Dedekind completeness: the least-upper-bound property

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

(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 $A\subseteq {\bigcup }_{A^{\mathrm{\prime }}\in S}{A}^{\mathrm{\prime }}=C$, 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.