Proving S Is Bounded Above: Supremum & Infimum | Asif

[asif zaidi]

I am trying to prove the following. I have a solution below. Can you tell if I am on the right track. P.S. I am doing calculus after 14 yrs so I am very rusty and probably sound stupid


1- Let T be a non-empty subset of R. Assume T is bounded below. Consider the set S = -T = {-t|t is an element of T}. Show that S is bounded above

Solution

a- Let -a= inf(T)
b- -(-a) is also an element of S (because it is a mapping)
c- Let b element of S

And this is where I am getting stuck at.
Intuitively, I know that a > b and it will be the supremum in S but I cannot prove it.

Thanks

Asif

 

[SiddharthM]

Don't bother with sups and infs if you are trying to give a general upper bound for -T.

Take the lower bound of T that is assumed to exist, call it B. We know that B<x for all x inside T. What can you say about -B in relation to -x? Now what is the set -T?

 

[asif zaidi]

If B is the lower bound in T and B<x in T
then -B>-x in all S (as S=-T which is given).

Is it this simple.
So in S, would B not be the least upper bound?

Thanks Siddharth for help

Asif

 

[SiddharthM]

B would be supS IF unless of course B=inf(T), in which case -B would be the least upper bound of S (prove it). But above we only assume that B was a lower bound of T not the GREATEST lower bound of T.