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

1. Sep 19, 2007

### ShengyaoLiang

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....?

2. Sep 19, 2007

### ShengyaoLiang

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?

3. Sep 19, 2007

### cks

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.

4. Sep 20, 2007

### HallsofIvy

Staff Emeritus
Notice that a "set of real numbers" is not a "subset of the set of real numbers" since the latter might be empty! It is more common for a theorem like this to start "Let A be a non-empty subset of R."

To prove your theorem- keep multiplying by -1!

For example, sup(A) is, by definition, an upper bound on A.

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.