# Prove that $\inf(A^{-1}) = \sup(A)^{-1}$

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 $$\sup(A) < 0$$.

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.

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 $$\sup(A) > 0$$?

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.