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.
Supremum 5, infimum 1
Supremum 1/2, infimum -1
Supremum 1, infimum -1
Supremum \(\sqrt 5\), infimum \(-\sqrt 5\)
Supremum 1, infimum 1/2
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, \(\inf(-A) = -\sup A \Leftrightarrow \sup A = -\inf(-A). \)
Give examples of;
A sequence of irrational numbers converging to a rational number
For instance, \((s_n) = (\pi/n)\).
A sequence of rational numbers converging to an irrational number
For instance, \((s_n) = ((1+\frac 1n)^n)\).
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.
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 \(\lim a_n = 2/3\).
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 \(\lim (\cos n)/n = 0\).
Let \((s_n)\) be a sequence in \(\mathbb R\).
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.
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.