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## 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##