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MATH 4181 Problem Set 3 Solutions 1 MATH 4181 001 Fall 1999 Problem Set 3 Solutions 1.(Exercise 6, page 46) Let aandbbe distinct points in a metric space X. Prove that there are neighborhoods NaandNbofaandbrespectively such that Na∩Nb=∅. Sinceaandbare distinct points, d(a,b) =r > 0. LetNa=Bd(a;r/3) andNb= Bd(b;r/3). Now, clearly, Na∩Nb=∅. 2.Let{xn}be a sequence in the metric space Xand assume that{xn}converges to x∈X. Prove that this limit is unique, i.e., if limxn=xandlimxn=Lthenx=L. We will use the result from above to prove this. Assume that x/negationslash=L. Then there are disjoint neighborhoods NxandNLofxandL, respectively. In fact, if we let r=d(x,L)/negationslash= 0 (by assumption), then we may take each neighborhood to be the ball of radiusr/3 centered at each respective point. Now, since the sequence converges to xwe know that for any /epsilon1 >0 there is a positive integer N/epsilon1so that for any n > N/epsilon1 xn∈Bd(x;/epsilon1). This must be true for /epsilon1=r/3. Thus, there is a positive integer Nso that for all n > N ,xn∈B(x;r/3). Applying the same analysis to the limit Lof the sequence{xn}gives us a positive integer Mso that for all n>M x n∈B(L;r/3). LetP= max{M,N}. Now, ifn > P we must have that xn∈B(x;r/3) andxn∈ B(L;r/3), orxn∈B(x;r/3)∩B(L;r/3) =∅. This is impossible, so we must have x=L. 3.Show that a finite subset of a metric space has no limit points and is therefore a closed set. LetA={x1,...,xn}be a finite set in the metric space ( X,d). Lety/negationslash∈A, then let dk=d(y,xk). Sincey/negationslash∈A dk/negationslash= 0 fork= 1,...,n . Letr= min{d1,...,dn}. Then for any/epsilon1<r ,B(y;/epsilon1) does not contain any elements of A, because they are all further away fromythan/epsilon1. Thus,yis not a limit point of A. Ify∈Awe use the same technique to construct a ball about ywhich contains no other points of A. Thus showing that y is not a limit point of A. Therefore, Ahas no limit points, A/prime=∅. This then means thatAcontains all of its limit points, since ∅⊂A. Therefore by our theorem, Amust be closed. 4.Let(X,d)be a metric space with the discrete metric. Prove (a)Every subset of Xis open. Note thatB(x; 1/2) ={x}is an open set. Thus, each singleton set is an open set. Every set is a union of its elements and therefore is a union of open sets. Therefore every set is open. (b)Every subset of Xis closed. For any set F, we know that X\Fis open from above. Thus, by definition, Fis closed. c/circlecopyrtDavid C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 3 Solutions 2 (c)No subset of Xhas a limit point. Using the same open set as the first part shows that each point has an open neighborhood that contains no other point of any other set. Thus, each point cannot be a limit point of any set. Thus, no subset of Xhas a limit point. c/circlecopyrtDavid C. Royster Introduction to Topology For Classroom Use Only