Unformatted text preview: s 0, Ý .
7. Suppose that f is a uniformly continuous function from a subset S of a metric space X into a metric space Y.
a. Prove that if x n is a Cauchy sequence in the set S then the sequence f x n is a Cauchy sequence in the
Suppose that x n is a Cauchy sequence in the set S, and, to show that f x n is a Cauchy
sequence in the space Y, suppose that 0. Using the fact that f is uniformly continuous on S,
choose 0 such that whenever t and x belong to S and d t, x , we have d f t , f x .
Using the fact that x n is a Cauchy sequence, choose an integer N such that, whenever m N
and n N, we have d x m , x n . Then, whenever m N and n N, we have
d f xm , f xn .
b. Prove that if the space Y is complete then whenever a sequence x n in the set S converges to a point
x X the sequence f x n converges in the space Y.
We assume that Y is complete and that x n is a sequence in S that converges to a point x of X.
Note that we cannot simply claim that f x n
f x because we have not said that the point x
must belong to the set S. However, the sequence x n must be a Cauchy sequence and
therefore, by part a, the sequence f x n must be a Cauchy sequence in Y, and must therefore
converge in Y because Y is complete.
c. Prove that if x S and if x n and t n are two sequences in the set S both of which converge to x and if
the space Y is complete then
lim f x n n Ý f t n .
We assume that x n and t n are sequences in the set S and that both of these sequences
converge to x. We now consider the sequence x 1 , t 1 , x 2 , t 2 , . More precisely, we define
un t n/2 if n is even x if n is odd. n1 /2 Since the sequence u n also converges to x, it is also a Cauchy sequence. Therefore the
sequence f u n is convergent in the space Y. Since both lim n Ý f x n and lim n Ý f t n are
partial limits of the sequence f u n we deduce that
lim f x n n Ý f t n n Ý f u n .
d. Given that Y is complete and that x S S, explain how we can use Part c to extend the definition of the
function f to the point x in such a way that the extension is continuous on the set S Þ x .
The continuity of f at the point x is really very simple. If y n is any sequence in the set S Þ x
converging to x then it is clear that f y n
f x . This part of the exercise also follows at once
from part e that we shall prove below.
e. Prove that if Y is complete then there exists a uniformly continuous function g from the set S into Y such
that g x f x for every point x S.
For each point x S S we define f x by choosing a sequence x n in S that converges to x
f x n Ý f xn .
We shall now show that this extension of the function f to S is uniformly continuous on S.
Suppose that 0.
First Proof (a bit messy, perhaps). Using the fact that f is uniformly continuous on S, choose
0 such that whenever u and v belong to S and d u, v we have d f u , f v /3. Now
suppose that u and v are any points of S satisfying the inequality d u, v /3. Choo...
View Full Document