Exercise 4.4.1
(a) It is already shown in Example 4.3.5.
(b) Let and , then we have
On the other hand, we have
Hence we can choose in Theorem 4.4.5. It follows from Theorem 4.4.5 that is not uniformly continuous on .
Lemma: If a function is uniformly continuous on a set , then it is also uniformly continuous on any subset of .
Proof: Since is uniformly continuous on , for any , there exists such that for all , , we have .
Now for any , , as is a subset of , we also have , , therefore we have .
(c) Let be a bounded subset of . Then there exists such that . The closed interval is a compact set. As is continuous on , by Theorem 4.4.7, it follows that is uniformly continuous on . Now it follows from the lemma above that is uniformly continuous on .
Exercise 4.4.2
(a) No, let and , thenHence by Theorem 4.4.5, is not uniformly continuous on .
(b) Yes. Note that is continuous on . Since is bounded and closed, is compact. By Theorem 4.4.7, is uniformly continuous on . Therefore, is also uniformly continuous on .
(c) Yes. We extend the function by defining , then it is clear that is continuous on . Since is bounded and closed, is compact. By Theorem 4.4.7, is uniformly continuous on . Therefore, is also uniformly continuous on .
Exercise 4.4.3
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.2.
Exercise 4.4.4
(a) See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.4.
(b) True. We argue it by contradiction. Suppose is not bounded. Then there exists a sequence such that and for all (this sequence can be obtained inductively). Since is bounded, by Theorem 2.5.5, then we have a convergent subsequence . Let , then butHence by Theorem 4.4.5, is not uniformly continuous, which is a contradiction. Therefore, we are done.
(c) False. Define a function as follows,Then is always a finite set and hence compact. But is not continuous on .
Exercise 4.4.5
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.7.
Exercise 4.4.6
(a) Let for . Then is continuous on . Let , then is Cauchy. But , so the sequence is not Cauchy.
(b) Impossible. Since is uniformly continuous on , for any , there exists such that provided . Since is Cauchy, there exists such that whenever . Hence for , we haveThis implies that is Cauchy.
(c) Impossible. Since is Cauchy, is bounded by Lemma 2.6.3. Let where for all . Then we can consider the restriction of on the interval . Because is compact, by Theorem 4.4.7, is uniformly continuous on . Using the same argument as in (b), you can show that is Cauchy.
Exercise 4.4.7
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.8 (b).
Exercise 4.4.8
(a) Impossible. Because is continuous and is compact, by Theorem 4.4.1, the range is compact as well. But is not compact.
(b) Let , see Example 4.2.6.
(c) Let
Exercise 4.4.9
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.9.
Exercise 4.4.10
The functions and are necessarily uniformly continuous.
is uniformly continuous. This follows from the triangle inequality. Let . Because and are uniformly continuous on , there exist and such that for all and on . Let . Then for all and , we haveThus is uniformly continuous.
may bot be uniformly continuous. Let , then is not uniformly continuous. Because we can choose and , thenBy Theorem 4.4.5, is not uniformly continuous.
may bot be uniformly continuous. Let and defined on , then is not uniformly continuous, see Exercise 4.4.2.
is uniformly continuous. Let . Since is uniformly continuous, there exists such that for all and we have Since is uniformly continuous, there exists such that for all and we have Therefore, if and , we have and therefore Thus is uniformly continuous.
Exercise 4.4.11
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.11.
Exercise 4.4.12
(a) (b) (c) may be false. Consider the example for all . Let , then which is clear not finite, not compact, not bounded.
(d) is true. To show is closed, it suffices to show that is open. We show that . Then because is open, by definition of Exercise 4.4.11, is also open.
Now we show . First we show . Let , then , i.e. . Therefore , hence .
Then we show . Let , then . Thus and . Therefore , this implies .
Now we have .
Exercise 4.4.13
See Understanding Analysis Instructors’ Solution Manual Exercise 4.4.13.