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.
okay. So taking your advice I have:
An upper bound: If a set A⊆R a real number s is an upper bound for A if for all x$\displaystyle \in$A, x$\displaystyle \leq$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$\displaystyle \leq$b.
So am I showing that there is a y such that y>s-$\displaystyle \epsilon$ or is that still just the definition. I'm sorry I think it is the wording of the question that is throwing me off.
Is this a step in the right direction?
If t is another upperbound for A, then t$\displaystyle \geq$ x for any x $\displaystyle \in$ A. Since s $\displaystyle \in$ A, we must have that t $\displaystyle \geq$ s. So therefore s is the smallest upper bound or least upper bound.