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Infimums and Darboux sums

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    If b>a, find the infimum of b - a for arbritrary b,a.


    2. Relevant equations
    -


    3. The attempt at a solution
    So this is for a darboux sum but I'm stuck on proving that 0 is the inf. It seems really easy but I can't seem to think straight.

    First thing is 0 is a lower bound of b - a.
    b - a > 0
    b > a

    Next is to prove it's the greatest lower bound. I tried to do this by supposing x > 0 is the greatest lower bound and find some contradiction but I can't figure it out.
     
  2. jcsd
  3. Feb 18, 2010 #2

    Dick

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    Re: Infimum

    Ok, go ahead. Suppose x>0 is a lower bound. Pick b=x/2 and a=0. Is x a lower bound for b-a?
     
  4. Feb 18, 2010 #3
    Re: Infimum

    Does that not only prove it then when a = 0? I guess I might as well say what I'm doing. Basically, I'm trying to get the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. I got the lower darboux sum and then when I try to do the upper darboux sum, I get [tex]\Sigma[/tex](xj+1 - xj)sup f. So if I expand this sum, I get 0 for every interval except the interval that contains 0 ie [xk,xk+1]. The upper darboux sum is xk+1 - xk. Clearly, the inf is 0 but I need to prove this.
     
  5. Feb 18, 2010 #4

    Dick

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    Re: Infimum

    Well, no. The assumption that was that x>0 was a lower bound of b-a for ALL b>a. If it's not a lower bound for SOME b>a, then x>0 is NOT a lower bound. But you are thinking about this way too generally. Of course you can make the upper sum as small as you want by picking x_(j+1)-x_j as small as you want. You might make the interval containing 0 be [-1/n,1/n]. So sure, you can make that as close to zero as you want by picking n large enough.
     
  6. Feb 22, 2010 #5
    Re: Infimum

    Thanks for that. Got it all sorted out now.
     
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