- #1

- 395

- 14

## 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.

## Homework Equations

## 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.