The research paper published by IJSER journal is about Generalized X-Closed Sets and (X, Y)* -Continuous Functions 1
ISSN 2229-5518
Alias B. Khalaf , Sarhad F. Namiq
(, )* -
Abstract— In this paper we introduce the concept of -open set and by using this set we define generalized -closed set we obtain some of its properties and also we define (, )* -continuous function and study some of its basic properties.
Index Terms— s-operation, -open, generalized -closed, -T space , (, )*
-continuous function .
—————————— ——————————
HE study of semi open sets and semi continuity in topo- logical spaces was initiated by Levine [5]. Analogous to
the concept of generalized closed sets introduced by Lev-
ine [6], Bhattacharya and Lahiri [3] introduced the concept of
Definition 3.1. [1] Let (X , )
be a topological space and
semi generalized closed sets in topological spaces. Kasahara
: SO (X ) P (X )
be an s-operation, then a subset A of X is
[4], defined the concept of an operation on topological spaces and introduced the concept of closed graphs of a function. Ahmad and Hussain [2], continued studying the properties of operations on topological spaces.
called a -open set if for each x A there exists a semi open
set U such that x U and (U ) A .
The complement of a -open set is said to be -closed. The family of all -open ( resp., -closed ) subsets of a topologi-
In this paper, we introduce new classes of sets called -
open and generalized -closed sets in topological spaces and
study some of their properties. By using these sets
cal space (X , ) is denoted by SO (X , )
SC (X , ) or SC (X ) ).
or SO (X )
( resp. ,
we define -T space and introduce the concept of
(, )* -continuous functions and study some of their basic
properties.
(X , ), SO (X ) SO (X ).
The following examples show that the converse of the
above proposition may not be true in general.
{a,b ,c},
and {,{a},X }.
We define an
s-operation
: SO (X ) P (X )
as (A ) A if
b A and
Throughout, X denote topological spaces. Let A be a subset of X , then the closure and the interior of A are denoted by
(A ) X otherwise. Here, we have {a,c} is semi open set but it is not -open.
Cl (A )
and
Int (A ) respectively. A subset A of a topological
space (X , ) is said to be semi open [5] if
A Cl (Int (A )) . The
Definition 3.4. Let (X , ) be a space, an s-operation is said
complement of a semi open set is said to be semi closed [5].
to be s-regular if for every semi open sets U and V of
x X ,
The family of all semi open (resp. semi closed) sets in a topo-
there exists a semi open set W containing x such that
logical space (X , ) is denoted by
SO (X , )
or SO (X )
(resp.
(W ) (U ) (V ).
SC (X , ) or SC (X ) ). We consider as a function defined on
SO (X )
into
P (X ) and
: SO (X ) P (X ) is called an s-
Definition 3.5. Let (X , ) be a topological space and let A be a
operation if
V (V ) for each non-empty semi open set V . It
subset of X . Then:
is assumed that ( ) and
(X ) X for any s-operation . .
(1) The -closure of A ( Cl (A ) ) is the intersection of
all -closed sets containing A
(2) The -interior of A ( Int (A ) ) is the union of all -
open sets of X contained in A
x X , is said to be a -limit point of A if
————————————————
Author is currently one of the staff members at Department of Mathemat- ics, College of Science, University of Duhok, Kurdistan-Region, Iraq.
E-mail address: aliasbkhalaf@gmail.com
Co-Author is currently one of the staff members at Department of Mathe-
matics, Faculty of Education and School of Science, University of Garmian,
every -open set containing x contains a point of
A different from x , and the set of all -limit points
of A is called the -derived set of A denoted by d (A ).
Proposition 3.6. For each point x X , x Cl (A ) if and only
Kurdistan-Region, Iraq
E-mail address: sarhad1983@gmail.com
if V A , for every V SO (X ) such that
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x V .
The research paper published by IJSER journal is about Generalized X-Closed Sets and (X, Y)* -Continuous Functions 2
ISSN 2229-5518
Proposition 3.7. Let A I
be any collection of -open
Theorem 3.14. Let A ,B be subsets of X. If
sets in a topological space (X , ) , then
A is a -open set.
: SO (X ) P (X ) is an s-regular s-operation, then:
A then there exist I I such that x A ,
I 0
0 (1)
Cl (A B ) Cl (A ) Cl (B ).
since A is a -open set for all I implies that there exists
a semi open set U such that
(U ) A
0
I
A . There-
Int (A B ) Int (A ) Int (B ).
fore
I
A is a -open subset of (X , ).
The following example shows that the intersection of two
-open sets need not be -open.
X {a,b ,c} and P (X ) . We define an s-
operation
: SO (X ) P (X )
as (A ) A if
A {a},{b}
and
1/ 2
(A ) X otherwise. Now, we have {a,b}
and {b ,c}
are -
In this section, we define a new class of sets called generalized
open sets but {a,b} {b ,c} {b} is not -open.
Proposition 3.9. Let be an s-regular s-operation. If A and
-closed set and we give some of its properties.
Definition 4.1. A subset A of a topological space (X , ) is
said to be generalized -closed ( briefly. g- -closed) if
B are -open sets in X , then A B is also a -open set.
Cl( A) U ,
whenever A U and U is a -open set in
x A B then x A
and x B . Since A and
(X , ).
B are -open sets, there exists semi open sets U and V such
We say that a subset B of X is generalized -open (brief-
that x U
and (U ) A ,
x V
and (V ) B . Since
ly. g- -open) if its complement
X \ B is generalized -
is a s-regular s-operation, this implies there exists a semi open
closed in (X , ).
set W of X such that
This implies that A B
(W ) (U ) (V ) A B .
is -open .
In the following proposition we show every -closed sub- set of X is g- -closed.
Proposition 3.10. Let (X , )
be a topological space and
Proposition 4.2. Every -closed set is g- -closed.
A X .
Then A is a -closed subset of X if and only if
Proof. A set A X
is -closed if and only if
d (A ) A.
Cl( A) A. Thus
U SO ( X ) containing
Cl( A) U for every
A .
A , B of a topological space
The reverse claim in the above proposition is not true in
(X , ), the following statements are true.
general. Next we give an example of a g- -closed set which is
A Cl (A ).
Cl (A ) is -closed set in X .
not -closed.
X {a,b ,c},
and P (X ).
We define
Cl (A ) is smallest -closed set which contain A .
an s-operation
: SO (X )
P (X ) as
(A ) A if
A {a}
and
(A ) X
otherwise. Then, if we let
A {a,b},
and
(4) A is -closed set if and only if A Cl (A ).
since the only -open supersets of A is X, so A is g- -closed
Cl () and Cl (X ) X .
but it is not -closed.
(6) If A and B are subsets of space X with A B .
Then Cl (A ) Cl (B ).
(7) For any subsets A , B of a space (X , ) ,
Proposition 4.4. The intersection of a g- -closed set and a - closed set is always g- -closed.
Proof. Let A be g- -closed and F be -closed. Assume that
Cl( A) Cl(B) Cl( A B).
U is -open set such that
A F U ,
set
G X
\ F .Then
A U G ,
since G is -open, then
(8) For any subsets A , B of a space (X , ) ,
Cl( A B) Cl( A) Cl(B).
Proposition 3.12. Let (X , ) be a topological space and
A X . Then Cl (A ) A d (A ).
Proposition 3.13. For a subset A of a topological space
U G is -open and since A is g- -closed, then
Cl( A) U G. Now by Proposition 3.8,
Cl( AF ) Cl (A) Cl (F )
Cl (A) F (U G) F= (U F ) (G F ) =
(U F ) U .
The union of two g- -closed sets need not be g- -closed,
as it is shown in the following example:
(X , ),
Int (A ) A \ d (X
\ A ).
X {a,b ,c},
and P (X ).
We define
an s-operation : SO (X )
P (X ) as:
Proposition 3.13. For any subset A of a topological space
(A ) A
, if A or {a,b} or {a,c} or {b,c}
X . The following statements are true.
(A ) X , otherwise
X \ Int( A) Cl( X \ A).
Then, if A {a} and
B {b}.
So, A and B are g- -
Cl( A) X \ Int( X \ A).
closed, but A B {a,b} is not a g- -closed, since {a,b}
is -open and Cl({a, b}) X.
X \ Cl( A) Int( X \ A).
: SO (X ) P (X ) is a s-regular s-operation.
Int( A) X \ Cl( X \ A).
Then the finite union of g- -closed sets is always a g- -
closed set.
Proof. Let A and B be two g- -closed sets, and let
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ISSN 2229-5518
A B U ,
where U is -open. Since A and B are g- -
implies
A X
\ F.
Since A is g- -closed and
X \ F is
closed sets, therefore Cl( A) U and Cl(B) U
implies
-open set, therefore
Cl( A) X
\ F ,
that
that Cl( A) Cl(B) U. But by Theorem 3.11, we have
Cl( A) Cl(B) Cl( A B). Therefore
is F X \ Cl( A).
Hence F Cl( A) X \ Cl( A) . This shows that
Cl (A B ) U . Hence we get A B is g- -closed set.
F
which is a contradiction. Hence
Cl( A) \ A
does
The intersection of two g- -closed sets need not be g- -
closed, as it is shown in the following example:
not contains any non empty -closed set in X .
{a,b ,c},
and P (X ). We define
Lemma 4.13. Let A be a subset of a topological space (X , ) .
an s-operation
: SO (X ) P (X ) as:
If d ( A) U
for U is -open, then
d (d ( A)) U ,
(A ) A
, if A {a} and (A ) X
, otherwise .
where is s-regular.
Then the sets
A {a,b}
and B {a,c}
are g- -closed
x d (d ( A)) but
x U . Then
and so
A U V X \ {y} A V {x }. It fol-
Theorem 4.8. If a subset A of X is g- -closed and
A B Cl( A) , then B is a g- -closed set in X .
lows that x U
which is contradiction.
Proof. Let A be g- -closed set such that
A B Cl( A).
Theorem 4.14. If is s-regular s-operation, then the -
Let U be a -open set of X such that B U . Since A is g-
derived set is g- -closed.
-closed, we have
Cl( A) U.
Now
Cl( A)
Proof. If A is any subset of a topological space (X , )
with
Cl(B) Cl(Cl( A)) = Cl( A) U. That
d ( A) U
for U is -open. Then by Lemma 4.13
is Cl(B) U , where U is -open. Therefore B is a g- - closed set in X .
The converse of the Theorem 4.8 need not be true as seen from the following example.
Cl(d ( A)) d (d ( A)) d ( A) U.
Theorem 4.15. A subset A of a topological space (X , )
-open if and only if F Int( A) whenever F A
is g- and
X {a,b ,c},
with
F is -closed in (X , ).
{,{a},{c},{a,c},{b ,c}, X }. Let
: SO (X ) P (X ) be
Proof. Let A be g- -open and F A where F is -
a -identity s-operation. If
A {a} and B {a,b} . Then
closed. Since X
\ A is g- -closed and X
\ F is a -open
A and B are g- -closed sets in (X , ).
But
set containing
X \ A implies
Cl( X \ A) X \ F. By
A B Ú Cl( A).
Theorem 4.10. Let : SO (X ) P (X ) be an s-operation. Then
Proposition 3.10,
F Int( A).
X \ Int( A) X \ F.
That is
for each
x X , {x } is -closed or X \{x } is g- -closed
Conversely, suppose that F is -closed and
F A ,
im-
in (X , ).
plies that
F Int( A).
Let
X \ A U , where U is -
Proof. Suppose that{x }is not -closed, then
X \{x }is not
open. Then
X \U A ,
where
X \U is -closed. By hy-
-open. Let U be any -open set such that
pothesis
X \ U Int( A). That is
X \ Int( A) U
and
X \ {x } U , then U
X . Therefore
Cl( X \ {x}) U.
then by Proposition 3.10,
Cl( X \ A) U. This implies
Hence X \{x }is g- -closed.
X \ A is g- -closed and A is g- -open.
The union of two g- -open sets need not be g- -open. As
Proposition 4.11. A subset A of (X , )
is g- -closed if and
it is shown in the following example:
only if Cl({x}) A
, holds for every x Cl( A).
{a,b ,c},
and P (X ).
We define
Proof. LetU be a -open set such that A U and let
an s-operation
: SO (X ) P (X )
as (A ) A if
x Cl( A).
By assumption, there exists a point
A {b} and
(A ) X
if A {b}. If
A {a} and
z Cl({x}) and
z A U . It follows from Proposition
B {c},
then A and B are g- -open sets in X, but
3.6, that U {x } , hence xU, implies Cl( A) U.
Therefore A is g- -closed.
A B {a,c} B {a,c} is not a g- -open set in X .
Conversely, suppose that
x Cl( A) such
: SO (X ) P (X )
be a s-regular s-
that Cl({x}) A .
Since,
Cl({x})
is -closed.
operation and let A and B be two g- -open sets in a space
Therefore,
X \ Cl({x})
is -open set in
X . Since
X , then A B
is also g- -open.
A X \ Cl ({x }) and A is g- -closed implies that
Proof. If A and B are g- -open sets in a space
X . Then
Cl (A ) X \ Cl ({x }) holds and hence x Cl (A ) a contradic-
X \ A and X
\ B are g- -closed sets in a space X . By The-
tion. Therefore Cl({x}) A .
orem 4.6, X
\ A X
\ B is also g- -closed set in X . That is
X \ A X
\ B X
\ (A B )
is a g- -closed set in X .
Theorem 4.12. If a subset A of X is g- -closed set in
Therefore A B is a g- -open set in X .
X . Then
Cl( A) \ A
does not contain any non empty -
closed set in X .
Proof. Let A be a g- -closed set in X . We prove the result
Theorem 4.18. A set A is g- -open if and only if
Int( A) X \ A G and G is -open implies G X .
by contradiction. Let F be a -closed set such that
Proof. Suppose that A is g- -open in
X . Let G be -open
F Cl( A) \ A
and
F . Then
F X \ A
which
and
Int( A) X \ A G.
This implies
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ISSN 2229-5518
X \ G X \ (Int (A ) X \ A ) X \ Int (A ) A.
That is
X \ G
B in Y , f
1 (B ) is -open in X .
(X \ Int (A )) \ (X \ A ). Thus X \G Cl (X \ A ) \ (X \ A ), since
Proof. Let f be a (, )* -continuous and B SO
(Y ), let
X \ Int( A)
C l( X \
A) . Now,
X \G is -closed
A f
1 (B ). We show that A is -open in X . For this, let
and X
\ A is g- -closed, by Theorem 4.12, it follows that
x A ,
then it implies that f (x ) B . Hence, by hypothe-
X \G . Hence G X .
sis, there exists
A SO (X ) such that
1
x A and
1
Conversely, let Int( A) X \ A G and G is -open, this
f (Ax ) B .
Then
Ax f
(f (Ax )) f
(B ) A .
implies that
G X .
Let U be a -open set such that
Thus A
{Ax : x A}.
It follows that A is -open in
X \ A U . Now Int( A) X \ A Int( A) U
which X .
is clearly, -open and so by the given condition
Conversely, let x X
and B SO (Y )
such
Int( A) U X ,
which implies that
Cl( X \ A) U.
that f (x ) B . Let
A f
1 (B ).
By hypothesis, A is -
Hence X
\ A is g- -closed, therefore A is g- -open.
open in X and also we have
x f
1 (B ) A as
f (x ) B .
Thus,
f (A ) f (f
1 (B )) B .
Hence f is
Theorem 4.19. Every singleton set in a space X is either g- - open or -closed.
Proof: Suppose that{x }is not g- -open, then by definition
(, )* -continuous.
: (X , ) (Y , ) be a function. Then
X \ {x } is not g- -closed. This implies that by Theorem
4.10, the set {x }is -closed.
the following statements are equivalent:
(1) f is (, )* -continuous.
Int( A) B A
and A is g- -open,
(2) The inverse image of each -closed set in Y is a -
then B is g- -open.
closed set in X .
Int( A) B A
implies
Cl (f
1 (V )) f
1 (Cl (V )), for every V
Y .
X \ A X \ B X \ Int( A).
That is,
X \ A
X \ B Cl( X \ A) by Proposition 3.10. Since X
\ A is g-
f (Cl (U )) Cl (f (U )), for every U
X .
-closed, by Theorem 4.8, X
\ B is g- -closed and B is
Bd (f
1 (V )) f
1 ( Bd (V )), for every
-open.
V Y .
Theorem 4.21. Let (X , ) be a topological space
f (d (U )) Cl (f (U )), for every U
X .
(X , ) and
: SO (X ) P (X ) be an s-operation. The
f 1 ( Int (V )) Int (f
1 (V )),
for every
space (X , )
is -T1/ 2
if and only if Each singleton
V Y .
{x } of X is either -closed set or -open set.
be -closed. Since f is
(, )* -continuous,
f 1 (Y
\ F ) X \ f
1 (F )
is -open.
Proof. Suppose {x } is not -closed. Then by Proposition
Therefore, f
1 (F ) is -closed in X .
4.10,
X \ {x } is g- -closed. Now since (X , ) is
(2) (3): Since Cl (V ) is -closed for every V
Y , then
-T1/ 2 , X
\ {x } is -closed i.e. {x }is -open.
f 1 (Cl (V ))
is -closed. Therefore
Conversely. Let A be any g- -closed set in (X , )
and
Cl (f
1 (V )) Cl (f
1 (Cl (V ))) f
1 (Cl (V )).
x Cl( A).
By (2) we have{x }
is -closed or -open.
and
If {x }is -closed then x A will imply
x Cl( A) \ A
f (U ) V . Then Cl (f
1 (V )) f
1 (Cl (V )).
Thus
which is not possible by Proposition 4.12. Hence
Cl (U ) Cl (f 1 (f (U ))) f 1(Cl (f (U ))) then we get
x A . Therefore,
Cl( A) A,
i.e. A is -closed. So
f (Cl (U )) Cl (f
(U )).
(X , ) is -T1/ 2 . On the other hand, if {x } is -open then
be a -closed set and
as x Cl( A), {x }A .
Hence
x A .
So A is -
U f
1 (W ).
This implies that
closed.
f (Cl (U )) Cl (f
(U )) Cl (f (f
1 (W ))) Cl (W ) W .
* * Thus
Cl (U ) f
1 (f (Cl (U ))) f
1 (W ) U . So
5 (, )
-Continuous and (, )
-Open Functions
U is -closed.
In this section, some types of continuous functions via s-
(2) (1): Let V Y
be an -open set, then Y
\V is -
operations are introduced and investigated. Several properties
closed. Hence,
f 1 (Y
\V ) X
\ f 1 (V )
is -closed in
of these functions are obtained.
X and so f
1 (V ) is -open in X .
Throughout, (X , ), (Ζ , )
and (Y , )
are topological
spaces and , and are s-operations on the family of semi
Bd ( f 1 (V )) f 1 ( Bd (V )) . This implies that
open sets of the topological spaces respectively.
f 1 (V ) \ Int (f
1 (V )) f
1 (V
\ Int (V )) f
1 (V ) \ f
1( Int (
: (X , ) (Y , ) is said to be
Hence we get f
1 ( Int (V )) Int (f
1 (V )).
(, )* -continuous, if for each x of X and each -open
set V of Y containing f (x ) , there exists a -open set U of
f 1 ( Int (V )) Int (f
1 (V )).
Implies that
X such that x U
and f (U ) V .
f 1 (V ) \ Int (f
1 (V )) f
1 (V ) \ f
1 ( Int (V ))
then
: (X , ) (Y , )
be a function, then
Bd (f
1 (V )) f
1 ( Bd (V )).
f is (, )* -continuous if and only if for each -open set
(, )* -continuous and by (4), we
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ISSN 2229-5518
have
f (Cl(U )) Cl( f (U ))
for each
U X . So
f (x ) f (U x ) f (A )
and by hypothesis,
f (d (U )) Cl (f (U )).
f (U x ) SO (Y ).
Hence
f (x ) Int (f (A )).
Thus
(6) (1): Let V be a -closed subset of Y and let
f (Int (A )) Int (f (A )).
f 1 (V ) W ,then by hypothesis,
Conversely, let U SO
(X ).
Then by hypothesis, we get
f (d (W )) Cl (f (W )). Thus
f (Int (U )) Int (f (U )).
Since Int (U ) U
as U is
f (d (f
1 (V ))) Cl (f (f
1 (V ))) Cl (V ) V .
-open. Also
Int (f (U )) f (U ).
Hence
Hence,
d (f
1 (V )) f
1 (V )
so by Proposition 3.4,
f (U ) Int (f (U )).
Thus
f (U )
is -open in Y . So
f 1 (V ) is -closed set. Therefore, by part (2) of this theo-
f is (, )* -open.
rem f is (, )* -continuous.
(1) (7): Let V Y , then f
1 ( Int (V ))
is -open set in
: (X , ) (Y , ) is
(, )* -
X . Thus
open if and only if
Int( f 1 (B)) f 1 ( Int(B)) for all
f 1 ( Int (V )) Int f
1 ( Int (V )) Int (f
1 (V )). The
B Y .
refore, f
1 ( Int (V )) Int (f
1 (V )).
Proof. Let B Y
, since Int (f
1 (B ))
is -open set in
(7) (1): Let V Y
be an -open set. Then
X and f is (, )* -open function , so
f (Int (f
1 (B )))
f 1 (V ) f 1 ( Int(V )) Int( f 1 (V )).
Therefore,
is -open set in
Y . We have
f 1 (V ) is -open. Hence f is (, )* -continuous.
f (Int (f
1 (B ))) f (f
1 (B )) B .
Hence f (Int (f
1 (B ))) Int (B ) by hypothesis. There-
: (X , ) (Ζ , ) is
fore Int (f
1 (B )) f
1 ( Int (B )).
(, )* -continuous and
g : (, ) (Y , )
is (, )* -
Conversely, let A X , then f (A ) Y . Hence by hypoth-
continuous, then
g f : (X , ) (Y , )
is (, )* -
esis, we obtain
continuous.
V SO (Y ).
Then g
(V ) SO () and
Int (A ) Int (f
plies that
1 (f (A ))) f
1 ( Int (f (A ))).
Im-
f 1 (g 1 (V )) SO
(X ).
This implies that
f (Int (A )) f (f
1 ( Int (f (A )))) Int (f (A )). Thus
Therefore,
f (Int (A )) Int (f (A )), for all
A X .
Hence, by
g f : (X , ) (Y , ) is (, )* -continuous.
Theorem 5.7, f is (, )* -open.
: (X , ) (Y , ) is said to be
: (X , ) (Y , )
is (, )* -
(, )* -open ( (, )* -closed ), if for any -open ( -
open if and only if
f 1 ( Cl(B)) Cl( f 1 (B))
for every
closed ) set A of (X , ),
f (A ) is -open ( -closed ).
subset B of Y .
Proof. Let B Y and let
x f 1 ( Cl(B)) , then
f : (X , ) (Y , ) is
f (x ) Cl (B ). Let U SO (X )
such that x U . By
(, )* -continuous and (, )* -closed function, then:
hypothesis,
f (U ) SO
(Y )
and
f (x ) f (U ).
Thus
(1) For every g - -closed set A of (X , ) the image
f (U ) B and hence
U f
1 (B ) .
Therefore,
x Cl (f
1(B )).
So we obtain
f (A ) is a g - -closed set.
f 1 (Cl (B )) Cl (f
1 (B )).
(2) For every g - -closed set B of (Y , ) the inverse
Conversely, let B Y , then (Y
\ B ) Y .
By hypothesis,
f 1 (Cl (Y
\ B )) Cl (f
1 (Y
\ B )).
Implies that
set f
1 (B ) is a g - -closed set.
X \ Cl (f
1 (Y
\ B )) X
\ f 1 (Cl (Y
\ B )).
Hence
Proof. (1) Let V be any -open set in (Y , )
such that
X \ Cl (X
\ f 1 (B )) X
\ f 1 (Y
\ Int (B )). Then
f (A ) V . Then by Theorem 5.2,
f 1 (V ) is -open. Since
Int (f
1 (B )) f
1 ( Int (B )).
Now by Theorem 5.8, it
A is
g - -closed and
A f
1 (V ),
we have
follows that f is (, )* -open.
Cl( A) f 1 (V ) and hence we get f (Cl (A )) V . By
assumption
f (Cl (A ))
is a -closed set. Therefore,
f : (X , ) (Z , )
and
Cl (f (A )) Cl (f (Cl (A ))) f (Cl (A )) V .
g : (Z , ) (Y , )
be two functions such that
This implies that
f (A ) is g - - closed.
g f : (X , ) (Y , ) is (, )* -continuous. Then:
(2) Let U be any -open set such that
f 1 ( B ) U . Let * *
H Cl (f
1 (B )) (X
\U ).
Then H is -closed in
(1) If g is a (, )
-open injection, then f is (, ) -
(X , ). This implies f
(H ) is -closed set in
Y . Since
continuous.
f (H ) f (Cl (f
1 (B )) X
\U ) Cl (B ) f (X
\U ) C(2l)(BIf) f
(iYs a\(B),. )* -open surjection, then g is (, )* -
This implies that
f (H )
and since f is a function,
continuous.
hence H
. Therefore,
Cl (f
1 (B )) U .
This im- *
plies f
1 (B ) is g - -closed.
Proof. (1) Let U SO (Z ).
Since g is
(, )
*
-open, then
g (U ) SO (Y ).
Also since g f is (, )
-continuous.
: (X , ) (Y , )
is (, )* -
Therefore, we have (g f )1 (g (U ))
SO (X ). Since g is
open if and only if
f (Int( A)) Int( f ( A))
for all
an injection function, so we have
(g f )1 (g (U )) (f 1
g 1)(g (U )) (f 1)(g 1(g (U ))) f 1(U ).
A X .
Consequently f
1 (U ) is -open in
X . This proves that
Proof. Let A X
and let
x Int (A ). Then there exists
f is (, )* -continuous.
U x SO (X )
such that
x U x
A .
So (2) Let V
SO
(Y ). Then (g f )1 (V ) SO
(X ) since
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g f is
(, )* -continuous. Also f is
(, )* -open,
1 1
f ((g f )1 (V )) is -open in Y .
Since f is surjec-
Int (f
(B )) f
( Int (B )) for all
B Y .
tive, then:
Proof.(1) (2). Obvious. Follows from Theorem 5.3 and
f ((g f )1 (V )) (f
(g f )1 )(V ) (f
(f 1
g 1 ))(V )The((ofremf 5.112).
g 1 )(V ) g 1 (V ).
Hence g is (, )* -continuous.
(1) (3). Follows from Theorem 5.3 and Theorem 5.9. (1) (5). Follows from Theorem 5.3 and Theorem 5.8.
f : (X , ) (Z , )
and
(1) (4). We have
Int (A ) X
\ Cl (X
\ A ) . Thus
g : (Z , ) (Y , ) are
(, )* -closed ( resp. open ) and
f (Int (A )) Y
\ Cl (f (X \ A )) Y
\ Cl (Y
\ f (A )) Int (f (A )).
(, )* -closed (resp. open) respectively. Then the composi- *
tion function g f
*
: (X , ) (Y , ) is a (, )* -closed (
f : (X , ) (Y , )
*
be a
(, ) -
resp.,
(, )
-open ) function.
continuous and (, )
-closed function. Then:
: (X , ) (Y , )
is (, )* -
(1) If f is injective and (Y , ) is a -T1/ 2 space, then
(X , ) is a -T1/ 2 space.
closed if and only if Cl (f (A )) f (Cl (A )),
subset A of X .
for every
(2) If f is surjective and (X , ) is a -T1/ 2 space, then
Proof. Suppose f is a
(, )* -closed function and A is an
(Y , ) is a -T1/ 2 space.
arbitrary subset of
X . Then f
(Cl (A )) is -closed set in
Proof. (1) Let A be a
g -- closed set in (X , ). To show that
Y . Since
f (A ) f (Cl (A )),
we obtain
A is
- closed. By Theorem 5.6, we have
f (A ) is
Cl (f (A )) f (Cl (A )).
g - - closed. Since (Y , )
is -T1/ 2 ,
f (A ) is a - closed
Conversely, suppose F is an arbitrary -closed set in X . By
set. Since f is injective and
(, )* -continuous,
hypothesis, we ob-
f 1 (f (A )) A
is a - closed set in X . Hence (X , ) is
tain f (F ) Cl (f (F )) f (Cl (F )) f (F ).
a -T1/ 2 space.
Hence Cl (f (F )) f (F ). Thus
f (F )
is -closed in Y .
(2) Let B be a
g - - closed set in (Y , ).
By Theorem 5.6,
It follows that f is (, )* -closed.
f 1 (B ) is
f 1 (B )
g --
-
closed. Since (X , )
f
is a
-T1/ 2 space,
is closed. Since
is surjective and (, )* -
: (X , ) (Y , )
be a bijective func-
continuous,
f (f
1 (B )) B
is a
- closed set in Y.
tion. Then the following statements are equivalent:
(1) f is (, )* -closed.
(2) f is (, )* -open.
Therefore (Y , ) is -T1/ 2 .
f 1 is ( , )* -continuous.
Proof. (1) (2): Let
U SO (X ).
Then
X \U is -
closed in
X . By(1),
f (X
\U )
is -closed in Y .
But
f (X
\U ) f (X ) \ f (U ) Y
\ f (U ). Thus f
(U ) is -
open in Y . This shows that f is (, )* -open.
(2) (3): Let A be a subset of
X . Since f is (, )* -open,
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f (Int (A )) Int (f (A )) for all
A X .
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