Proof that $\sup \frac{1}{A} = \frac{1}{\inf A}$

Question

I know this question was already asked a couple of times, but I didn't understand any of the answers (or they were solved without $ \epsilon $ which is want I want to practice according to my book), so I want to ask it again. Here is the question from a text book:

Let $A$ be a nonempty $\subset \mathbb{R}$ such that $\alpha:= \inf A > 0.\ $ Define $\frac{1}{A} := \left\{ \frac{1}{a} : a \in A \right\}.\ $ Prove that $\sup \frac{1}{A} = \frac{1}{\inf A}.$

This is what I got so far: $ 0 < \alpha \leq a \implies \frac{1}{\alpha} \geq \frac{1}{a} $, so $\frac{1}{\alpha}$ is an upper bound for $ \frac{1}{A} $. The next step would be to show that it is the smallest upper bound, so there always exists $a_0$ such that $ \frac{1}{a_0} > \frac{1}{\alpha} - \epsilon $ for all $\epsilon > 0$

This is where I'm stuck. A hint in the text book is to take a look at $ \frac{1}{\alpha} - \frac{1}{a} $ which is obviously $ \geq 0 $ but I don't see how that is supposed to help me.

Answer 1

Suppose $\frac{1}{\alpha}$ is not the least upper bound, then there is some other upper bound, call it $\frac{1}{a'} $ for which

$$0<\frac{1}{a'}<\frac{1}{\alpha} $$

Then since $\frac{1}{a'}$ is an upper bound of $\frac{1}{A}$, we have that

$$ \forall_{a\in A}[\frac{1}{a'} \geq \frac{1}{a}] $$

Or reversing the ratios,

$$ \forall_{a\in A}[a'\leq a] $$

So $a'$ is a lower bound for $A$

Then since $\frac{1}{a'}<\frac{1}{\alpha}$ we get that $ a'>\alpha$ which means there is some lower bound for $A$ which is larger that $\alpha$, in contradiction to $\alpha$ being the greatest lower bound.

Answer 2

As requested in the comments, this is how I would prove it.

Well, we need to prove that $\sup \frac{1}{A} = \frac{1}{\alpha}.$ By the definition of $\sup,$ this means that we need to prove two things:

$(1)\quad \frac{1}{\alpha}$ is an upper bound of $\frac{1}{A}.$

$(2)\quad $ If $x < \frac{1}{\alpha},\ $ then $x$ is not an upper bound of $\frac{1}{A}.$

Well, where do we start? Well, we already have that $A\subset \mathbb{R}$ such that $\alpha = \inf A>0.$ that is;

(i) $\alpha>0$ is a lower bound of $A,\ $ and

(ii) If $\beta > \alpha,$ then $\beta$ is not a lower bound of $A.$

(i) says that $0<\alpha \leq a\ \forall\ a\in A,\ \implies 0 < \frac{1}{a} \leq \frac{1}{\alpha}\ \forall a\in A,\ $ i.e. $\frac{1}{\alpha}\ $ is an upper bound of the set $ \frac{1}{A}.$ This proves $(1).$

Next, we try to prove $(2).$ So suppose $x < \frac{1}{\alpha}.\ $

We now split into two cases,$\ x \leq 0\ $ and $\ x>0.$

Consider $\ x \leq 0.\ $ Since $\ A\ $ is nonempty and $\ \inf A > 0,\ \exists\ a\in A\ $ such that $a>0.$ Therefore $\ x\ $ is not upper bound of $A,\ $ and so $(2)$ holds if $\ x \leq 0.$

So suppose $\ x>0,\ $ so that we now have $\ 0<x < \frac{1}{\alpha}.\ $ This implies $\ \frac{1}{x} > \alpha (>0),\ $ and so by (ii), $\ \frac{1}{x}\ $ is not a lower bound of $A.$ Therefore $\ \exists\ \mu\in A\ $ such that $\ (0 < ) \alpha \leq \mu < \frac{1}{x}.\ $ Since $\ \mu \in A,\ \frac{1}{\mu} \in \frac{1}{A}.\ $ This together with the fact that $\ \frac{1}{\mu} > x\ $ implies that $\ x\ $ is not an upper bound of $\ \frac{1}{A}.$ This proves $(2)$ and completes the proof.