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Journal of Al-Nahrain University Science Vol.17 (1), March, 2014, pp.160-166 Minimal and Maximal Beta Open Sets Qays Rashid Shakir Operations Management Techniques Department-Technical College of Management Baghdad, Baghdad-Iraq. E-mail: [email protected]. Abstract The present paper deals and discusses new types of sets all of these concepts completely depended on the concept of Beta open set. The importance concepts which introduced in this paper are minimal -open and maximal -open sets. Besides, new types of topological spaces introduced which called Tminand Tmaxspaces. Also we present new two maps of continuity which called minimal -continuous and maximal -continuous. Additionally we investigated some fundamental properties of the concepts which presented in this paper. Keywords: minimal -open set, maximal -continuous -open set, minimal -continuous and maximal contained in A is called the interior of A and denoted by AO and the intersection of all closed subset of X which contain A is called the closure of A and denoted by A. Introduction Minimal and maximal sets play an important role in the researches of generalized topological spaces, Nakaoka and Oda introduced these concepts in [1] and [2] and they used them to investigate many topological properties. In this paper we introduced the notion of minimal -open and maximal -open and their complements. Definition (6) [5]: A subset A of a space X is called a o -open set if A A .The complement of a -open set is defined to be a -closed set. Definition (7) [5]: Let X and Y be topological spaces and f:XY is a map then f is called a Definition (1) [1]: A proper nonempty open subset O of a topological space X is said to be minimal open set if any open set which is contained in O is or O. -continuous function if f1(A) is a -open set in X for every open set A in Y. Minimal and Maximal Definition (2) [2]: A proper nonempty open subset O of a topological space X is said to be maximal open set if any open set which is contains O is O or X. -open sets Definition (8): A proper -open subset B of a topological space X is said to be a minimal -open set if any -open set which is contained in B is or B. Definition (3) [3]: A proper nonempty closed subset O of a topological space X is said to be minimal closed set if any closed set which is contained in O is or O. Definition (9): A proper nonempty -open subset B of a topological space X is said to be a maximal -open set if any -open set which contains B is X or B. Definition (4) [3]: A proper nonempty closed subset O of a topological space X is said to be maximal closed set if any closed set which is contains O is O or X. Definition (10): A proper nonempty -closed subset F of a topological space X is said to be a minimal -closed set if any -closed set which is contained in F is or F. Definition (5) [4]: Let A be a subset of a topological space X, then the union of all open subset of X which 160 Qays Rashid Shakir XDF and this contradict being F is minimal -closed. let F be a -closed subset of X, suppose that there is a -closed K such that KF thus XFXKbut X-K is proper -open set. Contradiction to the assumption of being X-F is maximal -open.■ Definition (11): A proper nonempty -closed subset F of a topological space X is said to be a maximal -closed set if any -closed set which contains F is X or F. Remarks (12): (1) The family of all minimal -open (resp. minimal -closed) sets of a topological space X is denoted by MiO(X) (resp. MiC(X) ). (2) The family of all maximal -open (resp. maximal -closed) sets of a topological space X is denoted by MaO(X)(resp. Theorem (16): Let U and V be maximal -open subsets of a Topological space X, then UUV X or U=V. Proof: If UUV X then the proof is complete. If not, i.e. UUV X so we have to show that U=V. Since UUV X so U UUV and V UUV . But U is maximal -open set, so UUV X or UUV U Thus UUV U and so VU. Now since V UUV and V is maximal f-open set, so UUV X or UUV V ,but UUV X so UUV V and hence UV Therefore U=V.■ MaC(X)). Remark (13): The concept of minimal -open, maximal -open, minimal -closed and maximal f-closed are independent of each other as in the following example. Example (14): let X = {a, b, c} and {a, b}, X} so = { , {a}, O(X){,{a},{a,b},{a,c},X}, MiO(X) {{a}}, MiC(X) {{c},{b}}, MaO(X) {{a,b},{a,c}}, MaC(X) {{b,c}} Theorem (17): Let U be a maximal -open and V be a -open subsets of a Topological space X then UUV X or VU. So the following table show that the new sets are independent each other. Table (1). Proof: If UUV X then the proof is complete. If UUV X so U UUV and V UUV . Since U is maximal -open and U UUV so by definition of Maximal -open we have that UUV X or UUV Ubut UUV X so UUV U and hence VU.■ Theorem (15): Let F be a subset of a topological space X, then F is a minimal -closed if and only if XF is maximal -open set. Theorem (18): Let U be a maximal -open subset of a Topological space X with xX/ Uthen X/ UV for any -open subset of X with xV. Proof: Let xX/ Uand xV, so VU, thus by (17) we have that Proof: let F be a minimal -closed, so X-F is -open. We have to show that X-F is maximal -open suppose not, so there is a -open subset D of X such that XFD hence UUV X X \UI X \V X\ UV. ■ 161 Journal of Al-Nahrain University Science Vol.17 (1), March, 2014, pp.160-166 V I W V I X I W Set Theory V I WI UUV by (2.9 ) V I WI UUWI V Set Theory V I WI UUV I WI VSet Theory UI VUV I W sinceUI V W Theorem (19): Let F be a minimal -closed and K be a -closed subsets of a topological space X then FI K or FK. Proof: If FI K then the proof is complete. If FI K then we have to show that FK. Since FI K then FI K F and FI K K. But F is minimal -closed, so we have FI K F or FI K . Thus FI K F So FK.■ UUWI V Set Theory X I V sinceU UW X Thus VI WV V implies VW but V is maximal -open therefore V=W or VUW X but VUW X so V=W. ■ Theorem (22): U, V and W be maximal -open subsets of a Topological space X which are different from each other, then UI V UI W Proof: Let UI V UI W Theorem (20): Let F and K be minimal -closed subsets of a topological space X then FI K or FK. UI VUWI V UI WUWI V UI WUV UI VUW X UV X UW V W But V is maximal -open and W is a proper Proof: If FI K then the proof is complete. If FI K then we have to show that FK. Since FI K so FI K F or FI K K. Since F is minimal -closed so we have FI K F or FI K . But FI K hence FI K F which means FK. Now since K is minimal -closed so we have FI K K or FI K . But FI K hence FI K K which means KF. Therefore F=K. ■ subset of X so V=U, this result contradicts the fact that U, V and W are different from each other. Hence UVUW ■ Theorem (23): Let F be a minimal -closed subset of a Topological space X , if xFthen FK for any -closed subset K of X containing x. Proof: Suppose xKand FK so FI KF and FI K since xFI K But F is minimal -closed so FI K F. or FI K . hence FI K F which contract the relation FI K F. Therefore FK.■ Theorem (21): Let U, V and W be maximal -open subsets of a Topological space X such that U V, if UI V W, then either U=W or V=W. Proof: Suppose that UI V W, if U=W then the proof is complete. If U Wwe have to show that V=W Theorem: Let F -closed and sets if F () be minimal F UF then there exists A OAsuch that FFO . 162 Qays Rashid Shakir Let X is Tmaxspace. Suppose that X is not Tmin, so there is a proper -open subset Proof: First we have to show that FI FO , FO X\ F and FI FO then F U F X \ F which suppose that so A is K of X which is not minimal, this mean there exist an -open subset of X with HK. Thus we get that H is not maximal which is contradict of being X is Tmax.■ a FI FO . and hence FI FO F and FI FO FO since FI FO F and F is minimal -closed then FI FO F or F I FO thus FI FO F and hence FO F. Now since FI FO FO and FO is minimal -closed then FI FO FO or FI FO . Thus FI FO FO and hence F FO . Therefore F FO .■ contradiction. So Theorem (29): A topological space X is Tmin space if and only if every nonempty proper -closed subset of X is maximal -closed set in X. Proof: let F be a proper -closed subset of X and suppose F is not maximal. So there exists an -closed subset K of X with KX such that FK. Thus XKXF. Hence X-F is a proper -open which is not minimal and this contradicts of being X is Tminspace. Tminand Tmaxspace Definition (24): A topological space X is said to be Tminspace if every nonempty proper Suppose U is a proper -open subset of X. thus X-U is a proper -closed subset of X, so X-U is maximal -closed subset of X. and by (15) U is minimal -open. thus X is Tmin -open subset of X is minimal -open set. Definition (25): A topological space X is said to be Tmax space if every nonempty proper -open subset of X is maximal -open set. space. ■ Theorem (30): A topological space X is Tmax space if and only if every nonempty proper -closed subset of X is minimal -closed set in X. Example (26): Let X={a, b, c} and {,{a,b},{c},X} thus O(X)= , it is clear that {a, b} and {c} are maximal and minimal -open sets thus the space X is both Tminand Tmax. Proof: let F be a proper -closed subset of X, suppose F is not minimal -closed in X, so there is a proper -closed subset of X such that KF Thus XFXKbut X-K is proper -open in X so X-F is not maximal in X. Contradiction to the fact X-F is maximal -open. let U be a proper -open subset of X, then X-U is a proper -closed subset of X and so it is minimal -closed set. By (15) we get that U is maximal -open. ■ Remark (27): Tminand Tmaxspaces are identical. Theorem (28): A space X is Tmin if and only if it is Tmax. Proof: Let X is Tmin space. Suppose that X is not Tmax, so there is a proper -open subset K of X which is not maximal, this mean there exist a -open subset of X with KH . Thus we get that H is not minimal which is contradict of being X is Tmax. Theorem (31): Every pair of different minimal sets of Tminare disjoint. 163 -open Journal of Al-Nahrain University Science Vol.17 (1), March, 2014, pp.160-166 Proof: Let U and V be minimal f-open subsets of Tminspace X such that UV to show Definition (35): Let X and Y be topological spaces, a map f : XYisn called maximal -continuous if that UI V suppose not i.e. UI V . So UI V U andUI V V . Since UI V Uand U is minimal -open then UI V U or UI V thus UI V U. Now since UI V V and V is minimal then UI V V orUI V -open thus UI V V . Hence we get that U=V this result contradicts the fact that U and V are different. Therefore UI V .■ f 1(U) is maximal -open in X for any open subset U of Y. Example (36): Let X = Y = {a, b, c} and f :(X,) (Y,)is the identity map, where ={ , {a}, {b},{a, b}, X} and ={ , {a, c}, Y} then f is maximal -continuous since the only proper open subset of Y is {a, c} and f 1({a, c}){a, c}is maximal X. Theorem (32): Union of every pair of different maximal -open sets in Tmaxspace X is X. Theorem (37): Every minimal -continuous. Proof: Let U and V be maximal -open subsets of Tmaxspace X such that UV to that UUV X suppose not i.e. UUV X . So U UUV andV UUV . Since U UUV and U is maximal -open then UUV U orUUV X . Thus UUV U… (1). Now since V UUV and V is maximal -open then UUV V or UI V X Thus UUV V … (2) Hence from (1) and (2) we get that U=V this result contradicts the fact that U and V are different. Therefore UI V X .■ Remark 38: The converse is not true in general as in the following example. Example (39): Let X = Y = {a, b, c} and is f :(X,) (Y,) the identity map, where ={ , {a}, {c}, {a, c}, X} and ={ , {a, c}, Y} then f is -continuous but f is not minimal -open -continuous since f1({a,c}){a,c} is not minimal -open since {a}O(X) and {a}{a,c} . f 1(U) is minimal -open in X for any open Theorem (40): Let X and Y be topological spaces, if f : XYis an -continuous onto map and X is Tmin space then f is minimal subset U of Y. Y = {a, b, c} -continuous. and f :(X,) (Y,)is the identity map, where ={ , {a}, {a, b}, X} and ={ , {a}, Y} then f is minimal -continuous since the only proper open subset of and f 1({a}){a}is minimal Y is map is f : XY be a minimal -continuous map and U be open subset of Y. then f 1(U) is minimal -open in X and so f 1(U) is -open subset of X.■ Definition (33): Let X and Y be topological spaces, a map f : XYis called minimal -continuous if Example (34): Let X = -continuous in Proof: Let show Continuity with Minimal and Maximal Sets -open Proof: It is clear that the inverse image of and Y are f-open subsets of X. So let U be a proper open subset of Y. Since f is f-continuous so {a} -open in X. 164 Qays Rashid Shakir f 1(U) is proper f-open subset of X, but X is Tminso f 1(U) minimal f-open.■ maximal -continuous since f 1({a}){a} is not maximal -open since {a,c} {a}. Remark (41): The converse is not true in general as in the following example. Remark (49): Minimal -continuous and maximal -continuous maps are independent of each other and the following examples show that. Example (42): In (34) f is minimal f-continuous but X is not Tmin. Example (50): In (36) f is maximal f1({a,c}){a,c}is -open minimal -continuous. Theorem (43): Let X and Y be topological spaces, if f : XYis a -continuous onto map and X Tminspace then f is maximal -continuous. so subset of X but maximal -open in Theorem (52): Let f : XYbe a map and X and Y be topological spaces, then f is maximal (resp. minimal) -continuous if and only if f 1(F) is minimal (resp. maximal) -closed subset of X for each closed subset F of Y. Proof: let F be a closed set in Y. thus Y-F is -open.■ Remark (44): The converse is not true in general as in the following example. Example (45): In (36) f is maximal -continuous but X is not Tminspace. -continuous but f is not since f1({b}){b} is not maximal X. f 1(U) is a proper -open X is Tmin so f 1(U) is Theorem (46): Every maximal -continuous. since Example (51): In (34) f is minimal -continuous but it is not maximal -continuous Proof: It is clear that the inverse image of and Y are -open subsets of X. So let U be a proper open subset of Y. Since f is -continuous -continuous f1(YF) is maximal (resp. 1 1 maximal) -open. but f (YF) Xf (F) 1 so f (F) is minimal (resp. maximal) open and so closed. map is Theorem (53): Let X,Y and Z be topological spaces, if f : XYis a minimal ( respect. maximal) -continuous map and g:YZ is a continuous map then gf : XZ is a minimal (resp. maximal ) -continuous map. Proof: Let f : XYbe a maximal -continuous map and U be open subset of Y. then f 1(U) is maximal -open in X and so f 1(U) is -open subset of X.■ Proof: Let U be an open subset of Z, since g is continuous so g1(U) is an open subset of Y. But f is minimal (respect. maximal) Remark (47): The Converse is not true in general as in the following example. -continuous thus f1(g1(U)) gf 1 is a minimal (respect. maximal) -open subset of Example (48): Let X = X.■ Y = {a, b, c} and f :(X,) (Y,)is the identity map, then where ={ , {a}, {a, c}, X} and ={ , {a}, Y} then f is -continuous but f is not Conclusion In this paper we get some theorems presented to reveal many various properties of 165 Journal of Al-Nahrain University Science Vol.17 (1), March, 2014, pp.160-166 the minimal -open and maximal -open and their complements and we defined two types of topological spaces and finally we defined continuity over the new sets which produced here. References [1] Nakaoka F., and Oda N., “Some applications of minimal open sets”, Int. J. Math. Math. Sci. 27-8, 471-476, 2001. [2] Nakaoka F., and Oda N., “ Some properties of maximal open sets”, Int. J.Math. Math. Sci. 21, 1331-1340, 2003. [3] Nakaoka F., and Oda N., “On minimal closed sets”, Proceeding of Topological Spaces and its Applications, 5, 19-21, 2003. [4] Munkers J.R., “Topology”, Pearson Education (India), 2001. [5] Abd El-Monsef M. E., El-Deeb S. N., Mahmoud R. A., “ –open sets and -continuous mappings”, Bulletin of the Faculty of Science, Assiut University, 12, pp.77-90, 1983. الخالصة يناقش البحث الحالي أنواعا جديدة من المجموعات وكل هذه المفاهيم تعتمد على مفهوم.ويتعامل معها المجموعة المفتوحة من النمط بيتا وأهم هذه المفاهيم التي قدمت في هذا البحث هي المجموعة المفتوحة بيتا االصغرية كما تم تقديم نوعين.و المجموعة المفتوحة بيتا االكبرية جديدين من الفضاءات التبولوجيا سميت الفضاء من النمط كذلك قمنا بتعريف. Tmax و الفضاء من النمط Tmin نمطين جديدين من الدوال المستمرة هما الدالة المستمرة االصغرية من النمط بيتا و الدالة المستمرة االكبرية من النمط باالضافة الى ذلك تم دراسة بعض الخواص االساسية،بيتا .للمفاهيم المطروحة في هذا البحث 166

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