 Open Access
 Total Downloads : 155
 Authors : Luay A. AlSwidi, Maryam A. AlEthary
 Paper ID : IJERTV3IS20480
 Volume & Issue : Volume 03, Issue 02 (February 2014)
 Published (First Online): 07032014
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
New Function with “GemSet” in Topological Space
1. Luay A. ALSwidi, 2. Maryam A. ALEthary
1, 2. Mathematics Department , College of Education For Pure sciences University of Babylon .
AbstractIn this paper we introduce a new class of maps called Amap , and under the idea of Gem set in topological spaces and study some of its basic properties and relations among them .

INTRODUCTION
The impression of ideals in topological spaces in treated in the standard text by Kuratowski [2] and Vaidyanathaswamy [5]
,O.Njastad was introduced the idea of compatible ideals in 1966 . This ideal was also called as super compact by R.Vaidyanathaswamy . Furher D.Jankovic and T.R.Hamlett are also worked in this area , Jankovic and Hamlett [1] investigated supplementary properties of ideal spaces . An ideal on a topological space , is a nonempty collection of subsets of which satisfies the following properties : (1)
and implies ; (2) and implies
. An ideal topological space is a topological space
, with an ideal on and is denoted by , , , For a subsets , , = ;
, is called the locall function of with respect to o I and [2] .In 2012 [4] , AlSwidi and AlSada introduced a new type of ideals for a one point and denoted by , they defined the is an ideal on a topological space , at point x is defined by = { } , where U is nonempty set of X . In a topical paper , AlSwidi and Al Naff [2013] [3] hare considered a new set in topological space namely Gem set and depending on the , and, and some its properties are
studied , define a new separation axioms by using the idea of Gemset namely – space and – space,
Definition (2.2) [3]: Let , is a topological space ,
, We defineb = , for each .
Definition (2.3) [3]: A subset A of a topological space
, is called perfected set if , for each , and called coperfect if is a perfected set .
Definition (2.4) [3]: A topological space , is called :

0 space if and only if for each pair if distinct points x,y of X, there nonempty subsets A,B of X such that .

1 space if and only if for each pair if distinct points x,y of X, there nonempty subsets A,B of X such that .
2 space if and only if for each pair if distinct points x,y of X, there nonempty subsets A,B of X such that
= ,

0 space if and only if for each pair if distinct points x,y of X, there nonempty subset A of X such that .

1 space if and only if for each pair if distinct points x,y of X, there nonempty subset A of X such that .

2 space if and only if for each pair if distinct points x,y of X, there nonempty subset A of X such that = , .
Theorem (2.5) [3]: For a topological space , , then the following properties hold :
1 Every 0 space is a 0 space. 2 Every – space is a – space.
i=0,1,2 . and defines two mapping is called and
to carry properties of GemSet from a space to
3 Every
1
2 space is a
1

2 space.
other space and give more properties for new separation axioms . The aim of this paper is to introduce and study the concepts of new class of maps namely , and and Study of the most important properties and the relationship between them, as well as connect the properties
of the separation axioms type of – space and –
4 Every 0 space is a 0 space. 5 Every 1 space is a 1 space. 6 Every 2 space is a 2 space.
Definition (2.6) [3] : A mapping , ( , ) is called – map if and only if , for every subset A of X ,
, ( ) = ( )() .
space, i=0,1,2 with the functions and their effect upon . For a sub set of , is the Gemset of at the point .


PRELIMINARIES :
We recall the folloing definitions and results :
Definition (2.1) [3] : Let , is a topological space ,
, , we defined with respect to space ,
as follows : = { ,
} = , A set is called
GemSet .
Definition (2.7) [3]: A mapping , ( , ) is called – map if and only if , for every subset A of Y ,
,1 = (1 ) 1( ) .

THE NEW FUNCTION AND ITS PROPERTIES:
In this section we introduce a new class of functions namely
, and , and study their properties and relationships .
Definition (3.1) :
A mapping : , , is said to be :

is at ,
. () .

Also is said to be A map on it is A
map at each point on .

1 A map
. 1 ( ) 1 .

is
. [1 ] .

is , 1
and is bijective .
Theorem (3.2) : A mapping from a topological space
, into a topological space , is at x X . iff for each there exist such that
() , () whenever ( ).
Proof: By definition of .
Theorem (3.3) : A mapping from a topological space
, into a topological space , is at x X . iff for each there exist such that
there exist
() whenever = .
Proof: By definition of & theorem (1.6) .
Theorem (3.4) : A mapping from a topological space
, into a topological space , is
1 1 1 .
Proof : Suppose that and let , Put
= 1 and = 1() , Since A map so we get that but is one to one hence
( 1 ) , it follows 1
1 .
Conversely : Let , Put = 1 and
= 1() , By hypotheses : 1 1 it follows that ( ) (1 ) = . Thus A map .
(2) In the function : , , is bijective , then
1 = 1( ) .
Proof : By bijective of the map , we accomplished
1 = { 1 } = { 1 }
= { 1 1 () (1 )) = 1( ) .
Theorem (3.8) : If a bijective mapping : , ,
.
Proof : By noti function of .
Theorem (3.9) : An injective map from a topological space , onto a topological space , is iff 1 .
Proof : Assume that be map ,such that for
and , we gotting that , there exists
such that 1 . But ,
1 ( ) 1 1 = 1 and 1( )
1 . Thus 1 .
Conversely: For & and by identifiable of
of 1 , we have there exist
suc that 1 ( ) 1 so we get that
[1 ] [1 ] and by surjective of , we getting that [1 ] . Then is . Theorem (3.10) : A bijective function isits .
proof : By definition of and theorem (3.9) .
Theorem (3.11) : Every injective function is
.
Proof : Firsthand from definition of function .
Theorem (3.12) : A bijective is function
if is continuous function .
Proof : By continuity of and definition of and used the note (3.7) we can prove this theorem .
Theorem (3.13) : Every is if the function is injective .
Proof : Let , Put = & = 1()
[Since is injective ] , So by , we get that( ) = = then ( 1( ) ) . Hence is .
Theorem (3.14) : Every is if the
Theorem (3.5) : An injective map from a topological
space , onto a topological space , is iff
function is bijective .
Proof : Let , Put =
1
& = ()
. (1 ) 1( ) whenever
.
Proof: By definition of and identifiable of
.
Theorem (3,6) : A function : , , ,
if and only if ( ) ( ) for
and .
Proof : By definition of and identifiable of
.
Note (3.7) :
(1) Let : , , , if is injective map , then
= () .
Proof: Since the map is injective , then for =
= =
[Since is bijective ] , So by , we get that1 = [1 ] 1( ) = then (1 ) =
( ) so we get that ( ) . Hence is
.
Theorem (3.15) : If : , , is , open. and bijective map then is .
Proof : Let , , Put 1 = , 1 = . Since is so ( ) .Now we must prove that ( ) .Let ( ) so there exists
= and .Therefore
, .The ( ) it follows that ( )
() ( ) .By note (3.7)(1) we have = () ,
= since is bijective .So we get that ( )
(()) = () .
() for ( ) , Hence . Therefore ( )
. Thus is .
Theorem (3.16) : Let : , , be bijective function . Then is iff is .
Proof : By definition of and . Theorem (3.17) : Let : , , be bijective function . Then is iff is .
Proof : Straight from definition of and
.
Remark (3.18) : For a function : , , , the following diagram holds :
Diagram (3.1) : The relationship between the maps .
Bijective . Injective .
Bijective and open . Bijective and continuous .
Diagram (3.1) : The relationship between the maps .
Remark (3.17): The converse of theorems need not true as seen from the following examples .
Example(3.18) (1) : Let : , , if is
then need not .
Let = , , , , : , , =
, , , , , , , , {, , } , = , , =
, = , = , = .
but is not since let = , , =
= , , =
= = , ( ) = , , ,
( ) .
(2) : Let : , , if is then need not
.
Let = , , : , , =
, , , , , = , = , = .
but not since , let =
= , 1 = , .
1 = 1 = , So we get that
1 1 .
Let
= , , , = , , : ,
, = , , , , ,
= {, , , , , , , } = , = , =
. but is not since =
, = , , 1 , = , , [1 ]1() =
{} = {} , {, } .
(5) : Is the same example (3.18) (1) but is not
since if = , =
1 = = , = , , 1 = , , so 1 and such that 1 . (6) : Is the same example (3.18) (3) 1 is but is not since if = so such that ( )
() .
Theorem (3.18) : Let : , , be A map
: , , A map ,
: , , A map at .
Proof: Let .
b of . But
then there exist by of
. so we get that it follows that
so ( )
. Thus : , , A map . Theorem (3.19) : Let : , , be Then for , the following statements are true :
1)) The inclusion function :
(A,) , is .
2)) : A, , is .
3)) , , is always .


THE FUNCTION AND SEPARATION AXIOMS :
In this section we study the relationships between separations axioms _ , = 0,1,2 ; _ ,
= 0,1,2 and _ , = 0,1,2 ; by using the maps . Theorem (4.1) : Let : , , be bijective open mapping . Then :
(1) if (, ) is strongly and _ then
, _ .
(2) if (, ) is strongly and _ 1 then
, 1_ .
(3) if (, ) is strongly and _ 2 then
, 2_ .

: Let
:
,
11
,
if is then
Proof: Closely from use the definition of strongly space ,
open map and , = 0 ,1 ,2 , we evidential this
is .
theorem .
Let = , , : , 1 , 2 1 =
, , , , ,
2 = , , , , , , , {} , = , =
, = . so 1 is , by theorem (3.9) we get that . = = ,
= , , = [ ] = , , , So we get that [ ] . Therefore .
Remark (4.2): The converse of theorem need not true as seen from the following examples .
Example (4.3) :
(1) Let = , , and : , 1 , 2 s.t
1 = , , , , & 2 =
, , , , , , , and = , =
, = .Then , .Let =
2 0
(4) : Let : , , if is then is
.
= , , is not open set in , 1 , So we get that , 1 is not strongly space .
(2) Let = , , and : , 1 , () s.t
1 = , , , , and = , =
, = .Then , () 1 .Let
= = , , is not open set in , 1
So we get that , 1 is not strongly space .
(3) Let = , , and : , 1 , () s.t
1 = , , , , and = , =
, = .Then , () 2 .Let
= = , , is not open set in , 1
So we get that , 1 is not strongly space .
Corollary (4.4) : Let : , , is bijective and open maping , Then :
1 if (, ) is strongly and then
, _ .
2 if (, ) is strongly and 1 then
, 1_ .
3 if (, ) is strongly and 2 then
, 2_ .
Proof: By theorem (2.5) and (4.1) .
Theorem (4.5) : Let : , , is bijective and continuous mapping , Then :
Let = = , , is not open set in
, 1 , So we get that , 1 is not strongly space . Corollary (4.8) : Let : , , is surjective continuous maping , Then :
1 if , is strongly and then
, _ .
2 if , is strongly and 1 then
, 1_ .
3 if , is strongly and 2 then
, 2_ .
Proof: By theorem (2.5) and (4.5) .
Theorem (4.9): Let : , , be bijective open mapping , Then :
(1) if (, ) is strongly and _ then
, _ .
(2) if (, ) is strongly and _ 1 then
, 1_ .
(3) if (, ) is strongly and _ 2 then
, 2_ .
Proof: By definition of strongly space , open map and
, = 0 ,1 ,2 .
(1) if , is strongly & _ 0 then
(, ) 0_ .
(2) if , is strongly & _ 1 then
(, ) 1_ .
(3) if , is strongly & _ 2 then
(, ) 2_ .
Proof: By definition of strongly space , continuous map &
, = 0 ,1 ,2 .
Remark (4.10): The converse of theorem need not true as
seen from the same example (4.3) .
Theorem (4.11) : Let : , , is bijective continuous mapping , Then :
(1) if , is strongly and _ 0 then
(, ) 0_ .
(2) if , is strongly and _ 1 then
(, ) 1_ .
Remark (4.6): The converse of theorem need not true as seen
from the following example .
Example(4.7) :
(1) Let = , , and : , , , s.t
(3) if , is strongly and _ 2 then
(, ) 2_ .
Proof: By definition of strongly space , continuous map and
, = 0 ,1 ,2 .
1 2
1 = , , , , , and
2 = { , , , , , } , = , =
, = . . & 1 1 , Then , 1 0
. Let = = , is not open set in , 2 , So we get that , 2 is not strongly space
.
(2) Let = , , and : , () , , 1
s.t 1 = , and = , = , = .
. & 1 1 , Then , () 1 . Let = = , , is not open set in
, 1 ,So we get that , 1 is not strongly space .
(3) Let = , , and : , () , , 1
s.t 1 = , and = , = , = .
. & 1 1 , Then , () 2 .
Remark (4.12): The converse of theorem need not true as seen from the same example (4.7) .
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