1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Sup A < sup B

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data
    If sup A<sup B, then show that there exists an element b[tex]\in[/tex]B that is an upper bound for A.

    2. Relevant equations

    3. The attempt at a solution
    This one is really frustrating me. I'm having trouble even beginning.
    I know sup A implies s[tex]\leq[/tex]b where c is an upper bound for A.
    Similarily sup B implies s[tex]\leq[/tex] d where d is an upper bound for B.
  2. jcsd
  3. Sep 5, 2010 #2
    [tex]sup B= sup A+ \epsilon[/tex]

    Remember what sup b is and it's closeness to elements of B.
    Last edited: Sep 5, 2010
  4. Sep 5, 2010 #3
    I have no clue how you got to that.
  5. Sep 5, 2010 #4
    Well if x < y that means y= x+b where b= y-x.
  6. Sep 5, 2010 #5
    so x<x+b
    So that statement alone shows that b is an upper bound?
  7. Sep 5, 2010 #6
    I made an error while writing the latex. Please review my edit.
  8. Sep 5, 2010 #7
    [tex]sup B- \epsilon= sup A[/tex]
    Since sup B is the least upper bound what can we say about
    [tex]sup B- \epsilon_{0}[/tex] ?
    Is it in B?

    The point it that if we pick [tex] \epsilon_{0} [/tex] small enough we can find a number that is both in B and is an upper bound of A.
  9. Sep 6, 2010 #8
    [tex]sup B- \epsilon_{0}[/tex] must be an upper bound also?
  10. Sep 6, 2010 #9
    Do you know the definition of [tex] sup(B)[/tex] ?

    Given that sup(B) is the least upper bound this means that it is the smallest number that is an upper bound of the set B.

    That means given any [tex]\epsilon_{0}[/tex], no matter how small, there exist [tex]b_{0}\inB[/tex] such that
    [tex]b_{0} > sup(B) -\epsilon_{0}[/tex].

    Lets start again.

    [tex] sup(B) - sup(A) = \alpha >0[/tex]

    If we pick any [tex]\epsilon_{0}[/tex] such that [tex]\epsilon_{0} < \alpha[/tex] we can find an element in B such that
    [tex]b_{0} > sup(B) -\epsilon[/tex]. So [tex]b_{0}[/tex] is an upper bound of A.
  11. Sep 6, 2010 #10


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You could prove this by contradiction as well. Assume there is no element in B that is an upper bound for A, and then show it leads to the conclusion sup(A)>sup(B), which contradicts the condition sup(A)<sup(B).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook