# Least Upper Bound Property Proof

• Sep 7th 2008, 11:23 AM
Least Upper Bound Property Proof
Prove that an nonempty set S that is bounded above has a least upper bound.

Proof.

For this problem, we must use the following way to show the proof:

Let $u_0 \in S$ and $N_0$ be an upper bound of S.

Let $v_{0} = \frac {u_0 + N_0 }{2}$.

If $v_0$ is an upper bound, let $U_1 = v_0$ and $u_1 = x_0$

If $v_0$ is not an upper bound, then let $N_1 = N_0$ and $u_1 > v_0 \ \ \ u_1 \in S$

Repeat the process to obtain sequence $N_n$ and $u_n$, we have to show that they both converge to the LUB (S).

To be honest, I'm a bit lost on how to generate $N_n$ and $u_n$, so if $v_0$ is not an upper bound, that means $N_0$ is still the least upper bound since $v_0 \in S$, right? Suppose it is not true, then we let $v_1$ be something lower, or closer to the sequence, but how should the sequences $N_n$ and $u_n$ look like?

Thank you very much!
• Sep 7th 2008, 11:57 AM
Plato
Quote:

Prove that an nonempty set S that is bounded above has a least upper bound.

There is a real problem is offering any help in this case.
This statement is usually known as the Completeness Axiom.
If you are asked to prove what is an axiom in one development then there must be an axiom in the other development that helps prove the statement. Can you tell us what axioms you have been given that deal with bounded sets?
What are your axioms to this point?
• Sep 7th 2008, 12:16 PM
So far we know that:

A convergent sequence is bounded.

A montone sequence that is bounded converges.

And for the set S in the problem, S is a nonempty set in the set of real numbers.
• Sep 7th 2008, 01:38 PM
Plato
Quote:

So far we know that:A convergent sequence is bounded. A montone sequence that is bounded converges. And for the set S in the problem, S is a nonempty set in the set of real numbers.

It is clear that you will use “A montone sequence that is bounded converges.” However, I still do not know what else you have proven about the structure of the real numbers. Here is a fact that may help recall what you have proved about the real numbers. If S contains a point of U, then that point is the LUB of S. So if there is no LUB of S both S & U can be shown to be open. Have you done anything with connectivity?
• Sep 7th 2008, 05:38 PM
We haven't done anything with connected sets yet (at least not in this course). We went over field axioms and order axioms, well-ordered property, completeness property, and we talked about dense and equivalent classes. Haven't talk about open, close, or connected.
• Sep 8th 2008, 03:32 AM
Plato
Quote:

We haven't done anything with connected sets yet (at least not in this course). We went over field axioms and order axioms, well-ordered property, completeness property, and we talked about dense and equivalent classes. Haven't talk about open, close, or connected.

Well, I think that is what I asked to bengin with: "What is the statement of the completeness property?"
Because this is equivalent to that.
• Sep 8th 2008, 05:34 AM
The statement is: "An ordered field is said to be complete if it obeys the monotone sequence property".

I was working on this problem, and I understand the idea of the proof now.

So if a is an upper bound, then M would change; if a is not an upper bound, then x would change. In other words, $M_n$ is a sequence that is appraoching sup(S) from the positive side while $X_n$ does the same from the negative side.

Since, both sequences are monotonic, M being decreasing while X increasing, and they are both bounded by the sup(S), since if they move outside of their respective area they stop moving, sup(S) is the GLB of M and the LUB of x.

It is just that, I don't know how to write this correctly, since I do know what excatly $M_n$ and $x_n$ equals to. But here is what I think:

So I have $M_n = \frac { M_{n-1} + x_0 }{2}$

On the other hand, I have: $x_n > \frac { x_{n-1} +M_0 }{2}$

Now, by construction, M has to be decreasing and x has to be increasing, and they are both bounded by sup(S), but how do I show that?

Any hints would be appreicated, thank you!