Proving that the Archimedean axiom is true

[major_maths]

Homework Statement

Show that the Archimedean axiom O5 follows from the Least Upper Bound Property O6, together with the other axioms for the reals.

Homework Equations

O5 = [if a,b > 0, then there is a positive integer n such that b<a+a+a+...+a (n summands)] or [if a,b > 0, then b < na or b/a < n]

O6 = if A is any nonempty subset of R that is bounded above, then there is a least upper bound for A.

The Attempt at a Solution

My teacher told us to do this as a proof by contradiction so that's the format I'll be doing.

Suppose the Archimedean axiom is false towards a proof by contradiction. Therefore, there exists some a,b > 0 such that b \geq na, or b/a \geq n.
Then the set, say N, is bounded above by b/a and so sup(N) exists. Write sup(N) = S.

And then I can't figure out how to finish this proof.

 

[micromass]

That S=\sup(\mathbb{N}) means that there is a natural number n that is close to S. But then n+1>S...

Try to formalize this.

 

[major_maths]

If n+1>S, then there exists a natural number not bounded above by S.
This is a contradiction as the set N is the set of whole positive integers and adding 1 would not exclude any n previously in the set N.

Is that right?

 

[micromass]

If n+1>S, then there exists a natural number not bounded above by S.
This is a contradiction as the set N is the set of whole positive integers and adding 1 would not exclude any n previously in the set N.

Is that right?

Yes, that is correct. How would you choose n though?

 

[major_maths]

I would choose n to be close to S, or S-1<n<S.

So for any n bounded above by S but greater than S-1, if n+1>S, then there exists a natural number not bounded above by S.

And so on, and so on. Is that specific enough?

 

[micromass]

I would choose n to be close to S, or S-1<n<S.

So for any n bounded above by S but greater than S-1, if n+1>S, then there exists a natural number not bounded above by S.

And so on, and so on. Is that specific enough?

Yes, but you need to state why such an n exists. You probably know it, but I want to make sure.

 

[major_maths]

Oh. Um, it exists because S is the supremum of the set?

 

[micromass]

Oh. Um, it exists because S is the supremum of the set?

Yes, do you understand why?

 

[major_maths]

S is the supremum of the set because the set is bounded above by b/a, which is what sup(N[/]) is defined as at the beginning of the proof (b/a ≥ n).