Greatest lower bound proof

  • Thread starter cragar
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  • #1
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Homework Statement


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

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.
 

Answers and Replies

  • #2
Dick
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Well, yes, that would be true. What's your question?
 
  • #3
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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.
 
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  • #4
Dick
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Im guessing my statement isn't enough to prove it
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?
 
  • #5
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ok sweet, thanks for the help.
 

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