# Harrison Chapman

## Math 317: Homework 2

Due: Friday, September 13, 2019
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

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

3. Give examples of;

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

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

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

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

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 $$\lim a_n = 2/3$$.

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

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.

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.