= Approximation property of a supremum
{title2=$\forall\varepsilon>0\;\exists a\in A:\ s-\varepsilon<a\leq s$}
Let $A\subseteq\mathbb R$ be nonempty and bounded above. An <upper bound> $s$ equals $\sup A$ exactly when, for every $\varepsilon>0$, some $a\in A$ satisfies $s-\varepsilon<a\leq s$. If no such member existed, $s-\varepsilon$ would be a smaller <upper bound>. Conversely, the condition defeats every proposed <upper bound> $u<s$ by taking $\varepsilon=s-u$. <Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl/1.2 The set of real numbers>[Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl; Lebl, Basic Analysis I, §1.2: approximation of a supremum]{ref}.
For $(0,1)$ and $\varepsilon=0.2$, $a=0.9$ is one witness: $0.8<0.9<1$. Choosing $a_n\in A$ with $s-1/n<a_n\leq s$ produces a <sequence> converging to $s$ by the <squeeze theorem>. The choices need not form a <monotone sequence>.
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