# Thread: Prove that s is the sup of A.

1. ## Prove that s is the sup of A.

Could some one point me in the right direction with this proof to get me started? I'm not sure where to begin.

If a set A ⊆ R contains one of its upper bounds s, prove that s is the supremum of A.

2. Write out explicitly:

1. The definition of an upper bound.
2. The definition of a supremum.

3. Originally Posted by Mazerakham
Write out explicitly:

1. The definition of an upper bound.
2. The definition of a supremum.

An upper bound: If a set A⊆R a real number s is an upper bound for A if for all x $\in$A, x $\leq$s.

a supremum: a real number s is least upper bound for A if s is an upper bound for A having the property that, if b is also an upper bound for A then s $\leq$b.

So am I showing that there is a y such that y>s- $\epsilon$ or is that still just the definition. I'm sorry I think it is the wording of the question that is throwing me off.

4. Is this a step in the right direction?

If t is another upperbound for A, then t $\geq$ x for any x $\in$ A. Since s $\in$ A, we must have that t $\geq$ s. So therefore s is the smallest upper bound or least upper bound.

5. Originally Posted by alice8675309
Is this a step in the right direction?

If t is another upperbound for A, then t $\geq$ x for any x $\in$ A. Since s $\in$ A, we must have that t $\geq$ s. So therefore s is the smallest upper bound or least upper bound.

Right on the nose, and thus s is in fact the maximum of A as A contains it.

Tonio

6. Yep. I sketch could have helped too. Never try to do a symbolic proof without a picture in your head! The statement is of course intuitively true if you visualize it, and the proof should be connected to that intuition. Nice job.