . Since f is g -continuous
f (U ) and f 1 (V ) are g -open sets in X.
Further
is
injective,
f
1
f 1 (U ) f 1 (V ) = f 1 (U V ) = f 1 ( ) =
. Thus, for two disjoint points x1 , x2 of X there
exist disjoint
f 1 (V )
g -open sets
such
that
f 1 (U ) and
and
x1 f 1 (U )
x2 f (V ) . Therefore X is g T2 space.
1
f : X Y is g -irresolute
injective function and Y is g T2 space, then X
is g T2 space.
Theorem 3.9 Let
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Imperial Journal of Interdisciplinary Research (IJIR)
Vol-3, Issue-1, 2017
ISSN: 2454-1362, http://www.onlinejournal.in
Proof. Let x1 , x2 be pair of distinct points in X.
f is injective there exist distinct points y1
and y 2 of Y such that y1 = f ( x1 ) and
y2 = f ( x2 ) . Since Y is g T2 space, there
exist disjoint g -open sets U and V in Y such that
y1 U and y2 V . That is x1 f 1 (U ) and
x2 f 1 (V ) . Since f is g -irresolute
1
1
injective f (U ) and f (V ) are disjoint g open sets in X. Thus for two disjoint points x1 , x2 of
1
X there exist disjoint g -open sets f (U ) and
such
that
and
f 1 (V )
x1 f 1 (U )
1
x2 f (V ) . Therefore X is g T2 space.
Since
Theorem 3.10 In any topological space the
followings are equivalent.
(i) X is g T2 space.
(ii) For each x y , there exists a g -open set
U such that
x in U and y g cl (U ) .
x X , {x} = {g cl (U ) : U
is g -open set in X and x U }
..
(iii)For each
Proof. (i) (ii ) : Assume (i) holds. Let x X
and x y , then there exist disjoint g -open
y V . Clearly,
X - V is g -closed set. Since U V = ,
There
fore
U X V .
g cl (U ) g cl ( X V ) = X V .
Now y X V implies y g cl (U ) .
(ii ) (iii ) : For each x y , there exists a
g -open set U such that x U and
So
y g cl (U ) .
y {g cl (U ) : U is g -open set in X
and x U } = {x} .
(iii ) (i) : Let x , y X and x y . By
hypothesis there exists a g -open set U such that
x U and y g cl (U ) . This implies there
exists a g -closed set V such that y V .
Therefore y X V and X V is g -open
set. Thus, there exist two disjoint g -open sets U
and X - V such that x U and y X V .
Therefore X is g T2 space.
sets U and V such that x U and
Imperial Journal of Interdisciplinary Research (IJIR)
4. g -Regular Space
In this section, we introduce and study g regular spaces and some of their properties.
Definition 4.1 A topological space X is said to be a
g -regular if for each closed set F and each
x F , there exist disjoint g -open sets U
and V such that x U and F V .
point
Theorem 4.1 Every g -regular T0 space is
g T2 .
x , y X such that x y Let X be a
T0 - space and V be an open set which contains x
but not y . Then X - V is a closed set containing y
but not x . Now by g -regularity of X there
exist disjoint g -open sets U and W such that
and
Since
x U
X V W .
y X V , y W . Thus for x, y X with
x y , there exist disjoint open sets U and W such
that x U and y W . Hence X is g T2 .
Proof. Let
Theorem 4.2 In a topological space X, the following
conditions are equivalent:
(i) X is g -regular.
(ii) For every point x X and open set V
containing x there exists a g -open set U such
that
x U g cl (U ) V .
(iii)
For
every
closed
set
F,
F = {g cl (V ) : F V and V is g open set of X } .
(iv) For every set A and an open set B such that
A B , there exists g -open set O such
that A O and g cl (O) B .
(v) For every nonempty set A and closed set B such
that A B , there exist disjoint g -open
sets L and M such that
A L and B M .
Proof. (i) (ii ) Let V be an open set containing
x . Then X - V is closed set not containing x . Since
X is g -regular, there exist g -open sets L
x U , X L L and
U L . This implies U X L . Therefore,
g cl (U ) g cl ( X L) = X L ,
because X - L is g -closed. Hence
x U g cl (U ) X L V . That is
x U g cl (U ) V .
and
U
such
that
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Imperial Journal of Interdisciplinary Research (IJIR)
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(ii ) (iii ) Let F be a closed set and x F . Then
X - F is an open set containing x. By (ii) there is an
set
U
such
that
g -open
x U g cl (U ) X F . And so,
F X g cl (U ) X U . Consequently
X - U is g -closed set not containing x . Put
V = X g cl (U ) . This implies F V
and V is
g -open set of X and
x g cl (V ) , implies
{g cl (V ) : F V and V is g -open
set of X } F (i)
.
But F is closed and every closed set is g -closed.
Therefore
F {g cl (V ) : F V and V is g open set of X } (ii )
is always true. From (i) and (ii),
F = {g cl (V ) : F V and V is g open set of X } .
(iii ) (iv ) Let A B and B is open. Let
x A B . Then X - B is a closed set not
containing x. By (iii), there exists a g -open set
V of X such that
and
X B V
x g cl (V ) . Put O = X g cl (V ) ,
then O is g -open set of X, x A O and
g cl (O) g cl ( X V ) = X V B
. Hence g cl (O) B .
(iv ) (v) If A B = , where A is non empty
and B is closed, then A ( X B) and
X B is open. Therefore by (iv), there exists
g -open set L such that A L and
L g cl ( L) X B .
(v) (i) Let F be a closed set such that x F ,
then {x} F = . By (v), there exist disjoint open
sets L and M such that {x} L and F M ,
which implies x L and F M . Hence, X is
g -regular.
Theorem 4.3 If f : X Y is continuous bijective,
g -open function and X is a regular then Y is
g -regular.
Proof. Let F be a closed set in Y and y F . Take
y = f ( x) for some x X . Since f is
1
continuous, f ( F ) is closed set in X such that
x f 1 ( F ) . Now X is regular, there exist disjoint
Imperial Journal of Interdisciplinary Research (IJIR)
x U and
f ( F ) V . That is, y = f ( x) f (U ) and
F f (V ) . Since f is g -open function
f (U ) and f (V ) are g -open sets in Y and
is
bijective
f
f (U ) f (V ) = f (U V ) = f ( ) = .
Therefore, Y is g -regular.
open sets U and V such that
1
Theorem 4.4 If
f : X Y is -continuous
bijective, g -open function and X is
space, then Y is g -regular.
*
-regular
y F . Take
y = f ( x) for some x X . Since f is 1
continuous, f ( F ) is -closed set in X and
x f 1 ( F ) . Now X is -regular, there exist
disjoint -open sets U and V such that x U and
f 1 ( F ) V . That is, y = f ( x) f (U ) and
F f (V ) . Since f is g * -open function
f (U ) and f (V ) are g -open sets in Y and
is
bijective
f
f (U ) f (V ) = f (U V ) = f ( ) = .
Therefore, Y is g -regular.
Proof. Let F be a closed set in Y and
Theorem 4.5 If f : X Y is continuous
surjective, strongly g -open (resp. quasi g open) function and X is g -regular space then Y
is g -regular (resp. regular).
y F . Take
y = f ( x) for some x X . Since f is
1
continuous surjective, f ( F ) is closed set in X
1
and x f ( F ) . Now since X is g -regular,
there exist disjoint g -open sets U and V such
that
x U and f 1 ( F ) V . That is,
y = f ( x) f (U ) and F f (V ) . Since f is
strongly g -open (resp. quasi g -open) and
bijective, f (U ) and f (V ) are disjoint g open (resp. open) sets in Y. Therefore, Y is g Proof. Let F be a closed set in Y and
regular (resp. regular).
Theorem 4.6 If f : X Y is g -continuous,
closed, injection and Y is regular, then X is g regular.
Proof. Let F be a closed set in X and x F . Since
f is closed injection f (F ) is closed set in Y such
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f ( x) f ( F ) . Now Y regular, there exist
disjoint open sets G and H such that f ( x) G and
f ( F ) H . This implies x f 1 (G) and
F f 1 ( H ) . Since f is g -continuous,
f 1 (G) and f 1 ( H ) are g -open sets in X.
1
1
Further f (G) f ( H ) = . Hence X is
g -regular.
that
Theorem 4.7 If f : X Y is g -continuous,
-closed, injection and Y is -regular, then X is
g -regular.
*
Proof. Let F be a closed set in X and x F . Since
f is -closed injection f (F ) is -closed set in
Y such that f ( x) f ( F ) . Now Y is -regular,
there exist disjoint -open sets G and H such that
f ( x) G and f ( F ) H . This implies
1
1
x f (G) and F f ( H ) . Since
f is
g -continuous, f (G) and f ( H ) are
sets
in
X.
Further
g -open
1
1
f (G) f ( H ) = . Hence X is g *
1
1
regular.
Theorem 4.8 If f : X Y is g -irresolute,
closed, injection and Y is g -regular, then X is
g -regular.
Proof. Let F be a closed set in X and x F . Since
f is closed injection f (F ) is closed set in Y such
that f ( x) f ( F ) . Now Y is g -regular, there
exist disjoint g -open sets G and H such that
f ( x) G and f ( F ) H . This implies
x f 1 (G) and F f 1 ( H ) . Since f is
g -irresolute, f 1 (G) and f 1 ( H ) are
sets
in
X.
Further
g -open
1
1
f (G) f ( H ) = . Hence X is g regular.
5 g -Normal Spaces
In this section, we introduce and study g normal spaces and some of their properties.
Definition 5.1 A topological space X is said to be
g -normal if for every pair of disjoint closed sets
Imperial Journal of Interdisciplinary Research (IJIR)
E and F of X there exists disjoint g -open sets U
and V such that E U and F V .
Theorem 5.1 The following statements are
equivalent for a topological space X.
(i). X is g -normal.
(ii). For each closed set A and for each set U
containing A, there exist a g -open set V
containing A such that g cl (U ) U .
(iii). For each pair of disjoint closed sets A and B
there exists a g -open set U containing A such
g cl (U ) B = .
Proof. (i) (ii ) Let A be closed set and U be an
open set containing A. Then A ( X U ) = and
that
therefore they are disjoint closed sets in X. Since X is
g -normal, there exists disjoint g -open sets
A U , X U W that is
Now
V W = , implies
V and W such that
X W U .
Therefore
V X W .
g cl (U ) g cl ( X W ) = X W ,
because X - W is g -closed set. Thus
A V g cl (V ) X W U . That is
A V g cl (U ) U
(ii ) (iii ) Let A and B be disjoint closed sets in
X, then A X B and X B is an open set
containing A. By (ii) there exists a g -open set U
such that A U and g cl (U ) X B ,
which implies g cl (U ) B = .
(iii ) (i) Let A and B be disjoint closed sets in X.
By (iii) there exists a g -open set U such that
g cl (U ) B =
and
or
A U
B X g cl (U ) .
Now
U
and
X g cl (U ) are disjoint g -open sets
of
X
such
that
and
A U
B X g cl (U ) . Hence X is g normal.
Theorem 5.2 If X is -normal and F A = ,
where F is -closed set and A is g -closed
set, then there exist disjoint -open sets U and V in
X, such that F U and A V .
Proof. Let X be an -normal space and
F A = , since F is -closed set and A is
g -closed set such that A X F and
-open.
is
Therefore
X F
cl ( A) X F , implies cl ( A) F = .
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Now F is -closed, hence -closed, which
implies F and cl (A) are disjoint -closed sets
and X is -normal, therefore there exist disjoint open sets U and V of X such that F U and
cl ( A) V . That is, U and V of X are such that
F U and A V .
Theorem 5.3 If X is -normal, the following
statements are true.
(i). For each -closed set A and every g open set B such that A B there exists an -open
set U such that A U cl (U ) B .
(ii). For every g -closed set A and every open set containing A there exists an -open set
containing U such that A U cl (U ) B .
Proof. (i). Let A be a -closed set and B be a
g -open set such that A B . Then
A ( X B) = . Since A is -closed set and
X B be a g -closed, by theorem 5.2, there
exist -open sets U and V such that A U ,
and
Thus
U V = ,
X B V
A U X V B . Since X - V is -closed,
Therefore,
cl (U ) X V .
A U cl (U ) B .
(ii) Let A be a g -closed set and B be a open set such that A B . Then X B X A .
Since X is -normal and X - A is a g -open set
containing -closed set X - B, by (i) there exists
an
set
G
such
that
-open
X B G cl (G) X A .
That
is
A X cl (G) X G B .
Let
U = X cl (G) , then U is -open set and
A U cl (U ) B .
Theorem 5.4 If f : X Y is continuous
bijective, g -open function from normal space X
on to a space Y, then Y is g -normal.
Proof. Let E and F be disjoint closed sets in Y. Since
f is continuous bijective f 1 ( E ) and f 1 ( F )
are disjoint closed sets in X. Now X is normal, there
exist disjoint open sets U and V such that
f 1 ( E ) U and f 1 ( F ) V . That is,
E f (U ) and F f (V ) . Since f is g open function f (U ) and f (V ) are g -open
f
sets
in
Y
and
is
injective
Imperial Journal of Interdisciplinary Research (IJIR)
f (U ) f (V ) = f (U V ) = f ( ) = . Thus,
Y is g -normal.
Theorem 5.5 If
f : X Y is -continuous
bijective, g -open function from -normal
space X on to a space Y, then Y is g -normal.
Proof. Let E and F be disjoint closed sets in Y. Since
*
f is -continuous bijective f 1 ( E ) and
f 1 ( F ) are disjoint -closed sets in X. Now X is
-normal, there exist disjoint -open sets U and V
( E ) U and f 1 ( F ) V . That is,
E f (U ) and F f (V ) . Since f is g * open function f (U ) and f (V ) are g -open
is
injective
sets
in
Y
and
f
f (U ) f (V ) = f (U V ) = f ( ) = . Thus,
Y is g -normal.
such that f
1
Theorem 5.6 If f : X Y is g -continuous,
closed, injection and Y is normal, then X is g normal.
Proof. Let E and F be disjoint closed sets in Y. Since
f is closed injection f (E ) and f (F ) are
disjoint closed sets in Y. Now Y is normal, there exist
disjoint open sets G and H such that f ( E ) G
f ( F ) H . This implies, E f 1 (G) and
F f 1 ( H ) . Since f is g -continuous
1
1
function f (G) and f ( H ) are g -open
1
1
sets in X. Further f (G) f ( H ) = . Hence,
X is g -normal.
and
Theorem 5.7 If f : X Y is g -continuous,
-closed, injection and Y is -normal, then X is
g -normal.
Proof. Let E and F be disjoint closed sets in Y. Since
f is -closed injection f (E ) and f (F ) are
disjoint -closed sets in Y. Now Y is -normal,
there exist disjoint -open sets G and H such that
f ( E ) G and f ( F ) H . This implies,
*
E f 1 (G) and F f 1 ( H ) . Since f is
g * -continuous f 1 (G) and f 1 ( H ) are
sets
in
X.
Further
g -open
1
1
f (G) f ( H ) = . Hence, X is g normal.
Page 1879
Imperial Journal of Interdisciplinary Research (IJIR)
Vol-3, Issue-1, 2017
ISSN: 2454-1362, http://www.onlinejournal.in
Theorem 5.8 If f : X Y is g -irresolute,
closed, injection and Y is g -normal, then X is
g -normal.
Proof. Let E and F be disjoint closed sets in Y. Since
f is closed injection f (E ) and f (F ) are
disjoint closed sets in Y. Now Y is g -normal,
there exist disjoint g -open sets G and H such
f ( E ) G and f ( F ) H . This implies,
E f 1 (G) and F f 1 ( H ) . Since f is
g -irresolute f 1 (G) and f 1 ( H ) are
sets
in
X.
Further
g -open
1
1
f (G) f ( H ) = . Hence, X is g -
that
normal.
Theorem 5.9 If f : X Y is continuous bijective,
strongly g -open (resp. quasi g -open)
function from a g -normal space X onto a space
Y, then Y is g -normal (resp. normal).
Proof. Let E and F be disjoint closed sets in Y. Since
f is continuous bijective f 1 ( E ) and f 1 ( F )
are disjoint closed sets in X. Now X is g normal, there exist disjoint g -open sets U and V
1
1
such that f ( E ) U and f ( F ) V . That is,
E f (U ) and F f (V ) . Since f is strongly
g -open (resp. quasi g -open) function
f (U ) and f (V ) are g -open (resp. open) sets
f
in
and
is
injective
Y
f (U ) f (V ) = f (U V ) = f ( ) = . Thus,
X is g -normal (resp. normal).
6. Acknowledgements
The authors are grateful to the University Grants
Commission, New Delhi, India for its financial
support under UGC-SAP-DRS- III(F.510/3/DRS-III /
2016 (SAP-I) dated: 29-02-2016) to the Department
of Mathematics, Karnatak University, Dharwad,
India. Also this research was supported by the
University Grants Commission, New Delhi, India.
under No.F.4-1/2006(BSR)/7-101/2007(BSR) dated:
20th June, 2012.
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