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

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.