Let $A$ be a nonempty set of positive real numbers, and consider the set
$$B = \{ \tfrac1a \mid a\in A \}.$$
I have to prove the following:
1) The set $B$ is bounded above if and only if $\inf(A)> 0$.
2) If $\inf(A)> 0$, then $\sup(B)=\frac1{\inf(A)}$
If $B$ is bounded above, there is a real $M$ such that $a \geq \frac{1}{M}$ for all $a \in A$.
If $\inf(A) > 0$, this means that there is a positive lower bound $m$ for $A$.
If $A$ is bounded below, there is a real $m$ such that $a \geq m$ for all $a \in A$.
How do I conclude from here?
Best Answer
Hints:
(1)($\Rightarrow$) if $B$ is bounded above, there is a positive real $M$ such that $\frac{1}{a}\leq M$ for all $a\in A$. This means that $\frac{1}{M} \geq a$ for every $a \in A$. Since $1/M > 0$ we have a positive lower bound of $A$. By definition $\inf A$ is the biggest lower bound of $A$, so we conclude that ...
($\Leftarrow$) If $\inf(A)>0$, this means that there is positive lower bound $m$ for $A$ (we can take for instance $m=\inf A$). This means that $m \leq a$ for all $a\in A$. Hence $\frac{1}{m} \ldots \frac{1}{a}$ (fill in inequality, here we use the fact that $m>0$). So we have found that ...
(2) The supremum $\sup B$ is the smallest upper bound of $B$. So you have to prove that