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Supremum & inifimum

  1. Oct 25, 2007 #1
    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


    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.


  2. jcsd
  3. Oct 25, 2007 #2
    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?
  4. Oct 25, 2007 #3
    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

  5. Oct 25, 2007 #4
    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.
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