New Function with “Gem-Set” in Topological Space

New Function with “Gem-Set” in Topological Space

1. Luay A. AL-Swidi, 2. Maryam A. AL-Ethary

1, 2. Mathematics Department , College of Education For Pure sciences University of Babylon .

AbstractIn this paper we introduce a new class of maps called A-map , and under the idea of Gem- set in topological spaces and study some of its basic properties and relations among them .

1. 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 non-empty 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] , Al-Swidi and Al-Sada 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 non-empty set of X . In a topical paper , Al-Swidi 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 Gem-set 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 non-empty subsets A,B of X such that .

   • 1- space if and only if for each pair if distinct points x,y of X, there non-empty subsets A,B of X such that .

     2- space if and only if for each pair if distinct points x,y of X, there non-empty subsets A,B of X such that

     = ,

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

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

   • 2- space if and only if for each pair if distinct points x,y of X, there non-empty 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 Gem-Set 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 Gem-set of at the point .

2. 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

   Gem-Set .

   Definition (2.7) [3]: A mapping , ( , ) is called – map if and only if , for every subset A of Y ,

   ,1 = (1 ) 1( ) .

3. 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 is

   its .

   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 that

   1 = [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 .

4. 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_ .

1. : 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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