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:

$$ \frac{1}{\inf(A^{-1}) + \epsilon} < a \leq \sup{A} $$


$$ \frac{1}{\sup(A)} \geq \inf{A^{-1}} $$


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 $\sup(A) > 0$?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.