Many uncountable dense linear orders

Question

A theorem of Shelah (which has been quoted several times on this website) states that if $T$ is a countable unstable theory and $\kappa$ is an uncountable cardinal, then $T$ has $2^\kappa$ non-isomorphic models of cardinality $\kappa$. I was wondering if there is an elementary way ($\mathit{i.e.}$ requiring very little model and set theory) to illustrate this theorem in the special case of dense linear orders. In particular, I would like to construct explicitly $2^{\mathfrak{c}}$ pairwise non-isomorphic dense linear orders of cardinality $\mathfrak{c}$. My starting point was to consider the real line (“horizontally”) and to attach to every real number on the line a “vertical” fiber, which would be either $\mathbb{R}$ or $\mathbb{Q}$, and then order the whole subset of $\mathbb{R}^2$ obtained this way by the lexicographic order. Given any $A\subseteq\mathbb{R}$, I would then consider the dense linear order $\mathcal{M}_A$ which has $\mathbb{R}$ as a vertical fiber over every $x\in A$ and $\mathbb{Q}$ as a vertical fiber over every $x\notin A$. I have produced $2^{\mathfrak{c}}$ models of cardinality $\mathfrak{c}$, which unfortunately are not pairwise non-isomorphic (for example, whenever $A$ and $B$ are singletons, $\mathcal{M}_A$ and $\mathcal{M}_B$ are isomorphic). So I'm not sure this is the way to go. Any thoughts?

Answer 1

Start with $\kappa$ and replace each ordinal with either $(\omega_1^*+\omega_1)\times \mathbb{Q}$ or $(1+\omega_1^*+\omega_1)\times \mathbb{Q}$. This gives $2^{\kappa}$ many nonisomorphic dense orders.