Generalized γ-Axiom for Advanced Engineering Applications

Generalized -Axiom for Advanced Engineering Applications

S.MANIMEKALAI1, R.ANANDHI2

1(Mathematics, Dr.N.G.P Arts and Science College, India)

2(Mathematics, Dr.N.G.P Arts and Science college , India)

ABSTRACT

.

In this paper we discuss the properties, characterization of – open sets , relations between regular open set , – open

set and semi open set, – adherent, -closure of a subset A of a topological space, – irresolute function between topological spaces and equivalence relation – correspondence on the set of a topologies of a set X .The inter relationship between -axiom and various Separation axioms.

AMS(2000) Subject Classification No:primary:81T45,secondary:57N17 Keywords -adherent, -axiom, generalized -open set, -irresolute, -open set

A subset A of a topological space (X, ) is said to

1. INTRODUCTION

A subset A of a topological space ) is said to be -open iff there exists a regularly open set R such

that RAcl(R)The Paper is organized as follows.In section (2) Properties and characterization of -open sets and

relations between regular open sets, -opensets and semi open sets.

It is interesting to note that

1. The complement of a -open set is again a -open set.

2. Neither the union nor the intersection of two -open sets is -open. In Section (3) ,-adherant, -closure of a subset A of a topological space,-irresolute function between topological space and equivalence relation -correspondance on the set

of a topologies of a set X.Section (4) is devoted to the study of -axiom due to Sharma[27]. Separation axioms T0, T1 and T2

which are generalizations of separation axioms T0,T1 and

T2 respectively are studied. It is interesting to note

in a topological space all the three axioms T0, T1 and T2 coincide. Therefore these axioms are referred as

µ-axiom.The interrelationship between µ-axiom and various separation axioms semi T2, semi T1, semi T0 and

T0 are discussed.

2.-open sets Definition : 2.1

A subset A of a topological space (X, ) is said to be -open iff there exist a regularly open set R such that RAcl(R).

Theorem :2.2

be -open iff there exist a regularly closed set F such that int(F)AF.

Proof :

Assume A is a -open set.

Since A is -open, there exist a regularly open set R such that RAcl(R).

R=int(cl(R) cl(R) = cl(int(cl(R))) F= cl(int(F)).

we getAF

Hence int(F)AF.

Hence there exist a regularly closed set F such that int(F)AF

Let R=int(F)

Consider int(cl(R))=int(cl(int(F)))=int(F)=R, Hence RAcl(R)

Hence A is -open.

Theorem :2.3

A necessary and sufficient condition for a set A in a topological space (X,) is said to be -open is, int(cl(A))Acl(int(A)).

Proof :

Assume A is a -open set in (X, ).

Since A is -open,there exists a regularly open set R such that RAcl(R).

int(cl(R)) int(cl(A)) R int(cl(A))

Acl(R) int(cl(A)) int(cl(R))=R Hence int(cl(A))R

Hence int(cl(A))A

RA, Rint(A) cl(R)cl(int(A))

Acl(R), Acl(int(A))

Hence int(cl(A))Acl(int(A))

Conversely assume int(cl(A))Acl(int(A)) Let R= int(cl(A))

Then R= int(cl(A))Acl(int(A)) =cl(R). Therefore RAcl(R).

Theorem :2.4

A is -open in a topological space (X, ) iff A is semiopen as well as semi closed in (X, ).

Proof :

Assume A is -open, we get int(cl(A))Acl(int(A)) (by theorem 2..3)

int(cl(A))A and Acl(int(A)).

Acl(int(A)), for R=cl(A), we have RAcl(R) int(cl(A))A, for R=cl(A), we have int(R)Acl(A)=R

Conversely assume A is semi open as well as semi closed.

RA cl(R)cl(int(A)) Hence Acl(R)cl(int(A))

int(R)AR for some closed set R AR cl(A)R

Then int(cl(A))int(R)A Hence int(cl(A))Acl(int(A))

Hence by theorem 2.3, we have A is -open. Corollary :2.5 A set A in a topological space (X, ) is

-open iff A=s-cl-s-int(A) and A=s-int-s-cl(A). Remark : 2.6 By definition of a -open set it follows that every regularly open set is a -open set and every -open set is semi open set,

i.e., R O (X, ) V O(X, ) S O (X, )

However the converse of the above statements are not true in general as shown by the following examples.

Example : 2.7

Let X={a,b,c} and ={,X,{a},{b},{a,b}} Then

{a,c} is a -open set but not regularly open.

Example : 2.8

Let X={a,b,c} & ={,X,{a}} then the set

{a,b} is a semi open which is not -open.

Theorem :2.9

A -open set A is regularly open if Aint(cl(A))

Proof :

Assume A is –open, there exists a regularly open set O such that OAcl(O) and Aint(cl(A))

int(cl(A))OAint(cl(A))A Hence A is regularly open.

Theorem :2.10 A semi open set A is -open if int(cl(A)) A

Proof :

Since A is semi open, there exists an open set O such that OAcl(O)

Hence Acl(O)cl(int(A)) Hence Acl(int(A))

By our assumption, int(cl(A))A Hence int(cl(A))Acl(int(A))

Hence by theorem 2.3, we get A is -open. Theorem :2.11 A semi closed set A is – open if Acl(int(A))

Remark : 2.12

Neither the union nor the intersection of two – open sets is -open.

Example : 2.13

Let X={a,b,c} and ={,X,{a},{b},{a,b}} then

{a,c} and {b,c} are -open sets. But

{a,c}{b,c}={c} is not a -open sets.

Example : 2.14

Let X={a,b,c} and ={,X,{a},{b},{a,b}} then

{a} and {b} are -open sets but {a}{b}={a,b} is not a -open set.

Theorem :2.15

The complement of -open set is again a -open

set.

Proof :

Assume A is -open, there exists a regularly open set R such that RAcl(R)

Therefore we have X-R X-A X-cl(R) Hence int(X-R)X-AX-R

R is regularly open, X-R is regularly closed.

Hence by Theorem 2.2, and from (1), we have A is -open.

Theorem :2.16 If A is a -open set then

1. int(cl(A))=int(A)

2. cl(R)=cl(A), where R is a regularly open set such that RAcl(R).

Theorem :2.17

If A and R are regularly open sets and S is -open such that RScl(R), then AS=, implies AS=.

Proof : Assume A and R are regularly open sets and S is –open such that

RScl(R)

Since AR=, therefore RX-A. Hence X-A is closed.

Therefore RX-A implies cl(R) X-A Scl(R), we get SX-A.

Hence AS=.

Theorem :2.18

Intersection of a -open set S and regularly open set U is a -open set.

Proof :

Since S is -open, there exists a regularly open set R

Case(i): SU =

Let SU =

It is obvious that then SU is a -open set.

Case(ii): SU

we get RU (by Theorem 2.17) Since RS, we get RU SU

If SU, then either x belongs to U and R (or) x belongs to U and S-R.

Case (a): x belongs to U and R. Then URcl(UR) Hence URSUcl(UR) Therefore, SU is a -open set.

Case (b): x belongs to U and S-R.

Then x belongs to U and x is a limit point of R, since Scl(R).

Let N be a -neighbourhood of x.

Then NU is a -neighbourhood of x and (NU)R

which implies N(UR)

Hence x belongs to cl(UR) Hence RUSUcl(RU)

Hence SU is a -open set.

Theorem :2.19 If B is a subset of a topological space (X, ) such that ABcl(A). Then B is – open if A is -open.

Hence the claim.

Definition : 2.20

Let (X, ) be a topological space. Then the set f all regular open set forms a base for a topology

* on X called the semi regularization topology of

X such that * . The space (X, *) is called the

semi regularization space of (X, ).

Definition : 2.21

A topological space (X, ) is said to be semi regular iff = *.

Lemma : 2.22

R O (X,)=int V O (X, ), where int V O (X, ) denotes the collection of the interior of -open sets in a topological space (X, ).

Proof :

Let (X, ) be a topological space. Claim: R O (X,)=int V O (X, ) Let A int V O (X, ) .

Then A=int(B) for B int V O (X, ). A=int(B)=int(cl(B)) (by theorem 2.1.16(a)) Therefore A R O (X, ) HenceintVO(X,)RO(X,) ………..(1)

Then R V O (X,), (by Remark 2.1.6) Therefore R int V O (X, )

Hence R O (X, ) int V O (X, )

………..(2)

From (1) and (2), we have R O (X, )=int V O (X, )

Remark : 2.23

By Lemma 2.22, we get that the collection of -open sets generates a topological space on (X, ). Theorem :2.24

In a semi regular space (X, ), int V O (X, ) generates topology on X.

Proof :

Assume (X, ) is a semi-regular space.

To prove int V O (X, ) generates topology on X. By Lemma 2.22 collection of int V O (X, ) forms a base for a topology * on X such that * .

Since X is semi-regular , = *

Hence int V O (X, ) generates the topology .

Theorem :2.25

Let AYX, where Y is a regularly open subspace of a topological space (X, ). Then A V O (X, ) implies A V O (Y, y).

Proof :

Since A is -open in X, there is a regularly open set R in X such that RAclxR.

Hence RY AY (clxR)Y Hence A V O (Y, y).

Theorem :2.26

If R is a regularly-open subspace of a topological space (X, ) and V a -open set in X, then RV is -open in R.

Proof :

Assume R is a regularly open set in a topological space (X, ) and V is -open set in X. By Theorem 2.18, we get RV is a -open set in X.

Since R is a regularly open subspace of X, we get RV is a -open set in R

( by theorem 2.18).

Theorem :2.27

Let Y be a subspace of a topological space (X, ) and A V O (Y, y). Then V O (X, ) iff Y is -open in X.

Proof :

Since A V O (Y, y), we get A V O (X, ) Hence Y is regularly open.

Hence Y is a -open in Y. i.e., Y V O (Y, y). Y V O (X, ).

Hence Y is -open in X.

Conversely assume Y is -open in X.

Since A V O (Y, y), there exists a regularly open set R in Y such that RAclyR.

Let R1 be a regularly open set in X such that R1Y=R.

Then by Theorem 2.18, R is -open in X. Hence

R1YAcly(R1Y)={clx(R1Y)}Yclx(R1) implies

RAclxR (since R= R1Y)

By applying Theorem 2.19, we have A V O (X, ).

Hence A V O (X, ).

Definition : 2.28

A subset M of a topological space (X, ) is called a -neighbourhood of xX if there exists a -open set V in X such that VM

Theorem : 2.29

Let Y be a subspace of a X and A be a – neighbourhood of x in Y. Then A is a – neighbourhood of x in X iff Y is -open in X.

3.-irresolute functions

Definition : 3.1

A function f:XY is said to be -irresolute iff for any -open set V of Y, f1(V) is -open in X. Theorem : 3.2

An almost-continuous and almost-open mapping f:XY is -irresolute.

Theorem : 3.3

An identity mapping on a topological space (X, ) is -irresolute.

Definition : 3.4

A point x in a topological space (X, ) is said to be -adherent of a subset G of X if every -open set containing x has a non empty intersection with G.

Definition : 3.5

The set of all -adherent points of a set G is called -closure of G or the intersection of all – open sets containing G is known as -closure of G and denoted as -cl(G).

Theorem : 3.6

If A is a subset of a topological space (X, ) , then

A s-cl(A) -cl(A) -cl(A).

Proof :

Since every regularly open set is -open and every -open set is semi open, the Theorem follows.

Definition : 3.7

A subset A in (X, ) is said to be v – closed

if A= -cl(A).

Remark : 3.8

Two different topologies 1 and 2 on a set X may have same class of

-open sets. This leads to define the notation of – correspondence.

Definition : 3.9

Two topologies 1 and 2 on a set X are said to be correspondent if V O (X, 1) = V O (X, 2).

Example : 3.10

Let X={a,b,c} and 1 = {,{a},{b,c},X} and

2 = {,{a},{c},{a,c},{b,c},X}, then V O (X, 1) = V O (X, 2).

Theorem : 3.11

The relation of -correspondence on the collection of all topologies on a set X is an equivalence relation.

4.-axiom: Definition : 4.1

A topological space (X, ) is said to be T0 if for each pair of distinct points x,y of X there exists a -open set G containing x but not y (or) a -open set T containing y but not x.

Definition : 4.2

A topological space (X, ) is said to be T1 if for each pair of distinct points x,y of X there exists a -open set G containing x but not y (or) a -open set H containing y but not x.

Definition : 4.3

A topological space (X, ) is said to be T2 if for each pair of distinct points x , y of X there exists a -open set G and H such that G, H and G H =

Theorem : 4.4

For a topological space (X, ) the following are equivalent.

1. X is T0

2. X is T1

3. X is T2

Proof:

Let (X, ) be topological space.

Let x , y X such that x y.

Since X is -T0 , i.e., x G and y G x X-G and y X-G

Let X-G be H.

Then y H also G H =

Hence for x y X there exists -open set G containing x but not y and a -open set H containing y but not x.

Hence X is T1. Assume X is T1

Let x , y X such that x y.

Since X is T1 , there exists a -open set G containing x but not y and a

-open set H containing y but not x.

Since x G and G ,we get x X-G and y X-G.

Let H=X-G.

Then H is a –open set since G is a open set. Also G H =

Hence there exists -open sets G and H such that G, y H and G H=

Therefore X is T2.

As T2 axiom is stronger than T0 axiom, the result follows.

Remark : 4.5

The three axioms T0, T1, T2 coincide. Hence we shall refer the three axioms as -axiom. Definition : 4.6

A topological space (X, ) is said to be – space if for each pair of distinct points x , y of X there exists a -open set G containing x but not y. Proposition : 4.7

A -space X is semi T2.

Proof:

Since every -open set is semi open, we have X is semi T2.

Example : 4.8

Let X={a,b,c,d}, and ={, {a}, {b},

{c} ,{a,b}, {a,c}, {b,c}, {b,d}, {a,b,c}, {a,b,d},

{b,c,d}, X}. Then X is a semi T2 space but not a – space.

Remark : 4.10

1. Since every -space is semi T2 and semi T2 implies semi T1 , we get every -space is semi T1.

2. Every semi T1 space need not be a -space.

Example : 4.11

Let X={a,b,c,d}, and ={, {a}, {c}, {a,c},

{c,d}, {a,c,d},{b,c,d},X}. Then the space X is a semi T1 space which is not a -space.

Example : 4.12

Let X={a,b,c} and ={,{a},X}. Then X is a semi T0 space but not a -space.

Proposition :4.13

Every rT0 space is a -space.

Proof:

Let X be a rT0 space.

Then for each pair of distinct points x , y of X there exists a regularly open set G containing x but not y.

Since every regularly open set is a -open set , G is a -open set.

Then for each pair of distinct points x , y of X there exists a -open set G containing x but not y.

Hence X is a -space.

Hence every rT0 space is a -space.

Remark : 4.14

The converse of proposition 4.13 is not true.

Example : 4.15

Let X={a,b,c,}, and ={,{a},{b},{a,b}, X}. Then X is a -space but not rT0.

Remark : 4.16

The concepts of a space being a -space and T0-space are mutually independent.

Example : 4.17

Let X={a,b,c}, and ={,{a},{a,b}, X}.

Then X is a T0-space but not a -space.

Example : 4.18

Let X={a,b,c,d}, ={, {a}, {b}, {a,b}, X}.

Then X is a -space but it is not a T0-space.

Remark : 4.19

The axiom of -space is independent of the T1- axiom.

Example : 4.20

Let X be countable set, ={,X,A} where A X such that X-A is finite.

Example : 4.21

Let X={a,b,c,d} and ={ ,{a}, {b},

{a,b},{d}, {a,d}, {b,d}, {a,b,c}, {a,b,d},X}. Then X is a -space but not T1.

Theorem : 4.22

A space X is a -space iff each singleton set in X is v-closed.

Proof:

Let X be a -space.

Let x,y X such that x y.

Since X is a -space, there exists a -open set containing y but not x.

Then y -cl{x}.

Hence -cl{x} {x} Therefore {x} = -cl{x}. Hence {x} is v -closed. Conversely let x y X.

Then by the hypothesis, {x} and {y} are v- closed.

i.e., {x} = -cl{x} and {y} = -cl{y}. Therefore y -cl{x} and x -cl{y}.

Hence there exists a -open set containing x but not y.

Hence X is a -space.

Theorem : 4.23

The necessary and sufficient condition for a space X to be a -space is that for each x X there exist a -open set U of X containing x such that the subspace U is a -space.

Theorem : 4.24

Every regularly open subset Y of a -space X is a -space.

Proof:

Let Y be regularly open subset of a -space X.

Claim: Y is a -space.

Let x,y X such that x y. Then x y X.

Since X is a -space, there exist a -open set G containing x but not y.

Hence G Y is a -open set in Y (by Theorem 2.26), containing x but not Y.

Hence Y is a -space.

Theorem : 4.25

If f is a -irresolute function from a space X to a -space Y.Then X is a -space.

Theorem :4.26

The product of two -spaces is a -space.

CONCLUSION

This paper is an attempt to generalize -open sets due to Sharma [25] to fuzzyTopological spaces.

-open sets and -axiom due to Sharma[25,26] are analyzed. -open sets, its properties

and characterizations are analyzed .

-adherent, -closure of a subset A of a topological space, -irresolute function between topological

spaces and the relation -correspondence on the set of the topologies on a set X are analyzed.

separation axioms T0, T1 and T2 and their equivalence in topological spaces are discussed. Properties and characterizations of -spaces are analyzed.

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International Journal of Engineering Research & Technology (IJERT)

ISSN: 2278-0181

Vol. 1 Issue 5, July – 2012