Let $\varnothing\neq A\subseteq \mathbb R^+$, such that $\inf A>0$, then $$\sup\left (\dfrac{1}{A}\right )=\dfrac{1}{\inf A}.$$
My Try: Notice that $\inf A\leq a$ for all $a\in A$. Then, $1/a\leq 1/\inf A$, so $\sup(1/A)\leq 1/\inf A$.
On the other hand, we have $p\leq \sup(1/A)$ for all $p\in \dfrac{1}{A}$. Then, $\dfrac{1}{\sup(1/A)}\leq 1/p$, where $1/p\in A$. So $\dfrac{1}{\sup(1/A)}\leq \inf A$. Then, $\dfrac{1}{\inf A}\leq \sup(1/A)$.
That way we have to $\sup(1/A)\leq \dfrac{1}{\inf A}$ and $\dfrac{1}{\inf A}\leq \sup(1/A)$ Consequently, $\sup(1/A)=\dfrac{1}{\inf A}$. My proof is correct?