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

[cragar]

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.

 

[Dick]

Well, yes, that would be true. What's your question?

 

[cragar]

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.

 

[Dick]

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?

 

[cragar]

ok sweet, thanks for the help.