# Greatest lower bound problem - Rudin POMA Ch1 Exercise 5

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

## 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##

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jbunniii
Homework Helper
Gold Member
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 $\gamma < -\alpha$, then $\alpha < -\gamma$, so by (2), there exists $y \in A$ such that $\alpha \leq y < -\gamma$. But $y \in A$ iff $-y \in -A$, and $\alpha \leq y < -\gamma$ iff $\gamma < -y \leq -\alpha$, so this shows that there is an element of $-A$, namely $-y$, which is larger than $\gamma$. Therefore...