[proofplan] We prove both implications directly from the least-upper-bound definition of $\sup A$. If $s=\sup A$, then $s$ is an upper bound, and every number strictly below $s$ fails to be an upper bound; applying this to $s-\varepsilon$ gives an element of $A$ within $\varepsilon$ below $s$. Conversely, if $s$ is an upper bound and every interval $(s-\varepsilon,s]$ meets $A$, then every number $t<s$ fails to be an upper bound by choosing $\varepsilon=s-t$. Hence $s$ is the least upper bound of $A$. [/proofplan] [step:Derive the epsilon witnesses from the least-upper-bound property] Assume $s=\sup A$. By the definition of supremum, $s$ is an upper bound for $A$, so for every $a \in A$ one has $a \leq s$. Let $\varepsilon \in \mathbb{R}$ satisfy $\varepsilon>0$. Since $\varepsilon>0$, we have \begin{align*} s-\varepsilon < s. \end{align*} Because $s$ is the least upper bound of $A$, no real number strictly smaller than $s$ is an upper bound for $A$. Therefore $s-\varepsilon$ is not an upper bound for $A$. By the negation of the definition of upper bound, there exists $a \in A$ such that \begin{align*} s-\varepsilon < a. \end{align*} Since $s$ is an upper bound for $A$ and $a \in A$, we also have $a \leq s$. Thus \begin{align*} s-\varepsilon < a \leq s. \end{align*} Since $\varepsilon>0$ was arbitrary, the required epsilon condition holds. [guided] Assume $s=\sup A$. The meaning of this statement is twofold: first, $s$ is an upper bound for $A$; second, $s$ is the least among all upper bounds of $A$. From the upper-bound part, every element $a \in A$ satisfies \begin{align*} a \leq s. \end{align*} Now fix an arbitrary real number $\varepsilon>0$. We want to find an element of $A$ that lies above $s-\varepsilon$. The reason to look at $s-\varepsilon$ is that it is strictly smaller than $s$: \begin{align*} s-\varepsilon < s. \end{align*} Since $s$ is the least upper bound, any real number strictly below $s$ cannot itself be an upper bound for $A$. Therefore $s-\varepsilon$ is not an upper bound for $A$. Unpacking the negation of upper boundedness at the specific number $s-\varepsilon$, there must exist some $a \in A$ such that \begin{align*} s-\varepsilon < a. \end{align*} The same element $a$ also satisfies $a \leq s$, because $s$ is an upper bound for all of $A$. Combining the two inequalities gives \begin{align*} s-\varepsilon < a \leq s. \end{align*} Since the choice of $\varepsilon>0$ was arbitrary, this proves the epsilon condition. [/guided] [/step] [step:Use the epsilon witnesses to prove leastness of the upper bound] Assume that $s$ is an upper bound for $A$ and that for every $\varepsilon>0$ there exists $a \in A$ such that \begin{align*} s-\varepsilon < a \leq s. \end{align*} To prove $s=\sup A$, it remains to prove that no real number strictly smaller than $s$ is an upper bound for $A$. Let $t \in \mathbb{R}$ satisfy $t<s$. Define $\varepsilon := s-t$. Then $\varepsilon>0$. By the assumed epsilon condition, there exists $a \in A$ such that \begin{align*} s-\varepsilon < a \leq s. \end{align*} Since $\varepsilon=s-t$, we have $s-\varepsilon=t$, and hence \begin{align*} t<a. \end{align*} Thus $t$ is not an upper bound for $A$. Since every $t<s$ fails to be an upper bound, while $s$ is an upper bound for $A$, $s$ is the least upper bound of $A$. Therefore \begin{align*} s=\sup A. \end{align*} [/step]