Download On sp-gpr-Compact and sp-gpr-Connected in Topological Spaces

Indian Journal of Mathematics and Mathematical Sciences Vol. 7, No. 2, (December 2011) : 161-167
ON sp-gpr-COMPACT AND sp-gpr-CONNECTED IN
TOPOLOGICAL SPACES
D. Christia Jebakumari & M. Mariasingam
Abstract
Compact spaces and connected spaces constitute the most important classes
of topological spaces. They find very active role in abstract analysis. In this
paper, by using sp-gpr-closed and sp-gpr-open sets, we introduce two concepts
“sp-gpr-compact” and “sp-gpr-connected” in topological spaces. We study
their relations with other compact and connected spaces.
Keywords: sp-gpr-open, sp-gpr-compact, sp-gpr-connected etc.
1. INTRODUCTION AND PRELIMINARIES
Levine [7] introduced the concept of generalized closed sets as a generalization of
closed sets in topological spaces. Using pgpr-closed sets and pgpr-open sets, Anitha
and Thangavelu [3] defined and studied pgpr-compact and pgpr-connected spaces.
In this paper we have introduced sp-gpr-compact and sp-gpr-connected spaces.
Using these new types of spaces, several characterizations and its properties have
been obtained.Throughout this paper (X, τ), (Y, σ) denote the topological spaces.
For any subset A of (X, σ) cl A, int A denote the closure of A and interior of A
respectively. We recall some definitions and results that are useful in the sequel.
Definition 1.1: A subset A of a space (X, τ) is regular open [10] if A = int cl A
and regular closed if A = cl int A; pre-open [8] if A ⊆ int cl A and pre-closed if
cl int A ⊆ A; semi-pre-open [1] if A ⊆ cl int cl A and semi-pre-closed if int cl int A ⊆ A.
The pre-closure (resp. semi-pre-closure) of a subset A of X is the intersection
of all pre-closed (resp. semi-pre-closed) sets containing A and is denoted by
pcl A (resp. spcl A). The semi-pre-interior of a subset A of X is the union of all
semi-pre-open sets contained in A and is denoted by sp int A.
Definition 1.2: A subset A of (X, τ) is
(i)
g-closed [7] if cl A ⊆ U, whenever U is an open set containing A.
(ii) rg-closed [9] if cl A ⊆ U whenever U is a regular open set containing A.
(iii) pre-generalized pre-regular closed (briefly pgpr-closed) [2] if pcl A ⊆ U
whenever A ⊆ U and U is rg-open.
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(iv) semi-pre-generalized pre- regular closed (briefly sp-gpr-closed) [5] if
spcl A ⊆ U whenever A ⊆ U and U is rg-open.
A subset B of a space X is g-open if X \B is g-closed. The concepts of rg-open,
pgpr-open, sp-gpr-open are analogously defined.
Definition 1.3 [5]: A function f : (X, τ) → (Y, σ) is said to be
(i)
sp-gpr-continuous if f – 1(V) is sp-gpr-closed subset of X for every closed
set V in Y.
(ii) sp-gpr-irresolute if f – 1(V ) is sp-gpr-closed subset of X for every
sp-gpr-closed subset V in Y.
Definition 1.4 [3]: A topological space X is pgpr-compact if every cover of X
by pgpr-open subsets of X has a finite sub cover.
Definition 1.5 [3]: A topological space (X, τ) is said to be pgpr-connected if X
cannot be written as the disjoint union of two non-empty pgpr-open sets in X.
Lemma 1.6 [5]:
(i)
Every closed set is sp-gpr-closed.
(ii) Every open set is sp-gpr-open.
(iii) Every pgpr-closed set is sp-gpr-closed.
(iv) Every pgpr-open set is sp-gpr-open.
Lemma 1.7 [5]: A set A ⊆ X is sp-gpr-open if and only if F ⊆ sp int A whenever
F ⊆ A, F is rg-closed.
Lemma 1.8 [3]: For a topological space X, the following are equivalent:
(i)
X is pgpr-connected.
(ii) The only subsets of X which are both pgpr-open and pgpr-closed are the
empty set φ and X.
(iii) Each pgpr-continuous function of X into a discrete space Y with at least
two points is a constant map.
Lemma 1.9 [9]: Suppose that B ⊆ A ⊆ X, B is rg-closed relative to A and that
A is a g-closed open subset of X. Then B is rg-closed relative to X.
2. sp-gpr-COMPACT SPACE
Definition 2.1: A topological space X is sp-gpr-compact if every cover of X by
sp-gpr-open subsets of X has a finite sub cover.
On sp-gpr-Compact and sp-gpr-Connected in Topological Spaces
163
Definition 2.2: A subset S of a topological space X is sp-gpr-compact
relative to X ifΑfor every collection {Aα : α ∈ Ω} of sp-gpr-open subsetsΑof X such
that S ⊆  α there exists a finite subset ∆ of Ω such that S ⊆  α .
α∈Ω
α∈∆
Theorem 2.3: A sp-gpr-closed subset of a sp-gpr-compact space X is
sp-gpr-compact relative to X.
Proof: Let A be sp-gpr-closed subset of a sp-gpr-compact space X. Then X \ A
is sp-gpr-open. Let ξ be a cover for A by sp-gpr-open subsets of X. Then ξ ∪ {X \A}
is a cover for X by sp-gpr-open subsets of X. Since X is sp-gpr-compact, by
Definition 2.1 it has a finite sub cover, say {P1, P2, …, Pn} = ℘. If X \ A ∈ ℘, then
℘ is a finite sub cover of ξ for A. If X \ A ∈ ℘, then ℘ \ {X \ A} is a finite sub cover
of ξ for A. Thus A is sp-gpr-compact relative to X.
Theorem 2.4: Let f : (X, τ) → (Y, σ) be a surjective, sp-gpr-continuous function.
If X is sp-gpr-compact, then Y is compact.
Proof: Suppose X is sp-gpr-compact. Let {Aα : α ∈ Ω} be an open cover of Y.
Since f is sp-gpr-continuous by Definition 1.3(i) {f –1(Aα) : α ∈ Ω} is a cover for X
by sp-gpr-open sub sets of X. Since X is sp-gpr-compact by Definition 2.1 it has a
finite sub cover say {f –1(A1), f –1(A2), …, f –1(An)}. Since f is a surjection, {A1, A2, …, An}
is an open cover of Y and hence Y is compact. This proves the theorem.
Theorem 2.5: Suppose f : (X, τ) → (Y, σ) is sp-gpr-irresolute. Let S ∈ X be
sp-gpr-compact relative to X. Then the image f (S) is sp-gpr-compact relative to Y.
Proof: Let {Aα : α ∈ Ω} be a collection of sp-gpr-open sets in Y such that
Α
Α
–1
f ( S ) ⊆  α . Then S ⊆  f ( α) . Since f is sp-gpr-irresolute, by
α∈Ω
α∈Ω
Definition 1.3(ii), f –1(Aα) is sp-gpr-open in X for each α. Since S is sp-gprcomptact relative to X, by Definition 2.2 there exists
a finite sub collection
Α
Α
{f –1(A1), f –1(A2), …, f –1(An)} such that S ⊆
n

α =1
f −1 (
α ) . That is, f ( S ) ⊆
n

α =1
α
.
Hence by Definition 2.2 f (S) is sp-gpr-compact relative to Y.
Theorem 2.6: Every sp-gpr-compact space is pgpr-compact.
Proof: Let {Aα : α ∈ Ω} be a collection of pgpr-open sets which covers X.
By Lemma 1.6(iii) each Aα is sp-gpr-open. Since X is sp-gpr-compact by
Definition 2.1 there exists a finite sub-cover X. By Definition 1.4 X is pgpr-compact.
Theorem 2.7: Let (X, τ) be a topological space. X is sp-gpr-compact if and
only if any family of sp-gpr-closed sets with finite intersection property has
non-empty intersection.
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Proof: Suppose X is sp-gpr-compact. Let {Aα} be a family of sp-gpr-closed subsets
of X with finite intersection property. We claim that  Αα ≠ φ . Suppose  Αα = φ .
Α
X
α
α
Then X \  α = . Therefore,  ( X \ Aα ) = X . Also, since each Aα is sp-gpr-closed,
(
α
)
α
X \ Aα is sp-gpr-open. Therefore, {X \ Aα} is a cover for X by sp-gpr-open subsets
of X. Since X is sp-gpr-compact, by Definition 2.1 this cover has a finite
n
subcover Αsay {XX\ A1, X \ A2, …, X \ An}. Therefore,  ( Χ \ Αα ) = X . Therefore,
n
α =1
 n

X \   α  = which implies  Αα = φ . Which is a contradiction to the finite
 α =1 
α =1


intersection property. Therefore,  Αα ≠ φ . Conversely suppose that each family
α
of sp-gpr-closed sets in X with finite intersection property has non-empty
intersection. We wish to prove that X is sp-gpr-compact. Let {Aα : α ∈ I} be a cover
of X by sp-gpr-open subsets of X. Then,  Aα = X , that implies X \   Aα  is
 α∈I 
α∈I
Α


an empty set. Therefore,  ( X \ α ) = φ . Since Aα is sp-gpr-open, X\ Aα is sp-gprα∈I
closed for each a Therefore, {X \ Aα : α ∈ I} is a family of sp-gpr-closed sets whose
Α
intersection is empty. Hence by hypothesis there exists a finite sub collection
of
n
sp-gpr-closed subsets of X say {X \ A1, X \ A2, …, X \ An} such that  ( X \ α ) = φ ,
α =1
n
 n

that implies X \   Αα  = φ , which implies  Αα = X . This proves that, X is
 α =1 
α =1


sp-gpr-compact.
3. sp-gpr-CONNECTEDNESS
Definition 3.1: A topological space (X, τ) is said to be sp-gpr-connected if X cannot
be written as the disjoint union of two non-empty sp-gpr-open sets in X.
Definition 3.2: A subset S of X is sp-gpr-connected if it is sp-gpr-connected as
a subspace that is if S cannot be written as the disjoint union of two non-empty
sp-gpr-open sets in S.
Definition 3.3: A subset S of a topological space (X, τ) is said to be sp-gprconnected relative to X if S cannot be written as the disjoint union of two non-empty
sp-gpr-open sets in X.
Theorem 3.4: For a topological space X, the following are equivalent:
(i)
X is sp-gpr-connected.
(ii) The only subsets of X which are both sp-gpr-open and sp-gpr-closed are
the empty set φ and X.
On sp-gpr-Compact and sp-gpr-Connected in Topological Spaces
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(iii) Each sp-gpr-continuous function of X into a discrete space Y with atleast
two points is a constant map.
Proof: (i)  (ii): Suppose S ⊆ X is a pre-open subset which is both sp-gpropen and sp-gpr-closed. Then by Definition 1.2 its complement X \ S is also
sp-gpr-open and sp-gpr-closed. Then X = S ∪ (X \ S), a \ disjoint union of two
non-empty sp-gpr-open sets which contradicts (i). Hence S = φ or X.
(ii)  (i): Suppose X is not sp-gpr-connected. Then X = A ∪ B where A ∩ B = φ,
A ≠ φ, B ≠ φ and A and B are sp-gpr-open. Since A = X \ B, A is sp-gpr-closed. If (ii)
holds A = φ or X. A = φ is not possible by our choice. If A = X then B = φ that is also
not possible by our choice. This proves that X is sp-gpr-connected.
(ii)  (iii): Let f : X → Y be a sp-gpr-continuous function where Y is a discrete
space with atleast two points. Then f –1({y}) is sp-gpr-closed and sp-gpr-open for
each y ∈ Y. By (ii) f –1({y}) = φ or X. If f –1({y}) = φ for all y ∈ Y, f will not be a
function. So f –1({y}) = X. This proves that f is constant.
(iii)  (ii): Let S be both sp-gpr-open and sp-gpr-closed in X. Suppose S ≠ φ.
Let a, b in Y and a ≠ b. Let Y be a discrete space. Fix y0 and y1 in Y and y0 ≠ y1.
Define f : X → Y by f (x) = y0 for x ∈ S and f (x) = y1 for x ∈ X \ S. Then f is sp-gprcontinuous function. By (iii), f is constant. Therefore f = y0 or f = y1 . If f = y0 then
S = X and if f = y1 then S = φ. This proves the theorem.
Theorem 3.5:
(i)
Let f : X → Y be sp-gpr-continuous and onto and X be sp-gpr-connected.
Then Y is connected.
(ii)
Let f : X → Y be sp-gpr-irresolute and onto and X be sp-gpr-connected.
Then Y is sp-gpr-connected.
Proof: Suppose Y is not connected. Then Y = A ∪ B where A ∩ B = φ, A ≠ φ, B ≠ φ
and A and B are open in Y. If f is sp-gpr-continuous and onto, X = f –1(A) ∪ f –1(B)
where f –1(A) and f –1(B) are disjoint non-empty sp-gpr-open subsets of X. This
contradicts the fact that X is sp-gpr-connected. Thus (i) is proved.
Suppose Y is not sp-gpr-connected. Then Y = A ∪ B where A ∩ B = φ, A ≠ f,
B ≠ f and A and B are sp-gpr-open in Y. Since f is sp-gpr-irresolute and onto,
X = f –1(A) ∩ f –1(B) where f –1(A) and f –1(B) are disjoint non-empty sp-gpr-open
subsets of X. This contradicts the fact that X is sp-gpr-connected. Thus (ii) is proved.
Theorem 3.6: Every sp-gpr-connected space is connected.
Proof: Let X be a sp-gpr-connected space. Then by Theorem 3.4 the only
subsets of X which are both sp-gpr-open and sp-gpr-closed are the empty set φ and
X. Suppose X is not connected. Then there exist a proper non-empty subset B of X
which is both open and closed in X. Since by Lemma 1.6(i)every closed set is
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D. Christia Jebakumari & M. Mariasingam
sp-gpr-closed, B is a proper non-empty subset of X which is both sp-gpr-open and
sp-gpr-closed in X. Then by Theorem 3.4, X is not sp-gpr-connected. This proves
the theorem.
The following example shows that the converse is not true.
Example 3.7: Let X = {a, b, c}, τ = {φ, {a}, {b}, {a, b}, X }. The topological
space (X, τ) is connected. However, since {a, c} and {b} are both sp-gpr-open but
X is not sp-gpr-connected.
Theorem 3.8: Every sp-gpr-connected space is pgpr-connected.
Proof: Let X be a sp-gpr-connected space. By Theorem 3.4 the only subsets
of X which are both sp-gpr-open and sp-gpr-closed are the empty set φ and X.
Suppose X is not pgpr-connected. Then by Lemma 1.8 there exist a proper
non-empty subset B of X which is both pgpr-open and pgpr-closed in X. By
Lemma 1.6(iii) B is both sp-gpr-open and sp-gpr-closed in X. Then by Theorem 3.4
X is not sp-gpr-connected. This proves the Theorem.
The following example shows that the converse is not true.
Example 3.9: Let X = {a, b, c}, τ = {φ, {a}, {b}, {a, b}, X}. The topological
space (X, τ) is pgpr-connected. However, since {a, c} and {b} are both sp-gpropen but X is not sp-gpr-connected.
Lemma 3.10: Suppose Y ⊆ X, Y is clopen. If C is sp-gpr-open in X, then C ∩ Y
is sp-gpr-open in Y.
Proof: Let F ⊆ C ∩ Y and F be rg-closed in Y. Since Y is g-closed and open
relative to X, by Lemma 1.9 F is rg-closed in X. Since F ⊆ C, by Lemma 1.7
F ⊆ sp int (C) that implies F ⊆ Y ∩ sp int (C) = sp intY (C ∩ Y ). Again by Lemma 1.7,
C ∩ Y is sp-gpr-open in Y.
Theorem 3.11: Let Y ⊆ X be clopen in X. Suppose X = C ∪ D where C and D
are two disjoint non-empty sp-gpr-open subsets of X. If Y is a sp-gpr-connected
subspace of X, then Y lies entirely within either C or D.
Proof: Since C and D are both sp-gpr-open in X, by Lemma 3.10 the sets
C ∩ Y and D ∩ Y are sp-gpr-open in Y. These two sets are disjoint and their union
is Y. If they were both non-empty, then Y is not sp-gpr-connected. Therefore, one of
them is empty. Hence Y must lie entirely in C or in D.
Theorem 3.12: Let {Aα : α ∈ Ω} be a locally finite family of clopen sets in X
such that they have a common point. If each Aα is a sp-gpr-connected subspace of
X then their union is a sp-gpr-connected subspace of X.
Proof: Let p be a point of ∩Aα. Let Y = ∪Aα. Then Y is clopen. Suppose that
Y = C ∪ D, where C and D are two disjoint non-empty sp-gpr-open subsets of Y.
On sp-gpr-Compact and sp-gpr-Connected in Topological Spaces
167
The point p is in one of the sets C or D. If p ∈ C then Aα ⊆ C for every a, so that
∪ Aα ⊆ C. This shows that D = φ. Therefore Y = C is sp-gpr-connected.
Theorem 3.13: Let (X, τ) and (Y, σ) be any topological spaces. Let f : X → Y be
a sp-gpr-irresolute function and X be sp-gpr-connected subset of Y. Then f (X) is
sp-gpr-connected subset of Y.
Proof: Let f (X) = S so that f is a function from X onto S. Suppose S is not
sp-gpr-connected. Then there is a proper non-empty subset B of which is both
sp-gpr-open and sp-gpr-closed in S. Since f is sp-gpr-irresolute, by Definition1.3(ii)
f –1(B) is proper non-empty subset of X which is both sp-gpr-open and sp-gpr-closed
in X. Then Theorem3.4, X is not sp-gpr-connected. This shows that S is
sp-gpr-connected.
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D. Christia Jebakumari
Department of Mathematics,
Sarah Tucker College,
Tirunelveli-627007, India.
E-mail: [email protected]
M. Mariasingam
Department of Mathematics,
V.O. Chidambaram College,
Thoothukudi-628008, India.