Let $A \subseteq (0,\infty$) such that $\inf A>0$. Let $\frac{1}{A}=\{\frac{1}{a},a\in A\}$. Prove that $\sup \frac{1}{A}=\frac{1}{\inf A}$.

Question

Let $A \subseteq (0,\infty$) such that $\inf A>0$. Let $\frac{1}{A}=\{\frac{1}{a},a\in A\}$. Prove that $\sup \frac{1}{A}=\frac{1}{\inf A}$.

My attempt:

For $a\in A$ we have $$a\geq \inf A>0 \Rightarrow \frac{1}{a}\leq\frac{1}{\inf A}$$ $\Rightarrow \frac{1}{\inf A}$ is an upper bound of the set $\frac{1}{A}$. Now we have to prove that it is the least upper bound. So we have to prove that $(\forall \varepsilon>0)(\exists\frac{1}{a}\in\frac{1}{A})$ such that $\frac{1}{\inf A} -\varepsilon<\frac{1}{a}$. How do I continue from here?

Answer 1

Note $\inf A=\alpha$ is the infimum of $A$ if and only if

(1). For $\forall a\in A$, $a\ge \alpha$;

(2). For $\forall \epsilon>0$, there is $a\in A$ such that $\alpha\le a\le \alpha+\epsilon$.

Now from (1), one has

(1)' For $\forall \frac1a\in \frac1A$, $a\in A$ and hence $a\ge \alpha$ or $\frac1a\le\frac1\alpha$.

From (2), one has

(2)'. For $\forall \epsilon\in(0,\frac1\alpha)$, let $\epsilon'=\frac{\epsilon\alpha^2}{1-\epsilon\alpha}$. There is $a\in A$ such that $\alpha\le a\le \alpha+\epsilon'=\frac{1}{\frac1\alpha-\epsilon}$. Therefore $$\frac{1}{\alpha}-\epsilon<\frac1a\le\frac1\alpha. $$ From (1)'(2)', one can conclude that $\sup \frac1A=\frac1\alpha$.