# Completeness axiom/Achimedean principle and sup set.

1. Nov 19, 2012

### Max.Planck

1. The problem statement, all variables and given/known data
Let $$B=\left\{{\frac{1}{2},\frac{2}{3},\frac{3}{4},...}\right\}$$
Prove sup B = 1

2. Relevant equations
Archimedean principle:
Let a<b and a>0 $$\exists n \in{N}$$ such that an > b.

3. The attempt at a solution
Its trivial to see that 1 is an upper bound for B and B is nonempty, so B must have a finite supremum by the completeness axiom.

I want to show that 1 is the smallest upper bound. So I want to try to do a proof by contradiction:
Let M be an upper bound for B, and M<0.

I know each element in B is a/b, with natural numbers a and b and a<b. But how to continue further??

Last edited: Nov 19, 2012
2. Nov 19, 2012

### jbunniii

Well, surely you know more than that:
$$\frac{1}{2} = 1 - \frac{1}{2}$$
$$\frac{2}{3} = 1 - \frac{1}{3}$$
$$\frac{3}{4} = 1 - \frac{1}{4}$$
Does this suggest a formula for the elements of B?

3. Nov 20, 2012

### Max.Planck

Yes, each element in B is of the form:

$$1-\frac{1}{k} , k>1, k\in{N}$$

So intuitively its easy to see you can choose a k such that |1-(1-1/k)|<ε, for ε>0.

4. Nov 20, 2012

### Max.Planck

I want to prove that there exists a k, such that 1-1/k > M, when M<1.

5. Nov 20, 2012

### Dick

Pick ε=1-M>0. Then your Archimedean principle tells you there is a k such that kε>1. So ε>1/k. So?

6. Nov 20, 2012

### Max.Planck

Yes thanks, I think this is correct.

So the proof will look like this.

To prove, sup B = 1

Proof:
(1) Let M be an upper bound for B and M<1.

(2) 1-M > 0 (follows from 1)

(3) Exists natural number n s.t. : n(1-M) > 1 (Archimedean principle)

(4) (1-M) > 1/n (First multiplicative propery)

(5) -M > 1/n -1 (Follows from 4)

(6) M < 1 - 1/n (Contradiction, since M is an upper bound for B and 1-1/n is in B)

(7) Conclusion: 1 must be the least upper bound, therefore sup B = 1

7. Nov 20, 2012

### Dick

That looks ok to me. Since you know 1 is an upper bound, it must be the least upper bound.