Linear orders with dense isomorphic subsets Let $X$ and $Y$ be two complete (connected) linearly ordered sets.  Suppose $X'$ and $Y'$ are dense subsets of $X$ and $Y$, and $X'$ and $Y'$ are order isomorphic in their inherited orderings.  Is $X$ isomorphic to $Y$?
 A: Yes.  Indeed, $X$ must be canonically isomorphic to the Dedekind completion (with endpoints) of $X'$, and similarly for $Y$.  To be precise, let us say a Dedekind cut in $X'$ is a downward closed subset $L\subseteq X'$ which contains its supremum, if the supremum exists (note we do not require $L$ to be nonempty or a proper subset, to allow for cuts corresponding to endpoints).  Every element $x\in X$ determines a Dedekind cut $$L(x)=\{y\in X':y\leq x\},$$ and if $L(x)=L(x')$ then $x=x'$ since otherwise there would be an element of $X'$ between $x$ and $x'$ by density. (Note that here I take "dense" to mean dense in the order sense, not the topological sense, but they are equivalent given that $X$ is connected, which implies the open interval $(x,x')$ in $X$ must be nonempty.)  Moreover, every Dedekind cut $L\subseteq X'$ comes from an element of $X$ in this way: since $X$ is complete, $L$ has a supremum $x$ in $X$, and then $L=L(x)$.
So, $x\mapsto L(x)$ is a bijection between $X$ and the set of Dedekind cuts on $X'$, and it is moreover order-preserving when you order the Dedekind cuts by inclusion.  So, $X$ is isomorphic to the completion of $X'$.
