Dedekind completeness: the least-upper-bound property

(C1) $C$ is nonempty and proper: $S$ has a member $A_0$ with $A_0 \ne \emptyset$ and $A_0 \subseteq C$, so $C \ne \emptyset$; and every $A \in S$ satisfies $A \subseteq B$, so $C = \bigcup_{A \in S} A \subseteq B$ with $B \ne \mathbb{Q}$, hence $C \ne \mathbb{Q}$.

(C2) $C$ is downward closed: if $p \in C$ then $p \in A$ for some $A \in S$; for $q < p$, downward closure of $A$ gives $q \in A \subseteq C$.

(C3) $C$ has no greatest element: if $p \in C$ then $p \in A$ for some $A \in S$; as $A$ has no greatest element there is $r \in A$ with $r > p$, and $r \in C$.

$C$ is an upper bound for $S$: every $A \in S$ satisfies $A \subseteq \bigcup_{A' \in S} A' = C$, i.e. $A \le C$.

$C$ is below every upper bound: if a cut $D$ satisfies $A \le D$ for all $A \in S$, then $A \subseteq D$ for all $A$, so $C = \bigcup_{A \in S} A \subseteq D$, i.e. $C \le D$.

$C$ is a Dedekind cut.

Therefore $\sup S$ exists and equals $C = \bigcup_{A \in S} A$: $\mathbb{R}$ has the least-upper-bound property. With The Dedekind reals form a totally ordered field, $\mathbb{R}$ is a complete totally ordered field, the order-completeness of the Dedekind construction, the exact counterpart of Cauchy-sequence completeness.