TITLE: Non-isomorpic uncountable dense linear orderings (a textbook example)
QUESTION [5 upvotes]: We know that two countable dense open orderings are isomorphic, and we know that this is generally not true for uncountable structures. Now I quote from the textbook, that is "Model-Theoretic Logics": 
"there are easy examples of non-isomorphic dense open orderings even of the same cardinality $\aleph_\alpha$, for every $\alpha\geq1$. Take, for instance $\aleph_\alpha$ many copies of the rationals and order them according to $\aleph_\alpha$ or inversely."
I guess I can't picture this particular example to work for me in any way. Could you please elaborate? Thanks.

REPLY [5 votes]: The basic idea is to take $\mathbb{Q} \times \kappa$ (where $\kappa$ is an uncountable cardinal), and consider the following two orderings on this set:
$$
\langle p , \alpha \rangle \leq_1 \langle q , \beta \rangle \Leftrightarrow \begin{cases}
\alpha < \beta, \;\text{or} \\
\alpha = \beta, p \leq q
\end{cases} \\
\langle p , \alpha \rangle \leq_2 \langle q , \beta \rangle \Leftrightarrow \begin{cases}
\alpha > \beta, \;\text{or} \\
\alpha = \beta, p \leq q
\end{cases} \\
$$
So in each we are taking an anti-lexicographic order, but in the first we are going according to the usual ordering on $\kappa$ (the copy of $\mathbb{Q}$ indexed by $\beta < \kappa$ comes after the copies of $\mathbb{Q}$ indexed by all $\alpha < \beta$), and in the second we are going according to the opposite order on $\kappa$ (the copy of $\mathbb{Q}$ indexed by $\beta < \kappa$ comes before the copies of $\mathbb{Q}$ indexed by all $\alpha < \beta$).
One important difference between these two examples is that there is an order preserving injection from $\omega_1$ (with the usual order) into $\leq_1$, but not in $\leq_2$.

Clearly the map $\alpha \mapsto \langle 0 , \alpha \rangle$ is order preserving with respect to $\leq_1$.
Recall that there is no order preserving injection $\omega_1 \to \mathbb{Q}$.  So if $f : \omega_1 \to \langle \mathbb{Q} \times \kappa , \leq_2 \rangle$ were an order preserving injection, then there must be an uncountable $A \subseteq \omega_1$ such that the second coordinates of the $f(\alpha)$ for $\alpha \in A$ are distinct.  But this means that for $\alpha < \beta$ in $A$, since $f(\alpha) \leq_2 f(\beta)$ it follows that the second coordinate of $f(\beta)$ is strictly less than the second coordinate of $f(\alpha)$. Thus we have a infinite (uncountable!) strictly decreasing sequence in $\kappa$, which contradicts that $\kappa$ is well-ordered!