# Supremum, Infimum (Is my proof correct?)

1. Apr 11, 2016

### Incand

1. The problem statement, all variables and given/known data
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that
$\inf A = -\sup(-A)$.

2. Relevant equations
Definition:
Suppose $S$ is an ordered set, $E\subset S$, and $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:
(i) $\alpha$ is an upper bound of $E$.
(ii) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E$.
Then $\alpha$ is called the supremum of $E$ and we write $\alpha = \sup E$.
(Equivalently for infimum)

3. The attempt at a solution
From the definition of supremum $\exists \alpha > y, \forall y \in -A$ or equivalently $\exists \alpha > -x, \forall x \in A$.
Then $-\alpha < x, \forall x \in A$, hence $-\alpha = -\sup(-A)$ is a lower bound of $A$.

It's left to show that if $\gamma > -\alpha$ then $\gamma$ is not an lower bound of $A$.
Suppose $\gamma$ is a lower bound of $A$ with $\gamma > -\alpha$. Then $\gamma < x \forall x\in A$ or equivalently $-\gamma > -x \forall x \in A$. But this means that $\gamma$ is an upper bound of $-A$ and since $\sup(-A) = \alpha$ we have that $-\gamma \ge \alpha$ or equivalently $\gamma \le -\alpha$ a contradiction! Hence $\inf A = -\sup(-A)$.

Is the above correct? Anything I could do to improve it? I'm quite new to proofs so I'm not sure if I'm doing this right.

2. Apr 11, 2016

### Samy_A

Maybe I'm nitpicking, but it seems that you set $\alpha=-\sup(-A)$, and then prove that $-\alpha=\inf A$.
Shouldn't it be the other way around? You are given that $A$ is bounded below, meaning $\inf A$ exists. Set $\alpha=\inf A$ and prove that $\alpha = -\sup(-A)$.
Or, you could do it as you did, but then first prove that if $A$ has a lower bound, $-A$ has an upper bound.

3. Apr 11, 2016

### Incand

Good point! I'm going to go with trying to prove that $-A$ have an upper bound. It seemed easier to start on the more complicated side.
$A$ is bounded below, that is $\exists \xi \in \mathbf R$ such that $\xi < x, \forall x\in A$. But this means that $-\xi > -x, \forall x\in A$ that is $-\xi > y, \forall y \in -A$, hence $-A$ is bounded above since $-\xi \in \mathbf R$.

If I put that in before the start of my earlier proof, would that do it?

Edit: I'm also using the proposition that
If $x<0$ and $y<z$ then $xy>xz$
And others, but I guess I don't have to write thing like this out for every step.

4. Apr 11, 2016

Looks fine.