# Help, analsysis proof - inf (-A)=-sup(A) [-A= {-a : a∈A}]

help...

prove:
let　A be a set of real numbers bounders above.
then, inf (-A)=-sup(A) [-A= {-a : a∈A}]

really...thanks......i have no idear..how to prove it....?

the question didn't mention if A is a non-empty set...
so do i need to prove A is a non-empty set?
Or it said is real numbers, then it means is non empty set?

You don't have to prove A is non-empty.

A is a set of real numbers already means there exists an element x which is a real number and belongs to A. Since there exists such an element, thus A is non-empty.

HallsofIvy
Let b be any member of -A. Then b=-a for some a in A. We must have $a\le sup(A)$ so, multiplying by -1, $-a= b\ge -sup(A)$. That tells you that -sup(A) is a lower bound on -A. I'll leave the rest to you.