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Unformatted text preview: of a metric space X, d and that H is a closed subset of the metric
space S, d , prove that H must be closed in the metric space X, d .
Using Exercise 11 and the fact that S H is open in the space S, d we choose a set V that is open
in the space X such that
S HS V
and we observe that
HS
X V.
Thus H, being the intersection of two sets that are both closed in the space X, d , must be closed
in the space X, d .
15. Skip this exercise if you are not familiar with the concept of an uncountable set. Prove that every nonempty
open subset of R k is uncountable.
The proof follows at once from an earlier theorem about equivalence to R of subsets of R that
include intervals.
16. Suppose that X, d is a given metric space and that we have defined
x, y d x, y if d x, y 1 1 if d x, y 1 . a. Prove that the function is also a metric on the set X.
b. Prove that the metric spaces X, d and X, have exactly the same open sets.
We saw in an earlier exercise that the function is a metric. Part b follows from the fact that
whenever x X and 0 r 1, the ball with center x and radius r n the metric space X, d is exactly
the same as the ball with center x and radius r in the metric space X, .
17. This exercise refers to the optional topic of a convex hull that appears in the reading material on convexity.
Prove that if U is an open subset of R k then the convex hull of U is open.
We assume that U is an open subset of R k . Suppose that x is any member of the convex hull co U
n
of U and choose a positive integer n and numbers r j
0, 1 for j 1, 2, , n such that j1 r j 1
and
n x rjxj.
j1 We may assume that the coefficient r 1 is not zero. Now using the fact that U is open and that
x 1 U we choose 0 such that B x 1 ,
U. We see that if y is any member of the ball
B x 1 , r 1 , since 120 n yy x rjxj
j1 r1 y x n r1 x1 x1 B x1, rjxj
j2 and since
y x
r1 we see that y U co U . 18. The purpose of this exercise is to exhibit an example of a closed subset H of R 2 whose convex hull fails to
be closed. We define
1
1 x
1
1x 1 fx 1 if x 0 if x 0 0.8 0.6 0.4 0.2 4 2 0 2 x 4 and we define
H x, y
0 y fx .
Prove that the set H has the desired properties.
Given any positive integer n and points A j x j , y j in the set H for j 1, 2, , n and nonnegative
n
numbers r j for which j1 r j 1 we have
n n n rj xj, yj rjAj
j1 j1 n rjxj,
j1 rjyj
j1 n and since j1 r j y j 1 we know that the point 0, 1 does not belong to co H . On the other hand,
whenever p 0 the point
p2
1
1
0, 2
1 p, 1
1 p, 1
2
1p
2
1p
p
1
must belong to co H and so the point 0, 1 is close to co H . Therefore co H is not closed.
We can, of course, do much better. In fact
co H 0, 0 Þ x, y
R2 0 y 1 . Exercises on Closure
1. Suppose that
S 0, 1 Þ 1, 2 .
a. What is the set of interior points of S in the metric space R?
The set of interior points of S is 0, 1 Þ 1, 2 .
b. Given that U is the set of interior points of S, evaluate U. 121 0, 1 Þ 1, 2 0, 2...
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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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