Approximation property of a supremum ($\forall \epsilon >0\exists a\in A:s-\epsilon <a\le s$)

Let $A\subseteq \mathbb{R}$ be nonempty and bounded above. An upper bound $s$ equals $\sup A$ exactly when, for every $\epsilon >0$, some $a\in A$ satisfies $s-\epsilon <a\le s$. If no such member existed, $s-\epsilon$ would be a smaller upper bound. Conversely, the condition defeats every proposed upper bound $u<s$ by taking $\epsilon =s-u$. ^[ref].

For $(0,1)$ and $\epsilon =0.2$, $a=0.9$ is one witness: $0.8<0.9<1$. Choosing $a_n\in A$ with $s-1/n<a_n\le s$ produces a sequence converging to $s$ by the squeeze theorem. The choices need not form a monotone sequence.