Available online through
ISSN 2229 – 5046A STRONG FORM OF COMPACT SPACES
*
C. Duraisamy &
**R. Vennila
*
Department of Mathematics, Kongu Engineering College, Perundurai, Erode-638052, India
**
Department of Mathematics, Sriguru Institute of Technology, Varathaiyangar Palayam, Coimbatore-641110, India
(Received on: 19-07-12; Accepted on: 05-08-12)
ABSTRACT
I
n this paper, we introduce and study the notion of N-λ
-open sets as a generalization ofλ
-open sets.Keywords: Topological spaces, λ-open sets,
λ
-compact spaces.2000 Mathematics Subject Classification: 54C10, 54D10.
1. INTRODUCTION AND PRELIMINARIES
Throughout this paper
(X,
τ)
and(Y,
σ)
stand for topological spaces with no separation axioms assumed, unlessotherwise stated. Maki [3] introduced the notion of
Λ
-sets in topological spaces. AΛ
-set is a set A which is equal to its kernel, that is, to the intersection of all open super sets of A. Arenas et.al. [1]Introduced and investigated the notion of λ-closed sets by involvingΛ
-sets and closed sets. Let A be a subset of a topological space (X,τ). The closure and the interior of a set A is denoted by Cl (A), Int (A) respectively. A subset A of a topological space (X,τ) is said to be λ -closed [1] if A = B ∩ C, where B is aΛ
-set and C is a closed set of X. The complement of λ-closed set is called λ -open [1]. A point x∈X in a topological space (X,τ) is said to be λ-cluster point of A [2] if for every λ-open set U of X containing x,A
∩
U
≠
φ
. The set of all λ-cluster points of A is called the λ-closure of A and is denoted by Clλ(A) [2]. A point x∈X is said to be the λ-interior point of A if there exists a λ-open set U of X containing x such that U⊂ A. The set of all λ-interior points of A is said to be the λ-interior of A and is denoted by Intλ(A). A set A is λ-closed(resp.λ-open) if and only if Cl λ(A) = A (resp. Intλ(A) = A) [2]. The family of all λ-open (resp. λ-closed) sets of X is denoted by λO(X) (resp. λC(X)). The family of all λ-open (resp. λ-closed) sets of a space (X,τ) containing the point x∈X is denoted by λO(X, x) (resp. λC(X, x)).
2.
N-λ
-OPEN SETSDefinition 2.1: A subset A of a topological space X is said to be N-
λ
-open if for everyx
∈
A,
there exists aλ
-opensubset
U
x∈
X
containing x such thatU
x−
A
is finite set. The complement of an N-λ
-open subset is said to beN-
λ
-closed.The family of N-
λ
-open (resp.N-λ
-closed) subsets of a space(X, )
τ
is denoted byN
λO(X)
(resp.N
λC(X)
).Proposition 2.2: Every
λ
-open set is N-λ
-open. Converse not true.Example 2.3: Let
X
=
{
a,
b,
c
}
,τ
=
{
φ
,
{ }
a
,
X
}
, then{ }
b
is N-open but not N-λ
-open (since X is a finite set).Corollary 2.4: Every open set isN-
λ
-open, but not conversely.Proof: Follows from the fact that every open set is
λ
-open. Corresponding author:**Lemma 2.5: A subset A of a topological space X is N-
λ
-open if and only if for everyx
∈
A
, there exists aλ
-opensubset
U
xcontaining x and a finite subset C such thatU
x−
C
⊂
A
.Proof: Let A be N-
λ
-open andx
∈
A
, then there exists aλ
-open subsetU
xcontaining x such thatU
x−
A
is finite. LetC
=
U
x−
A
=
U
x∩
(
X
−
A)
. ThenU
x− ⊆
C
A
. Conversely, letx
∈
A
. Then there exists aλ
-open subsetU
xcontaining x and a finite subset C such thatU
x− ⊆
C
A
. Thus,U
x− ⊆
C
A
andU
x−
A
is a finite set.Theorem 2.6: Let X be a topological space and
F
⊆
X.
If F is N-λ
-closed, thenF
⊆ ∪
K
C
for someλ
-closed subset K and a finite subset C.Proof: If F is N-
λ
-closed, thenX
−
F
is N-λ
-open and hence for everyx
∈ −
X F
, there exists aλ
-open set U containing x and a finite set C such thatU C
− ⊆ −
X F
.ThusF
⊂ −
X (U C)
−
= −
X (U
∩
(X C))
−
=
(
X U
− ∪
)
C
. LetK
=
X
−
U
. Then K is aλ
-closed set such thatF
⊆ ∪
K
C
.Lemma 2.7: The union of any family of N-
λ
-open sets is N-λ
-open.Proof: If
{
U :
ii
∈
I
}
is a collection of N-λ
-open subsets of X and1
U
n i i
x
=∈
.Thenx
∈
U
jfor somej
∈
I
. Thisimplies that there exists a
λ
-open subset V of X containing x such thatV
−
U
j is finite. Since1
V
U
V U ,
n
i j i=
−
⊆ −
then
1
V
U
n i i=
−
is finite. Therefore,U
λO(X)
1
N
∈
=
in i .The intersection of all N-
λ
-closed sets of X containing A is called the N-λ
-closure of A and is denoted byN
C
l
λA)
. And the union of all N-λ
-open sets of X contained in A is called the N -λ
-interior and is denotedby
N
Int
λ(A)
.The proof of the following lemma is obvious and hence omitted.
Lemma 2.8: Let A be a subset of a topological space
(X,
τ)
. Then(i)
x
∈
N
Cl
λ(A)
if and only ifA
∩
U
≠
φ
for everyU
∈
N
λO(X,
x
)
; (ii) A is N-λ
-closed if and only ifA
=
N
Cl
λ(A)
;(iii)
N
Cl
λ(A)
is N -λ
-closed.Corollary 2.9: The intersection of an N-
λ
-open set with an open set is N-λ
-open.Question: Does there exist an example for the intersection of N-
λ
-open sets is N-λ
-open?3.
λ
- COMPACT SPACESDefinition 3.1: A collection
{
U
α:
α
∈
∆
}
ofλ
-open sets in a topological space X is called aλ
-open cover of a subset B of X ifB
⊂
{
U
α:
α
∈
∆
}
holds.Theorem 3.3: If X is a topological space such that every
λ
-open subset isλ
-compact relative to X, then every subset isλ
-compact relative to X.Proof: Let B be an arbitrary subset of X and let
{
U :
ii
∈
I
}
be the cover of B byλ
-open sets of X. Then the family{
U :
ii
∈
I
}
is aλ
-open cover of theλ
-open set∪
{
U :
ii
∈
I
}
. Hence by hypothesis there is a finite subfamily{
U :
N
0}
j
i
j
∈
which covers∪
{
U :
ii
∈
I
}
. This subfamily is also a cover of the set B.Theorem 3.4: A subset A of a topological space is
λ
-compact relative to X if and only if for any cover{
V
α:
α
∈
∆
}
of A by N-λ
-open sets of X, there exists a finite subset∆
0of∆
such thatA
⊆
∪
{
V
α:
α
∈
∆
0}
.Proof: Let
{
V
α:
α
∈
∆
}
be a cover of A andV
α∈
N
λO(X)
.For eachx
∈
A
, there existsα
( )
x
∈
∆
such thatx
∈
V
α( )x . SinceV
α( )x is N-λ
-open, there exist aλ
-open setU
α( )x such thatx
∈
U
α( )x andU
α( )x−
V
α( )x is finite. The family{
U
α( )x:
x
∈
A
}
is aλ
-open cover of A. Since A isλ
-compact relative to X, there exists a finitesubset
x
1,
x
2,...
x
n, such thatA
⊆
∪
{
U
( ):
i
∈
F
}
i
x
α , where
F
=
{
1,2,...
n
}
. Now, we have,( ) ( )
(
)
( )(
)
FU
V
V
A
∈
∪
−
⊆
i
x
α
x
α
x
α i i i
(
( ) ( ))
( )F F
V
V
U
∈ ∈
∪
−
=
i x
α
i
x
α
x
α i i i . For each
x
i,U
α( )xi−
V
α( )xi is a finite set and there exists a finite subset∆
( )
x
i of∆
such that(
α( )x α( )x)
{
αα
( )
x
i}
i
i
−
V
∩
A
⊆
∪
V
:
∈
∆
U
. Therefore, wehave
(
{
( )
}
)
( )
∪
∪
∈
∆
⊆
∈
∈
FV
:
FV
A
i x
α
i
i
α
α
x
i . Hence A isλ
-compact relative to X.Corollary 3.5: For any topological space X, the following properties are equivalent: (i) X is
λ
-compact.(ii) Every N-
λ
-open cover of X admits a finite subcover.Theorem 3.6: A topological space X is
λ
-compact if and only if every proper N-λ
-closed set isλ
-compact with respect to X.Proof: Let A be a proper N-
λ
-closed subset of X. Let{
U
α:
α
∈
∆
}
be a cover of A byλ
-open sets of X. Now for eachx
∈ −
X
A
, there is aλ
-open setV
x such thatV
x−
A
is finite . Then{
U :
αα
∈∆ ∪
} {
V :
xx
∈ −
X A
}
isa
λ
-open cover of X. Since X isλ
-compact, there exist a finite subset∆
1 of∆
and a finite number of points,say,
x
1,
x
2,...
x
n inX
−
A
such thatX
=
(
∪
{
U
α:
α
∈
∆
1}
)
∪
(
∪
{
V
xi:
1
≤
i
≤
n
}
)
hence{
}
(
U
:
1)
A
⊂
∪
αα
∈
∆
(
{
i
n
}
)
i
x
≤
≤
∩
∪
∪
A
V
:
1
since
A
∩
V
xi is finite for each i, there exists a finitesubset
∆
2 of∆
such that(
{
i
n
}
)
i
x
≤
≤
∩
∪
A
V
:
1
⊂
{
U
α:
α
∈
∆
}
. Therefore, weobtain
A
⊂
∪
{
U
α:
α
∈
∆
1∪
∆
2}
. This shows that A isλ
-compact relative to X. Conversely let{
V
α:
α
∈
∆
}
be anyλ
-open cover of X. We choose and fix oneα
0∈
∆
.Then∪
{
V
α:
α
∈
∆
−
{ }
α
0}
is aλ
-open cover of a N-λ
-closed set
X
V
{
V
:
0}
0
⊂
∪
∈
∆
−
α αα
.Therefore,
X
=
∪
{
V
α:
α
∈
∆
0∪
{ }
α
0}
. This shows that X isλ
-compact.Theorem 3.7: Let
(X,
τ)
be a topological space such that .Then isλ
-compact if and only if is compact.Proof: Let
{
V
α:
α
∈
∆
}
be anyλ
-open cover of(
X,
N
λO(X)
)
. For eachx
∈
X
, there existsα
( )
x
∈
∆
such that( )x
x
∈
V
α . SinceV
α( )x is N-λ
-open, there exist aλ
-open setU
α( )x such thatx
∈
U
α( )x andU
α( )x−
V
α( )x issubset
x
1,
x
2,...
x
n, such thatX
=
∪
{
U
α( )xi:
i
∈
F
}
, whereF
=
{
1,2,...
n
}
. Now, we have( ) ( )
(
)
( )(
)
FU
V
V
X
∈∪
−
=
i x α x α xα i i i
(
( ) ( ))
( )F F
V
V
U
∈ ∈∪
−
=
i x α i x α xα i i i . For each
x
i,U
α( )xi−
V
α( )xi is a finite set and there exists a finite subset∆
( )
x
i of∆
such that(
α( )x α( )x)
{
αα
( )
x
i}
i
i
−
V
∩
X
⊆
∪
V
:
∈
∆
U
. Therefore, wehave
(
{
( )
}
)
( )
∪
∆
∈
∪
=
∈ ∈
F FV
:
V
X
i x α i iα
α
x
i . HenceN
λO(X)
is compact. Conversely, let U be aλ
-opencover of
(X,
τ)
. ThenU
⊆
N
λO(X)
. Since(
X,
N
λO(X)
)
is compact, there exists a finite subcoverof
U
⊆
N
λO(X)
. Since(
X,
N
λO(X)
)
is compact, there exists a finite subcover of U for X. Hence(X,
τ)
isλ
-compact.Theorem 3.8: An N -
λ
-closed subset of aλ
-compact space X isλ
-compact relative to X.Proof: Let A be a N-
λ
-closed subset of X. Let{
U
α:
α
∈
∆
}
be a cover of A byλ
-open sets of X. Now for eachX
A
x
∈ −
there is aλ
-open setV
x such thatV
x−
A
is finite. Since{
U :
αα
∈ ∆ ∪
} {
V :
xx
∈ −
X
A
}
is aλ
-open cover of X and X isλ
-compact, there exist a finite subcover{
U :
N
} {
V :
N
}
i
i
xii
α
∈
∪
∈
.Since
(
)
N
A
V
∈∩
ixi is finite, so for each
x
j∈
∪
(
V
xi∩
A
)
there isU
α(xi)∈
{
U
α:
α
∈
∆
}
such that) (
U
i x jx
∈
α andj
∈
N
. Hence{
U
αi:
i
∈
N
}
∪
{
U
α(xi):
i
∈
N
}
is a finite subcover of{
U
α:
α
∈
∆
}
and itcovers A. Therefore, A is
λ
-compact relative to X.Corollary 3.9: If a topological space X is
λ
-compact and A isλ
-closed, then A isλ
-compact relative to X.4. PRESERVATION THEOREMS
Definition 4.1: A function
f
:
(X,
τ)
→
(Y,
σ)
is said to be N-λ
-continuous (resp.λ
-continuous [1]) if the inverseimage of every open subset of Y is N-
λ
-open in X.It is clear that every
λ
-continuous function is N -λ
-continuous but not conversely.Example 4.2: Let
X
=
{
a,
b,
c
}
,τ
=
{
φ
,
{ }
a
,
X
}
andσ
=
{
φ
,
{ }
a,
c
,
X
}
. Clearly the identity functionσ)
(X,
τ)
(X,
:
→
f
is N -λ
-continuous but notλ
-continuous.Theorem 4.3: A function
f
:
(X,
τ)
→
(Y,
σ)
is N-λ
-continuous if and only if for each point x in X and each openset V in Y with
f
( )
x
∈
V
, there is an N-λ
-open set U in X such thatx
∈
U
, andf
( )
U
⊆
V.
Proof: Let V be an open set in Y and let
x
∈
f
−1( )
V
. Thenf
( )
x
∈
V
and thus there existU
x∈
N
λO(X)
such thatx
∈
U
x andf
( )
U
x⊆
V
. Thenx
∈
U
x⊆
f
−1( )
V
=
∪
{
U
x:
x
∈
f
−1( )
V
}
Then by Lemma 2.7f
−1( )
V
isN-
λ
-open. Conversely, letx
∈
X
and V be an open set of Y containingf
( )
x
. Thenx
∈
f
−1( )
V
∈
N
λO(X)
since f is N-
λ
-continuous. LetU
=
f
−1( )
V
. Thenx
∈
U
andf
( )
U
⊆
V
.Theorem 4.4: Let
f
:
(X,
τ)
→
(Y,
σ)
be a N-λ
-continuous function. If X isλ
-compact, then Y is compact.Proof: Let
{
V
α:
α
∈
∆
}
be an open cover of Y. Then,{
f
−1( )
V
α:
α
∈
∆
}
is a N -λ
-cover of X. Since X isλ
-compact, by Corollary 3.9 there exist a finite subset∆
0of∆
such thatX
=
∪
{
f
−1( )
V
α:
α
∈
∆
0}
hence{
V
:
0}
Definition 4.5: A function
f
:
X
→
Y
is said to be stronglyλ
-open if the image of eachλ
-open subset of X isλ
-open in Y.Proposition 4.6: If
f
:
X
→
Y
is stronglyλ
-open, then the image of an N-λ
-open set of X is N-λ
-open in Y.Proof: Let
f
:
X
→
Y
be stronglyλ
-open and W an N-λ
-open subset of X. For anyy
∈
f
( )
W
, there existx
∈
W
such thatf
( )
x
=
y
. Since W is N-λ
-open, there exists aλ
-open set U suchthat
x
∈
U
andU
−
W
=
C
is finite. Since f is stronglyλ
-open,f
( )
U
isλ
-open in Y such thaty
=
f
( ) ( )
x
∈
f
U
and
f
( ) ( )
U
−
f
W
⊆
f
(
U
−
W
)
=
f
( )
C
is finite. Therefore,f
( )
W
is N-λ
-open in Y.Definition 4.7: [2] A function
f
:
X
→
Y
is said to beλ
-irresolute if the inverse image of eachλ
-open subset of Yis
λ
-open in X.Proposition 4.8: If
f
:
X
→
Y
is aλ
-irresolute injection and A is N-λ
-open in Y, thenf
−1( )
A
is N-λ
-open in X.Proof: Assume that A is an N-
λ
-open subset of Y. Letx
∈
f
−1( )
A
. Thenf
( )
x
∈
A
and there exists aN-
λ
-open set V containingf
( )
x
such thatV
−
A
is finite. Since f isλ
-irresolute,f
−1( )
V
is aλ
-open set containing x. Thusf
−1( )
V
−
f
−1( )
A
=
f
−1(
V
−
A
)
and it is finite. It follows thatf
−1( )
A
is N-λ
-open in X.Definition 4.9: A function
f
:
X
→
Y
is said to be N-λ
-closed iff
( )
A
is N-λ
-closed in Y for eachλ
-closed set A of X.It is clear that every strongly
λ
-closed function is N-λ
-closed but not conversely. The function f in Example 4.2 isN -
λ
-closed but not strongly N-λ
-closed.Theorem 4.10: If
f
:
X
→
Y
is an N -λ
-closed surjection such thatf
−1(
y
)
isλ
-compact relative to X foreach
y
∈
Y
and Y isλ
-compact, then X isλ
-compact.Proof: Let
{
U
α:
α
∈
∆
}
be anyλ
-open cover of X. For eachy
∈
Y
,
f
−1(
y
)
isλ
-compact relative to X and there exists a finite subset∆
( )
y
of∆
such thatf
−1(
y
)
⊂
∪
{
U
α:
α
∈
∆
( )
y
}
.Now, we put( )
{
y
}
y
)
=
∪
U
α:
α
∈
∆
(
U
andV
(
y
)
=
Y
−
f
(
X
−
V
(
y
)
)
. Then, since f is N-λ
-closed,V
(
y
)
is an N-λ
-open set in Y containing y such that
f
−1(
V
(
y
)
)
⊂
U
(
y
)
. Since{
V
(
y
)
:
y
∈
Y
}
is an N-λ
-open cover of Y, byCorollary 3.9 there exists a finite subset
{
y
k:
1
≤
k
≤
n
}
⊆
Y
such that
( )
n
k k
y
1
V
Y
=
=
. Therefore,( )
n(
( )
)
k
k
y
f
y
f
1 1 1
V
X
= −
−
=
=
n( )
k k
y
1
U
=
⊆
n{
( )
}
k
k
y
1
:
U
=
∆
∈
=
αα
. This shows that X isλ
-Compact.Definition 4.11: A function
f
:
X
→
Y
is said to be N-λ
-continuous if for eachx
∈
X
and eachλ
-open set V ofY containing
f
( )
x
, there exist an N-λ
-open set U of X containing x such thatf
( )
U
⊆
V
.Theorem4.12: Let
f
:
X
→
Y
be a N-λ
-continuoussurjection from X onto to Y. If X isλ
-compact, then Y isλ
-compact.{
x
k:
1
≤
k
≤
n
}
⊆
X
such that
( )n
k x
α k 1
U
X
=
=
. Therefore,( )
( )
=
=
=
n kxk
f
f
1
U
X
Y
α
( )n
k x
α k 1
V
=
⊆
. This showsthat Y is
λ
-Compact.REFERENCES
[1] Arenas F.G., J. Dontchev and M. Ganster, On λ-closed sets and dual of generalized continuity, Q&A Gen.Topology, 15, (1997), 3-13.
[2] Caldas M., S. Jafari and G. Navalagi, More on λ-closed sets in topological spaces, Revista Columbiana de Matematica, 41(2), (2007), 355-369.
[3] Maki H., Generalized