(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Proving inf A=-sup(-A)

**Physics Forums | Science Articles, Homework Help, Discussion**