Question

Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x is in A. Prove that
inf(A)=-sup(-A)

Answer 1

Let \(y=\inf(A)\). Since \(y\) is the greatest lower bound, you have \(y<w\) for all \(w\in A\).

Since \(-A=\left\{-x~|~x\in A\right\},\) you have \(-y<-w\) for all \(w\in A\); in other words, you have \(y>w\) for all \(-w\in -A,\) which means ...

Answer 2

question: what is inf(A) and sup (A) ?

Answer 3

Thank you @SithsAndGiggles, I don't actually know how to give medals, I'll figure it out in a moment and give you one.

Answer 4

You're welcome!

@Loser66, http://en.wikipedia.org/wiki/Infimum and http://en.wikipedia.org/wiki/Supremum

Answer 5

thanks for the link, I got it

Answer 6

@SithsAndGiggles But wouldn't it be less than or equal to?

Answer 7

@SithsAndGiggles

Why would you have ..."you have −y<−w for all w∈A"

Answer 8

@Zarkon, I must have made some jumps in reasoning. Should it be "you have -y>-w" instead? I'm not sure myself, I've only read the wiki pages to find out what inf and sup mean.

Answer 9

you have y<w therefore -y>-w..which is what you want

Answer 10

Ah, I see. I messed up with the dividing-both-sides-of-an-inequality-by-a-negative thing. Thanks for pointing out the error!