Homepage › Solution manuals › Walter Rudin › Principles of Mathematical Analysis › Exercise 4.11
Exercise 4.11
Exercise 11: Suppose is a uniformly continuous mapping of a metric space into a metric space and prove that is a Cauchy sequence in for every Cauchy sequence in . Use this result to give an alternative proof of the theorem stated in Exercise 13.
Answers
Let and let such that if . Let be a Cauchy sequence in . Then there is an integer such that if and . Hence if and , so is a Cauchy sequence in .
Let be a dense subset of a metric space , and let be a uniformly continuous mapping of into a complete metric space . We can extend to a function on all of as follows. Let and be a sequence in which converges to . Then is a Cauchy sequence in which converges to since is complete. This sets up a well-defined function from to since if is another sequence in converging to , then converges to . Since is uniformly continuous on , for there is such that if with then . There is an integer such that and for all . Then for such we have
Also, there is an integer such that for all we have and . Hence, for large enough we have
Considering the constant sequence for , we see that extends to all of . To see that is uniformly continuous, let . Since is uniformly continuous on , there is such that if with then . Let such that , and let and be sequences in which converge to and , respectively. Let be a large enough integer so that and . Then
so that . Also, if is large enough we have and . Hence, for large enough we have
Hence is uniformly continuous on .