156 Solutions and Hints for Selected Exercises SECTION 1.5 Exercise 1.5.4. Let us frst show that A+ B is bounded above. Since A and B are nonempty and bounded above, by the completeness axiom, supA and supB exist and are real numbers. In particular, a ≤ supA for all a ∈ A and b ≤ supB 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 ≤ supA+supB, which shows that A + B is bounded above. We will now show that supA + supB is the supremum of the set A+ B by showing that supA + supB satisfes conditions (1’) and (2’) of Proposition 1.5.1. We have just shown that supA + supB is an upper bound of A + B and, hence, supA + supB satisfes 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 ε ε supA − < a and supB − < b. 2 2 It follows that supA + supB− ε < a+ b. This proves condition (2’). It follows from Proposition 1.5.1 applied to the set A + B that supA + supB = sup(A + B) as desired. SECTION 1.6 1 Exercise 1.6.2. Let x = . By Theorem 1.6.2(iv), there exists m ∈ Z such that r 1 m − 1 ≤ < m. r 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 1 < r ≤ . n + 1 n SECTION 2.1 Exercise 2.1.16. (a) Suppose that limn →∞ an = ℓ. Then by Theorem 2.1.6, lim a2n = ℓ and lim a2n+1 = ℓ. (5.13) n→∞ n→∞ Now suppose that (5.13) is satisfed. Fix any ε > 0. Choose N1 ∈ N such that |a2n − ℓ| < ε whenever n ≥ N1, and choose N2 ∈ N such that |a2n+1 − ℓ| < ε whenever n ≥ N2.
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