Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document 8937022
Document related concepts
Transcript
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[1] introduced a new class of g-closed sets in a topological space in 1970 andM.K.R.S.Veerakumar[3] 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)[1] if cl(A) U whenever A U and U is open in (X, τ). 2) generalized star closed set (briefly g*-closed)[3] if cl(A) U whenever A U and U is g-openin R S. Publication (rspublication.com), [email protected] Page 124 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[3]: 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[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] Page 125 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. Remark 3.4:1) g*- additive g*-countably additive. 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 xg*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 contradiction. 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] Page 128 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 is contradiction. Corollary 4.12:Let {An} be a of g*-open, g*-connected subsets of X such that AnAn+1, for all n. Then An is g*-connected. R S. Publication (rspublication.com), [email protected] Page 129 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 AA 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 BC.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 AAA B. Since is arbitrary, AB 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. R S. Publication (rspublication.com), [email protected] Page 130 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. R S. Publication (rspublication.com), [email protected] Page 131 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 [1]N.Levine, Rend. Cire. Math. Palermo, 19(1970), 89 – 96. [2]James R.Munkres, Topology, Ed-2.,PHI Learning Pvt.Ltd.New Delhi,2010. [3]M.K.R.S.Veerakumar, Between closed sets and g-closed sets, Mem. Fac. Sci. Kochi Univ.(Math), 21(2000), 1-19. R S. Publication (rspublication.com), [email protected] Page 132
