# Supremum Proof

1. Sep 7, 2009

### jgens

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

Consider the open interval $(a,b)$. Prove that $\mathrm{sup}{(a,b)} = b$.

2. Relevant equations

N/A

3. The attempt at a solution

I'm terrible at these proofs so I would appreciate it if someone could verify (or correct) my solution.

Proof: Clearly $b$ is an upper bound since $\forall{x} \in (a,b)$ we have the strict inequality $a < x < b$. Now, suppose that $b$ is not the least upper bound. Letting $c = \mathrm{sup}{(a,b)}$ there must be some real $\varepsilon > 0$ such that $b - \varepsilon = c$. However, since $\varepsilon > 0$ this implies that $\frac{\varepsilon}{2} > 0$ and similarly that $b > b - \frac{\varepsilon}{2} > b - \varepsilon = c$ which contradicts the fact that $c = \mathrm{sup}{(a,b)}$. This completes the proof.

I know that my proof is definately wordy, but is it correct? Thanks for any suggestions!

2. Sep 7, 2009

### LeonhardEuler

You may have the basics of a proof there, but it is definitely unclear. Why dose $b > b - \frac{\varepsilon}{2} > b - \varepsilon = c$ contradict the assumption that c is the supremum? If you clear that up, then the proof should be fine. Remember, there are 2 properties c should satisfy to be the supremum: It should be an upper bound, and every other upper bound should be greater than it. Which, if any, of these properties have you shown c not to satisfy?

3. Sep 7, 2009

### JG89

The proof seems fine to me.

4. Sep 7, 2009

### snipez90

Yeah it looks clear to me. I mean the OP demonstrates that b - e/2 is in the open interval and (a,b) is greater than the supposed sup (a,b), so there's really not much else to say...

5. Sep 7, 2009

### LeonhardEuler

But he doesn't. He doesn't explicitly say it and he doesn't restrict e to eliminate the possibility that b-e/2 < a so that it is not in the interval.

6. Sep 7, 2009

### JG89

He did say it. He said "which contradicts the fact that $$c = sup{(a,b)}$$ . This completes the proof. ".

Also, it's pretty obvious that b - e is not less than a (because it's the LUB for the set), and so b - e/2 cannot be less than a.

I see what you mean, but in my opinion, being that wordy would just make the proof longer than it has to be, without really contributing anything.