Let $H$ be a nonempty set and let $f:H\to\mathbb R$ be such that $f(H)=\{f(h): h \in H\}$ is a bounded subset of $\mathbb R$. Prove that ...

Question

Let $H$ be a nonempty bounded subset of $\mathbb R$ such that $\inf H>0$.

Show that $\frac1{\inf H} = \sup\{\frac1h : h\in H\}$

Proof:

$\sup\{\frac1h : h\in H\} = \frac1{\inf H}$

By definition, if $H$ is bounded there exists $a,b\in H$ such that $a\leq a_1\leq...x\leq....\leq b$.

\begin{align}\sup\{\frac1h : a\leq x\leq b\} = \frac1{\inf H}\\ \frac1a = \sup\{\frac1h:a\leq x\leq b\}\\ \frac1a = \sup\{\frac1h:a\leq x\leq b\} = \frac1{\inf H}\end{align}

$\inf H = a$ since by definition any $k$ such that $a>k$ implies $k\notin H$.

$\frac1a=\sup\{\frac1h:a\leq x\leq b\} = \frac1{\inf H}=\frac1a$

Answer 1

Let $$A=\{\frac 1h , h\in H\}$$

$$a\in A \implies$$ $$\exists h\in H \;: a=\frac 1h .$$

$$ h\in H \implies h\ge \inf H>0 \implies$$ $$ a\le\frac{1}{\inf H} $$ thus $\frac{1}{\inf H}$ is an upper bound for $A$.

$\inf H$ is limit of a sequence $(h_n)$ of elements in $H$ with $h_n\ge \inf H>0$.

but $$\frac{1}{\inf H}=\lim_{n\to+\infty} \frac{1}{h_n}$$

thus $\frac{1}{\inf H}$ is an upper bound and limit of a sequence of elements in $A$ . therefore it is the $\sup$.