S-open set in Bitopological space
Samah Abd Al-hadi Abbass
Babylon University College of science For women
Computer Department
Abstract
The primary purpose of this paper is to introduce and study a new types of open sets called S-open sets , continuous , separation axioms are study with respect to the new open set .
ةص?خلا
-S ىمسـت (X,P1,p2) يجوـلوبتلا يئاـنثلا ءاضفلا يف ةحوتفملا تاعومجملا نم ديدج عون ةسارد وه ثحبلا اذه نم فدهلا
ةحوتفم
Introduction
A triple (X,p1,P2) where X is anon empty set and P1,P2 are topologies on X , is called abitopological space and kelly [Kelly ,1963] initiated the study of such spaces. after that more mathematician working on the new space bitopological spaces deferent sets are defined and study in it .
if A be a subset of X, the interior (resp. closure) of A with respect to the topology pi (i = 1, 2) will be denoted by int pi (A) (resp. cl pi (A)).
The purpose of this paper is to define S-open sets and define separation axioms and continuous function associated S-open sets in bitopological spaces and investigate some of their properties and relations and its effects on some theorems and properties .
1-preminaries
in this section we define s-open set and we give some basic remark on it . Definition(1-1):
Let (X,p1),(X,p2) are two topologicl on X asubset A of X is said to be S-openset in
(x,p1,p2) iff A= int p1(U?V) such thatU,V are p2-openset .the complement of S-open set is called S-closed set
Example(1-1)
Let X={a,b,c} ,p1={x , ?,{a},{c},{a,c}},p2={X, ?,{b },{a,c} S.O(X)={X, ?,{a,c}}
Theorem(1-1):
The family of all S-open set is a topological space Proof:
a) ? and X is clearly that S-open set
b) Let G1,G2 two S-open sets
Then G1=int p1(U?V) , G2=intp1(U1?V1) , Where U,V,U1,V1 are p2-open sets
then G1?G2 = intp1(U?V) ?int p2(U1?V1) = int p1(W?H) where W=U?U1
H=V?V1
Then G1?G2 is S-open set
434 c)Let ? be any index and G? is S-open set for every ??? ,then G?= intp1(U??V?) fior some U? and V? be p2-open sets then
U G?=U int p1(U??V?) =intp1(U(U??V?) )=intp1[(U???(U?)) ? (U???(V? ))] Since U???(U?) and (U???(V?) be p2-open sets
Then U??? G? is s-open set Remark(1-1):
1/If Ais S-open set then it is P1-open set .and the convers is not true
2/ If A is S-open set and B is P1-open set thenA?B and AUB are P1-open set 2- S-separation axioms and continuous function
In this section we will define separation axioms with respect to s-open sets. Definition(2-1):
Abitopological space is said to be S-T0-space iff for each two distinct points x,y in X there exist s-open sets U such that x?U , y?U or there exist S-open set y?W , x?W .
Definition(2-2):
Abitopological space is said to be S-T1-space iff for each two distinct points x,y in X there exist two s-open sets U and W such that x?U , y?U, y?W , x?W .
Theorem(2-1):
let (X, P1, P2) be a bitopological space then X is S- T1-space iff {x} is S-closed set for each x?X.
Proof: see [sharma ,1963] Definition(2-3):
Abitopological space is said to be S-T2-space iff for each two distinct points x,y in X there exist two s-open sets U and W such that x?U , y?W ,and U? W=?
Definition(2-4):
Abitopological space is said to be S-regular-space iff for each points x in X ands S- closed set H such that x? H there exist two s-open sets U and W such that x?U , H?W ,U?V=?
Definition(2-5):
Abitopological space is said to be S-T3-space iff the bitoplogical space is S-T1 and S- regular
Definition(2-6):
Abitopological space is said to be S-normal-space iff for each two S-closed sets H and F , such that H?F=? there exist two s-open sets U and W such that F?U , H?W , and U?W=?
435Journal of Babylon University/Pure and Applied Sciences/ No.(2)/ Vol.(20): 2012 Definition(2-7):
Abitopological space is said to be S-T4-space iff the bitoplogical space is S-T1 and S- normal .
Theorem(2-2): let (X, P1, P2) be abitopological space
If (X, P1, P2) is S-Ti-spase, i=0,1,2 then(X, P1) is Ti-space where ,i=0,1,2 Proof:
Exist by definition
- the following diagram explain that the relation between the above separation axioms
P1-T1
P1- T0 ?
P1-T2 P1-T3 P1-T4
?
?
?
?
S-T0? S-T1? ST2? ?? S-T3 ?S-T4 Examp(2-1 ):
Let X={ a,b,c} and P1=D , D is the discrete topology on X which is Ti-space for i=0,1,2. P2={X, ?,{a},{b},{a,b}}, S.O(X)=P2 then (X,P1,P2) is not S-Ti-space for i=0,1,2.
Example(2-2) : let X={a,b,c,d} and
P1={X,?,{a},{c},{b},{a,c},{a,b},{b,c,d},{c,d},{b,c},{a,b,c},{a,c,d}} P2={X,?, {{a},{c},{b},{c,d},{a,b},{a,c},{b,c},{b,c,d},{a,c,d}}
S.O(X)={X,?,{a},{b},{c},{a,c},{a,b},{b,c},{c,d},{b,c,d},{a,b,c},{a,c,d}} Then clearly that (X,P1,P2) is S-T2 but not S-T3
Example(2-3) : let X={a,b,c,d} and P1=D
P2={X,?, {a},{c},{d},{a,d},{c,d},{a,b},{a,c},{b,c},{a,b,d}{a,b,c},{a,c,d}} S.O(X)={ X,?,{c},{a},{d},{a,b},{a,c},{a,d},{c,d},{a,d,b},{a,b,c}{a,c,d}}
Then clearly that (X,P1,P2) is S-T3 but not S-T4
Example (2-4) : let X={a,b,c,} and P1={X,?,{a},{c},{a,c}}, P2={X,?, {{a},{c},{a,c},{b,c}}
S.O(X)={X,?,{a},{c},{a,c}}
Then clearly that (X,P1,P2) is S-To but not S-T1 3- S-continuous functions
Now we define S-continuous function with some theorems on it we study . Definition(3-1):
A mapping f:( X, P1, P2) ? (Y, P*1 , P *2) is said to be S-continuous iff the inverse image of each S-open set in Y is P1-open set in X
Theorem(3-1):
If f: (x, P1)?(Y, P*1) is continuous then f:( X, P1, P2) ? (Y, P*1 , P *2) is S-continuous
436Proof
Let A is S-open set in Y then A is P*1-open set and since f is continuous then f-1(A) is P1-open set and there for f is S-continuous.
Example (3-1)
Let X={a,b,c} , P1={X,?,{a},{b,c}} ,P2={X,?,{a},{a,c}} Y={1,2,3} ,P1*={Y,?,{1},{2},{1,2}}
P2 *={Y,?,{2}}
Let g: ( X, P1, P2) ? (Y, P*1 , P *2) defined by g(a)=2 , g(b)=1 , g(c)=3 Then clearly that g is S-continuous but not P1-P1*- continuous Remark(3-1):
The composition of two S-continuous functions is not S-continuous functions since not every P1-open set is S-open set
Definition(3-2):
Amapping f:( X, P1, P2) ? (Y, P*1 , P *2) is said to be
1/ S-open map iff f(u) is S-open set in Y for each S-open set U in X
2/ S-homeomorphism map iff f is 1-1,onto, S-continuous and S-open map Definition (3-3):
Amapping f:( X, P1, P2) ? (Y, P*1 , P *2) is said to be s-inn map iff the inverse image of each S-open set in Y is S-open set in X.
Theorem (3-2): if f:( X, P1, P2) ? (Y, P*1 , P *2) is S-inn map then f is S-continuous map:
Proof: let U is S-open set in Y then f-1(U) is S-open set in X and since every S-open set is p1-open set then f-1(U) is p1-open set then f is S-continuous .
Example (3-2)
Let X={a,b,c} , P1={X,?,{a},{c}, {a,c}} ,P2={X,?,{a},{a,c}} Y={1,2,3} ,P1*=D , D is the discrete topology on X .
P2 *={Y,?,{2},{1},{1,2}}
Let g: ( X, P1, P2) ? (Y, P*1 , P *2) defined by g(a)=1 , g(b)=3 , g(c)=2 Then clearly that g is S-continuous but not S-inn map
Theorem(3-3): let (X, P1, P2) be a bitopological space if X is
S-Ti where ,i=0,1,2 then X is a topological properties with respect to S-Ti Proof:
Exist by definition Remark(3-2)
let (X, P1, P2) be a bitopological space
1- S-Ti ,i=3,4 is not topological properties
2- if f:( X, P1, P2) ? (Y, P*1 , P *2) is S-inn map then Ti-space ,i=3,4 is a topological property.
Definition(3-4) : let ( X, P1, P2) be abitopological space and Y be asubset of X then TY is the collection given by
P iY= {D?Y : D is S–open set} then (Y,P1Y ,P2Y) is called subspace of ( X, P1, P2) .
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