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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?

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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

  1. $\frac{1}{\inf A}$ is an upper bound of $B$.
  2. If $M$ is an upper bound of $B$, then $\frac{1}{\inf A} \leq M$.
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  • $\begingroup$ Ok, I will take a look at it. $\endgroup$ – Peter van de Berg Nov 15 '18 at 13:59
  • $\begingroup$ I am aware that these are just mere hints, but I am not going to give the whole answer for free. I guess you understand that. But if you have any extra questions, or if these hints turn out to be insufficient, feel free to ask extra questions (preferably in you original post, so that everyone can read them). $\endgroup$ – Ernie060 Nov 15 '18 at 14:01
  • $\begingroup$ Thanks for the hint. I will ask you when I need more help. $\endgroup$ – Peter van de Berg Nov 15 '18 at 14:03
  • $\begingroup$ I am still not able to conclude from your hints. I understand your hints but do not know how to give a conclusion. $\endgroup$ – Peter van de Berg Nov 15 '18 at 17:14
  • $\begingroup$ No problem, I have added a few lines. Now you should only fill in the conclusion. I have also given a hint for part (2). $\endgroup$ – Ernie060 Nov 15 '18 at 17:37

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