Non-isomorpic uncountable dense linear orderings (a textbook example)

Question

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.

Answer 1

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!