Supremum/Infimum proof: $\sup \{1/x; x\in A\} = 1/\inf(A)$

Question

Assume that $\inf(A)>0$ and let $A'=\left\{\frac{1}{x} : x\in A\right\}$. I need to show that $\sup(A') = \dfrac{1}{\inf(A)}$.

I think this is quite simple, $\sup(A')$ must be $\dfrac{1}{\inf(A)}$ since $\inf(A)$ is the smallest number to divide by, that is $x > \inf(A)$ for all $x\in A$.

Maybe I made a mistake, but if not is this sufficient? Or, how could I make this more formal?

EDIT I could probably make this formal by proving this by contradiction...

Answer 1

What you’ve said is a reasonable intuitive explanation of the result, but it’s not a proof. In order to show that $\sup A'=\dfrac1{\inf A}$, you must show two things:

1. $x\le\dfrac1{\inf A}$ for each $x\in A'$, and
2. if $u<\dfrac1{\inf A}$, then there is an $x\in A'$ such that $x>u$.

In other words, you must show that $\dfrac1{\inf A}$ is an upper bound for $A'$, and that no smaller number is an upper bound for $A'$. Neither of these is difficult, but you need to do them in order to have a proof.

For (1), for instance, if $x\in A'$, then $x=\dfrac1a$ for some $a\in A$, and $a\ge\inf A>0$, so $$x=\frac1a\le\frac1{\inf A}\;.$$

I’ll let you take a stab at (2) on your own.

Answer 2

Here is an abstract way to approach the problem. As a warm-up, try showing that if $f$ is an order-preserving function $\mathbb{R}^+ \to \mathbb{R}^+$ then $f(\inf A) = \inf f[A]$ where $f[A]$ denotes the pointwise image of $A$ under $f$. Roughly speaking this is because the notion of infimum just has to do with the ordering and not with any algebraic structure, but you can prove it rigorously from the definition of infimum. Once you have done this there is an easy modification for order-reversing functions like $x \mapsto 1/x$.

Answer 3

Here, we complete the answer of Brian M.Scott.

For (2): First, we note that since $\inf(A)>0$ then $A\subset \mathbb{R}^*_+$ so if we take $u\leq 0$ s.t $\displaystyle u<\frac{1}{\inf A}$ then any $a\in A$ verify $a\geq u$.

Now suppose that there's $u>0$ s.t $\displaystyle u<\frac{1}{\inf A}$ then $\displaystyle \frac{1}{u}>\inf A$ and by characterization of the infimum of $A$ there's $a\in A$ s.t $\displaystyle a\leq \frac{1}{u}$ so $x=\displaystyle \frac{1}{a}\in A'$ and $x\geq u$ and this allows us to conclude.