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.