Proving the Greatest Lower Bound of a Set Using the Archimedean Property

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Homework Help Overview

The discussion revolves around proving the greatest lower bound of the set \(\{\frac{1}{n}:n\in\mathbb{N}\}\) using the Archimedean property of the real numbers. Participants explore the implications of the Archimedean principle in relation to the set's elements and their bounds.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the greatest lower bound and whether the initial statements provide sufficient proof. There are considerations about the implications of choosing a real number close to zero and the necessity of demonstrating that no positive number can serve as a lower bound.

Discussion Status

The conversation indicates that some participants are questioning the adequacy of their reasoning and whether additional assumptions are needed. There is an acknowledgment of the Archimedean property as a key aspect of the argument, but no consensus has been reached on the completeness of the proof.

Contextual Notes

Participants are navigating the requirements of the homework task, specifically focusing on the definitions and properties related to bounds and the Archimedean principle. The discussion reflects uncertainty about the assumptions that can be made in the proof process.

cragar
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Homework Statement


Use the Archimedean property of \mathbb{R} to prove that
the greatest lower bound of {\frac{1}{n}:n\in\mathbb{N}}=0
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.
 
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Well, yes, that would be true. What's your question?
 
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:
cragar said:
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?
 
ok sweet, thanks for the help.
 

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