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Generalized -Closed Sets and

Continuous Functions

Alias B. Khalaf , Sarhad F. Namiq

(,)* -

AbstractIn 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 Termss-operation, -open, generalized -closed, -T space , (,)*

-continuous function .

—————————— ——————————

1 INTRODUCTION

HE study of semi open sets and semi continuity in topo- logical spaces was initiated by Levine [5]. Analogous to

3 - OPEN SET

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.

Proposition 3.2. For a topological space

(X ,), SO(X )  SO (X ).

Proof. Obvious

The following examples show that the converse of the
above proposition may not be true in general.

2 PRELIMINARIES

Example 3.3. Let X

 {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

(3) A point

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

Proof. Straightforward.

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x V .

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Proposition 3.7. Let AI

be any collection of  -open
Theorem 3.14. Let A ,B be subsets of X. If

sets in a topological space (X ,) , then


Ais a -open set.

: SO (X )  P (X ) is an s-regular s-operation, then:

Proof. Let x

A then there exist I I such that x A ,

  I 0

 0 (1)

Cl (A B )  Cl (A )  Cl (B ).

since Ais a -open set for all I implies that there exists
a semi open set U such that

(U )  A

0

  I

A. There-

(2)

Int (A B )  Int (A )  Int (B ).

fore

  I


Ais a -open subset of (X ,).

Proof. Obvious.

The following example shows that the intersection of two

-open sets need not be -open.

Example 3.8. Let

X  {a,b ,c} and  P (X ) . We define an s-

4 GENERALIZED -CLOSED SET AND -T SPACE

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

Proof. Let

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.

Proof. Obvious.

Cl( A)  A. Thus

U SO( X ) containing

Cl( A)  U for every

A .

Proposition 3.11. For subsets

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

(1) (2)

A Cl (A ).

Cl (A ) is -closed set in X .

not -closed.

Example 4.3. Let

X {a,b ,c},

and P (X ).

We define

(3)

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

(5)

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).

Proof. Obvious.

Proposition 3.12. Let (X ,) be a topological space and

A X . Then Cl (A )  A d (A ).

Proof. Obvious.

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 ).

Example 4.5. Let

X {a,b ,c},

and P (X ).

We define

Proof. Obvious.

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

(1)

X \ Int( A)  Cl( X \ A).

Then, if A  {a} and

B  {b}.

So, A and B are g- -

(2)

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.

(3)

X \ Cl( A)  Int( X \ A).

Theorem 4.6. If

: SO (X )  P (X ) is a s-regular s-operation.

(4)

Int( A)  X \ Cl( X \ A).

Then the finite union of g- -closed sets is always a g- -

Proof. Obvious.

closed set.
Proof. Let A and B be two g- -closed sets, and let

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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 .

Example 4.7. Let 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

Proof. Suppose

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

Example 4.9. Let

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).

Example 4.16. Let X

{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

Theorem 4.17. Let

: 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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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.

Theorem 5.3. Let f

: (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.

Theorem 4.20. If

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 .

Proof.

Int( A)  B A

implies

(3)

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-

(4)

f (Cl (U ))  Cl (f (U )), for every U

X .

-closed, by Theorem 4.8, X

\ B is g- -closed and B is

(5)

Bd (f

1 (V ))  f

1 (Bd (V )), for every

-open.

V Y .

Theorem 4.21. Let (X ,) be a topological space

(6)

f (d (U ))  Cl (f (U )), for every U

X .

(X ,) and

: SO (X )  P (X ) be an s-operation. The

(7)

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.

Proof. (1) (2): Let F Y

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.

(3) (4): Let U X

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

(4) (2): Let W Y

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

(5) (7): Let V Y , then by hypothesis,

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 (

Definition 5.1. A function f

: (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

(7) (5): Let V Y , then by hypothesis,

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

Theorem 5.2. Let f

: (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

(1) (6): Since f is

(,)* -continuous and by (4), we

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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

Theorem 5.8. A function f

: (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-

Proposition 5.4. If the functions f

: (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.

Proof. Let

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.

Definition 5.5. A function f

: (X ,)  (Y ,) is said to be

Theorem 5.9. A function f

: (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

Theorem 5.6. Suppose that

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,

Theorem 5.10. Let

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.

Theorem 5.7. A function f

: (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-

(5)

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.

Theorem 5.11. Let

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 (

Theorem 5.16. Let

f : (X ,)  (Y ,)

*

be a

(,) -

resp.,

(,)

-open ) function.

continuous and (,)

-closed function. Then:

Proof. Obvious.

Theorem 5.12. A function f

: (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 (,)* -

Theorem 5.13. Let f

: (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 .

(3)

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,

REFERENCES

so by Theorem 5.12,

f 1 (Cl (f (A )))  Cl (f 1 (f (A ))). This

implies that

Cl (f (A ))  f (Cl (A )).

Thus

[1] Alias B. Khalaf and Sarhad F. Namiq, New types of continuity and

Cl ((f

1 )1 (A ))  (f

1 )1 (Cl (A )),

for all

A X .

separation axiom based operation in topological spaces, M. Sc. Thesis,

Then by Theorem 3.1.6, it follows that continuous.

f 1 is

(, )* -

University of Sulaimani (2011).

[2] B. Ahmad, S. Hussain, Properties of -operations in topological

(3) (1): Let A be an arbitrary -closed subset of X . Since

spaces, Aligarh Bull. Math., 22, No. 1 (2003), 45-51.

f 1

is a

(, )* -continuous. Then by Theorem 3.1.6,

[3] P. Bhattacharyya and B. K. Lahiri, Semi-generalized closed set in to-

(f 1 )1 (A )

is -closed in Y .
But

(f 1 )1 (A )  f (A ).

pology, Indian J. Math., 29(1987). no. 3, 375-382. MR 90a:54004. Zbl

This means that f is (, )* -closed.

687.54002.

[4] S. Kasahara, Operation-compact spaces, Math. Japonica, 24 (1979), 97-

Definition 5.14. A function f

*

: (X ,)  (Y ,)

is said to

*

105.

be (,)

-homeomorphism if it is bijective,

*

(,) -

[5] N. Levine, Semi-open sets and semi-continuity in topological spaces,

continuous and (,)

-open.

Amer. Math.Monthly, 70 (1)(1963), 36-41.

Corollary 5.15. If f

: (X ,)  (Y ,) is a bijective function,

[6] N. Levine, Generalized closed sets in topology, Rend. Circ. Math. Pa-

then the following statement are equivalent.

(1) f is (,)* -homeomorphism.

lermo, 19, pp.89- 96, (1970).

(2)

f (Cl (A ))  Cl (f (A )) for all

A X .

(3)

Cl (f

1 (B ))  f

1 (Cl (B )) for all B Y .

(4)

f (Int (A )) Int (f (A )) for all

A X .

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