Introduction to Mathematical Analysis I - Second Edition

146 Solutions and Hints for Selected Exercises SECTION 1.5 Exercise 1.5.4 . Let us first show that A + B is bounded above. Since A and B are nonempty and bounded above, by the completeness axiom, sup A and sup B exist and are real numbers. In particular, a ≤ sup A for all a ∈ A and b ≤ sup B for all b ∈ B . For any x ∈ A + B , there exist a ∈ A and b ∈ B such that x = a + b . Thus, x = a + b ≤ sup A + sup B , which shows that A + B is bounded above. We will now show that sup A + sup B is the supremum of the set A + B by showing that sup A + sup B satisfies conditions (1’) and (2’) of Proposition 1.5.1 . We have just shown that sup A + sup B is an upper bound of A + B and, hence, sup A + sup B satisfies condition (1’) . Now let ε > 0. Using ε 2 in part (2’) of Proposition 1.5.1 applied to the sets A and B , there exits a ∈ A and b ∈ B such that sup A − ε 2 < a and sup B − ε 2 < b . It follows that sup A + sup B − ε < a + b . This proves condition (2’) . It follows from Proposition 1.5.1 applied to the set A + B that sup A + sup B = sup ( A + B ) as desired. SECTION 1.6 Exercise 1.6.2 . Let x = 1 r . By Theorem 1.6.2( d) , there exists m ∈ Z such that m − 1 ≤ 1 r < m . Since 1 / r > 1, we get m > 1 and, so, m ≥ 2. It follows that m − 1 ∈ N . Set n = m − 1 and then we get 1 n + 1 < r ≤ 1 n . SECTION 2.1 Exercise 2.1.12 . (a) Suppose that lim n → ∞ a n = ` . Then by Theorem 2.1.9 , lim n → ∞ a 2 n = ` and lim n → ∞ a 2 n + 1 = `. (5.2) Now suppose that ( 5.2 ) is satisfied. Fix any ε > 0. Choose N 1 ∈ N such that | a 2 n − ` | < ε whenever n ≥ N 1 , and choose N 2 ∈ N such that | a 2 n + 1 − ` | < ε whenever n ≥ N 2 .