Epsilon characterisation of the supremum

For the forward implication assume $u = \sup S$, that is, $u$ is an upper bound of $S$ that is $\le$ every upper bound of $S$, and let $\varepsilon > 0$ be arbitrary.

For the converse implication assume that $u$ is an upper bound of $S$ such that for every $\varepsilon > 0$ there exists $s \in S$ with $u - \varepsilon < s$, and let $u'$ be an arbitrary upper bound of $S$.

Since $u - (u - \varepsilon) = \varepsilon > 0$, we have $u - \varepsilon < u$.

By totality either $u \le u'$ or $u' < u$; in the second case put $\varepsilon_0 := u - u'$, so that $\varepsilon_0 > 0$ and $u - \varepsilon_0 = u'$.

The element $u - \varepsilon$ is not an upper bound of $S$: if it were, the leastness of $u$ among upper bounds would give $u \le u - \varepsilon$, which contradicts $u - \varepsilon < u$ by trichotomy.

In that second case the hypothesis applied to $\varepsilon_0$ yields $s_0 \in S$ with $u' = u - \varepsilon_0 < s_0$, so $s_0 \le u'$ fails, contradicting that $u'$ is an upper bound of $S$; the second case is therefore impossible and $u \le u'$.

Failing to be an upper bound of $S$ means precisely that some $s \in S$ does not satisfy $s \le u - \varepsilon$, and by totality that says $u - \varepsilon < s$; since $\varepsilon > 0$ was arbitrary, the forward implication is proved.

Since $u'$ was an arbitrary upper bound of $S$, we get $u \le u'$ for every upper bound $u'$; as $u$ is itself an upper bound, $u$ is a least upper bound of $S$, hence $u = \sup S$ by uniqueness, which proves the converse implication.

Both implications hold, so for an upper bound $u$ of a nonempty set $S$ bounded above, $u = \sup S$ if and only if for every $\varepsilon > 0$ there is $s \in S$ with $u - \varepsilon < s$.