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```International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
g*-ConnectedSpace
Pauline Mary Helen M, Associate Professor in Mathematics, Nirmala College for Women, Coimbatore
Vakithabegam R, M.Phil Scholar, Nirmala College for Women, Coimbatore
Abstract: In this paper, g*- Connected space is introduced and few of itscharacteristics
are discussed.
Key words: g*-neighbourhood, g*-limit point, g*-closure, g*-interior, g*-multiplicative
space, g*-connected space.
1. Introduction
N.Levine introduced a new class of g-closed sets in a topological space in 1970
andM.K.R.S.Veerakumar introduced g*- closed sets in 1991. In this paper g*
connected space is defined and their properties are investigated.
Throughout this paper X and Y represent topological spaces. For a subset A of a space X,
cl(A) and int(A) denote the closure of A and interior of A.
2. Preliminaries
Definition 2.1:A subset A of a topological space (X, τ) is called a
1)generalized closedset (briefly g-closed) if cl(A)  U whenever A  U and U is
open in (X, τ).
2) generalized star closed set (briefly g*-closed) if cl(A)  U whenever A  U and U
is g-openin
R S. Publication (rspublication.com), [email protected]
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
(X,τ). The complement of g*-closed set is said to be g*-open. The collection of all g*open sets and g*closed sets in X will be denoted by G*O(X) and G*C(X) respectively.
Definition 2.2: A function f : (X,)  (Y,) is called
1) g*-irresolute if f
-1
(V) is a g*-closed set of (X,) for every g*-closed set V of (Y,).
2) g*-continuous if f -1 (V) is a g*-closed set of (X,) for every closed set V of (Y,).
3) strongly g*-continuous if f -1(V) is a closed set of (X,) for every g*-closed set V of
(Y,).
4) g*-resoluteif f (V) is g*-closed in Y whenever V is g*-closed in X.
Remark 2.3: 1)ɸ and X are g*-open in any topological space (X,).
2)Every closed sets isg*-closed but the converse is not true.
3)Union of two g*-closed sets is g*-closed.
4) Arbitrary union of g*-closed sets need not be g*-closed.
5)Intersection of g*-closed sets need not be g*-closed.∴ G*O(X) is not atopology in X.
3. g*-multiplicative spaces
Definition 3.1:A topological space (X,) is defined to be
1) g*-finitely multiplicative if finite intersection of g*-closed sets is g*-closed.
2) g*-countably multiplicative if countable intersection of g*-closed sets is g*-closed.
3) g*-multiplicative if arbitrary intersection of g*-closed sets is g*-closed.
4) g*-countablyadditive if countable union of g*-closed sets is g*-closed.
5) g*-additive if arbitrary union of g*-closed sets is g*-closed.
R S. Publication (rspublication.com), [email protected]
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
Example 3.2: Any indiscrete space (X,) is g*-additive, g*-countably additive,g*multiplicative, g*-finitely multiplicative, g*-countably multiplicative.
Example 3.3: Any infinite set with finite complement topology is not g*additive,notcountably g*-additive, g*-multiplicative, g*-finitely multiplicative, g*countably multiplicative.
2) g*-multiplicative  g*-countably multiplicative  g*-finitely multiplicative.
Remark 3.5: If (X,) is g*-multiplicative then G*O(X) is a topology in X.
Definition 3.6: A space (X,) is said to be g*-discrete space if every subset is g*-open.
Note 3.7: Every discrete topological space is g*-discrete since every open set is g*open.But the converse need not be true.
Example 3.8: An indiscrete space (X,) is g*-discrete but not discrete. Since in this space
all subsets are g*-open but not open.
Definition 3.9: For a subset A in a topological space (X,), define g*-closure of A and g*interior of A as follows:g*cl(A)= intersection of all g*-closed sets containing A and
g*int(A) = union of all g*-open sets contained in A.
Remark 3.10:1)Since intersection of g*-closed sets need not be g*-closed, g*cl(A)need
not be g*-closed. 2) If A is g*-closed then g*cl(A) = A.But A = g*cl(A) does not imply A
isg*-closed.3) If (X,) is g*-multiplicative then g*cl(A) is g*-closed.4) If (X,) is g*multiplicative then g*cl(A) = A A is g*-closed.5)Since union of g*-open sets need not
be g*-open, g*int(A) need not be g*-open.6) If A is g*-open then g*int(A) = A.7)But A =
g*int(A) need not imply A is g*-open.8) If (X,) is g*-multiplicative then A=g*int(A) 
A is g*-open.
R S. Publication (rspublication.com), [email protected]
Page 126
International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
Definition 3.11: Let (X,) be a topological spaceand x ∈X. Every g*-open set containing
x is called g*-neighbourhood of x.
Definition 3.12: If Ais a subset ofa topological space X and x is a point of X, we say that
x is a g*-limit point of A if every g*-neighbourhood of xintersects A in some point other
than x.
Theorem 3.13:Let A be a subset of a topological space X. Let A' be the set of all g*-limit
points of A. Then g*cl(A) = A A'.
Proof: Let x  A.Then every g*-neighbourhood of x intersects A. Suppose x  g*cl(A),
there exists a g*-closed set F containing A and x  F.Then (X–F) is a g*-open
setcontaining x and (X−F)A =  which is a contradiction.x g*-cl(A). On the other
hand,Let x  g*-cl(A). If x A then g*cl(A)  A A. If x  A then xg*cl(A)  every
g*-closed set containing A contains x.x  A. Hence g*cl(A)  A A.g*cl(A) = A
A.
4.g*-ConnectedSpace
Definition 4.1:Let X=A∪B is said to be a g*-separation of X if A and B are non empty
disjoint and g*-open sets.If there is no g*-separation of X then X is said to be g*connected. Otherwise it is said to be g*-disconnected.
Note :If X=A∪B is a separation then Ac=B and Bc =A and hence A and B are g*-closed.
Remark4.2:(X,τ) is g*-connected if and only if the only subsets which are both g*-open
and g*-closed are X and Ф.
Proof :Let (X,τ) be g*-connected.Suppose there exists a subset A which is both g*-open
and g*-closed then X=AUAC is a g*-separation, which is a contradiction. Conversely let
the only subset which is both g*-open and g*-closed be X and Ф.Suppose X=A∪Ac is a
R S. Publication (rspublication.com), [email protected]
Page 127
International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
separation of X then A is both g*-open and g*-closed.Therefore A = X or Ф which is a
Example 4.3:An Infinite set with cofinite topology is g*-connected since it is impossible
to find two disjoint g*-open sets.
Proof : Suppose A and B are disjoint g*-open sets then AC U BC =X.AC and BC areg*closed implies AC and BC are finite. Hence X is finite which is a contradiction.
Example 4.4: Any indiscrete topological space (X,τ) with more than one point is not g*connected since every subset is g*-open.
Proof : Suppose A is a proper subset of X. Then A and AC are g*-open. Therefore
X = A∪AC is a g*-separation.
Theorem 4.5:Every g*-connected space is connected but not conversely.
Proof:Obvious, since every open set is g*-open.
Example 4.6:Any indiscrete topological space (X,τ) is connected but not g*-connected.
Defnition 4.7:Let Y be a subset of X .Then Y=A∪B is said to be g*-separation of Yif A
and B are non-empty disjoint g*-open sets in X. If there is no g*-separation of Y then Y
is said to be g*-connected subset of X.
Theorem 4.8:Let (X,τ) be a topological space and X=A∪B is a g*-separation of X. IfY
is ag*-open, g*-connected subset of X thenYis completely contained in either A orB.
Proof:X = A∪B is a g*-separation of X. Suppose Y intersects both A and B then Y =
(A∩Y)∪(B∩Y) is a g*-separation of Y which is a contradiction.
Theorem 4.9:Let (X,τ) and (Y,σ) be two topological spaces andf : (X,τ) → (Y,σ) be a
bijection function. Then
i)
f is g*-continuous and X is g*-connected  Y is connected.
ii) f is continuous and X is g*-connected  Y is connected.
R S. Publication (rspublication.com), [email protected]
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
iii) f is strongly g*- continuous then X is connected  Y is g*-connected.
iv) f is g- open then Y is g*-connected  X is g*-connected.
v)
f is open then X is connected  Y is g*-connected.
vi) f is g*- irresolute then X is g*-connected  Y is g*-connected.
vii) f is g*- resolute then Y is g*-connected  X is g*-connected.
Proof: (i) Suppose Y=A∪B is a separation of Y then X = f-1(Y).X= f-1(Y) = f-1(A) ∪f1
(B)is a g*- separationof X , which is a contradiction.Therefore Y is connected.
Proof for (ii) to (vii)are similar to the above proof.
Theorem 4.10:A topological space (X,τ) is g*-disconnected if and only if there exists a
g*-continuous map of X onto discrete two point space Y={0,1}.
Proof:Let (X,τ) be g* -disconnected and Y={0,1} is a space with discrete topology.Let
X = A∪B is a g*-separation of X.Define f:X→ Y such that f(A)=0 andf(B)=1. Obviouslyf
is onto, g*- continuous map.Conversely, let f:X→ Y be g*- continuous, onto map thenX =
f-1(0) ∪f-1(1) is a g*-separation of X.
Theorem 4.11:Let Aα be an arbitrary family if g*- open, g*-connected subset of X with
the common point p then Aα isg*-connected.
Proof: Let A= B∪C be a g*-separation of A.Then B and C are disjoint non empty
g*-open sets in X.p A p B or p  C. Assume that p  B.Thenby theorem
(4.8),A is completely contained in B for all (since p  B). Therefore C is empty,which
Corollary 4.12:Let {An} be a of g*-open, g*-connected subsets of X such that
AnAn+1, for all n. Then An is g*-connected.
R S. Publication (rspublication.com), [email protected]
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
Proof : This can be proved sequence by induction on n. Using above theorem, the result
is true for n= 2. Assume that the result to be true when n = k.Now to prove the result
k
k
i 1
i 1
when n = k + 1. By the hypothesis  Ai is g*-connected, g*-open.Now ( Ai )  Ak 1   .
k 1
Thereforeby the above theorem  Ai is g*-connected. By induction hypothesis the result
i 1
is true for all n.
Corollary 4.13:Let {A}A be an arbitrary collection of g*-open g*-connected subsets
of X. Let A be a g*-open g*-connected subset of X. If AA for all  thenA  (A)
is g*-connected.
Proof:Suppose that A  (A) = B  C be a g*-separation of the subset A
(A).Since A  BC.By theorem (4.8), A  B or A  C.Without loss of generality
assume that A  B. Let  A be arbitrary. Then A B C,by theorem (4.8), A B or
A C. But AAA B. Since  is arbitrary, AB for all . Therefore A 
(A)  B which implies C =  which is a contradiction. Therefore A  (A) is g*connected.
*
*
Definition 4.14:A space(X,τ) is said to be Totally g -disconnected if its only g -
connected subsets are one point sets.
Example 4.15:Let(X,τ) be an indiscrete topological space with more than one point.
Here all subsets are g*-open.If A = { x1, x2,} then A = {x1}{x2} is a g*- separation of
A. Therefore any subset with more than one point is g*-disconnected. Hence (X,τ) is
totally g*- disconnected.
Example 4.16:An infinite set with finite complement topology is not totally g*disconnected, because it is impossible to find disjoint g*-open sets.
Remark 4.17:Totally g*-disconnectednessimplies g*-disconnectedness.
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
Definition 4.18:A point x  X is said to be g*-boundary of A (g* Bd(A)) if everyg-*open
set containing x intersects both A and X-A.
Theorem 4.19:Let (X,τ) be a topological space and let A be a subset of X. If Cisg*-open,
g*-connected subset of X that intersects both A and X-A then C intersectsg* Bd(A).
Proof:It is given that C∩A   and C∩Ac   .Now C=(C∩A)(C∩Ac) is non-empty
disjoint union.Suppose both are g*-open then it is a contradiction to the fact that C isg*connected.Hence either C∩A or C∩ACis not g*-open.Suppose that C∩A is not g*open.Then there exists x  C∩Awhich is not a g*-interior point of C∩A.Let U be ag*open set containing x.Then C∩U is a g*-open set containing x and hence (C∩U )∩(C∩A
c
)   .This implies U intersects both A and Ac and x  g* Bd(A).HenceC∩g*Bd(A)  
.
Next we prove the intermediate value theorem of calculus suitably generalized.
Theorem 4.20:(Generalisation of Intermediate Value theorem )Let f :X →R be a g*continuous map where X is a g*- connected space and R with usualtopology. If x, yare
two points of X and b= f(y) then every real number r between a and b is attained at a
point in X.
Proof:Assume the hypothesis of the theorem. Suppose there is no point c  X such that
f(c)= r.Then A=(-∞, r) and B= (r,∞) are disjoint open sets in R and X= f-1(A)f-1(B)
which is a g*- separation of X.This is a contradiction to the fact that X is g*connected.Therefore there existsc  X such that f(c)=r.Hence proved.
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International Journal of Computer Application
Available online on http://www.rspublication.com/ijca/ijca_index.htm
Issue 4, Volume 3 (May-June 2014)
ISSN: 2250-1797
References
N.Levine, Rend. Cire. Math. Palermo, 19(1970), 89 – 96.
James R.Munkres, Topology, Ed-2.,PHI Learning Pvt.Ltd.New Delhi,2010.
M.K.R.S.Veerakumar, Between closed sets and g-closed sets, Mem. Fac. Sci. Kochi
Univ.(Math), 21(2000), 1-19.
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