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 $\epsilon >0$ be arbitrary.

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

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

By totality either $u\le u^{\mathrm{\prime }}$ or $u^{\mathrm{\prime }}<u$; in the second case put ${\epsilon }_0:=u-u^{\mathrm{\prime }}$, so that ${\epsilon }_0>0$ and $u-{\epsilon }_0=u^{\mathrm{\prime }}$.

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

In that second case the hypothesis applied to ${\epsilon }_0$ yields $s_0\in S$ with $u^{\mathrm{\prime }}=u-{\epsilon }_0<s_0$, so $s_0\le u^{\mathrm{\prime }}$ fails, contradicting that $u^{\mathrm{\prime }}$ is an upper bound of $S$; the second case is therefore impossible and $u\le u^{\mathrm{\prime }}$.

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

Since $u^{\mathrm{\prime }}$ was an arbitrary upper bound of $S$, we get $u\le u^{\mathrm{\prime }}$ for every upper bound $u^{\mathrm{\prime }}$; 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 $\epsilon >0$ there is $s\in S$ with $u-\epsilon <s$.