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2014, Vol.5(6): Pg.490-499
On α Regular ω-Open Sets in Topological Spaces
R. S. Wali1 and Prabhavati S. Mandalgeri2
1
Department of Mathematics,
Bhandari Rathi College, Guledagud, Karnataka State, INDIA.
2
Department of Mathematics,
K.L.E’S , S.K. Arts College & H.S.K. Science Institute,
Hubli, Karnataka State, INDIA.
(Received on: November 15, 2014)
ABSTRACT
The aim of this paper is to introduce & study the αrωopen sets in topological space and obtain some of their properties.
Also we introduce αrω-interior , αrω-neighbourhood, αrω –limits
points in topological spaces., Applying αrω-closed sets we
introduce αrω-closure and discuss some basic properties of thiese.
Keywords: Topological spaces, αrω-closed sets, αrω-open sets.
1. INTRODUCTION
Regular open sets and rw-open sets
have been introduced and investigated by
Stone11 and Benchalli and Wali2 respectively.
Levine6, Nagaveni9, Pushpalatha10 introduced
and investigated semi open sets, generalized
closed sets, weakly generalized closed sets,
respectively. Maki et al.7 introduced and
studied generalized α-closed sets and αgeneralized closed sets. S. S. Benchalli et
al.3 studied ωα-closed sets in topological
spaces. R.S.Wali and P S. Mandalgeri13
introduced and studied αrω-closed sets.
We introduce and study the αrωopen sets in topological space and obtain
some of their properties. Also we introduce
αrω-interior, αrω-neighbourhood, αrω–limits
points in topological space.
2. PRELIMINARIES
Throughout this paper space (X, )
and (,Y, σ) (or simply X and Y) always
denote topological space on which no
separation axioms are assumed unless
explicitly stated. For a subset A of a space
X, Cl(A), Int(A) and Ac denote the Closure
of A, Interior of A and Compliment of A in
X respectively.
Definition 2.1 -A subset A of a topological
space (X, ) is called.
(1) a regular open set [11] if A = int(clA))
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and a regular closed set if A = cl(int(A)).
(2) Regular semi open set [4] if there is a
regular open set U such that U ⊆ A ⊆ cl(U).
Definition 2.2 Let (X, ߬) be topological
space and A ⊆X
The intersection of all semi closed
(resp pre-closed, α-closed and semi-preclosed ) subsets of space X containing A is
called the Semi closure (resp pre-closure ,
α-closure and Semi-pre-closure) of A and
denoted by sCl(A) (resp pCl(A),αCl(A),
spCl(A)
Definition 2.4 A subset A of a topological
space (X, ߬ ) is called
1) a generalized semi-closed set(briefly gsclosed)[1] if scl(A) ⊆ U whenever A ⊆
U and U is open in X.
2) an α-generalized closed set(briefly αg closed)[7] if αcl(A)⊆U whenever A⊆U
and U is open in X.
3) a generalized semi pre regular closed
(briefly gspr-closed) set [10] if spcl(A )
⊆ U whenever
A⊆ U and U is
regular open in X.
4) generalized pre regular closed set(briefly
gpr-closed)[2] if pcl(A) ⊆ U whenever
A ⊆ U and U is regular open in X.
5) a weakely generalized closed set
(briefly, wg-closed)[9] if cl(int(A))⊆ U
whenever A ⊆ U and U is open in X.
6) a regular weakly generalized closed set
(briefly, rwg-closed)[9] if cl(int(A))⊆ U
whenever A ⊆ U and U is regular open
in X.
7) an α-generalized regular closed (briefly
αgr-closed) set [12] if αcl(A)⊆ U
whenever A ⊆ U and U is regularopen in X.
8) ωα-closed set [3] if αcl(A) ⊆ U
whenever A⊆ U and U is ω-open in X.
9) regular ω- closed (briefly rω -closed) set
[2] if cl(A)⊆ U whenever A⊆ U and U
is regular semi- open in X.
10) generalized pre closed (briefly gpclosed) set [8] if pcl(A)⊆U whenever
A⊆U and U is open in X.
The compliment of the above
mentioned closed sets are their open sets
respectively.
3. α Regular ω-CLOSED SETS IN
TOPOLOGICAL SPACE
Definition : 3.1 [13] A subset A of
topological space (X, ߬) is called a α regular
ω-closed set (briefly αrω-closed set) if
αCl(A)⊆U whenever A ⊆ U and U is rωopen in (X, ߬).
Results3.2 : From [13]
1) Every closed set is αrω-closed set in X
2) Every α-closed set in X is αrω-closed
set
3) Every regular closed set is αrω-closed
set in X.
4) Every αrω-closed set is αg-closed(also
αgr-closed, ωα-closed, gs-closed, gsprclosed, rwg-, closed, gp-closed, wgclosed, gpr-closed ) set in X
5) The Union of two αrω -closed subsets
of X is αrω – closed set.
4.
α REGULAR ω-OPEN SETS
In this section, we define αrω-open
sets in topological spaces and obtain some of
their properties.
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492
Definition 4.1 A subset A of a topological
space (X , ߬) is called α regular ω-open
(briefly αrω-open) set in X if Ac is αrωclosed in X
X A ⊆ X U and X U is rω-open set of
(X, ߬). This implies that αcl(X A) ⊆ X U.
But αcl(X A)=X αint(A). Thus X
αint(A) ⊆ X U, so U ⊆ αint (A)
The following theorem is the analogue of
results 3.2 ( 1) to (4)
Conversely
Theorem 4.2 For any topological spaces
(X, ߬) we have the following
(i) Every open (resp α-open ,regular-open)
set is αrω-open.
(ii) Every αrω-open set is αg –open set.
(iii) Every αrω-open set is αgr-open ( resp
ωα-open gs-open,gspr-open , ωg-open, rωgopen,gp-open and gpr-open) set.
Theorem 4.3 If A and B are αrω-open sets
in space X, then A∩B is also an αrω-open
in X.
Suppose U ⊆ αint (A) whenever U is rωclosed and U A. To prove that A is αrωopen set. Let F be rω-open set of (X, ߬) s.t
X A ⊆ F. Then X F ⊆ A. Now X F is rωclosed set containing A, So X F ⊆ αint(A) ,
X − αint(A) ⊆ F but αcl(X A)= X −
αintሺAሻ ⊆ F. Thus αcl(X A)⊆ F i.e X A is
αrω-closed set & hence A is αrω-open set.
Theorem4.7 If αint(A) ⊆ B ⊆ A and A is
αrω-open set, then B is αrω-open set.
Proof: Let A and B be two αrω-open sets
in X. Then Ac and Bc are αrω-closed sets
in X by Results 3.2 (5), Ac ∪ Bc is also
αrω-closed set in X. that is Ac ∪ Bc
=(A∩B)c is αrω-closed set in X.
Therefore A∩B is an αrω-open set in X.
Proof: Let αint(A) ⊆ B ⊆A , Thus X A ⊆
X B⊆ X αint(A) , i.e X A
X B ⊆
cl(X A), Since X A is αrω-closed set, by
Theorem3.20 X B is αrω-closed set.
Therefore B in αrω-open set.
Remark 4.4 The union of αrω-open set in
X is generally not an αrω-open set in X.
Theorem 4.8 If A X is αrω-closed then
αcl(A) −A is αrω-open set
Example 4.5 Let X={a,b,c,d}, ߬ ={X,ϕ,{a},
{c,d},{a,c,d}} If A={c} B={a} Then A & B
are αrω-open set in X but A ⋃ B={a,c} is
not an αrω-open set in X.
Proof: Let A be αrω-closed. Let F ⊆
αcl(A) −A, where F is rω-closed by
Theorem 3.17 F=ϕ Therefore F ⊆ αint
(αcl(A) −A) and Theorem4.6 αcl(A) −A is
αrω-open set.
Theorem 4.6 A subset of A of a topological
space X is αrω-open iff U ⊆ αint(A),
whenever U is rω-closed and U ⊆ A
Proof: Assume that A is αrω-open set in X
and U is rω-closed set of (X, ߬) s.t U ⊆ A.
Then X-A is a αrω-closed set in (X, ߬). Also
The reverse implication does not hold good
Example 4.9:Let X={a,b,c,d}, ߬ ={X,ϕ,{a},
{c,d},{a,c,d}} A={b,c} αcl(A)={b,c,d}
αcl(A) –A={d} which is αrω-open set in X,
but A is not αrω-closed.
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Theorem 4.10 A set A is αrω-open set in
(X, ߬) if and only if U=X whenever U is
rω-open and
αint(A) (X A) ⊆ U.
Proof: Suppose that A is αrω-open set in X.
Let U be rω-open and αint(A) ∪ (X-A) ⊆ U
Uc ⊆ (αint(A) ∪Ac)c = (αint(A))c ∩ A i.e Uc
⊆ (αint(A))c Ac (because A B = A∩Bc)
Thus Uc ⊆ αcl(Ac)–Ac (because (αint(A))c
= αcl(Ac))
Now Ac is also αrω-closed and Uc is rωclosed then by Theorem--- it follows Uc=ϕ
then U=X
Conversely: Suppose F is αrω-closed and
F⊆A. Then αint(A) ∪ (X-A) ⊆ αint(A) ∪
(X-F). It follows that αint(A) ∪ (X) –F)=X
Theorem 4.11 If A and B be subsets of
space (X, ߬). If
B is αrω-open and
αint(B)⊆A then A∩B is αrω-open.
Proof: Let B is αrω-open in X. αint(B) ⊆ A
and αint(B) ⊆ B is always then αint(B) ⊆
A∩B also αint(B) ⊆ A∩B ⊆ B and B is
αrω-open set by Theorem 4.7 , A∩B is also
is αrω-open set in X.
5. αrω-CLOSURE AND αrω-INTERIOR
In this section the notation of αrωclosure and αrω-Interior is defined and some
of its basic properties are studied.
Definition 5.1 For a subset A of (X, ߬), αrωclosure of A is denoted by αrωcl(A) and
defined as αrωcl(A)= ∩{G : A⊆G, G is αrω–
closed in (X, ߬) } or ∩{G : A⊆G, G ߳
αRωC(X)}
Theorem 5.2 If A and B are subsets of space
(X, ߬) then
i) αrωcl (X) = X , αrωcl(ϕ)=ϕ
ii) A ⊆ αrωcl(A)
iii) If B is any αrω-closed set containing A,
then αrωcl(A) ⊆ B
iv) If A ⊆ B then αrωcl(A) ⊆ αrωcl(B)
v) αrωcl(A)= αrωcl (αrωcl (A))
vi) αrωcl(A B) = αrωcl (A) ∪ αrωcl (B)
Proof:
i)
By definition of αrω=closure, X is only
αrω-closed set containing X. therefore
αrωcl(X) = Intersection of all the αrωclosed set containing X =∩{X} =X
therefore αrωcl (X) = X and again by
definition of αrω-closure αrωcl(ϕ) =
Intersection of all αrω-closed sets
containing ϕ
= ϕ ∩ any αrω-closed set containing ϕ
= ϕ
Therefore αrωcl(ϕ)=ϕ
ii) By definition of αrω-closure of A, it is
obvious that A ⊆ αrωcl(A)
iii) Let B be any αrω-closed set containing
A. Since αrωcl(A) is the intersection of
all αrω-closed set containing A ,
αrωcl(A) is contained in every αrωclosed set containing A. Hence in
particular αrωcl(A)⊆B
iv) Let A and B be subsets of (X, ߬) such
that A⊆B by definition αrω-closure ,
αrωcl(B)=∩{F : B ⊆F ߳ αRωC(X)} If
B ⊆ F ߳ αRωC(X) , then αrωcl(B) F.
since A ⊆ B , A ⊆ B ⊆F ߳ αRωC(X), we
have αrωcl(A) ⊆F , αrωcl(A) ⊆∩{F : B
⊆F ߳ αRωC(X)} = αrωcl(B)
Therefore αrωcl(A) ⊆ αrωcl(B)
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v) Let A be any subset of X by definition
of αrω-closure , αrωcl(A)= ∩{F : A ⊆F
߳ αRωC(X)}
vi) If A ⊆ F ϵ αRωC(X) then αrωcl(A) F ,
since F is αrω-closed set containing
αrωcl(A) by (iii) αrωcl(αrωcl(A))
F
Hence αrωcl(αrωcl(A)) = ∩{F : A
⊆F ߳ αRωC(X)} = αrωcl(A). Therefore
αrωcl(αrωcl(A)) = αrωcl(A)
Let A and B be subsets of X , Clearly
A⊆A B, B⊆A B from (iv) αrωcl(A) ⊆
αrωcl(A B)αrωcl(B) ⊆ αrωcl(A B)
hence αrωcl(A) ∪αrωcl(B) ⊆ αrωcl(A B)
(1)
Now we have to prove that αrωcl(A B) ⊆
αrωcl(A) ∪αrωcl(B)
Suppose x ∉ αrωcl(A) ∪αrωcl(B) then
αrω-closed set A1 and B1 with AcA1 , B⊆B1
and x ∉ A1 B1
We have A B ⊆A1 B1 and A1 B1 is the
αrω-closed set by theorem --- such that x ∉
A1 B1
Thus x αrωcl(A B) hence αrωcl(A B) ⊆
αrωcl(A) ∪αrωcl(B)
(2)
From (1) and (2) we have αrωcl(A B) =
αrωcl(A) ∪αrωcl(B)
Theorem 5.3 If A ⊆ X is αrω-closed set
then αrωcl(A) =A
Proof : Let A be αrω-closed subset of X. we
know that A ⊆ αrωcl(A) –(1)
Also A⊆A and A is αrω-closed set by
theorem5.2 (iii) αrωcl(A) ⊆A –(2)
Hence αrωcl(A) =A
The Converse of the above need not
be true as seen from the following example.
494
Example5.4 Let X={a,b,c,d}, ߬ ={X,ϕ,{a},
{c,d},{a,c,d}} A={b,c} αrωcl(A)={b,c}=A
then A is not αrω-closed set.
Theorem 5.5 If A and B are subsets of
⊆
space
X
then
αrωcl(A B)
αrωcl(A) ∩αrωcl(B)
Proof: Let A and B be subsets of X, Clearly
A B⊆A , A B⊆B by theorem5.2(iv )
αrωcl(A B) ⊆ αrωcl(A), αrωcl(A B) ⊆
αrωcl(B)
hence
αrωcl(A B)
⊆
αrωcl(A) ∩αrωcl(B)
Remark5.6 In general αrωcl(A) ∩αrωcl(B)
αrωcl(A B) as seen from the following
example.
Example5.7 Consider X={a,b,c,d}, ߬ ={X,ϕ,
{a},{b},{a,b},{a,b,c}} , A={b,c} , B={c,d} ,
A∩B={c} αrωcl(A)={b,c,d} αrωcl(B)={c,d}
αrωcl(A∩B)={c} and
αrωcl(A) ∩αrωcl(B)={c,d} therefore
αrωcl(A) ∩αrωcl(B) αrωcl(A B).
Theorem 5.8 For an x ߳ X, x ߳ αrωcl(A) if
and if A V ≠ ϕ for every αrω-open set V
containing x.
Proof: Let x ߳ αrωcl(A) To prove A V ≠ ϕ
for every αrω-open set V containing x by
contradiction.
Suppose αrω-open set V containing x s.t A
V = ϕ then A ⊆ X-V , X-V is αrω-closed
set , αrωcl(A) ⊆ X-V This shows that x ∉
αrωcl(A) which is contradiction . Hence A
V ≠ ϕ for every αrω-open set V containing x.
Conversely:
Let A V ≠ ϕ for every αrω-open set
V containing x. To prove x αrωcl(A). we
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prove the result by contradiction. Suppose x
αrωcl(A) then there exist a αrω-closed
subset F containing A s.t x F. Then x
X-F is αrω-open. Also (X-F) A=ϕ which is
contradiction. Hence x αrωcl(A).
αRωC(X)} If A⊆F αRωC(X) then A⊆F
GPRC(X), because every αrω-closed is gprclosed i.e gprcl(A) ⊆F therefore gprcl(A)
⊆∩{F : A ⊆F
αRωC(X)}= αrωcl(A).
Hence gprcl(A) ⊆ αrωcl(A)
Theorem 5.9 If Ais subset of space X then
(i)
αrωcl(A) ⊆ cl(A)
(ii)
αrωcl(A) ⊆ αcl(A)
Theorem 5.12 αrω-closure is a Kuratowski
closure operator on a space X.
Proof: Let A and B be the subsets space X.
i) αrωcl (X) = X , αrωcl(ϕ)=ϕ ii)
A ⊆
αrωcl(A) iii) αrωcl(A)= αrωcl (αrωcl (A))
iv) αrωcl(A B) = αrωcl (A) αrωcl (B) by
theorem 5.2
Hence αrω-closure is a Kuratowski closure
operator on a space X.
Proof: (i) Let A be subset of space X by
definition of Closure Cl(A)= ⊆∩{F : A ⊆F
C(X)}
C(X) then A⊆F
αRωC(X)
If A⊆ F
because every closed set is αrω-closed that is
αrωcl(A)⊆F therefore αrωcl(A) ⊆∩{F :
A⊆F C(X)}=cl(A) Hence αrωcl(A) ⊆ cl(A)
(ii) Let A be subset of space X by definition
of α-closure αcl(A)= ⊆∩{F : A ⊆F αC(X)}
If A⊆ F αC(X) then A⊆F αRωC(X)
because every α-closed set is αrω-closed that
is αrωcl(A)⊆F therefore αrωcl(A) ⊆∩{F :
A⊆F αC(X)} = αcl(A) Hence αrωcl(A) ⊆
αcl(A)
Remark 5.10 Containment relation in the
above theorem5.9 may be proper as seen
from following example.
Example 5.11 Let X={a,b,c,d}, ={X,ϕ,
{a},{b},{a,b},{a,b,c}}, A={a} cl(A)={a,c,d}
αrωcl(A)={a,d}, αcl(A)={a,c,d} It follows
that αrωcl(A) cl(A) and αrωcl(A) αcl(A)
Theorem 5.12 If A is subset of space X then
gprcl(A)⊆αrωcl(A) where gprcl(A)= ⊆∩{F :
A ⊆F GPRC(X)}
Proof: Let A be a subset of X by definition
of αrω-closure, αrωcl(A)=∩{F : A ⊆F
Definition 5.13: For a subset A of (X, ) ,
αrω-interior of A is denoted by αrωint(A)
and defined as αrωint(A)= {G: G⊆A and G
is αrω-open in X} or {G: G⊆A and G
αRωO(X)}
i.e αrωint(A) is the union of all αrω-open
set contained in A.
Theorem 5.14 Let A and B be subset of
space X then
i) αrωint(X) = X , αrωint(ϕ)=ϕ
ii) αrωint(A) ⊆ A
iii) If B is any αrω-open set contained in
A, then B ⊆ αrωint(A)
iv) If A ⊆ B then αrωint(A) ⊆ αrωint(B)
v) αrωint(A)= αrωint(αrωint(A))
vi) αrωint(A∩B) = αrωint(A) αrωint(B)
Proof: i) and ii) by definition of αrω-interior
of A, it is obvious
iii) Let B be any αrω-open set s.t B ⊆ A.
Let x B , B is an αrω-open set contained in
A, x is an αrω-interior of A i.e. x
αrωint(A). Hence B ⊆ αrωint(A).
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iv) ,v), vi) similar proof as theorem5.2 and
definition of αrω-interior
Theorem 5.15 If a subset A of X is αrωopen then αrωint(A) = A
Proof: Let A be αrω-open subset of X. We
know that αrωint(A) ⊆ A –(1) Also A is
αrω-open set contained in A from Theorem
5.14 iii) A ⊆ αrωint(A) --(2) Hence From
(1) and (2) αrωint(A) = A
Theorem 5.16 If A and B are subsets of
space X then αrωint(A) αrωint(B) ⊆
αrωint(A B)
Proof: We know that A ⊆ A B and B⊆A
B, We have Theorem 5.14 iv) αrωint(A)⊆
αrωint(A B) and αrωint(B)⊆ αrωint(A B).
This implies that αrωint(A) αrωint(B) ⊆
A B Remark 5.17: The converse of the
above theorem need not be true as seen from
the following example.
Example 5.18: Let X={a,b,c,d}
={X,ϕ,
{a},{b},{a,b},{a,b,c}}, A={b,c}, B={a,d},
A B ={a,b,c,d} αrωint(A)={b,c} αrωint(B)
={a},
αrωint(A B)=X,
{αrωint(A)
αrωint(B)={a,b,c}, therefore
αrωint(A B) αrωint(A) αrωint(B)
Theorem 5.19 If A is a subset of X then i)
int(A) ⊆ αrωint(A) ii) αint(A) ⊆ αrωint(A)
Proof: i) Let A be a subset of a space X. Let
x int(A) => x
{G : G is open , G ⊆ A}
=> an open set G s.t. x G ⊆ A => an
αrω-open set G s.t. x G ⊆ A, as every open
set is an αrω-open set in X=> x {G : G is
496
αrω-open set in X}=> x αrωint(A). Thus x
int(A) => x αrωint(A)
Hence int(A) ⊆ αrωint(A).
ii) Let A be a subset of a space X. Let x
int(A) => x
{G : G is α-open , G ⊆ A}
=> an α-open set G s.t. x G ⊆ A => an
αrω-open set G s.t. x G ⊆ A, as every αopen set is an αrω-open set in X=> x {G :
G is αrω-open set in X}=> x αrωint(A).
Thus x α-int(A)
=> x
αrωint(A). Hence αint(A) ⊆
αrωint(A).
Remark 5.20: Containment relation in the
above theorem may be proper as seen from
the following example
Example 5.21 Let X={a,b,c,d} ={X,ϕ,{a},
{b},{a,b},{a,b,c}} A={b,c} , Int(A)={b},
αint(A)={b}, αrωint(A)={b,c}
therefore
int(A) ⊂ αrωint(A) and αint(A) ⊂ αrωint(A)
Theorem 5.22: If A is subset of X, then
αrωint(A) ⊆ gpr-int(A), where gpr-int(A) is
given by
gpr-int(A) = {G ⊆ X : G is gpr-open , G ⊆
A}
Proof: Let A be a subset of a space X. Let x
int(A) => x
{G : G is αrω-open , G
⊆ A}
=> an αrω-open set G s.t. x G ⊆ A, as
every αrω-open set is an gpr-open set in
X=> x {G : G is gpr-open , G ⊆ A }=> x
gprint(A). Thus x αrω int(A) => x gprint(A) Hence αrωint(A) ⊆ gpr-int(A).
Theorem 5.23: For any subset A of X (i) X
– αrωint(A) = αrωcl(X – A)
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(ii) X – αrωcl(A) = αrωint(X – A)
Proof: x ∈ X-αrω int(A) , then x is not in
αrωint(A) , i.e. every αrω-open set G
containing x s.t. G ⊆ A . This implies every
αrω-open set G containing x intersects (X –
A) i.e. G ∩ (X − A) ≠ φ . Then by theorem
5.8 x ∈ αrωcl(X − A)
∴ X − αrω int(A) ⊆ αrωcl(X − A)
(1)
and Let x ∈ αrωcl(X − A) , then every
x ∈ αrω-open set G containing x intersects
X – A i.e. G ∩ (X − A) ≠ φ , i.e. every
αrω-open set G containing x s.t. G ⊆ A .
Then by definition 5.13, x is not in
αrω int(A) , i.e. x ∈ X − αrω int(A) and so
αrωcl(X − A) ⊆ X − αrω int(A)
(2)
Thus X − αrω int(A) = αrωcl(X − A)
6. αrω-NEIGHBOURHOOD AND αrωLIMITS POINTS
In this section we define the notation
of αrω-neighbourhood , αrω-limits points
and αrω derived sets and show of their basic
properties and analogous to those for open
sets.
Defintion 6.1 Let (X, τ) be a topological
space and Let x ∈ X , A subset of N of X is
said to be
αrω-neighbourhood of x if there exists an
αrω-open set G s.t. x ∈ G ⊆ N .
Defintion 6.2 i) Let (X, τ) be a topological
space and A be a subset of X, A subset of N
of X is said to be αrω-neighbourhood of A
if there exists an αrω-open set G s.t.
A⊆G⊆N.
ii) The collection of all αrω-neighbourhood
of x ∈ X is called αrω-neighbourhood
system at x and shall be denoted by
αrω-N(x)
Definition 6.3 Let (X, τ) be a topological
space and A be a subset of X, then a point
x ∈ X is called a αrω- limit point of A iff
every αrω-neighbourhood of x contains a
point of A distinct from x i.e.
( N − {x}) ∩ A ≠ φ for each
αrω-neighbourhood N of x. Also
equivalently iff,
every αrω-open set G
containing x contains a point of A other than
x.
The set of all αrω limit points of the set A is
called derived set A and is denoted by
αrωd(A)
Theorem 6.4 : Every neighbourhood N of
x ∈ X is a αrω-neighbourhood of X
Proof: Let N be neighbourhood of point
x ∈ X . To prove that N is a αrωneighbourhood of x by definition of
neighbourhood, ∃ an open set G s.t.
x ∈ G ⊂ N . Hence N is αrω-neighbourhood
of x
Remark 6.5 In general, a αrω-nbhd N of x ϕ
X need not be a nbhd of x in X, as seen from
the following example.
Example 6.6 Let X = {a, b, c, d} with
topology τ = {X, φ, {a} , {b} , {a, b} , {a, b,
c}}. Then
αRrωO (X) = {X, φ, {a} , {b} , {c} , {a, b} ,
{b,c} , {a, c} , {a, b, c}}. The set {c,d} is
αrω-nbhd of the point c, since the αrω-open
sets {c} is such that c ϕ {c} ϕ {c,d}.
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R. S. Wali, et al., J. Comp. & Math. Sci. Vol.5 (6), 490-499 (2014)
However, the set {c,d} is not a nbhd of the
point c, since no open set G exists such that
c ϕ G ϕ {c,d}.
Theorem 6.7. If a subset N of a space X is
αrω-open, then N is a αrω-nbhd of each of
its points.
Proof: Suppose N is αrω-open. Let x ϕ N.
We claim that N is αrω-nbhd of x. For N is
a αrω-open set such that x ϕ N ϕ N. Since x
is an arbitrary point of N, it follows that N is
a αrω-nbhd of each of its points.
Remark 6.8 The converse of the above
theorem is not true in general as seen from
the following example.
Example 6.9. Let X = {a, b, c, d} with
topology τ = {X, φ, {a} , {c,d} , {a, c,d}}.
Then αRωO (X) = {X, φ, {a} , {b} , {c} , {c,
d}, {a, c, d}}. The set {a, b} is a αrω-nbhd
of the point a, since the αrω-open set {a} is
such that a ϕ {a} ϕ {a,b}. Also the set {a, b}
is a αrω-nbhd of the point {b}, Since the
αrω-open set {b} is such that b ϕ {b} ϕ {a,
b}. That is {a, b} is a αrω-nbhd of each of
its points. However the set {a, b} is not a
αrω-open set in X.
Theorem 6.10. Let X be a topological space.
If F is a αrω-closed subset of X, and x ϕ Fc.
Prove that there exists a αrω-nbhd N of x
such that N∩F = φ.
Proof: Let F be αrω-closed subset of X and
x ϕ Fc. Then Fc is αrω-open set of X. So by
theorem 6.7 Fc contains a αrω-nbhd of each
of its points. Hence there exists a αrω-nbhd
N of x such that N ϕ Fc. That is N ∩ F = φ.
Theorem 6.11. Let X be a topological space
and for each x ϕ X, Let αrω-N (x) be the
498
collection of all αrω-nbhds of x. Then we
have the following results.
(i) ∀ x ∈ X, αrω-N (x) _≠ φ.
(ii) N ϕ αrω-N (x) ⇒ x ∈ N.
(iii) N ϕ αrω-N (x), M ⊃ N ⇒ M ∈ αrω-N
(x).
(iv) N ϕ αrω-N (x),M ϕ αrω-N (x) ⇒ N ∩M
ϕ αrω-N (x).
(v) N ϕ αrω-N (x) ⇒ there exists M∈ αrωN(x) such that M ϕ N and M ϕ αrω-N (y) for
every y ϕ M.
Proof. (i) Since X is a αrω-open set, it is a
αrω-nbhd of every x ϕ X. Hence there exists
at least one αrω-nbhd (namely - X) for each
x ϕ X. Hence αrω-N (x) ≠ ϕ for every x ϕ X.
(ii) If N ϕ αrω-N (x), then N is a αrω-nbhd
of x. So by definition of αrω-nbhd, x ϕ N.
(iii) Let N ϕ αrω-N (x) and M ϕ N. Then
there is a αrω-open set G such that x ϕ G ϕ
N. Since N ϕ M, x ϕ G ϕ M and so M is αrωnbhd of x. Hence M ϕ αrω-N (x).
(iv) Let N ϕ αrω-N (x) and M ϕ αrω-N (x).
Then by definition of αrω-nbhd there exists
αrω-open sets G1 and G2 such that x ϕ G1 ϕ
N and x ϕ G2 ϕ M. Hence x ∈ G1 ∩ G2 ⊂
N ∩M → (1). Since G1 ∩ G2 is a αrω-open
set, (being the intersection of two αrω-open
sets), it follows from (1) that N ∩ M is a
αrω-nbhd of x. Hence N ∩M ∈ αrω-N (x).
(v) If N ϕ αrω-N (x), then there exists a αrωopen set M such that x ϕ M ϕ N. Since M is
a αrω-open set, it is αrω-nbhd of each of its
points. Therefore M ϕ αrω-N (y) for every y
ϕ M.
Theorem 6.12. Let X be a nonempty set,
and for each x ϕ X, let αrω-N (x) be a
nonempty collection of subsets of X
satisfying following conditions.
(i) N ϕ αrω-N (x) ⇒ x ϕ N
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R. S. Wali, et al., J. Comp. & Math. Sci. Vol.5 (6), 490-499 (2014)
(ii) N ϕ αrω-N (x),M ϕ αrω-N (x) ⇒ N ∩M ϕ
αrω-N (x).
Let τ consists of the empty set and all those
non-empty subsets of G of X having the
property that x ϕ G implies that there exists
an N ϕ αrω-N (x) such that x ϕ N ϕ G, Then
τ is a topology for X.
Proof: (i) φ ϕ τ by definition. We now show
that x ϕ τ . Let x be any arbitrary element of
X. Since αrω-N (x) is nonempty, there is an
N ϕ αrω-N (x) and so x ϕ N by (i). Since N
is a subset of X, we have x ϕ N ϕ X. Hence
Xϕτ.
(ii) Let G1 V τ and G2 ∈ τ. If x ∈ G1 ∩ G2
then x ϕ G1 and x ϕ G2. Since G1 ϕ τ and
G2 ϕ τ , there exists N ϕ αrω-N (x) and M ϕ
αrω-N (x), such that x ϕ N ϕ G1 and x ϕ M
⊂ G2. Then x ∈ N ∩ M ⊂ G1 ∩ G2. But N
∩M ∈ αrω-N (x) by (2). Hence G1 ∩ G2 ∈
τ.
(iii) Let Gλ ∈ τ for every λ ∈ ∧ . If x ∈
∪ {Gλ : λ ∈ ∧ }, then x ∈ Gλx for some λx
V ∧ . Since Gλx V τ , there exists an N V
αrω-N (x) such that x ϕ N ϕ Gλx and
consequently x ∈ N ⊂ ∪ {Gλ : λ ∈ ∧ }.
Hence ∪ {Gλ : λ ∈ ∧ } ∈ τ. It follows that τ
is topology for X.
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