Mathematics 242 – Principles of Analysis Selected Solutions, Problem Set 2 September 22, 2000

§1.5/16. We want to show that if

A = \{x ∈ R : 0 < x < 1 and x = \sqrt{2}r for r ∈ Q\}

then lub(A) = 1. To do this we must show that 1 is an upper bound for A, and that no β < 1 is an upper bound for A. The fact that 1 is an upper bound for A is “built into” the definition of A, so to speak – all x ∈ A have x < 1. Hence we need to show that given any β < 1, there exists some x = \sqrt{2}r (an element of A) satisfying β < x < 1. Consider the interval (\frac{β}{\sqrt{2}}, \frac{1}{\sqrt{2}}). By Theorem 1.5.6 (or Problem 15), there exists a rational number r in this interval, so

\frac{β}{\sqrt{2}} < r < \frac{1}{\sqrt{2}}.

Multiplying through by \sqrt{2} > 0, the inequalities are preserved, so

β < r\sqrt{2} < 1.

Hence β is not an upper bound for A, since the number r\sqrt{2} ∈ A but it is larger than β. Therefore, lub(A) = 1.

§2.1/2a. By inspection, lim_{n→∞} \frac{16}{n^{3}} = 0. To prove this using the ϵ − n_{0} definition, we

proceed as follows. Let ϵ > 0 be any positive number. If n_{0} > \sqrt[3]{\frac{16}{ϵ}}, and n ≥ n_{0}, then

n > \sqrt[3]{\frac{16}{ϵ}}, so n^{3} > \frac{16}{ϵ}, so \frac{16}{n^{3}} < ϵ. Since |\frac{16}{ϵ} − 0| < ϵ for all n ≥ n_{0}, this shows that the definition of convergence is satisfied with limiting value a = 0.

§2.1/16 c. We must show that if lim_{n→∞} y_{n} = a, lim_{n→∞} z_{n} = a, and x_{2k} = y_{k}, x_{2k−1} = z_{k}, then lim_{n→∞}x_{n} = a also. To prove this, let ϵ > 0 be any positive number. Since lim_{n→∞} y_{n} = a, there is some k_{0} ∈ N such that for all k ≥ k_{0}, |y_{k} − a| < ϵ. If n = 2k is an even integer, this shows that |x_{n} − a| = |x_{2k} − a| = |y_{k} − a| < ϵ for all n = 2k ≥ 2k_{0}. Similarly, since lim_{n→∞} z_{n} = a, there is some ℓ_{0} ∈ N such that for all ℓ ≥ ℓ_{0}, |z_{ℓ} − a| < ϵ. If n = 2ℓ − 1 is an even integer, this shows that |x_{n} − a| = |x_{2ℓ−1} − a| = |z_{ℓ} − a| < ϵ for all n = 2ℓ − 1 ≥ 2ℓ_{0} − 1. If n ≥ max(2k_{0}, 2ℓ_{0} − 1), this shows that |x_{n} − a| < ϵ in all cases (i.e. whether n is even or odd). Therefore lim_{n→∞} x_{n} = a, since the definition is satisfied.