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 Authors : S. Manimekalai, R. Anandhi
 Paper ID : IJERTV1IS5122
 Volume & Issue : Volume 01, Issue 05 (July 2012)
 Published (First Online): 02082012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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

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

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
Âµaxiom.The interrelationship between Âµaxiom 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=sclsint(A) and A=sintscl(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 XR XA Xcl(R) Hence int(XR)XAXR
R is regularly open, XR 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

int(cl(A))=int(A)

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 RXA. Hence XA is closed.
Therefore RXA implies cl(R) XA Scl(R), we get SXA.
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 SR.
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 SR.
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 semiregular 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 semiregular , = *
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 regularlyopen 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 almostcontinuous and almostopen 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 scl(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.

X is T0

X is T1

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 XG and y XG
Let XG 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 XG and y XG.
Let H=XG.
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

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

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 T0space are mutually independent.
Example : 4.17
Let X={a,b,c}, and ={,{a},{a,b}, X}.
Then X is a T0space 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 T0space.
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 XA 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 vclosed.
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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