Let A be bounded above so that s = sup A exists. Show that s Є A bar.
Let $\displaystyle A$ be a bounded non-empty set. Let $\displaystyle s = \sup A$. To show $\displaystyle s \in \bar A$ it is sufficient to show $\displaystyle s \in \partial A$ i.e. is a boundary point. Let $\displaystyle \epsilon > 0$ then $\displaystyle s - \epsilon$ is not an upper bound which means there is $\displaystyle a\in A$ such that $\displaystyle s - \epsilon < a\leq s$, and $\displaystyle s+\epsilon$ is above the supremem thus if $\displaystyle b>s+\epsilon$ then $\displaystyle b\not \in A$. This means $\displaystyle (s-\epsilon,s+\epsilon)$ intersects both $\displaystyle A$ and $\displaystyle \mathbb{R} - A$ non-trivially. It follows that $\displaystyle a\in \partial A$