Mathematics 242 – Principles of Analysis Solutions for Exam 1 – February 21, 2014

Directions: Do all work in the blue exam booklet (do not place any work that you want to have considered for credit on this sheet). There are 100 regular and 10 extra credit points.

I. A) (10) Define: The real number a is a least upper bound of A ⊂ R, and state the Least

Upper Bound Axiom for R.

Solution: a is a least upper bound of A if (1) a ≥ x for all x ∈ A, and (2) if a′ ≥ x for all x ∈ A, then a′ ≥ a. The LUB Axiom states that if A is a nonempty subset of R that is bounded above, then A has a least upper bound in R.

B) (10) Let

A = ⋂_{n=1}^{∞} (1 + \frac{1}{n}, 4 − \frac{1}{n})

Explain why A is bounded and determine the least upper and greatest lower bounds for A.

Solution: When n = 1, we have (1 + \frac{1}{n}, 4 − \frac{1}{n}) = (2, 3). For n > 1, 1 + \frac{1}{n} < 2 and 4 − \frac{1}{n} > 3. Hence the intersection of all the intervals (1 + \frac{1}{n}, 4 − \frac{1}{n}) is equal to the interval for n = 1: A = (2, 3). Hence glb(A) = 2 and lub(A) = 3.

C) (15) Let A be a bounded subset of R with lub(A) = 2, and let B = {−4x+3 | x ∈ A}. What can be said about glb(B)? Prove your assertion.

Solution: glb)(B) must exist and equal −5. Proof: First, 2 is an upper bound for A, so for all x ∈ A, x ≤ 2. Hence −4x ≥ −8, and so −4x + 3 ≥ −5. Next if b is any lower bound for B, we have b ≤ −4x + 3 for all x ∈ A, hence \frac{b−3}{−4} = \frac{3−b}{4} ≥ x for all x in A. Since 2 = lub(A), this implies \frac{3−b}{4} ≥ 2, so 3 − b ≥ 8, and hence b ≤ −5. This shows glb(B) = −5.

II. (15) Let x_{n} be the sequence defined by the rules x_{1} = 1 and x_{n+1} = −\frac{2}{5}x_{n} + 1 for all n ≥ 1. Show by mathematical induction that

0 ≤ x_{n} ≤ 1 for all n ≥ 1.

Solution: The base case for the induction is n = 1 and x_{1} = 1 ≥ 0, so there is nothing else to be proved there. Now for the induction step, assume 0 ≤ x_{k} ≤ 1. Then the \frac{−2}{5} ≤ \frac{−2}{5}x ≤ 0, so \frac{−2}{5} + 1 ≤ \frac{−2}{5}x + 1 ≤ 1, which shows what we want since \frac{−2}{5} + 1 ≥ 0.

III. Let x_{n} = \frac{3n^{2}}{5n^{2}+3n+1} for all natural numbers n ≥ 1.

A) (10) Determine lim_{n→∞} x_{n} intuitively.

Solution: Dividing numerator and denominator by n^{2} have

x_{n} = \frac{3}{5 + \frac{3}{n} + \frac{1}{n^{2}}}

so we expect x_{n} → \frac{3}{5} as n → ∞. B) (20) Use the ε, n_{0} definition of convergence to prove that {x_{n}} converges to the number

you identified in part A. Solution: Let ε > 0. Since N is not bounded in the real numbers, there exists n_{0} in N with n_{0} > \frac{2}{5ε}, so \frac{2}{5n_{0}} < ε. Then for any n ≥ n_{0}, we have

|\frac{3n^{2}}{5n^{2} + 3n + 1} − \frac{3}{5}| = \frac{9n + 3}{25n^{2} + 15n + 5}

< \frac{10n}{25n^{2}}

= \frac{2}{5n}

≤ \frac{2}{5n_{0}}

< ε.

This shows x_{n} → \frac{3}{5} as n → ∞. IV. True-False. For each true statement, give a short proof or reason. For each false statement, give an explicit counterexample. A) (10) If A and B are two nonempty bounded sets of real numbers and lub(B) > lub(A),

then y > x for all y ∈ B and all x ∈ A. Solution: This is FALSE. A counterexample would be something like A = [0, 1] and B = [−1, 2] Then lub(A) = 1 and lub(B) = 2, but B also contains elements smaller than elements in A: −1 ∈ B but −1 < x for some x ∈ A (in fact all x in A, but that would not be necessary to get a counterexample).

B) Let

A = \{x ∈ R | 0 < x < 1, and x = r\sqrt{2} for some r ∈ Q\}.

Then lub(A) = 1. Solution: This is TRUE. First, x < 1 for all x ∈ A by definition. Now, let 0 < a < 1 be any real number. We claim that there are elements of A between a and 1, so that a cannot be an upper bound for A. To see this, note that by a theorem we proved in class, there must be rational numbers r in the interval (\frac{a}{\sqrt{2}}, \frac{1}{\sqrt{2}}). So for any such r,

a < r\sqrt{2} < 1. Hence r\sqrt{2} ∈ A. This shows that a cannot be an upper bound for A. Hence lub(A) = 1.

Extra Credit (10) Is it possible to produce a sequence x_{n} whose terms include all the positive and negative integers? If so, give an indication how to construct such a sequence. If not, give a reason why there cannot exist such a sequence.

Solution: There is such a sequence, and an example can be constructed like this: Let

x_{n} = \{ \begin{matrix}\frac{1−n}{2} & if n is odd \\ \frac{n}{2} & if n is even \end{matrix}

Listing out the first few terms we see

0, 1, −1, 2, −2, 3, −3, …

so it is clear that every integer will be in the set of terms of the sequence. The positive integers come from n even and the non-positive integers come from n odd.