Harrison Chapman

Math 317: Homework 2

1. For each subset of $\mathbb R$ below, determine both the supremum and the infimum. You do not need to give a rigorous proof of your answer.

   1. $$A = \{ 3, 5, 1, 4 \}$$
   2. $$B = \left\{\dfrac {(-1)^n}n \;:\; n \in \mathbb N \right\}$$
   3. $$C = \left\{\cos\left(\dfrac{n\pi}{3}\right) \;:\; n \in \mathbb N \right\}$$
   4. $$D = \{ r \;:\; r \in \mathbb Q,\; r^2 < 5 \}$$
   5. $$E = \left\{ 1 - \dfrac{1}{2^n} \;:\; n \in \mathbb N \right\}$$

   1. Supremum 5, infimum 1

   2. Supremum 1/2, infimum -1

   3. Supremum 1, infimum -1

   4. Supremum $\sqrt 5$, infimum $-\sqrt 5$

   5. Supremum 1, infimum 1/2

   Show solution

2. Let $A$ be a nonempty, bounded set of real numbers. Let $-A$ be the set $\{ -x \;:\; x \in A \}$. Prove that

   $$\sup A = -\inf(-A).$$

   Let $M = \sup A$, which exists and is real as $A$ is bounded. So $\forall a \in A, M \ge a$. Then $\forall y=-a \in -A$, $M \ge a \Rightarrow -M \le -a = y$. So $-M$ is a lower bound for $-A$.

   Suppose $L$ is any lower bound for $-A$. So $\forall a \in A$, $L \le -a \Leftrightarrow -L \ge a$ so $-L$ is an upper bound for $A$, and in particular $-L \ge M$ as $M$ is the least upper bound of $A$, and we see that $L \le -M$, so $-M$ is the greatest lower bound for $-A$.

   That is, $\inf(-A) = -\sup A \Leftrightarrow \sup A = -\inf(-A).$

   Show solution

3. Give examples of;

   1. A sequence of irrational numbers converging to a rational number

      For instance, $(s_n) = (\pi/n)$.

      Show solution

   2. A sequence of rational numbers converging to an irrational number

      For instance, $(s_n) = ((1+\frac 1n)^n)$.

      Show solution

4. For each of the following sequences, determine its limit (usual Calculus reasoning will be helpful here), then prove that the sequence does indeed converge to this limit.

   1. $$a_n = \frac {2n-3}{3n+4}$$

      Let $\epsilon > 0$. Consider $N = \frac 19 \left( \frac {17}{\epsilon} - 12 \right)$. Then for $n > N$,

      $$\left\vert \frac {2n-3}{3n+4} - \frac 23\right\vert = \frac {17}{9n + 12} < \frac {17}{9N + 12} = \epsilon.$$

      So $\lim a_n = 2/3$.

      Show solution
   2. $$s_n = \frac {\cos n}{n}$$

      Let $\epsilon > 0$. Consider $N = 1/\epsilon$ and $n > N$. Then

      $$\left\vert \frac {\cos n}n \right\vert = \frac{\vert \cos n \vert}{\vert n \vert} \le \frac 1{\vert n \vert} = \frac 1n < \frac 1N = \epsilon.$$

      So $\lim (\cos n)/n = 0$.

      Show solution

5. Let $(s_n)$ be a sequence in $\mathbb R$.

   1. Prove that $\lim{s_n} = 0$ if and only if $\lim{ s_n^2 } = 0$. (Hint: This is an “if and only if” statement, so you will need to prove both directions of the statement.)

      ($\Rightarrow$) Suppose $\lim s_n = 0$. Consider $\epsilon > 0$. Then there exists $N$ so that $n > N$ implies that $\vert s_n \vert < \sqrt{\epsilon}$.

      So then $\vert s_n ^2 - 0 \vert = s_n ^2 < \epsilon$. So $\lim s_n^2 = 0$.

      ($\Leftarrow$) The reverse direction is very similar to the forward direction.

      Show solution

   2. Prove that if $s_n = (-1)^n$ that $\lim{s_n^2}$ exists but $s_n$ diverges.

      We have that $s_n$ diverges by Example 4 in Section 8 (pp41-42). Also, $s_n^2 = 1$ which converges to $1$ as it is constant.

      Show solution