This is very similar to this question, whose answer I do not understand. I want to proved that for $A \subseteq \mathbb{R}$:
$$ \inf(A^{-1}) = \sup(A)^{-1} $$
where $A^{-1} = \{\frac{1}{a} \mid a \in A\}$ and it is given that $\sup(A) < 0$.
My first attempt was this:
Let $i := \inf(A^{-1})$, then we know:
$$ \begin{align} &(1) \quad \forall a \in A: \frac{1}{a} \ge i \\ &(2) \quad \forall \epsilon > 0 \in \mathbb{R}: \exists a \in A: \frac{1}{a} < i + \epsilon \end{align} $$
from which it follows that:
$$ \begin{align} &(1) \quad \forall a \in A: a \leq \frac{1}{i} \\ &(2) \quad \forall \epsilon > 0 \in \mathbb{R}: \exists a \in A: a > \frac{1}{i + \epsilon} \end{align} $$
Now if I could show that $\frac{1}{i + \epsilon} \geq \frac{1}{i} - \epsilon$ that would show that $\sup(A) = \frac{1}{i}$ and conclude the proof. But I don't know how to show that that's true.
My second attempt was this:
Choose an arbitrary $\epsilon > 0$, then pick an $a \in A$ such that $a > \sup(A) - \epsilon$. Then we have:
$$ \frac{1}{\sup(A) - \epsilon} > \frac{1}{a} \geq \inf(A^{-1}) $$
and if we let $\epsilon$ go to zero we get:
$$ \frac{1}{\sup(A)} \geq \inf(A^{-1}) $$
but I don't know how to show that it also holds that:
$$ \frac{1}{\sup(A)} \leq \inf(A^{-1}) $$
because if repeat the procedure, picking an $a$, such that $\frac{1}{a} < \inf(A^{-1}) + \epsilon$ then I also get:
$$ \frac{1}{\inf(A^{-1}) + \epsilon} < a \leq \sup{A} $$
i.e.
$$ \frac{1}{\sup(A)} \geq \inf{A^{-1}} $$
again.
How can fix one or both of these approaches? And where do I actually need to use $\sup(A) < 0$? Is the equality not true for $\sup(A) > 0$?
Here is a simple approach.
We know that for any $a$ in A, $$ a \leq Sup(A)$$ Or $$\frac{1}{Sup(A)} \leq \frac{1}{a}$$
So $Sup(A)^{-1}$ is a lower bound to $A^{-1}$. Since $Inf(A^{-1})$ is greatest lower bound to $A^{-1}$ we must have $Sup(A)^{-1} \leq Inf(A^{-1})$.
Now similarly $\frac{1}{Inf(A^{-1})}$ is an upper bound to the set $A$, $Sup(A)^{-1} \geq Inf(A^{-1})$. So we have your equality.