Dedekind Completeness is Self-Dual

Theorem
Let $\left({S, \preceq}\right)$ be a Dedekind complete ordered set.

If a subset $A \subseteq S$ is non-empty and bounded below, then $A$ admits an infimum in $S$.

That is, the dual of a Dedekind complete ordered set is also Dedekind complete.

Proof
Let $B \subseteq S$ be set of all lower bounds for $A$.

Then every element of $A$ is an upper bound for $B$.

Therefore, $B$ is non-empty and bounded above.

By the definition of Dedekind completeness, $B$ admits a supremum $x \in S$.

By the definition of supremum, it follows that every element of $A$ succeeds $x$.

That is, $x$ is a lower bound for $A$.

If $y \in S$ is a lower bound for $A$, then $y \in B$, and so $y \preceq x$.

Hence, $x$ is the infimum of $A$.