Thread: [SOLVED] Help finishing my proof for: Prove that inf A = -sup(-A)

Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
Prove that inf A = -sup (-A).

My work:

Proof: Let A be non empty subset of R and -A = {-x : x in A}. If -x is in -A then
-x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
-sup (-A) is a lower bound of A.


Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
Any help is greatly appreciated. Thank you for your time!

Originally Posted by ilikedmath

Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
Prove that inf A = -sup (-A).

My work:

Proof: Let A be non empty subset of R and -A = {-x : x in A}. If -x is in -A then
-x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
-sup (-A) is a lower bound of A.

your work is incomplete. showing it is a lower bound is not enough. you need to show it is the greatest lower bound. there are ways to do that elegantly. i will use the fact that for .

Let and be as defined. By the definition of infimum, we have that for all . so that for all . but that means is an upper bound for . thus, since the supremum is the least upper bound, we must have ........(1)

Also, for all by the definition of supremum. but that means for all , so that is a lower bound for the set . since the infimum is the greatest lower bound, we must have that ...........(2)

By (1) and (2) we have , as desired

Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
Any help is greatly appreciated. Thank you for your time!

yes, we can assume the supremum and infimum exist. they are and in the extreme cases

thanks for the help!