Exercise 1.8
Solution:
(a) Because , we have . Similarly, we have . We haveHence .
(b) We give a counterexample. Let . Set and , then and . But .
Exercise 1.10
Solution: We remark the following identityTherefore, we have
Another solution: We prove it by induction on .
If , the identity is clearly true.
Suppose the identity is true for , we prove it for . By induction hypothesis, we haveThen we haveHence the identity is true for .
By Mathematical Induction, we are done!
Exercise 1.20
Solution:
(a)Let and . Since , we have . By Theorem 1.24 and setting , there exists such that Hence is an upper bound of .
(b) We give an example. Let and . Then . But no element in is an upper bound of since they are all smaller than one.
Exercise 1.21
Solution: Since is bounded below, there exists such that for all . Hence is a lower bound of and . Let .
Since is non-empty, we have an element in A and hence an element in . This implies that .
Note that for all . Multiplying both sides by , we obtain for all . This implies that for all . Therefore is bounded above.
We then show that . We use Theorem 1.24. We have
(a) for all , and hence for all . This implies is an upper bound of ;
(b) for any , there exists such that , and thus . This implies that is not an upper bound of .
Therefore, according to Theorem 1.24, we have .
Exercise 1.22
Solution:
Exercise 1.23
Solution: Let , then each is open and non-empty. Moreover,We show that .
We argue it by contradiction. Suppose , then there exist . Hence for all , that is Since , we have . By the Archimedean Principle, we can find such that Therefore which contradicts with if we take .
Hence .
Exercise 1.24
Solution: We use Theorem 1.24. Let and . Then for any and , we have and . Adding them together, we conclude that for all and . Hence is an upper bound of . (This is part (a) of Theorem 1.24 we would like to show.)
By Theorem 1.24, for any , there exist and such thatAdd them together, we get thatHence we obtain part (b) of Theorem 1.24.
Using Theorem 1.24, we get that , i.e. .
Exercise 1.25
Solution: We give a counter example. Let , then we have We also haveHence , but .
Exercise 1.26
Solution: We shall use Definition 1.29 to show that the set is not dense in . Let Then and . We show that there is no element satisfing .
We argue it by contradiction. Suppose and , where , then we have , , andBecause , we have which is impossible. Hence cannot happen. By Definition 1.29, the set is not dense in .