1. Not finding help here? Sign up for a free 30min 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!

Prove Sup S = c

Tags:
  1. Sep 26, 2015 #1
    1. The problem statement, all variables and given/known data
    S≡{x|x∈ℝ,x≥0,x2 < c} Prove Sup S = c

    2. Relevant equations


    3. The attempt at a solution
    Since x in the set real numbers, there are two cases for x: x < 1 or 0 <= x <=1

    if 0 <= x <=1, then x < c + 1 since c is positive.
    if x < 1, then x^2 < c < x*c + x = x(c+1)
    thus x < c+1, by the completeness axiom, S has a least upper bound, let's called it b.

    I am stuck at this step, can anyone give me a hit or a suggestion to keep going. Thanks
     
  2. jcsd
  3. Sep 27, 2015 #2

    Mark44

    Staff: Mentor

    It would seem to me that the two cases are ##0 \le x \le 1## and ##x > 1##. Is there some reason you're not considering ##x > 1##?
     
  4. Sep 27, 2015 #3
    Set S consists of all real numbers ##x## such that ##x\geq 0## and ##x^2 < C##.
    We know ##C>0##, for if ##C<0##, there exists no real number that satisfies ##x^2 < C## and if ##C=0##, then ##x^2 < 0##, which is also impossible.
    Provided that ##C>0## then we can say set S is non-empty and then apply the axiom:
    If a non-empty subset has an upper bound, then it has a least upper bound.

    If we establish ##b## as the candidate for the least upper bound, then ##b## satisfies certain critera:
    1) ##x\leq b## for every ##x\in S## (in other words, ##b## has to be an upper bound)
    2)For every ##\varepsilon >0##, there exists an ##x_0## such that ##b-\varepsilon < x_0## (meaning ##b## is the least of upper bounds)
    EDIT:
    The second criterion could also be written as:
    2a)For every ##y\in\mathbb{R}## such that ##y < b##, there exists an ##x_0\in S## such that ##y < x_0##
    2 and 2a mean the same thing.

    We can agree that ##b## is an upper bound, now fix an arbitrary ##\varepsilon >0## and show that ##b## is indeed the least of upper bounds and in the finale it follows immediately after that ##\sqrt{C} = b##.
     
    Last edited: Sep 27, 2015
  5. Sep 27, 2015 #4
    Surely sup S = sqrt c not c ? Unless I'm missing something ?
     
  6. Sep 27, 2015 #5
    I meant ##\sqrt{C}## thanks for the correction :)
     
  7. Sep 27, 2015 #6
    ok then so the proof of the OP would proceed as follows:

    suppose Sup S = sqrt c + epsilon where epsilon > 0. Then (Sup S - epsilon /2)^2 > c (contradicting the definition of Sup S)
    suppose Sup S = sqrt c - epsilon. Then (sup S + epsilon/2)^2 <c (again contradicting the definition of Sup S)
    so Sup S = sqrt c
     
  8. Sep 27, 2015 #7
    Well, hurts my eyes without TeX, for one, but more importantly, do not post complete solutions. I merely referred to the definition in an attempt to hint at how to proceed.
     
  9. Sep 27, 2015 #8
  10. Sep 27, 2015 #9
    That is a typo
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted