Let $S$ be a bounded and nonempty subset of $\mathbb{R}^{+}$. Prove that $\sup(\frac{1}{S}) = \frac{1}{\inf S}, \inf S > 0$

Question

Define $$ \frac{1}{S} = \{\frac{1}{s} | s \in S\}$$.

Please check my attempt

Proof. Let $\frac{1}{s} \in \frac{1}{S}$. Thus $\inf S \leq s$, hence $\frac{1}{s} \leq \frac{1}{\inf S}$, meaning $\frac{1}{\inf S}$ is an upper bound of $\frac{1}{S}$ and $\sup(\frac{1}{S}) \leq \frac{1}{\inf S}.$

We now show that $\frac{1}{\inf S}$ is the least upper bound of $\frac{1}{S}$. Suppose $x$ is an upper bound of $\frac{1}{S}$. Then $\forall s \in S,$

$$\frac{1}{s} \leq x \implies \frac{1}{x} \leq s.$$

Thus $\frac{1}{x}$ is a lower bound of $S$ and $\frac{1}{x} \leq \inf S$. Thus $\frac{1}{\inf S} \leq x$, so $\frac{1}{\inf S}$ is the least upper bound of $\frac{1}{S}$, therefore $\sup(\frac{1}{S}) = \frac{1}{\inf S}. \square$

Answer 1

Try like this.

By $\inf S \leq s$ for $s$ in $S$ you have that $1/(\inf S) \geq 1/s $ for $s$ in $S$. Thus $1/(\inf S) $ is a majorant. But if $s_n \to \inf S $ then $1/s_n \to 1/(\inf S) $, thus $1/(\inf S) = \sup S^{-1}$ by definition.