18 100B Fall 2002 Homework 4 Solutions Was due by Noon Tuesday October 1 Rudin 1 Chapter 2 Problem 22 Let Qk Rk be the subset of points with rational coe cients This is countable as the Cartesian product of a nite number of countable sets Suppose that x x1 xk Rk By the density of the rationals in the real numbers given 0 there exists yi Qk such that xi yi k i 1 k Thus if y y1 y2 yk then x y k maxki 1 xi yi shows the density of Qk in Rk Thus Rk is separable 2 Chapter 2 Problem 23 Given a separable metric space X let Y X be a countable dense subset The product A Y Q is countable Let Ua a A be the collection of open balls with center from Y and rational radius If V X is open then for each point p V there exists r 0 such that B p r V By the density of Q in X there exists y Q such that p B y r 2 Moreover there exists q Q with r 2 q r Then x B y q Thus each point of V is in an element of one of the Ua s which is contained in V so Ua V Ua V It follows that the Ua a A form a base of X actually now more usually called an open basis 3 Chapter 2 Problem 24 By assumption X is a metric space in which every in nite set has a limit point For each positive integer n choose points x1 n x2 n successively with the property that d xj n xk n 1 n for k j After a nite number of steps no futher choice is possible Indeed if there were an in nite set of points E satisfying d x x 1 n for all x x in E then E could have no limit point since a limit point q X would have to satify d q pi 1 2n for an in nite number of di erent pi E and this would imply that d p1 p2 d p1 q d q p2 1 n which is a contradiction Let Y X be the countable subset as a countable union of nite sets consisting of all the xj n for all n Then Y is dense in X To see this given p X and 0 choose n 1 If p xj n for some j then it is in Y If not then for some j d p xj n 1 n otherwise it would be possible to choose another xj n contradicting the fact that we have chosen as many as possible Then d p q for some q Y which is therefore dense and X is therefore separable 4 Chapter 2 Problem 26 By assumption X is a metric space in which every in nite subset has a limit point By the problems above it is separable and hence has a countable open basis Ui Let Va a A be an arbitrary open cover of X Each Va is a union of Uj s by the de nition of an open basis For each j such that Uj is in one of these unions choose a Vaj which contains it Then for every b A Vb must be contained in a union of the Uaj s hence in the 1 2 union of the Vaj s which therefore form a countable subcover of the original open cover Va Consider the successive open sets N Vai i 1 If one of these contains X then we have found a nite subcover of the Va s So suppose to the contrary that FN X N Vai N i 1 1 The FN s are decreasing as N increases Let E X be a set which contains one point from each FN It must be an in nite set since otherwise some xed point would be in FN for arbitrary large hence all N but FN N N since together all the Vai do cover X By the assumed property of X E must have a limit point p For each N all but nitely many points of E lie in FN so p must be a limit point of FN for all N but each FN is closed so this would mean p FN for all N contradicting 1