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International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 S 1 -compact and S -closed spaces Alias B. Khalaf* and Nehmat K. Ahmed** Abstract - The objective of this paper is to obtain the properties of s -compact and s –closed spaces by using nets, filterbase and s -complete accumulation points. Index Terms— s -open sets, semi open sets, s -compact spaces, s –closed spaces, s -complete accumulation points. —————————— —————————— of X is said to be semi-open [14] (resp., pre-open [15], - 1 INTRODUCTION AND PRELIMINARIES open [16], -open [1] regular open [19] and regular -open I [22]) set if A cl int A , (resp., A int clA , A int cl int A , t is well- Known that the effects of the investigation of properties of closed bounded intervals of real numbers, spaces of continuous functions and solution to differential equations are the possible motivations for the formation of the A cl int clA , A int clA A int clA ). and complement of S -open (resp., semi-open, pre-open, The -open notion of, compactness. Compactness is now one of the most , -open, regular open, regular -open) set is said to be S - important, useful, and fundamental notions of not only closed (resp., semi-closed, pre- closed, general topology but also of other advanced branches of regular closed, regular -closed ). The intersection of all S - mathematics. Recently Khalaf A.B. et al. [13] introduced and closed (resp., semi-closed , pre- closed, -closed) sets of X investigated the concepts of S -space and S -continuity. A semi open subset A of a topological space ( X , ) is said to be S -open if for each that x F A x X there exists a -closed set F such . The aim of this paper to give some characterizations of S -compact spaces in terms of nets and filter-bases. We also introduce the notion of S -complete accumulation points by which we give some characterizations of S -compact spaces. Throughout the present paper, ( X , ) and (Y , ) or simply X and Y denote topological space . In [14] containing a subset A is called the -closed , -closed, S -closure (resp., semi- closure, pre-closure -closure) of A and denoted by S clA (resp., sclA, pclA , clA). The union of all S -open (resp., semi-open ,pre-open, -open) set of X contained in A is called the S -interior (resp., semi-interior, pre-interior, interior) of A and denoted by S intA (resp., sintA , pintA , intA). The family of all S -open (resp., semi-open, pre- open, -open -open, regular -open, regular open, S - Levine initiated semi open sets and their properties. closed, semi-closed , pre- closed, -closed , -closed , Mathematicians gives in several papers interesting and regular -closed , and regular closed) subset of a topological different new types of sets. In [1] Abd-El-Moonsef defined the space X is denoted by class of -open set. In [18] Shareef introduced a new class of O(X), O(X), R O(X), RO(X), semi-open sets called S P - open sets. We recall the following C(X), definitions and characterizations. The closure (resp., interior) called -open [21] if for each x A , there exists an open set B of a subset A of X is denoted by clA (resp., intA). A subset A such that x B int clB A . A subset A of a space X is called S O(X) (resp., SO(X), PO(X), S C(X), SC(X), PC(X), C(X) , R C(X) and RC(X)). A subset A of X is -semi-open [12] (resp., semi- -open [7] if for each x A, there exists a semi-open set B such that x B clB A (resp., ———————————————— *Department of Mathematics, Faculty of Science, University of Duhok, Duhok City, Kurdistan- Region, Iraq. E-mail: [email protected] ** Co-Author is currently pursuing Ph. D. degree program in Mathematics at the Department of Mathematics, College of Education, Salahaddin University. E-mail: [email protected] x B SclB A . Definition 1.1 [15]. A topological space ( X , ) is said to be : 1- Extremely disconnected if clV for every V . 2- Locally indiscrete, if every open subset of X is closed. IJSER © 2012 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 2 3- Hyper-connected if every non-empty open subset of X is F S clV ) for any S -open set V containing x and dense. every F . Theorem 1.2 [14] Let A be any subset of a space X. Then A SC( X ) if and only if clA cl int A . It is clear from the definition above, that S -converges (resp., S -accumulates) of a filter bases in a topological spaces The following results can be found in [13]. implies S - - converges (resp., S - - accumulates) but the Proposition 1.3: If a space X is T1 , then SO( X ) S O( X ) . converses are not true in general as shown in the following Proposition 1.4: If a topological space X is locally indiscrete, examples then every semi-open set is S open set. Example 2.3: Let X { a , b , c , d} and { , X ,{a , b},{ a , b , c}} we get Proposition1.5: A space X is hyper-connected if and only if S O( X) { , X ,{ a , b},{ a , b , c},{ a , b , d}} and let {X ,{c , d},{{b , c , d}} , S O( X ) { X , } . then Corollary 1.6: For any space X, S O( X , ) S O( X , ) converges to a because the set { a , b} S O( X ) contains a, but Proposition 1.7: If a topological space ( X , ) is -regular, there exists no B such that B {a , b} . Also S - - then S O( X ) accumulates to a point b, but Proposition 1.8: 1-Every Sp -open set is S -open . b, because the set { a , b} S O( X ) contains b, but there exist S - -converges to a point a, but does not does not S - S -accumulates to 2- An S -open set is regular -open . {c , d} such that { a , b} {c , d} . 3- A regular closed set is S -open . Theorem 2.4: Let ( X , ) be a topological space and let 4- Every Regular open set is S -closed . filter base on X. Then the following statements are equivalent: 1- Definition 1.9: A subset A of a space X is said to be S -open There exists a filter base finer than {Ux } , where {Ux } is the family of S -open sets of X containing x. [18] (resp., SC [4]) if for each x A SO( X) there exists a pre- 2- closed (resp., closed) set F such that x F A . Definition 1.10: A filter base be a There exists a filter base 1 finer than and S converges to x. in a space X is s-converges Proof: Let 1 be a filter base which is finer than both and [20] (resp., rc-converges [11] ) to a point x if for every semiopen (resp., regular closed) subset U of X containing x there {Ux } . Then S - converges to x since it contains {Ux } . exist B such that B clU (resp., B U ). Conversely, let 1 be the filter base which is finer than Definition 1.11: A filter base in a space X is s-accumulates and which converges to x. Then must contain {Ux } by [20] (resp., rc- accumulates [11] )to a point x if for every semi- definition. open (resp., regular closed) subset U of X containing x there Corollary 2.5: If is a maximal filter base in a topological exist B such that B clU (resp., B U ). space ( X , ) , then S -converges (resp., S - converges) to a point x X if and only if S -accumulates (resp., S - -accumulates) to a point x. 2 S -compact and S -closed spaces Proof: Let be a maximal filter base in X and S - In this section, we introduce new classes of topological accumulates to a point x X , and then by Theorem 2.4, there space called S -compact and S - closed spaces . exists a filter base 1 finer than and S -converges to x. Definition 2.1: A filter base is S -convergent (resp., S - convergent) to a point x X , if for any S -open set V Theorem 2.6: Let be a filter base in a topological space containing x, there exists F such that F V (resp., ( X , ) . If S - converges (resp., S - accumulates) to a point F S clV ) . x X , then rc-converges (resp., rc-accumulates) at a point Definition 2.2: A filter base accumulates) to a point But is maximal filter base. Thus it is S -convergent to x. is S -accumulates (resp., x X , if F V (resp., S - x X . Proof: Suppose that be a filter base S - converges to a IJSER © 2012 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 x X . Let V be any regular closed set containing x, by Theorem 2.1.15, then V is S -open set containing x. since point 3 F V (resp., F S clV ). Hence S -converges (resp., S - converges) to a point x X . shows that rc-converges (resp., rc-accumulates ) to a point be a filter base in a topological space ( X , ) and E is any -closed set containing x. If there exists F such that F E (resp., F SclE ), then xX . S - accumulates (resp., S - converges (resp., S - accumulates) to a point Theorem 2.11 : Let x X , There exists F such that F V (resp., F V ). This The following example Show that the converse of Theorem 2.6 is not true in general. xX . Proof: Similar to proof of Theorem 2.10. Example 2.7: Let X { a , b , c , d} and { , X ,{a , c}} , then the family RC ( X ) { , X } and S O( X) { , X ,{ a , c},{ a , b , c},{ a , c , d}} and = {{b},{a , b, c}, X } . Then S -accumulates) at a point is rc-converges (resp., rc- Definition 2.12: A topological space ( X , ) is said to be compact (resp., S -closed) if for every cover {V : } of X, S -open sets, there exists a finite subset 0 of such accumulates) to a point c X , but not S - converges (resp., by S - accumulates) to a point c X . that X {V : 0 } (resp., X {S clV : 0 } ) Theorem 2.8: Let It is clear that S -compact is S -closed, but not conversely be a filter base in a topological space ( X , ) . If S - converges (resp., S - accumulates ) to a point as shown in the following examples x X , then s- converges (resp., s- accumulates) at a point Example 2.13: Consider an countable space X with Co- xX . countable topology. Since X is Proof: Suppose that is S - S - converges to a point x X . Let T1 , then by Proposition 1.3 the family of open, semi-open set and S - open sets are identical. V be any semi-open set containing x, then by Theorem 1.2 Hence X is an S - closed because every open set in X is dense clV cl int V , so clV is regular closed , by Theorem , not S - compact [19] p-194. Proposion 1.8 clV is S -open set containing x. Since converges (resp., S - accumulates ) to a point is S - x X , then Theorem 2.14: If every -closed cover of a space has a finite subcover, then X is S -compact. there exists B such that B clV (resp., B clV ). This Proof: Let {V : } be any S - open cover of X, and implies that is s- converges (resp., s- accumulates) at a x X , then for each x V ( x ) and , there exists a point x X . The converse of Theorem 2.8 is not true in general as shown closed sets x X such that x F ( x) V ( x ) , so the family { F ( x ) : } is a -closed cover of X, then by hypothesis, this in the following example: Example 2.9: In Example 2.7 the family family has a finite subcover such that SO( X ) { , X ,{a , b},{ a , b , c},{a , b, d}} S O( X ) , considering the X { F ( xi ) ; i 1,2,...n} {V ( xi ) ; i 1,2,...n} . Therefore family = {{c},{ a, b , c}, X} . Then is filter base s-converges X {V ( xi ) : i 1,2,...n} . Hence X is S -compact. The following theorem shows the relation between S - (resp., s-accumulates) to a point b X , but not S - converges (resp., S - accumulates) to a point b X . compact and some other compactness Theorem 2.10: Let be a filter base in a topological space Theorem 2.15: Every semi-compact space, is S -compact ( X , ) and E is any -closed set containing x. If there exists spaces. F such that F E (resp., F S clE ). Then S - Proof: Let {V : } be any converges (resp., S - converges) to a point x X . {V : } is a semi--open cover of X. Since X is semi- Proof: Let V be any S -open set containing x. then V is semi- compact, there exists a finite subset 0 of such that open and for each x V , there exist a -closed set E such X {V : 0 } . Hence X is that x E V . By hypothesis, there exists F such that F E V (resp., F S clE S clV ). Which implies that - S - open cover of X. Then S -compact. The following example shows that the converse of Theorem 2.15 is not true in general. IJSER © 2012 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 Example 2.16: Let X= R with the topology Then ( X , ) { X , ,{0}} . 4 regular T1 space there exist Gx SO( X ) and x Gx SclGx V ( x ) . Then by Proposition 1.3, the family is not semi-compact, since the space X is hyperconnected, then by Proposion 1.5 {Gx : x X} is an S -open cover of X. Since X is S -closed S O (X ) = { , X }. Then ( X , ) is S -compact. space, then there exists a subfamily {G xi Theorem 2.17: If a topological space ( X , ) is T1 and S - T1 X is T1 and S -compact space. Let i 1 regularity in Theorem 2.22 can not be dropped Example 2.23: In Example 2.20 ( X , ) is neither s-regular nor X , there exist (x) such that x V ( x ) . Since by proposition 1.3, the family {V : } is i1 The following example shows that the condition of s- {V : } be any semi-open cover of X. Then for every x n that X S clGxi V ( xi ) . Thus X is compact. compact space , then it is semi-compact. Proof: Suppose that X is n : i 1,2,...n} such S -open compact but it is S -closed because the only non-empty S - open set in ( X , ) is X itself. cover of X. Since X is S -compact, there exists a finite subset Theorem 2.24: Let X be an almost- regular space if X is S - 0 compact, then it is nearly compact. of in X such that X {V : 0 } . Hence X is semi- compact. Proof: Let {V : } be any regular open cover of X, Since X Theorem 2.18: If a topological space ( X , ) is locally is almost- regular space then for each indiscrete. Then S -compact space X is semi-compact . x X and each regular open set V ( x ) , there exist an open set G x such that Proof: Follows from Theorem 1.4. x G x clG x V ( x ) . But clG x is regular closed for each In general S -compact spaces and compact spaces are not x X . Therefore X comparable as shown in the following examples. clGx xX : x X } is an V ( x ) This implies that the ( x ) Example 2.19: The one-point compactification of any discrete family {clG x spaces is not S-closed [10] corollary 4, therefore the space is S -compact, then there exists a subfamily not S -compact, but it is compact. {clG xi : i 1,2,...n} such that Example 2.20: Let X=(0,1) with the topology { X , ,G (0,1 n1 ), n 2,3,...} then ( X , ) is not compact [19], p. 76, but it is S -compact since the only S -open subset of . Theorem 2.21: Let ( X , ) be a topological space. If X is - of such that i 1 i 1 is nearly compact. Theorem 2.25: Let X be semi- regular s-closed space, then it is V is semi-open for each , since X is semi-regular for each x X and V ( x ) , there exists a semi-open set G x such that Proof: Let {V : } be any open cover of X. Since X is - 0 n Proof: Let {V : } be any S - open cover of X, Then regular and S -compact , then X is compact. cover of X. Since X is n X clGxi V ( xi ) . Thus X S -compact. ( X , ) are X and regular By Proposition 1.7, {V : } forms an S -open cover of X. Since X is S -open x Gx SclGx V ( x ) . Then the family {Gx : x X} is semi-open S -compact, there exists a finite subset cover of X. Since X is s-closed space, then there exists a subfamily {G xi S int F 0 S cl( X \ F 0 ) , hence X is compact. : i 1,2,...n} such that n n i1 i 1 X SclGxi V ( xi ) . Thus X is S -compact. Theorem 2.22: If X is an s- regular S -closed T1 space, then Theorem 2.26: For any topological space ( X , ) . The following X is compact. statements are equivalent: Proof: Let {V : } be any open cover of an s-regular and 1- ( X , ) is S -compact spaces (resp., S -closed) S -closed T1 space X, then for each and for each 2- For any S -open cover {V : } of X, there exists a finite x X there exists ( x) such that x V ( x ) . Since X is s- subset 0 of such that X {V : 0 } (resp., IJSER © 2012 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 5 X {S clV : 0 } ) a filter base on X. By hypothesis, S - accumulates (resp., 3- Every maximal filter base in X S - converges (resp. S - S - -accumulates) to some point x X . This implies that for - converges ) to some point x X . every S -open set V containing x, F V (resp., 4- Every filter base in X S - accumulates (resp., S - - F S clV ), for every F and every .Since accumulates) to some point x X . x F ,there exist 0 such that x F 0 . Hence X \ F 0 is 5- For every family { F : } of S -closed subsets of X such S -open set containing x and F 0 X \ F 0 (resp., that { F : } there exists finite subset 0 of such S int F 0 S cl( X \ F 0 ) ). Which contracting the fact that that { F : 0 } (resp., {S int F : 0 } ) S - accumulates to x (resp., S - -accumulates). So the proof: 1 2. Straightforward. 2 3 Suppose that for every S -open cover {V : } of X , there exist a finite subset 0 of such that 5 1. Let {V : } be S -open cover of X. Then { X \ V : } is a family of S -closed subsets of X such X {V : 0 } (resp., X {S clV : 0 } and let that { X \ V : } . By hypothesis, there exists a finite { F : } be a maximal filter base. Suppose that does not S -converges (resp., S - - converges ) to any point of X. Since is maximal, by Corollary 5.1.4 assertion in (5) is true. dose not S - subset 0 of such that { X \ V : 0 } (resp., {S int( X \ V ) : 0 } . Hence X {V : 0 } (resp., X {S clV : 0 } . This shows that X is S -compact (resp., accumulates (resp., S - -accumulates) to any point of X. S -closed) . This implies that for every x X there exists S -open set Theorem 2.27: If a topological space ( X , ) is S -closed and Vx and F ( x ) such that F ( x ) Vx (resp., T1 -space then it is nearly compact. F ( x ) S clV x ). The family {Vx : x X} is an S -pen cover of Proof: Let {V : } be any regular open cover of X. Then X and by hypothesis, there exists a finite number of points {V : } is a S -open cover of X. Since X is S -closed , there x1 , x 2 ,..., x n exists a finite subset 0 of such that X {V : 0 } . of X such that X {V( xi ) : i 1,2,...n} (resp., X {S clV( xi ) : i 1,2,...n} ). Since is a filter base on X, there Hence X is nearly compact. exists a F0 such that F0 { F ( xi ) : i 1,2,..., n} . Hence F0 V ( xi ) (resp., F0 S clV ( xi ) ) for i 1,2,...n which 3 Characterization of S compact spaces implies that F0 ({V( xi ) : i 1,2,...n}) (resp., Definition 3.1: A point x in X is said to be S -complete F0 ({S clV( xi ) : i 1,2,...n) F0 X Therefore, we accumulation point of a subset A of X if obtain F0 . Which is contradict the fact that Card ( A U ) Card(A) for each U S ( X , x) . Where Card(A) , thus denotes the cardinality of A . Definition3.2: In a topological space X, a point x is said to be is S - converges to some point x X 3 4. Let be any filter base on X. Then, there exists a maximal filter base 0 such that 0 . By hypothesis 0 is S -converges (resp., S - - converges) to some point x X . For every F and every S -open set V containing x, there exists F0 0 such that F0 V (resp., F0 S clV ), hence an S -adherent point of a filter on X if it lies in the S closure of all sets of . Theorem3.3 : A space X is S -compact spaces if and only if each infinite subset of X has S -complete accumulation point. Proof: Let the space X be S -compact and S an infinite subset F0 F V F (resp., S clV F . This shows that S - of X. Let K be the set of points x in X which are not S - accumulates at x (resp., S - -accumulates). complete accumulation points of S. Now it is obvious that for 4 5. Let { F : } be a family of S -closed subsets of X each point x in K , we are able to find U( x) S O( X , x) such such that { F : } . If possible suppose that every finite that Card(S U( x ) ) Card(S) . If K is the Whole space , then subfamily { F i : i 1,2,..., n} . Therefore A Y X form IJSER © 2012 http://www.ijser.org International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 6 {U( x ) : x X } is S -cover of X. By hypothesis X is S - (x) such that compact, so there exists a finite subcover {U( xi ) ; i 1,2,..., n} { x : ( x)} is a subset of X V( x) . Then the collection such that S {U( xi ) S ; i 1,2,..., n} , then {V( x ) : x X } is S - cover of X. By hypothesis of theorem, X Card(S) max{Card(U( xi ) S); i 1,2,...n} , which does not agree is S -compact and so with what we assumed. This implies that S has an S - {V( xi ) : i 1,2,...n} such that X {V( xi ) : i 1,2,...n} .Suppose that complete accumulation. Now assume that X is not S - the corresponding elements of V( x ) { x : ( x )} . This implies that has a finite subfamily compact and that every infinite subset S of X has an S - be { ( xi )} where i=1,2,…,n, since is well-ordered and { ( xi)} where i=1,2,…,n is complete accumulation point in X. it follows that there exists finite. The largest elements of an cover with no finite subcover. Set { ( xi)} exists. Suppose it is { ( xi )} . Then for { ( xi )} . We have min{Card( ): , wher is anS cov er of X} . Fix , for which Card( ) and {U : U } X . Let N denote the { x : } ( X V( xi ) ) X V( xi ) . Which is impossible. i 1 i 1 set of natural numbers, then by hypothesis Card( N ) [ By This shows that well-ordering of ] . By some minimal well-ordering “~” , (ii) suppose that U is any member of . By minimal well- has an S -complete accumulation point by utilizing above ordering “~”, we have theorem. Suppose that S X is an infinite subset of X. Card({V : V , V ~ U }) Card({V : V }) . Since can not According to Zorns Lemma, the infinite set S can be well- have any subcover with cardinality less that , then for each ordered. This means that we can assume S to be a net with a U we have X {V : V , V ~ U ] . For each U choose domain which is a well ordered index set. It follows that S has a point x(U ) X {V { x( V )} : V , V ~ U } . We are always S - adherent point z. Therefore is an S -complete able to do this, if not, one can choose a cover of smaller accumulation point of S this shows that X is S -compact. cardinality from . If H { x(U ):U } , then to finish the Theorem3.5 : A space X is S -compact if and only if each proof we will show that H has no S -complete accumulation family of S -closed subsets of X with the finite intersection point in X. Suppose that z is a point of the space X. Since S -cover of X, then z is a point of some set W in fact that U~V we have is . By the has at least one S -adherent point in X . (i): Now it is enough to prove that each infinite subset property has a non-empty intersection . Proof: Given a collection of subsets of X. let {X w : w } be the collection of their complements . x(U ) W . It follows that Then the following statements hold. T {U : U and x(U ) W } {V : V , V ~ W } . But i- Card(T ) . Therefore Card( H W ) . But is the collection of S -open sets if and only if is a Card( H ) Card( N ) , since for two distinct points U and W in collection of S -closed sets. , we have ii- the collection covers of X if and only if the intersection x(U ) x(W ) , this means that H has no S - of all the elements of complete accumulation point in X which contradicts our is non empty v assumptions. Therefore X is S -compact. iii- The finite sub collection { wn ,...wn } of Theorem3.4: For a topological space the following are only if the intersection of the corresponding elements equivalent: vi X wi of i- X is S -compact. The statement (i) is trivial . While the statement (ii) and (iii) ii- Every net in X with well-ordered directed set as its domain follows from De-Morgan Law X v = ( X v ) . The j proof of theorem now proceeds in two steps. Taking the Proof:(i) (ii): Suppose that X is S -compact and Assume that is empty. j accumulates to some point of X.. { x : } a net with a well-ordered set covers X if and contra positive of the theorem and the complement . as domain. . The statement X is S -compact is equivalent to: Given any has no S - adherent point in X. Then for each collection point x in X there exists V( x ) S O( X , x) and an IJSER © 2012 http://www.ijser.org of S -open subsets of X, if covers X, then International Journal of Scientific & Engineering Research Volume 3, Issue 4, April-2012 ISSN 2229-5518 some finite sub collection of covers X. This statement is is a filter base on X with no S -adherent point. By hypothesis equivalent to its contra positive, Which is the following. Given any collection 7 of S -open sets, if no finite sub covers X, then does not cover X . Letting be as earlier, the collection { X w : w } , and applying collection of S -converges to some point z in X. Suppose arbitrary element of exists an . Then for each F such that F is an V S O( X , z) , there F V . Since is a filter base such that F F F F V where F is a (i) to (iii), we see that this statement is in turn equivalent to there exists a the following. nonempty. This means that F V is nonempty for every Given any collection of S -closed sets , if every finite intersection of elements of V S O( X , z) and correspondingly for each is non empty. This is just the , z is a point of S clF . It follows that z S clF Therefore, z is condition of our theorem. Theorem3.6: A space X is S -compact if and only if each filter S adherent point of . Which is contradiction . This base in X has at least one S -adherent point. shows that X is S compact. 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