On Separation Axioms With Respect To Gem Set

On Separation Axioms With Respect To Gem Set

1Luay A. Al-Swidi 2Maryam S. AL-Rubaye

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

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

Abstract : In this paper, we create a new definition of space namely" _, _ and _ in topological space " and we do a new definition of separation axioms by using the idea of "Gem set''.

Keyword- New separation axioms, , _

and _ , Gem set and topological ideal .

1-INTRODUCTION AND PRELIMINARIES :

The epigram of ideal presented first by K. Kuratowski [1]. In general topological Hamlett and Jankovi´c [2, 3, 4,5] they introduced the application of topological ideal in generalization of most essential properties and the ideal as this form: An ideal I on a topological space (X,T) is a non-empty collection of subsets of X which satisfies the following properties: (1) A I and B A implies B I. (2) A I and B

I implies AB I and called the (X,T,I) ideal topological space in addition to K. Kuratowski [1] used the ideal to define space (X,T,I) and a subset A X, (I) ={ x X : U A I for every U T(x)} is called the local function of A with respect to I. We simply write Ainstead of ( I) ,F.G. Arenas

, J .Dontchev and M.L .Puertas [6] introduce some weak separation axioms under the concept ideal . In recent years Al- Swidi and AL-sadaa [7] they defined for each element had ideal by this form: let (X,T) be a topological space and x X

,we denote by Ix to an ideal {G X: x } , Where X is non- empty set ,again Al-Swidi and AL-Nefee [8] they use the idea of ideal defined new set namely "Gem set " which means that A subset B of a topological space (X,T) . Then they defined Bx with respect to space (X,T) as follows

: ={y : , for every G T(y)} where T(y)={G

: } .A set was called "Gem set" and define a new separation axioms by using Gem set namely it the"I-Ti- space "and "I-Ti- space " ,i=0,1,2. Through out this paper we defined anew separation axioms by benefit of Gem set namely "Si -space " i=0,1,2,3,4 and studied proprieties and the relationship between "I-Ti-space "and Ti-space. I=0,1,2. Also we define anew space called R_space , M_space and N_space and we study the proprieties and relation between them.

Definition1.1 [6].A topological space (X,T) is called

• I _T°_space if and only if for each pair of distinct point x

  ,y of X there exist nonempty subset A ,B of X such that yAx or xBy .

• I_T1_space if and only if for each pair of distinct point x, y of X there exist nonempty subset A ,B of X such that yAx and xBy .

• I _T2_space if and only if for each pair of distinct point x, y of X there exist nonempty subset A,B of X such that yAx and xBy with By Ax = Ø.

• I _T°_space if and only if for each pair of distinct point x ,y of X there exist nonempty subset A of X such that yAx or xAy .

• I_T1_space if and only if for each pair of distinct point x

  ,y of X there exist nonempty subset A of X such that x Ay and yAx .

• I _T2_space if and only if for each pair of distinct point x

,y of X there exist nonempty subset A of X such that x Ay and yAx with Ay Ax = Ø.

Definition1.2 [6] Let (X,T) be a topological space and A

X We define Prx A = Ax A, for each x X.

Proposition 1.3 [6] Let (X,T)be topological x A if and only if x Ax ,for each A X ,xX.

Definition1.4 [6] A mapping f :X Y is called I-map .if and only if , for every subset A of X ,x X f((Ax ) =

A (x).

Remark 1.5 [5] Let f: (X, ) (Y, ) be I-map .then

,f(prx (A))= pr(x)(A)) for every subset A of X and x Proposition 1.6 Let (X,T) be topological space, and A subset of X ,x X. If x A , Then Ax = Ø

Proof:- Let Ax Ø Then there exist at least one

element, say y Ax by definition of Gem set then A Gy Ix.Hence x A GY sox A which Contradiction ,then Ax = Ø

Definition1.7 Let (X,T) be a topological space ,for each x X

,anon empty subset A of X ,is called a strong set if and only if (Ax is open set and x A).

Definition 1.8 A topological space (X,T) is said strong space if every sub set of X is strong set .

2-"_ , _ and _ in topological space "

In this section, we offer new definitions of the spaces through Gem set call them R_space , M_space and N_space with study some results and properties.

Remark 2.1 . If a function f:(X, T) (Y, ) is one to one

,then1 Iy = I1(y)for each y Y.

Remark 2.2 . If a function f:(X, T) (Y, ) is bijection , then I1(y) = Iy for each y Y.

Theorem 2.3 . If a function f:(X, T) (Y, ) is continuous and one to one ,then1 By (1 B )1(y) for each y Y.

Proof:- Let a 1 By ,then there exist d By with a=1 d . And Hd B Iy for each Hd T(d) ,then

1 (Hd ) f 1(B) 1( ) ( by remark 2.1 ).Hence by continuity of f we get that a ((1 )1 .Then

1 By ((1 B )f1(y) .

Theorem 2.4 . If a function f:(X, T) (Y, ) is continuous, open and bijection , then (1 B )1 y 1 By for each yY.

a

Proof :- Let a (1 B )1 (y) then U 1 B

I1 y for each Ua T(a) thus (Ua ) B Iy (by Remark 2.2).Therefor by properties of open map ,we get that a (

1 By hence (1 B )f1(y) 1 By .

Corollary 2.5 If a function f:(X, T) (Y, ) is continuous, open and bijection . Then (1 B )1 y = 1 By for each yY.

Proof :- By using the theorem 2.3 and 2.4 , we get the prove. Definition 2.6 . A topological space (X,T) is called R_space if and only if for each x Xand a nonempty subset A of X such that x Prx (A) then there exist a nonempty subset B, C of X such that x Bx , Prx (A) Cx .

we noted that in definition of R_space that Bx and Cx are disjoint where if x C Cx = Ø this contradiction with Prx (A) Cx .

Example2.7. Let X = {x, y, z } , T ={X,Ø}Ix={Ø,{z},{y},{z,

y}},Iy = Ø, z , x , z, x andIz={Ø,{y},{x},{y, x}}

set A ={y},B ={x, y}and C ={z, x} then prx (A)={ y},Bx =

{x, y, z}and Cx ={x,z,y} it follows x Bx , prx ( A) Cx .

eroehTm 2.8. Every subspace of R_space is R_space .

Proof :- Let Y , y Yand A Y such that y pry (A) ,then there exist a subset H of X such that pry (A)= pry (H) Y ,so we get that y pry (H) Y, but X is R_space then there exist a nonempty subsets B,C of X such that pry (H) By and y Cy .Thus pry (H) Y By Yand y Cy Y .Put Dy = By Y and Sy = Cy Y it is follows that pry (A) Dy and y Sy .Hence Y is R_subspace.

Theorem 2.9. Let f be bijection open and continuous map from (X,T) space onto R- space (Y,). Then (X,) is – space ,if f is -map.

proof:-Let x X and A X such that x prx A . Since f is I-

map. then f(x) pr(x)( A ),but Y is R- space so there exist a nonempty sub sets B,C of Y such that pr(x)( A ) B(x) and f(x) C(x).So 1 (B(x)) and 1(C(x)) is a nonempty subset of X such that . 1(pr(x) f A

1 B x and x 1 C(x) = [1(C)]x ,( by corollary 2.5) prx A Bx and x [1()] .Hence (X ,T) is R- space.

Theorem 2.10. Let f be bijection open and continuous map fro (X,T) -space onto (Y,). Then (Y,) is -space ,if f

is -map. Definition 2.11.:A topological space (X,T) is called M_space if and only if for each x X and a nonempty subset A of X such that x Prx (A) then there exist a nonempty B,C of X , such that x Prx C and Prx (A) Prx (C),With

prx B Prx C = Ø.

Example 2.12. Let (X,T ) be a topological space such that X ={x, y, z } ,T={X,Ø,{y},{z},{y, z}},Ix

={Ø,{z},{y},{z ,y}} , Iy ={Ø,{z},{x},{z ,x}} and

Iz ={Ø,{x},{y},{x ,y}} . let x X and a nonempty subset set A={y} of X such that x prx (A)={y} then there exist a nonempty subset B ={x}and C ={z, y} of X then,prx (B)={x}and prx (C)={z,y} it follows x prx (B) , prx ( A) prx (C) with prx (B) prx (C)={x } z, y = Ø .

Theorem 2.13. Every subspace of M_space , is M_subs pace.

Theorem 2.14. Let f be bijection open and continues map from M- space (X,T),onto (Y,) space. Then (Y,) is M- space if f is -map .

Proof:-Let y Y and B Y such that y pry B (By Remark

1.5) then 1 () pr 1 (1 B ) since X is M- space then there exist a nonempty sub sets D,C of X such that pr1 y (1 B ) pr1 (D)and 1()

pr1 (C).with pr1( )(D) pr1( ) C = Ø. Now ( pr1( )(1 B ) (pr1( )(D)), (1())

(pr1( )(C)) since f is I-map ,then pry (B) pry ((D)), y pry ((C)) with ( prx (D) prx C ) = pry ( D ) pry (C ) = Ø = Ø then Y is M- space. Theorem 2.15 Let f be bijection _ and continues map from ( X,T) space onto M- space (Y,) space. Then (X,) is M- space if f is open map .

proof :- The same of above theorem .

Definition 2 .16 A topological space (X,T) is said to be N space if and only if for each x X and every tow subsets M,L of X then there exist a nonempty subsets B,C of X such that prx (M ) Bx and prx (L ) Cx .

Remark 2.17. We can not say Bx Cx = Ø is disjoint set if Bx Cx = Ø then x C then Cx = Ø this contradiction with prx (L ) Cx ,similar x B .

Theorem 2.18 Every subspace of N space is N

space.

Theorem 2.19 Let f be is a bijection and continues function of space (X,T) onto N- space (Y,) space. Then (X,) is N- space if f is open map.

Proof:-Let L,M pair sub set of X. so L , M is disjoint subset of Y since Y is N- space then there exist a nonempty sub sets B,C of Y such that pry L By and pry M Cy . so 1 By and 1 Cy is a

nonempty subset of X, thus 1 prX L 1(By ,

1 prX M 1(Cy (by corollary 2.5)

1(pry L = prx L (1 (B))x and prx M

(1(C))x).Hence (X ,T) is N- space.

" separation axioms"

In this section, we introduce the concept of new definition separation axioms called Si -spaces and investigate some of their properties and study the relationship between "I-Ti- space ","I-Ti-space and Ti-space.

Definition 3. 1 :-

1. A topological space (X,T) is said S°_space if and only if for each pair of distinct point x ,y of X there exist nonempty subset A of X and A contain at least one of them such that yAx or xAy .

2. A topological space (X,T) is said S1_space if and only if for each pair of distinct point x ,y of X there exist nonempty subset A of X and A contain at least one of them such that x Ay and yAx .

3. A topological space (X,T) is said S2_space if and only if for each pair of distinct point x ,y of X there exist nonempty subset A of X and A contain at least one of them such that x Ay and yAx with Ay Ax = Ø.

4. A topological space (X,T) is said S3_space if it was

   x .hence (X,T) is (_ 1_ and _ 2_

   )but not (1_ and 2_ )since Let x ,y

   such that there exist A={x} . ={x, y} so y

   Then (X,T) is not( 1_ and 2_).

   Example 3.7 Let X={x, y,z} T={X,Ø,{x},{x, y}}and =

   Ø, , , , , = Ø, , , , . Let x, y

   there exist A={z} and B={y}, = Ø = {, } , x and y ..Hence (X,T) is _ 1_ but not

   1_ since Let x, y ,A = {x}then y =

   {, , }then (X,T) is ( _ 1_and _ 2_) but not (1-space and 2-space ).

   Preposition 3.8 Let (X,T) be topological space if X strong space then we have.

   1. Every _ is _.

   2. Every 1_ is 1_.

   3. Every 2_ is 2_.

   4. Every __space is _

   M space and S1_space

   5. Every _1_space is 1_.

5. A topological space (X,T) is said S4_space if it was

6. Every _2_space is 2_.

N space and S1_space.

Theorem 3.2 For topological space (X,T), then the following properties hold:

1. Every T°_space is S°space

2. Every T1_space is S1space.

3. Every T2_space is S2_space.

4. Every S°_space is I_T_space.

5. Every S1_space is I_T1space.

6. Every S2_space is I _T2_space.

7. Every S°_space is I_T_space.

   7. Every – _ is _.

   8. Every _°_space is °_.

   9. Every _1_space is 1_.

   10. Every _2_space is 2_.

   11. Every _°_space is °_.

   12. Every _ _space is _.

8. Every S1_space is I_T1_space. 1 1

9. Every S2_space is I_T2_space.

13. Every _ _space is _.

2 2

Proof :- Straight forward .

Remark. 3.3. The converse of theorem need not be true as since from the following examples .

Example 3.4 Let(X,T) be a topological space such that X={x, y,z,h} T={ X,Ø,{z},{y,z}}and,

= Ø, , , , , , , , , , , , ,

and = Ø, , , , , , , , , , , . Let x, y

such that x then there exist a nonempty subset A={x} of X contain at least one element of them we say that x such that x = Ø. Then (X,T ) is °-space but not °- space.

Example 3.5 Let(X,T) be a topological space such that X={x, y, z} T={X,Ø,{x},{x ,z},{x ,y},{y}}and =

, , , , , = {,{x},{z},{x,z}} .Let x, y such that x then there exist a nonempty subset A={x} of X contain at least one element of them we say that x such that x = Ø.and y = {, }.Hence (X,T ) is (1- space and 2-space )but not (1-space and 2-space)

Example 3.6 Let X={x, y, z} T={X,Ø,{z}} =

Ø, , , , , = Ø, , , , , . Let x ,y

there exist A={z} . =Ø and = Ø so y and

Proof:- (1) Let x ,y such that x .Since (X,T) is

_ then there exist a nonempty subset A of X contain at least one element of them we say that x such that x

or y .Assume y since X is strong space then A is strong set so is open set and then x .Hence (X

,T)is _.

Proof:- (2) Let x ,y such that x .Since (X,T) is

1_then there exist a nonempty subset A of X contain at least one element of them we say that x such that x

and y since X is strong space then

and is open set with y

(by proposition 1.6)) x and y ,hence (X,T) is

1_.

Proof:- (3) The same way of proof (2)

Proof:-(4) Let x ,y such that x .Since (X,T) is –

°_ thenThere exist a nonempty subset A of X such that x or y .Since X is strong space then A is strong therefor x hence (X,T) is _.

Proof:-(5) Let x ,y such that x .And let (X,T) is –

1_ thenThere exist a nonempty subset A of X such that

x and y .Since X is strong space then A is strong therefor x hence (X,T) is 1_ .

Proof:-(6) The same way of proof (5)

Proof:-(7) Let x ,y such that x .And let (X,T) is

°_ then

There exist two a nonempty subset B, A of X such that x

or y assume x since X is strong space then A is strong set .So Then x .Hence (X,T) is _.

Proof 🙁 8) Let x ,y such that x .and let (X,T) is –

°_ then

There exist two a nonempty subset B, A of X such that x

or y ,since X is strong space then A ,B are strong sets .So , are open sets and x ,y .(by proposition 1.6 ) then x and y and we have x

or y .Hence (X,T) is °_.

Proof :- (9) Let x ,y such that x .And let (X,T) is –

1_ then

There exist two a nonempty subset B, A of X such that x

and y since X is strong space then A ,B are strong sets .so , are open sets and x ,y .(by proposition 1.6 ) then x and y and we have x

and y .Hence (X,T) is 1_.

Proof :- (10) The same way of proof (9)

Proof :- (11 ) Let x ,y such that x .Since (X,T) is –

°_ thenThere exist a nonempty subset A of X such that x or y .Since X is strong space then A is strong

therefor x (by proposition 1.6 ) x .Hence (X,T) is _ .

Theorem 3.14 Let f be is open and bijection from (X,T) onto Si_space (Y,) then (X,) is Si_space if f is

for each i=0,1,2

Proof:- Assume i=2 ,let x1, x2 X such that x1 x2 since f is one to one then(x1) (x2) but Y is S1-space then there exist a nonempty subset B of Y such that B contain at least one element we say that (x1) B with (x1) B(x2)and (x2) B(x1)such that Bf(x2) Bf(x1) = Ø so x1 1 B and 1 ((x1)) 1(Bx2))=x1

1 (B)x2 , x2 1(B)x1 with

1 (B)(x1)) 1(B) (x2)) = 1(B)x1 1(B)x2 = Ø.Then X is S2-space.

Proposition 3.15 The product space of 2-space is 2-space.

Proof :- Let X and Y are 2-space to prove X× is 2-space

.Let (1,1), (2,2) be two distinct point of X× either

1 2 or 1 2 take 1 2.since X is 2-space then there exist a nonempty sub set A of X such that contain at least one element we say that 1 and 1 2 and

2 1 with 2 1 = Ø . it follows 1 × and

2 × is a nonempty subset of X× then (1,1) 2 ×

and (2,2) 1 × with (1 × ) (2 × )

=(1 2 ) × = Ø × = Ø. Hence X× is 2-space. Corollary 3.16The product space X=×{ : } If is 2- space then X is 2-space.

Proof :- Let is 2-space to prove X =×{ : } is 2- space let x={ : : } and y={ : : } be two distinct point of X then for some where ,

.since is -space then there exist a nonempty sets in

Proof :- (12) Let x, y and let X is – _

2

1 and contain at least one element we say that such

then there exist a subset A of and

that

and

with = Ø

.since then ( ) since( ) = Ø.

Put = then = Ø then y and we have

.Since X is strong set then and .Hence (X,T)

_1_space.

Proof :- (13) The same way the proof (12).

Corollary 3.9 Let (X,T) be topological space if X is strong space and – space then we have .

1. Every 2_ is 3_.

2. Every -2_ is 3_.

Theorem 3.10 Let (X,T) is 3-space then _2_space

Proof:- By (Let (X,T) is 3-space then 2-space ) then X is

2-space and by 2-space is _2_space then (X,T) is _2_space.

Remark 3.11 Every 2 -space is a 1 space.

Remark 3.12 If X is door space then every 3 space is 3

space

Theorem 3.13 Let f be is bijection and I -map from Si_space (X,T) onto (Y,) space then (Y,) is Si_space if f is open and continuous for each i=0,1,2.

Proof :-Assume i=2,let y1, y2 Y such that y1 y2 since f is one to one then1(y1) 1(y2) because of X is S1-space

then there exist a nonempty subset A of X such that A contain at least one element we say that 1 (y ) A and 1(y )

since = and = it follows =

and = then x 1( ) and y 1( ) with 1 = 1 Ø = Ø , 1( ) and 1( ) are a nonempty sets in X

.Hence X is 2-space.

REFERENCE :

1. K. Kuratowski, Topologie I, Warszawa, 1933.

2. T.R. Hamlett and D. Jankovi´c, Ideals in Topological Spaces and the Set Operator ,Bollettino U.M.I., 7 (1990), 863874.

3. T.R. Hamlett and D. Jankovi´c, Ideals in General Topology, General Topology and Applications,(Middletown, CT, 1988), 115125; SE: Lecture Notes in Pure & Appl. Math.,123 (1990), Dekker, New York.

4. D. Jankovi´c and T.R. Hamlett, New topologies from old via ideals, Amer. Math.

   Monthly, 97 (1990), 295310.

5. D. Jankovi´c and T.R. Hamlett, Compatible Extensions of Ideals,

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6. Arenas , F.G . Dontchev , j .and Puertas M.L ."idealization some weak separation axioms "Acts Math .Hunger ,Vol .89,No.2,pp 47-53. [7]- AL-Swidi ,L.Aand AL-sadaa , A.B (2013):"Turing Point of Proper Ideal "in Archive Des Science (Impact factor 0.474), volume 65, No.7, ISSN 1661-464X, pp (213-220).

[8]-AL-Swidi,L.Aand AL-Nafee, A.B (2013): New separation

1 1 axioms using the ideal "Gem set" in topology space :Mathematical

2

Af1(y2 ) and 1(y ) A1(y1)) with A1(y2 )

Theory and Modeling;Vol.3, No.3,2013.www.iiste.org.

A1 (y1 )) = Ø so (1(y )) A = y

(A) since f is I –

1 1

map then we get that y1 (A)y2 and =y2 (A)y1 with

(A)y2 (A)y1 ) = Ø = Ø . Then Y is S1-space.