For each subset of 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 be a nonempty, bounded set of real numbers. Let be the set . Prove that
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 . Consider . Then for ,
So .
Let . Consider and . Then
So .
Let be a sequence in .
Prove that if and only if . (Hint: This is an “if and only if” statement, so you will need to prove both directions of the statement.)
() Suppose . Consider . Then there exists so that implies that .
So then . So .
() The reverse direction is very similar to the forward direction.
Prove that if that exists but diverges.
We have that diverges by Example 4 in Section 8 (pp41-42). Also, which converges to as it is constant.