Supremum not in interior of set $A$

Question

If $ A \neq \emptyset $ and is bounded above and $ s = \sup A $ then $ s \in \bar A $ and $ s \notin \operatorname{int}A$

I've got the first part, showing that $s$ is a limit point of $A$. Thus, $ s \in \bar A$. But, how to go about showing the second part?

Answer 1

If $s\in Int(A)$, then there is a $\delta>0$ s.t. $$(s-\delta,s+\delta)\subset A$$ and this is a contradiction with the fact that $s=\sup A$ (since if $x\in A$, by definition of the supremum, $x\leq s$, whereas $s+\delta>s$ and $s+\delta\in A$).

Answer 2

Actually, no interior point can be an upper bound, let alone the supremum.

Indeed, let $b$ be a point in the interior of $A$. Then there is $\varepsilon >0$ such that $(b-\varepsilon,b+\varepsilon) \subseteq A$. In particular, $b+\varepsilon/2 \in A$ and so $b$ cannot be an upper bound for $A$.