1873_solutions

# X 1 x 2 x 2 x 1 x 3 y 1 y2 y2 x 3 y

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Unformatted text preview: every cube I c, r and every cube I c, r is 2r k . Given any points x and y in the cube I c, r we have x y kx y 2r k . Ý Now suppose that 0  p  2r k . We define u to be the point u 1 , u 2 , , u k of R k whose coordinates u j are all equal to 1, we choose a number q between p and 2r k and we define q xc u 2k and q y  c u. 2k The points x and y lie in the cube I c, r and x y c q u 2k c q u 2k  q k u  q  p. 9. Prove that if d is the discrete metric on a set X and c X then the diameter of the ball B c, 1 is 0 and the diameter of the ball B c, 1 is 1. These assertions follow at once from the observation that B c, 1  c and B c, 1  X. Strictly speaking, the exercise should have stipulated that the set X shoud contain more than one point. 10. Given that c R k and r  0 prove that the sets B c, r , B c, r , I c, r and I c, r are all convex. Suppose that x and y belong to B c, r and that 0 t 1. We see that c 1 t x  ty  1 t c x  t c y 1 tc x  tc y  1 t r  tr  r. Thus B c, r is convex and we can argue in the same way that B c, r is convex. By replacing the Euclidean norm by the Ý-norm we obtain the analogous results for I c, r and I c, r . 11. Suppose that S is a nonempty subset of R k and that H is the convex hull of S. a. Prove that for every point a Since S R k we have sup a x x S  sup a y y H. H we have sup a x xS sup a y y H. on the other hand, if n is a positive integer and x 1 , , x n belong to S and r 1 , , r n are n nonnegative numbers and j1 r j  1 then 115 n a n rjxj n  n rja j1  rjxj j1 rj a j1 xj j1 n n rj a r j sup xj j1 a x x S j1  sup a x x S b. Prove that diam S  diam H . We see at once that diam S diam H . Now given any point x S we deduce from part a that sup x y y H  sup x y y S  diam S . Using part a again we deduce that if y H then sup x y x H  sup x y xS diam S . Some Exercises on Total Boundedness 1. Given that S is a subset of a metric space X, prove that S is totally bounded if and only if for every possible to find finitely many points x 1 , x 2 , , x n that belong to the set S such that  0 it is n  B xj, S . j1 Solution: The “if” part of this exercise is obvious. We assume that a subset S of a given metric space X is totally bounded and that  0. and we need to find finitely many points x 1 , x 2 , , x n in S such that n  B xj, S . j1 Using the fact that S is totally bounded we choose a positive integer k and points y 1 , y 2 , , y k in X such that k B S yj, j1 . 2 For each j, in the event that the set B yj, S 2 is nonempty, we choose a member of this set and call it x j . We can now show that S is included in the union of the balls B x j , where j runs through those integers for which x j has been defined: Suppose that x S and, using the fact that k B x yj, j1 2 choose j such that x B yj, 2 . Since B yj, we know that x j is defined and we have d x, x j 2 S d x, y j  d y j , x j  116 2  2 . 2. Given that S is an infinite set, find a bounded subset E of the metric space boun...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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