$\sup (1/A) = 1/\inf (A)$, for $\inf (A) > 0$

Question

Here is the textbook question:

$A$ is a nonempty set of positive real numbers.

Define $\frac{1}{A} = \{z = \frac{1}{x}\mid x \in A\}$.

Show that $$\sup (\frac{1}{A}) = \frac{1}{\inf (A)}$$

First forgive non math syntax, the question is posted from mobile

now, if $\inf(A) > 1$ it is pretty straight forward

since for every $x \in A$ $\inf (A) \le x$ by simple operations $z = \frac{1}{x} \ge \sup(A)$

but if $\inf(A)$ belong to $]0, 1[$ it is not possible

note that is the answer supplied by the textbook

Answer 1

I think you can just use a straightforward argument. Recall for a moment that $\inf A$ is characterized by the following two properties:

1. $\inf A \le a$ for all $a \in A$, and
2. for every $\epsilon > 0$ there exists $b \in A$ with the property that $b < \inf A + \epsilon$.

Now let $\epsilon > 0$ be given. Choose $b \in A$ (which by hypothesis satisfies $b > 0$) satisfying $b < \inf A + \epsilon$. Since $\dfrac 1b \in \dfrac 1A$ you may take the reciprocal of each side to find $$\frac{1}{\inf A + \epsilon} < \frac 1b \le \sup \frac 1A.$$ Now let $\epsilon \to 0^+$ to conclude $$\frac 1{\inf A} \le \sup \frac 1A.$$

The other direction can be proved in a very similar manner.