1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 xc 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 j1 r j  1 then 115 n a n rjxj n  n rja j1  rjxj j1 rj a j1 xj j1 n n rj a r j sup xj j1 a x x S j1  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 . j1 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 . j1 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, j1 . 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, j1 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...
View Full Document

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.

Ask a homework question - tutors are online