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Greatest lower bound proof

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the Archimedean property of [itex] \mathbb{R} [/itex] to prove that
    the greatest lower bound of [itex] {\frac{1}{n}:n\in\mathbb{N}}=0 [/itex]
    the archimedean principle says that for any number y there is a natural number
    such that 1/n<y for y>0

    3. The attempt at a solution
    since all of our numbers in our set are positive. I could pick a real number as close to zero as I wanted but there would still be a natural such that 1/n is smaller than the real I picked, there zero is the greatest lower bound of the set.
     
  2. jcsd
  3. Feb 2, 2012 #2

    Dick

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    Well, yes, that would be true. What's your question?
     
  4. Feb 2, 2012 #3
    Im guessing my statement isn't enough to prove it. Or should I assume that there is a real just to the right of zero and claim that this is the greater lower bound, but then there is always another number 1/n that is smaller and since all 1/n are in the set, that zero has to be the greatest lower bound.
     
    Last edited: Feb 2, 2012
  5. Feb 2, 2012 #4

    Dick

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    I don't see why not. If y>0 then y is not a lower bound of {1/n} by the Archimedean property. If y=0, then it is since 1/n>0 for all n. Doesn't that make y=0 the greatest lower bound?
     
  6. Feb 2, 2012 #5
    ok sweet, thanks for the help.
     
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