Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of lower bound of a nonempty set of real numbers

  1. Feb 16, 2008 #1
    1. Let [itex] A [/itex] be a nonempty set of real numbers which is bounded below. Let [itex] -A [/itex] be the set of numbers [itex] -x [/itex], where [itex] x \in A [/itex]. Prove that [itex] \inf(A) = -\sup(-A) [/itex].

    Intuitively this makes sense if you draw it on a number line. But I am not sure how to formally prove it.
     
  2. jcsd
  3. Feb 16, 2008 #2
    If -x is in -A then -x<sup(-A)
    this implies:
    x>-sup(-A)

    Therefor -sup(-A) is a lower bound of A.

    Now establish it is the least lower bound (probably via contradiction).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of lower bound of a nonempty set of real numbers
  1. Nonempty convex sets (Replies: 10)

Loading...