Is $\mathbb{R}$ a maximum Dedekind complete field?

Question

Let $\mathbb{R}\subseteq F$ be an ordered Dedekind complete Field (every Dedekind cut of $F$ is already in $F$), does this mean $\mathbb{R}=F$?

Answer 1

It is the unique Dedekind complete field, so in particular it is maximal.

To see that, note that a Dedekind complete field must be Archimedean, otherwise the cut defined by the natural numbers is not realized, then the rational numbers are dense in the field, so every cut is generated by a set of rationals, which is exactly to say that every cut is a real number.