$\sup (1/A) = 1/\inf (A)$, for $\inf (A) > 0$
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
$\endgroup$ 21 Answer
$\begingroup$I think you can just use a straightforward argument. Recall for a moment that $\inf A$ is characterized by the following two properties:
- $\inf A \le a$ for all $a \in A$, and
- 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.
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