# Dense linear orders embedded in the rationals

Let $$(\Omega,\leq)$$ be a dense linear order, with $$\Omega \subset \mathbb{Q}$$, and where "$$\leq$$" is induced by the usual order relation in $$\mathbb{Q}$$.

Is there an elegant general description for $$\Omega$$ (as a subset of $$\mathbb{Q}$$) ?

Yes, after possibly deleting its endpoints, it is isomorphic to $$\mathbb{Q}$$. That is, $$\Omega$$ will be isomorphic to one of the following:

• $$\mathbb{Q}$$,
• $$\mathbb{Q} \cup \{\infty\}$$,
• $$\{-\infty\} \cup \mathbb{Q}$$,
• $$\{-\infty\} \cup \mathbb{Q} \cup \{\infty\}$$.

I think the interpretation of $$-\infty$$ and $$\infty$$ should be clear.

The reason for this is that the theory of dense linear orders without endpoints is $$\omega$$-categorical. This is a well-known fact, and a proof of it appears in pretty much every book on model theory. It is not hard to see from that point that the other variants of the theory of dense linear orders (i.e. left endpoint, right endpoint or both) are also all $$\omega$$-categorical.

In particular, any dense linear order is going to be infinite, and since $$\Omega \subseteq \mathbb{Q}$$ it must be countable. Thus by $$\omega$$-categoricity it must be isomorphic to one of the models mentioned above.

Edit: you also ask "(as a subset of $$\mathbb{Q}$$)", unfortunately this is generally not possible I think. Although it might depend on what you call "an elegant description". For example, any union of open intervals in $$\mathbb{Q}$$ will be an example of such an $$\Omega$$. But also weirder sets like the set of all $$\frac{a}{b}$$ where this fraction is in reduced form and $$b$$ is odd.

Following up on Mark Kamsma's answer, there are $$2^{\aleph_0}$$-many suborders of $$\mathbb{Q}$$ which are isomorphic to $$\mathbb{Q}$$, even up to automorphisms of $$\mathbb{Q}$$. This suggests that there are too many to be "nicely classifiable".

To see this, let $$\eta\in 2^\omega$$ be any infinite binary sequence, and define for all $$n\in \omega$$: $$I_n = \begin{cases} (n,n+1) & \text{if }\eta(n) = 0 \\ (n,n+\frac{1}{2}) & \text{if } \eta(n) = 1\end{cases}$$ Let $$\Omega_\eta = \bigcup_{n\in \omega} I_n$$. Then $$\Omega_\eta$$ is a dense linear order without endpoints. Namely, it is the interval $$(0,\infty)$$ in $$\mathbb{Q}$$, with a sequence of gaps: $$\{n+1\}$$ is removed when $$\eta(n) = 0$$, and the entire interval $$[n+\frac{1}{2},n]$$ is removed when $$\eta(n) = 1$$. Any automorphism of $$\mathbb{Q}$$ which maps $$\Omega_\eta$$ to $$\Omega_{\eta'}$$ must map $$(0,\infty)$$ to $$(0,\infty)$$, and must map the $$n^\text{th}$$ gap in $$\Omega_\eta$$ to the $$n^{\text{th}}$$ gap in $$\Omega_{\eta'}$$. So these gaps must have the same type, and we conclude that $$\eta = \eta'$$.

Now consider the space of suborders of $$\mathbb{Q}$$ (by which I mean the Cantor space $$2^\mathbb{Q}$$, where an element encodes whether each $$q\in \mathbb{Q}$$ is in or out of the suborder). This is a Polish space, and the subspace of suborders which are themselves dense linear orders (without endpoint) is a $$G_\delta$$ subspace, hence also Polish. The Polish group $$\text{Aut}(\mathbb{Q})$$ acts on both spaces by automorphisms of $$\mathbb{Q}$$. So above I showed that this group action has $$2^{\aleph_0}$$-many orbits.

But there is one orbit which is much larger than all the others, in the sense that it is comeager in $$2^{\mathbb{Q}}$$. This is the orbit of any suborder of $$\mathbb{Q}$$ which is dense and codense in $$\mathbb{Q}$$. For example, the suborder of all dyadic rationals in $$\mathbb{Q}$$ is in this "generic" orbit.