# Proving inf A=-sup(-A)

## Homework Statement

Classic proof, I saw it before but I can't remember it.
A, a subset of R is bounded below. -A is -a where a is in A
I don't know why I'm having such a difficult time.

## The Attempt at a Solution

Because A is bounded below, there is some m such that m≤a for all a in A. Simply multiply by -1 to get -m≥-a for all -a in -A, proving that -A has an upper bound. By the completeness axiom, -A has a least upper bound. Lets call this u. u has two properties. u≥-a for all -a in -A. Also v≥u for all other upper bounds v. Multiply by -1. -u≤a for all a in A AND -v≤-u for all lower bounds -v. Therefore, -u is the inf of A. Therefore inf A=-sup(-A).

Is this okay? I feel uneasy.

## Answers and Replies

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HallsofIvy
Science Advisor
Homework Helper
First, any number in -A is of the form -x for some x in A. Then, a is a lower bound for A: for any x in A, $a\le x$. Therefore, $-a\ge -x$. -a is an upperbound on -A.

Further, a is the greatest lower bound of A. If b is another lower bound on A, b< a. Let p be an upper bound on -A. $p\ge x$ for all x in A. Then, again since any number in -A is of the form -a for some a in A, $p\ge -a$. Can you complete that?

First, any number in -A is of the form -x for some x in A. Then, a is a lower bound for A: for any x in A, $a\le x$. Therefore, $-a\ge -x$. -a is an upperbound on -A.

Further, a is the greatest lower bound of A. If b is another lower bound on A, b< a. Let p be an upper bound on -A. $p\ge x$ for all x in A. Then, again since any number in -A is of the form -a for some a in A, $p\ge -a$. Can you complete that?
We can't assume that A has a greatest lower bound, this is just one part of a proof of that.

Is there anything wrong with my proof? It seems very similar to yours.