Greatest lower bound problem - Rudin POMA Ch1 Exercise 5

[EdMel]

Homework Statement

5. 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
$$\inf A=-\sup(-A)\text{.}$$

Homework Equations

The Attempt at a Solution

Does the proof below look OK? I am a bit uneasy about just multiplying through (2) by -1 to get (4). Thanks in advance.

Proof: As $A$ is bounded below $\inf A$ exists, let's call this $\alpha$. Then we can say
(1) $\forall x\in A$, $x\geq\alpha$, and
(2) if $\gamma>\alpha$ then $\exists y\in A$ such that $\alpha\leq y<\gamma$ (in other words $\gamma$ is not a lower bound of A).
To get set $-A$ we must multiply each member of $A$ by $-1$. So if $x\geq\alpha$ is true then $-x\leq-\alpha$ is true, and as $-x$ is an arbitrary member of $-A$ it must be that
(3) $\forall-x\in-A$, $-x\leq-\alpha$,
so $-\alpha$ is an upper bound of $A$. Similarly, multiplying the inequalities in (2) by $-1$ means that
(4) if $-\gamma<-\alpha$ then $\exists-y\in-A$ such that $-\alpha\geq-y>-\gamma$,
so that $-\alpha$ is the least upper bound of $-A$. Thus we can write $\sup(-A)=-\alpha=-\inf(A)$ or $-\sup(-A)=\inf(A)$, which is what we needed to show. $\square$

 

[jbunniii]

Yes, your proof looks OK. If multiplying (2) by -1 to get (4) makes you uneasy, then try writing it in a bit more detail. For example:

If \gamma &lt; -\alpha, then \alpha &lt; -\gamma, so by (2), there exists y \in A such that \alpha \leq y &lt; -\gamma. But y \in A iff -y \in -A, and \alpha \leq y &lt; -\gamma iff \gamma &lt; -y \leq -\alpha, so this shows that there is an element of -A, namely -y, which is larger than \gamma. Therefore...