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

[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...?

 

[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?

 

[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.

 

[HallsofIvy]

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.