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Unformatted text preview: of this exercise.
Well, of course, such a choice of x n would make f x n
Ý which we know now to be
impossible. 7. a. Given that S is a set of real numbers, that a
for all x S S and that
f x x1a
S, prove that f is continuous on S but not uniformly continuous. Hint: Use the preceding exercise. Choose a sequence
f xn Ý as n x n in S a that converges to x. Note that Ý. b. Given that S is a set of real numbers and that S fails to be closed, prove that there exists a continuous
function on S that fails to be uniformly continuous on S.
Choose a number a S S and use Part a.
c. Is it true that if S is an unbounded set of real numbers then there exists a continuous function on S that
fails to be uniformly continuous on S?
No, as we remarked after Exercise 1, every function on the set Z of integers must be uniformly
continuous.
8. Is it true that the composition of a uniformly continuous function with a uniformly continuous function is
uniformly continuous?
Yes, this assertion is true. Suppose that f is a uniformly continuous function on a set S and that g is
a uniformly continuous function on a set T that includes the range f S of f. To show that the
function g f is uniformly continuous on S, suppose that 0.
Using the uniform continuity of g on the set T we choose a number 0 such that the inequality
g u g v  holds whenever u and v belong to T and u v  . Now, using the uniform
continuity of f on the set S we choose a positive number such that the inequality f s f t 
will hold whenever s and t belong to S and s t  . Then whenever s and t belong to S and
gft  .
s t  we have g f s
9. a. Suppose that f is uniformly continuous on a set S and that x n is a convergent sequence in S. Prove that
the sequence f x n cannot have more than one partial limit.
We know from the result proved in Exercise 6 d that the sequence f x n is bounded. We
choose a partial limit y of f x n and we want to prove that y is the only partial limit of f x n .
Suppose that z is any number unequal to y. We define
1 y z .
3
Choose a number 0 such that the inequality 203 f s f t 
holds whenever s and t belong to S and s t  /2.
We write the limit of x n as x and choose an integer N such that the inequality x n
holds whenever n N. Using the fact that y is a partial limit of the sequence f x n
an integer m N such that f x m
y  . Now given any integer n N, since
x n x m  x n x  x x m 
2
2
we must have
f xm 
f x n
and consequently
f xm  2
y f x n  y f x m  f x n y f xm z f xn Thus if n N, the number f x n cannot lie in the neighborhood z
that z fails to be a partial limit of the sequence f x n . x
2
we choose ,z of z and we conclude b. In Part a, did you assume that the limit of the sequence x n belongs to S? If so, go back and do the
problem again.
c. Prove that if f is uniformly continuous on a set S and x n is a convergent sequence in S then the sequence
f x n is also convergent. Do not assum...
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 Fall '08
 STAFF
 Math, Calculus

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