This preview shows page 1. Sign up to view the full content.
Unformatted text preview: h this exercise is quite difficult right now, it will
become considerably easier after we have studied the concept of a derivative.
I am resisting the urge to write a direct solution of this exercise. Those students who wish to
attempt it now will probably want to be left alone. All others can wait until after the mean value
theorem when the fact that f is constant will follow at once from the obvious fact that f x 0 for
every number x. Exercises on the Distance Function
1. Two subsets A and B of a metric space X are said to be separated from each other if
A BA B .
Prove that if two sets A and B are separated from each other then 225 A x B x 0 A x whenever x B x 0 A and whenever x B.
We know that if x A then A x 0, and that, since x does not belong to B, we have B x 0.
Therefore the inequality
A x B x 0
holds whenever x A and the same kind of argument shows that
A x B x 0
whenever x B.
2. Prove that if two sets A and B are separated from each other then there exist two open sets U and V that are
disjoint from each other such that A U and B V. Solution: Suppose that A and B are subsets of a metric space X and that A and B are separated from
each other. Define
U x X A x B x 0 V x X A x B x 0 and 3. Given two subsets A and B of a metric space X, prove that the following conditions are equivalent:
a. We have
B . A
b. There exists a continuous function f : X 0, 1 such that
0 if x A 1 if x fx B. The fact that condition a implies condition b is just Urysohn’s lemma. Now suppose that condition b
holds and choose a continuous function f : R 0, 1 such that
0 if x A 1 if x fx B. Since
and since the set x X A
x X fx 0
f x 0 is closed, we have
A
x X fx 0 . In the same way,
B
and therefore A x X fx 1 B . 4. Suppose that A, B and C are closed subsets of a metric space X, and that no two of these three sets intersect.
Prove that there exists a continuous function f : X R such that
1 if x 226 2 if x B. 3 if x fx A C Define
C
2
f 1 A A B .
C
B
AC
AB
5. Suppose that S is a subset of a metric space and that S fails to be closed. Prove that there exists a convergent
sequence x n in S and a continuous function f from S to R such that the sequence f x n fails to converge. Solution: We begin by choosing a point w S S and we choose a sequence x n in S that converges
to the point w. We define E to be the range of the sequence x n and, Using the fact that that the set E
must be infinite, we choose an infinite subset A of E such that the set B E A is also infinite. Since no
member of S can lie in the closures of both A and B, the function
f A
B
A
is continuous on the set S. Furthermore, since there must be infinitely many integers n for which x n A
and infinitely many integers n for which x n B the sequence f x n has no limit. Exercises on Continuous Functions on Compact Spaces
1. Give an example of a function f that is continuous from a closed set H of real numbers into R such that rang...
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.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details