# Thread: Least Upper Bond Proof. I have no Idea D:.

1. ## Least Upper Bound Proof. I have no Idea D:.

Let S be a bounded set that is not empty.
Prove that sup(S) is a boundary point of S

can someone help me complete this proof? i've been trying for a long time and i can't come up with anything other than

If S is a (not empty) set of real numbers and S has an upper bound y,
then there is a least upper bound of the set S.
That least upper bound is also called the supremum of S

2. I think that you need help with the idea of boundary points.
The point $b$ is a boundary point of $S$ if and only if $\left( {\forall \varepsilon > 0} \right)\left[ {\left( {\exists x \in S} \right)\left( {\exists y \notin S} \right)\left[ {\left\{ {x,y} \right\} \subseteq \left( {b - \varepsilon ,b + \varepsilon } \right)} \right]} \right]$.
In other words, in every $\varepsilon-\mbox{neighborhood}$ of $b$ contains a point of $S$ and a point not in $S$.

Using the definition of supremum, the proof is simple.

3. hmm. see i'm not very good with this. i'm just starting off, and i'm having a hard time with the writing of proofs. it's hard. thanks you though. the definition of a supremum is the LUB on a function over its domain right?kj

4. would it help if i were to plug anything in?>

5. errr, anyone?
i hate to post in a row lke this but i really need help