# Greatest lower bound proof

1. Feb 2, 2012

### cragar

1. The problem statement, all variables and given/known data
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

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. Feb 2, 2012

### Dick

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

3. Feb 2, 2012

### 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.

Last edited: Feb 2, 2012
4. Feb 2, 2012

### Dick

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. Feb 2, 2012

### cragar

ok sweet, thanks for the help.