Microsoft Word - 267-276 267 | Mathematics 2016) عام 1العدد ( 29مجلة إبن الهيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Weak N-open sets Amer I. l Al-Saeed Dept. of Mathematics/College of Science /Al-Mustansiriya University Received in:30/April/2015,Accepted in:6/December/2016 Abstract In this paper we introduce new class of open sets called weak N-open sets and we study the relation between N-open sets , weak N-open sets and some other open sets. We prove several results about them. Keywords: N-open, semi-open,  -open, b-open, β-open,  -closed. 268 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Introduction Let ),( X be a topological space (or simply, a space) and XA  . Then the closure of A and the interior of A will be denoted by )( Acl and )int( A , respectively. A subset XA  is called semi-open [1] if there exists an open set O such that )(OclAO  . Clearly A is semi-open if and only if ))(int( AclA  . The complement of a semi-open set is called semi-closed [1]. A is called preopen [2] (respt.  -open [3],b-open[4] ,β-open[5]) if ))(int( AclA  (respt. )))(int(int( AclA  , ))(int())(int( AclAclA  , )))(int(clAclA  ). A space ),( X is called anti-locally countable [8] if every non-empty open subset is uncountable. In [6], the concept of  -closed subsets was explored where a subset A of a space ),( X is  -closed if it contains all of its condensation points (a point x is a condensation point of A if UA  is uncountable for every open U containing x). The complement of an  -closed set is called  - open or equivalently for each Ax  , there exists U such that Ux  and AU  is countable.  -closure and  -interior, that can be defined in an analogous manner to )( Acl and )int( A , will be denoted by )( Acl and )(int A , respectively. For several characterizations of  -closed subsets, see [7,8]. Weak N-open sets Definition[8,9] 2.1: A subset A of a topological space (X, τ) is called: 1) pre- -open if ))((int AclA  . 2)  - -open if )))((int(int AclA  . 3) b- -open if ))((int))((int AclAclA   . 4)  - -open if )))(((int Aclcl  . 5) N-open subset of X if for any x in A there exists an open set U containing x such that U\A is finite. Remark[9] 2.2: Every open set is N-open and every N-open set is ω-open. The converse of above remark may be not true in general as seen in the following examples. Examples 2.3: (1) Let N be the set of natural numbers with topology defined on it by  = {Ui: Ui= {i, i+1, i+2,…}, iN} }{ , then U5  {3} is N-open since for any aU5  {3} there exists U2 containing a and U2\ (U5  {3})={2,4} is finite, but U5  {3} is not open. (2) Let (R, u ) be usual topological space and Let Q be rational numbers then R\Q is ω-open Since for any x R\Q there is open set (x- ,x+ ) containing x and (x- ,x+ )\(R\Q) is countable, but not N-open, since every open set A containing x implies A\(R\Q) is infinite. The family of all N-open sets in a topological space (X, ) will be denoted by NO(X), and it is clear form a topology N on X which is finer than . Remark 2.4: For any finite topological space N = P(X) (that is any subset of X is N-open). Definition [9] 2.5: Let (X,τ) be a topological space. And A  X, then the union of all N-open contained in A is called N-interior and we will denote it by Nint (A). Note that the N-interior is the maximal N-open which contains in A. Proposition 2.6: Let (X,τ) be a topological space. And A,B  X, then 269 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 (1) int(A)  Nint (A)  int (A)  A (2) A is N-open iff Nint (A) =A. (3) Nint (X)=X and Nint ( )= (4) Nint ( Nint (A))= Nint (A). (5) If A  B, then Nint (A)  Nint (B). (6) Nint (A)  Nint (B)  Nint (A  B). (7) Nint (A  B)= Nint (A)  Nint (B). Proof As in the usual case By the following example we note that in general A  Nint (A) and Nint (A  B)  Nint (A)  Nint (B) Example 2.7: Let (R, u ) be the usual topological space and Let A=(2,3] and B=(3,4], then Nint (A)=(2,3) , Nint (B)=(3,4), hence A  Nint (A) and Nint (A  B)=(2,4), and Nint (A)  Nint (B)=(2,4)\{3}, therefore Nint (A  B)  Nint (A)  Nint (B). Definition[9] 2.8: Let (X,τ) be a topological space. And A  X, then the intersection of all N- closed sets containing A is called N-closure of A and we will denote it by Ncl (A). Note that the N-closure is the smallest N-closed containing A. Proposition 2.9: Let (X,τ) be a topological space. And A,B  X, then (1) A  cl (A)  Ncl (A)  cl (A). (2) A is N-closed iff Ncl (A)=A. (3) Ncl (X)=X and Ncl ( )= . (4) Ncl ( Ncl (A)) = Ncl (A). (5) If A  B, then Ncl (A)  Ncl (B). (6) Ncl (A  B) = Ncl (A)  Ncl (B). (7) Ncl (A  B)  Ncl (A)  Ncl (B). (8) Nint (X\A) = X\ Ncl (A). (9) Ncl (X\A) =X\ Nint (A). Proof As in the usual case Proposition [9] 2.10: Let (X,τ) be a topological space. And A  X, then x )( AclN iff for every N-open set U containing x, then UA  . The following are new modified definition. 270 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Definition 2.11: Let (X,τ) be a topological space. A subset A of X is called (1)  - N-open if A  Nint ( cl ( Nint (A))). (2) pre-N-open if A  Nint ( cl ((A)). (3) b- N-open if A  Nint ( cl (A))  cl ( Nint (A)). (4)  - N-open if A  cl ( Nint ( cl (A))). The complement of  - N-open (resp. pre-N-open, b- N-open,  - N-open) set is called  - N-closed (resp. pre-N- closed, b- N- closed,  - N- closed) Proposition 2.12: In any topological space the following are satisfied (1) Every N-open is  - N-open. (2) Every  -N-open is pre-N-open. (3) Every pre-N-open is b-N-open. (4) Every b-N-open is  -N-open. Proof (1) suppose that A is N-open, then by Proposition (2.9(2)) A= Nint (A)  cl ( Nint (A)) and by Proposition(2.6(4,5)) A  Nint ( cl ( Nint (A))), hence A is  - N-open. (2) Suppose that A is  -N-open, then A  Nint ( cl ( Nint (A)))  Nint ( cl (A)) (by Proposition(2.6(1))), hence A is pre-N-open. (3) Suppose that A is pre-N-open, then A  Nint ( cl (A))  Nint ( cl (A))  ( cl ( Nint (A)) therefore A is b-N-open. (4) Suppose that A is b-N-open, then A  Nint ( cl (A))  ( cl ( Nint (A)) , if A  Nint ( cl (A)), then A  cl ( Nint ( cl (A))), and if A  Nint ( cl (A)), then A  cl ( Nint (A))  cl ( Nint ( cl (A)))( by Proposition(2.9(1))), that is A is  -N-open. The converse of proposition (2.12) may be not true in general to show that see the following examples Examples 2.13: (1) Let X be infinite set and A,B,C and D be subsets of X such that each of them is infinite and the collection {A,B,C,D}be a partition of X, define the topology on X by  ={ , X,A,B,{A,B},{A,B,C}},where {A,B,C}=A B  C and that similarly for {A,B},{A,B,D}and {B,C,D}. Then {A,B,D} is  -N-open but not N-open and {B,C,D} is b-N-open but not pre-N-open. (2) In example (2.7) let A= Q  [0,1], then A is  -N-open but not b-N-open. (3) In example (2.7) let A= Q, then A is pre-N-open but not  -N-open Proposition 2.14: In any topological space the following are satisfied (1) Every  -open set is  -N-open, and every  -N-open set is  - -open. (2) Every pre-open set is pre-N-open, and every pre-N-open set is pre- -open. (3) Every b-open set is b-N-open, and every b-N-open set is b- -open. (4) Every  -open set is  -N-open, and every  -N-open set is  - -open. Proof (1) suppose that A is -open set, that is A  (intint(cl (A))) and by proposition (2.9(1)) We have A  (intint(cl (A)))  Nint ( cl ( Nint (A ))) ,then A is  -N-open and if A is -N- open, Also by proposition (2.9(1)) we have A  Nint ( cl ( Nint (A )))  ((int(int  cl A))), then A is  - -open. similar proof for the other cases. Remark 2.15: The following examples show that the converse of some points of proposition (2.14) may be not true in general 271 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Examples 2.16: (1) In example (2.3(1)) Let A={1,2,3} ,then A is  - -open (since X=N is countable) but not  -N-open since U4 is open and X\U4=A then A is closed hence cl (A)=A and Nint (A)= since 1A and only open set containing 1 is X and X\A not finite hence A is not N-open and does not contain N-open set unless  , hence A  cl ( Nint ( cl (A)))=  , therefore A is not  -N-open. (2) Let X={1,2,3} and  ={X, , {a},{b},{a,b}}. Then {c} is an  -N-open but not  -open. Lemma [8] 2.17: Let (X,τ) be a topological space, then the following properties hold: 1. Every  -open set is  - -open. 2. Every  - -open set is pre- -open. 3. Every pre- -open set is b- -open. 4. Every b- -open set is  - -open. The following diagram explains the relation among the above concepts. open  -open pre-open b-open  -open N-open  -N-open pre-N-open b-N-open  -N-open  -open - -open pre- -open b- -open  - -open Definition 2.18[8]: A topological space (X,τ) is called door space if every subset of X is open or closed. Proposition 2.19: If (X,τ) is door space , then every pre-N-open set is N-open. Proof : Let A be pre-N-open. If A is open then it is N-open. Otherwise A is closed, then A  Nint ( cl (A)) = Nint (A)  A, then by proposition(2.6(2)) A is N-open. Proposition 2.20: Let (X,τ) be a topological space. And Let A be b-N-open such that Nint (A)= , then A is pre-N-open. Proof It is clear Lemma 2.21[10]: Let (X,τ) be a topological space. And U be an open set of X, then cl (U  A)= cl (U  cl (A)) and hence U  cl (A)  cl (U  A) for any subset A of X. Proposition 2.22: A subset U of a topological space (X,τ) is pre-N-open set iff there exists a pre-N-open set A such that U  A  cl (U). Proof since U  A  Nint ( cl (A)) , also cl (A)  cl ( cl (U))= cl (U), and by proposition(2.6(5)) we have Nint ( cl (A))  Nint ( cl (U)), that is U  Nint ( cl (U)), hence U is pre-N-open set. Conversely: suppose that U is pre-N-open. If we take A=U, then A is pre-N-open set such that U  A  cl (U). 272 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Proposition 2.23: A subset A of a topological space (X,τ) is semi open iff A is  -N-open and Nint ( cl (A))  cl ( int (A)). Proof Let A be semi open set, then A  cl ( int (A))  cl ( Nint (A))  cl ( Nint ( cl (A))) (by proposition (2.9(1)), hence A is  -N-open. Now, cl (A)  cl ( int (A)), then Nint ( cl (A))  cl ( int (A)) (by proposition (2.9(1)). Conversely: A  cl ( Nint ( cl ( A)))  cl ( cl ( int (A))) = cl ( int (A)), hence A is semi open set. Proposition 2.24: In any topological space the intersection of a  -N-open set and open set is  -N-open. Proof Let Ube an open set and A be a  -N-open since every open set is N-open, then by lemma (2.21) we have U  A  U  cl ( Nint (A))  cl (U  cl ( Nint ( cl (A))) = cl ( Nint (U)  Nint ( cl (A))) = cl ( Nint (U  cl (A))) (by proposition 2.9(7)) That is U  A  U  cl ( Nint (A))  cl ( Nint ( cl (U  A))). Hence U  A is  -N-open. Proposition 2.25: In any topological space the intersection of a b-N-open set and an open set is b-N-open. Proof Let A be a b-N-open and U be an open set, then U  A  U  [ Nint ( cl (A))  cl ( Nint (A))] =[U  Nint ( cl (A))]  [U  cl ( Nint (A))] = [ Nint (U)  Nint ( cl (A))]  [U  cl ( Nint (A))]  [ Nint (U  cl (A))]  [ cl (U  Nint (A))]  Nint ( cl (U  A))  cl ( Nint (U  A)). Hence U  A is b-N-open. Proposition 2.26: In any topological space the intersection of a  -N-open set and an open set is  -N-open. Proof: Let A be an  -N-open set and U be an open set, then U  A  Nint (U)  Nint ( cl ( Nint (A)))  Nint (U  cl ( Nint (A)))  Nint ( cl (U  Nint (A))  Nint ( cl ( Nint (U  A))). Therefore U  A is  -N-open. Remark 2.27: The intersection of two pre-N-open ( resp. b-N-open,  -N-open) set need not be pre-N-open ( resp. b-N-open,  -N-open) in general to show that for pre-N-open see the following example. Example 2.28: In example (2.7) let A=Q and B= (R\Q)  {1}. Then A and B are pre-N- open , b-N-open and  -N-open sets but A  B= {1} is not pre-N-open set since Nint ( cl ({1}))=  A  B is not b-N-open since Nint ( cl ({1}))  cl ( Nint ({1}))=    =  . A  B is not  -N-open since cl ( Nint ( cl ({1})))= cl ( Nint ( {1}))= cl ( )= . Proposition 2.29: In any topological space the union of any family of b-N-open ( resp. pre-N- open,  -N-open) set is b-N-open ( resp. pre-N-open,  -N-open). Proof: Let{Aα}  be a family of b-N-open sets, since Aα  Nint ( cl (Aα))  cl ( Nint ( Aα))  , then   Aα    [ Nint ( cl (Aα))  cl ( Nint (Aα))] 273 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016  [   Nint ( cl (Aα))]  [   cl ( Nint (Aα))]  Nint (   ( cl (Aα)))  cl (   ( Nint (Aα)))  Nint ( cl (   (Aα)))  cl ( Nint (   (Aα))). Hence   Aα is b-N-open.( A similar proof for the other cases). Contra N-continuous Definition 3.1: A function f from a topological space (X,τ) into a topological space (Y,   ) is Called: 1- Contra-continuous if the inverse image of each open subset of Y is closed subset of X [11]. 2-  -continuous if the inverse image of each open subset of Y is  -open subset of X [6]. 3- Contra- -continuous if the inverse image of each open subset of Y is  -closed subset of X [13]. 4- N-continuous if the inverse image of each open subset of Y is N-open subset of X [9]. 5- Contra N-continuous inverse image of each open subset of Y is N-closed subset of X. Since every closed set is N-closed, then every contra continuous is contra-N-continuous, and since every N-closed is  -closed, then every contra-N-continuous is co- -continuous. But in general: the converse of above may be not true, show the following examples Examples 3.2: 1- Let f be the identity function from the set of natural numbers with indiscrete topology onto itself with the discrete topology, then f is contra-N-continuous but not contra- continuous. 2- Let f be the identity function from the set of rational numbers with indiscrete topology onto itself with the discrete topology, then f is -continuous and contra- -continuous but f is not contra-N-continuous. 3- Let X={1,2,3}, Y={a,b}, X ={X,  ,{2}} and Y ={Y,  ,{b}}. Define f from X into Y by f(1)=a, f(2)=f(3)=b, then f is contra-N-continuous but not continuous. 4 -Consider the two functions f and g from usual topological space into space Y={0,1} with topology defined by Y ={Y,  ,{0}} defined by f(x)=      Ax ARx 1 \0 g(x)=      Ax ARx 0 \1 where A is finite set in R. Then f is continuous but not contra-N-continuous and g is contra- N-continuous but not  -continuous The following diagram explains the relation among the above concepts. Contra continuous Contra-N-continuous contra- -continuous 274 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 Continuous N-continuous  -continuous Definition [13] 3.3: Let A be a subset of a topological space (X,τ). The kernel of A is the set defined as }:{)ker( UAUA   Lemma [13] 3.4: The following properties hold for subsets A and B of a topological space (X,τ) 1- )ker( Ax  iff FA  for any closed subset F containing x. 2- )ker( AA  and )ker( AA  if A is open in X. 3- If BA  , then )ker()ker( BA  . Theorem 3.5: Let f be a function from topological (X,τ) into a topological space (Y, ). Then the following are equivalent:- 1- f is contra N-continuous. 2- For every closed subset F of Y, 1f (F) is N-open. 3- For each Xx  and each closed subset F of Y containing f(x), there exists an N-open U containing x such that f(U)  F. 4- f( (Ncl A)  ker (f(A)) for all A  X. 5- ))(ker())(( 11 BfBfclN   for all YB  . Proof (1  2) Let F  Y closed set. Then Y\F is open and since f is contra N-continuous, then )\(1 FYf  is N-closed in X, but )\(1 FYf  = )(\ 1 FfX  , that is, )(1 Ff  is N-open. (2  3)Let Xx  and F  Y such that Fxf )( , we have )(1 Ffx  which is N-open set Put )(1 FfU  , then we have FFffUf   ))(()( 1 , hence FUf )( . (3  4)Suppose that, there exists ))(( Aclfy N and ))(ker( Afy  for some subset A of X, then )( xfy  for some ))( Aclx N , hence there exists a closed set F  Y such that Fy  and )( AfF  , thus by (3) there exist NU  such that FUf )( , then )())(( 11 FfUffU   , hence AU  and )( Aclx N , that is ))(()( Aclfxfy N Which is contradiction. (4  5) Let YB  . Then )ker())((ker( 1 BBff  and by (4) we have )ker()))((ker()))((( 11 BBffBfclf N   , hence ))(ker())(( 11 BfBfclN   . (5  1) Let V be any open of Y, then by lemma and the assumption )())(ker())(( 111 VfVfVfclN   , therefore )())(( 11 VfVfclN   . Hence ))(()( 11 VfclVf N   and then f is contra N-continuous. 275 | Mathematics 2016) عام 1العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham J. for Pure & Appl. Sci. Vol.29 (1) 2016 References 1. Levine, N. (1963); Semi-open sets and semi-continuity in topological spaces, Math. 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