Theorem: Let X Be a Topological Space, and Let a Be a Subset of X Then a Is Closed Iff D(A) A
1 Metric Space: Open sets: Let (X,d) be a metric space . A subset G of X is said to be d-open iff to each x G there exist r >o such that S(x,r) G. n + Def : Let (X,d) be a metric space, and let xо X if r R then the set {x X; d(x,xо)< r} is called an open sphere (or open ball).the point xо is called the center and r the radius of the sphere. and we denoted by S(xо,r) or by B(xо,r): i.e. S(xо,r)={ xо X; d(x,xо)< r} Closed set is define and denoted by S[xо,r]={ xо X; d(x,xо)≤ r}. Ex: Let x R then a subset N of R is U nbd of x iff there exist a u-open set G such that xG N , but G is U-open and x G implies that there exist an >0 such that (x-,x+) G. Thus N is a U-nbd of x if N contains an open interval (x-,x+) for some >0. In particular every open interval containing x is a nbd of x. Ex: Consider the set R of all real numbers with usual metric space d(x,y) = x y and find whether or not the following sets are open. A= (0,1) ,B=[0,1), C= (0,1] ,D=[0,1], E= (0,1) (2,3) , F={1}, G={1,2,3} . Soln : A is open set Let x be appoint in A, we take r=min{x-0,1-x}, then it is evident that (x-r, x+r) A 1 1 1 3 For example consider 4 (0,1), then r=min{ 4 -0,1- 4 }=min{ , 4 }= 1 ( - , + ) =(0, 2 ) (0,1)=A.
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