Mathematics 242 – Principles of Analysis Selected Solutions for Problem Set 5, “B” portion March 18, 2011

§2.3/15. The idea of this problem is to show that if we assume that every bounded monotone sequence converges, then it follows that every nonempty set A that is bounded above has a least upper bound. (In other words, we will show that the statement of the Completeness Axiom follows from the Monotone Convergence Theorem. We proved the opposite implication in class, and both of these statements are also equivalent to the Nested Interval Theorem(!)) So we start with a set A ⊂ R that is bounded above. We construct two sequences x_{n} and y_{n} as described in the problem. We take x_{1} ∈ A an arbitrary element and we let y_{1} in R be any upper bound for A. Then assuming x_{n} and y_{n} have already been constructed, let z_{n} = \frac{x_{n}+y_{n}}{2} and consider two possible cases for determining x_{n+1} and y_{n+1}:

• Case 1: If z_{n} is an upper bound for A, then we let x_{n+1} = x_{n} and y_{n+1} = z_{n}.

• Case 2: If, on the other hand z_{n} is not an upper bound for A, then there exists some a′ ∈ A with a′ > z_{n}, so we let x_{n+1} = a′ and y_{n+1} = y_{n}.

We note the following:

• By construction, x_{n} ∈ A for all n and y_{n} is an upper bound of A for all n.

• The sequence {x_{n}} is monotone increasing and bounded above by y_{1} since all x_{n} ∈ A.

• The sequence {y_{n}} is monotone decreasing and bounded below by any x_{m}.

(a) It follows now that {x_{n}} and {y_{n}} are both convergent sequences, since our hypothesis is that every bounded monotone sequence in R converges.

(b) Since z_{n} is the midpoint of the line segment from x_{n} to y_{n} for each n, we note that for all n ≥ 1,

y_{n+1} − x_{n+1} ≤ \frac{1}{2}(y_{n} − x_{n}).

(We have equality in Case 1 and the strict inequality in Case 2 as above.) If we “unwind” this back to the beginning of the construction, we see

y_{n+1} − x_{n+1} ≤ \frac{1}{2}(y_{n} − x_{n})

≤ \frac{1}{2^{2}} (y_{n−1} − x_{n−1})

⋮

≤ \frac{1}{2^{n}} (y_{1} − x_{1}).

Hence if we let n → ∞, we see that

lim_{n→∞}(y_{n} − x_{n}) = 0,

since \frac{1}{2^{n}} → 0 as n → ∞. That implies

lim_{n→∞} y_{n} = lim_{n→∞} x_{n}

by part (a) and Theorem 2.2.5a.

(c) Call the limit of {x_{n}} and {y_{n}} α ∈ R. We want to show that α satisfies the properties of the least upper bound of A. First, since every y_{n} is an upper bound for A, we have x ≤ y_{n} for all x ∈ A and all n. It follows from Theorem 2.2.7 that x ≤ α for all x ∈ A. Hence α is an upper bound for A. On the other hand, if ε > 0, then since x_{n} → α, there exist n_{0} ∈ N such that α − ε < x_{n} ≤ α for all n ≥ n_{0}. Since the x_{n} ∈ A for all n, this implies that α − ε is not an upper bound for A. It follows that α = lub(A).

§2.5/12. Both parts of this are TRUE. For part (a) suppose that we have an sequence of positive numbers {x_{n}} with x_{n} → 0 as n → ∞. Then we want to show how to produce a strictly decreasing subsequence. We can take x_{n_{1}} = x_{1} to start. Now suppose we have found

x_{n_{1}} > ⋯ > x_{n_{k}}.

Letting ε = x_{n_{k}} in the definition of convergence we see that there exists some N_{0} such that |x_{n} − 0| = x_{n} < ε = x_{n_{k}} for all n ≥ N_{0}. Letting n_{k+1} be any one such integer that is also > n_{k} we get x_{n_{k+1}} satisfying

x_{n_{1}} > ⋯ > x_{n_{k}} > x_{n_{k+1}}.

By induction, this shows there exists a strictly decreasing subsequence.

Now for (b), if infinitely many x_{n} > 0, then we can apply the same reasoning as in (a) to get a strictly decreasing subsequence chosen from the positive terms in the whole sequence. If there are not infinitely many x_{n} > 0, then there must be infinitely many x_{n} < 0 and we can repeat essentially the same argument above with the negative terms. We can let n_{1} be any index such that x_{n_{1}} < 0. Then assuming we have found

x_{n_{1}} < ⋯ < x_{n_{k}},

let ε = |x_{n_{k}}| in the definition of convergence. we see that there exists some N_{0} such that |x_{n} − 0| = |x_{n}| < ε = |x_{n_{k}}| for all n ≥ N_{0}. Letting n_{k+1} be any one such integer that is also > n_{k} and such that x_{n_{k+1}} < 0, we get x_{n_{k+1}} satisfying |x_{n_{k+1}}| < |x_{n_{k}}|, which implies x_{n_{k+1}} > x_{n_{k}} (since both are negative). Hence

x_{n_{1}} < ⋯ < x_{n_{k}} < x_{n_{k+1}}.

By induction, this shows there exists a strictly increasing subsequence.

13. We suppose that {x_{n}} is a bounded sequence with the property that every convergent subsequence converges to the same limit a. Aiming for a contradiction, suppose x_{n} ↛ a.

This is equivalent to saying there exists some ε_{0} > 0 such that for all n_{0} ∈ N, |x_{n} −a| ≥ ε_{0} for some n ≥ n_{0}.

Let ε_{0} > 0 be any fixed number for which it is true that

(*) for all n_{0} ∈ N, |x_{n} − a| ≥ ε_{0} for some n ≥ n_{0}.

We claim that there is a subsequence {x_{n_{k}}} with the property that |x_{n_{k}} − a| ≥ ε_{0} for all k ∈ N. To see this, we apply (*) above with n_{0} = 1, and we get a first term x_{n_{1}} for n_{1} ≥ 1 satisfying |x_{n_{1}} − a| ≥ ε_{0}. Then assuming that x_{n_{1}}, …, x_{n_{k}} have been constructed, apply the statement (*) above with n_{0} = n_{k} + 1. We get n_{k+1} ≥ n_{k} + 1 such that |x_{n_{k+1}} −a| ≥ ε_{0}. Continuing in the same way (i.e. by induction) this shows our claim.

It is clear that the subsequence {x_{n}} constructed here cannot converge to a since all the terms are uniformly bounded away from a. Moreover, no subsequence of this subsequence can converge to a either, for the same reason.

However, since the original sequence was bounded, this subsequence is bounded too. Hence the Bolzano-Weierstrass theorem applies and shows that there must be a convergent subsequence of the {x_{n_{k}}}. That convergent subsequence is also a subsequence of the original {x_{n}}. Hence the assumption implies it must converge to a. This contradicts what we said in the last paragraph.

Hence x_{n} must converge to a. (Wow!)