Proving that Supremum of (a,b) is Equal to b

[jgens]

Homework Statement

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

Homework Equations

N/A

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 definitely wordy, but is it correct? Thanks for any suggestions!

 

[LeonhardEuler]

Homework Statement

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

Homework Equations

N/A

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 definitely wordy, but is it correct? Thanks for any suggestions!

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?

 

[JG89]

The proof seems fine to me.

 

[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...

 

[LeonhardEuler]

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...

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.

 

[JG89]

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.

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.