1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Upper and lower bounds

  1. Oct 9, 2007 #1
    I was recently looking at the solutions to a problem set I have in analysis and there was something I didn't understand. This is the problem in the problem set:

    1. The problem statement, all variables and given/known data

    Let A ⊂ R be nonempty. Define −A = {−x / x ∈ A}. Show that sup(−A) = − inf A and inf(−A) = − sup A.

    So I thought the definition of a, say, least upper bound, was that [tex]\alpha[/tex] is a least upper bound of E if it is an upper bound of E and if [tex]\gamma < \alpha[/tex] then [tex]\gamma[/tex] is not an upper bound of E. But in the solution, they almost seem to be using a different definition of least upper bound / greatest lower bound.

    3. (part of the solution)

    This is just part of the solution. I'm not really confused about the actual answer to the question, I'm just confused about how they define the least upper bound.

    In the solution, in a case where they assumed A had a lower bound:

    Let us define r := inf A. Then r ∈ R. Since r is a lower bound for
    A,we know that x ≥ r, ∀x ∈ A. It follows that: −x ≤ −r, ∀x ∈ A.
    Hence, −r is an upper bound for A. We must show that it is the least
    upper bound for A. To see this, let e > 0 be given. Since r = inf A,
    it follows that ∀ e > 0, ∃a ∈ A such that a < r + e. So, ∀ e > 0, ∃a ∈ A
    such that −a > −r −e . Hence, since such an element of −A can be
    found ∀ e > 0, it follows that −r is the least upper bound.

    ...What they were doing with e seems almost more like something with a limit point...So I got confused...
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted