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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
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 Ýnorm 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...
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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.
 Fall '08
 STAFF
 Math, Calculus

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