Order on the Dedekind reals

For Dedekind cuts $A, B \in \mathbb{R}$ (Dedekind cut, The real numbers $\mathbb{R}$ as Dedekind cuts), the order is set inclusion:

$A \le B \iff A \subseteq B, \qquad A < B \iff A \subsetneq B.$

That is, $A < B$ means $A \subseteq B$ and $A \ne B$.

A cut $A$ is called positive iff $0^{*} < A$, and nonnegative iff $0^{*} \le A$, where $0^{*} = \{\, r \in \mathbb{Q} : r < 0 \,\}$ is the cut of $0$. Negative and nonpositive cuts are defined symmetrically: $A < 0^{*}$ and $A \le 0^{*}$.

Positivity has a convenient rational restatement: $0^{*} \le A$ iff every rational $r < 0$ lies in $A$, and $0^{*} < A$ iff moreover $0 \in A$. Indeed if $0 \in A$ then downward closure (C2) forces every $r < 0$ into $A$, so $0^{*} \subseteq A$ and the inclusion is proper; conversely $0^{*} \subsetneq A$ supplies some $q \in A$ with $q \ge 0$, whence $0 \in A$ by (C2). Thus a cut is positive exactly when it contains $0$.

Inclusion is manifestly reflexive, antisymmetric, and transitive; what is not immediate is that it is total (any two cuts are comparable), which is Inclusion totally orders the Dedekind reals. Compatibility of this order with the field operations (translation invariance of $\le$ and closure of nonnegatives under products), making $\mathbb{R}$ a totally ordered field, is The Dedekind reals form a totally ordered field.