1. Use Theorem 3.17 to prove that for any two real sequences {an), lim sup an + bn < lim sup an +lim sup bn provided the sum on the right hand side is not of the form oo-oo Note: Other proofs exist, but you need to use Theorem 3.17 to receive credit for this problem 2. Let f be a real uniformly continuous function on a bounded set E C R. (a) Prove that f is bounded on E (b) Show that the conclusion is false if boundedness of E is omitted from the hypothesis 3. Show that if X, Y, Z are metric spaces, f : X → Y and g : Y → Z are uniformly continuous, then the composite function go f:g(f ( is uniformly continuous. 4. (a) Let {Pn] be a sequence in a metric space X such that every subsequence of (pn has a subsequence that converges to p in X. Then Pn converges to p in x. (b) Show that the statement is no longer true if the words to p are omitted
3.17 Theorem Let {sn} be a sequence of real numbers. Let E and s* have the same meaning as in Definition 3. 16. Then s* has the following two properties: (a) s* e E. (b) If x> s*, there is an integer N such that n � N implies sn < x. Moreover, s* is the only number with the properties (a) and (b). Of course, an analogous result is true for s * . Proof (a) If s* = + 00, then E is not bounded above; hence {sn} is not bounded above, and there is a subsequence {snk } such that snk ) + 00. If s* is real, then E is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28. If s* = - 00, then E contains only one element, namely - 00, and there is no subsequential limit. Hence, for any real M, Sn > M for at most a finite number of values of n, so that Sn ) - 00. This establishes (a) in all cases. (b) Suppose there is a number x> s* such that Sn � x for infinitely many values of n. In that case, there is a number y E E such that y � x > s*, contradicting the definition of s*. Thus s* satisfies (a) and (b). To show the uniqueness, suppose there are two numbers, p and q, which satisfy (a) and (b), and suppose p < q. Choose x such thatp < x < q. Since p satisfies (b), we have Sn < x for n � N. But then q cannot satisfy (a).
![1. Use Theorem 3.17 to prove that for any two real sequences {an),
lim sup an + bn < lim sup an +lim sup bn
provided the sum on the right hand side is not of the form oo-oo
Note: Other proofs exist, but you need to use Theorem 3.17 to receive credit for this problem
2. Let f be a real uniformly continuous function on a bounded set E C R.
(a) Prove that f is bounded on E
(b) Show that the conclusion is false if boundedness of E is omitted from the hypothesis
3. Show that if X, Y, Z are metric spaces, f : X → Y and g : Y → Z are uniformly continuous, then the
composite function go f:g(f (
is uniformly continuous.
4. (a) Let {Pn] be a sequence in a metric space X such that every subsequence of (pn has a subsequence
that converges to p in X. Then Pn converges to p in x.
(b) Show that the statement is no longer true if the words to p are omitted](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F105e5746-ccd5-4056-9254-8489b4131a73%2Ffa11c727-821f-4e18-b13b-12ec5fa16f8b%2Fzwzr358.jpeg&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 8 steps with 7 images
