Show that $\sup(\frac{1}{A})=\frac{1}{\inf A}$ 
Given nonempty set $A$ of positive real numbers, and define  $$\frac{1}{A}=\left\{z=\frac{1}{x}:x\in A \right\}$$ Show that $$\sup\left(\frac{1}{A}\right)=\frac{1}{\inf A}$$


let $\sup\left(\frac{1}{A}\right)=\alpha$ and $\inf A = \beta$. Apply the definition of supremum, $z<\alpha$, then there exist $z'\in 1/A$ such that for every $\epsilon>0$, $z'>\alpha-\epsilon$
And the definition of infimum, $x>\beta$ and there exists $x'\in A$ such that for every $\epsilon>0$, $x'<\epsilon+\beta$. 
At this step, I don't see how to relate $\beta$ with $\alpha$ which is $\alpha=\frac{1}{\beta}$, can anyone give me a hit or suggestion. Thanks. 

The crucial facts here are that if $x \in A$, then $x>0$ and the function
$x \mapsto {1 \over x}$ reverses order in $A$, that is, if $x,y \in A$, then $x<y$ iff${1 \over x } > {1 \over y}$.
We have $\sup_{x' \in A} {1 \over x'} \ge {1 \over x}$ for all $x \in A$.
Now let $x_n \in A$ such that $x_n \to \inf A$. Then this gives
 $\sup_{x' \in A} {1 \over x'} \ge {1 \over \inf A}$.
Since $x \mapsto {1 \over x}$ reverses order in $A$, we have
${1 \over \inf A} \ge {1 \over x}$ for all $x \in A$. Taking the
$\sup$ yields the desired answer,
${1 \over \inf A} \ge \sup_{x \in A} {1 \over x}$.
