156 Solutions and Hints for Selected Exercises SECTION 1.5 Exercise 1.5.4. Let us first show that A+Bis bounded above. Since Aand Bare nonempty and bounded above, by the completeness axiom, supAand supBexist and are real numbers. In particular, a≤supAfor all a∈Aandb≤supBfor all b∈B. For anyx∈A+B, there exist a∈Aandb∈Bsuch that x=a+b. Thus, x=a+b≤supA+supB, which shows that A+Bis bounded above. We will now show that supA+supBis the supremum of the set A+Bby showing that supA+ supBsatisfies 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 satisfies condition (1’). Now let ε >0. Using ε 2 in part (2’) of Proposition 1.5.1 applied to the sets AandB, there exits a∈Aandb∈Bsuch that supA− ε 2 <aand supB− ε 2 <b. It follows that supA+supB−ε <a+b. This proves condition (2’). It follows from Proposition 1.5.1 applied to the set A+Bthat supA+ supB=sup(A+B) as desired. SECTION 1.6 Exercise 1.6.2. Let x = 1 r . By Theorem 1.6.2(iv), there exists m∈Zsuch 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 2.1.16. (a) Suppose that limn→∞an =ℓ. Then by Theorem 2.1.6, lim n→∞ a2n =ℓ and lim n→∞ a2n+1 =ℓ. (5.13) Now suppose that (5.13) is satisfied. Fix anyε >0. Choose N1 ∈Nsuch that |a2n −ℓ| <ε whenever n≥N1, and choose N2 ∈Nsuch that |a2n+1 −ℓ| <ε whenever n≥N2.
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