Order types of subsets of the reals

Question

Is there a subset of $\mathbb{R}$ that has the order type $\omega + (\omega^*+\omega)\cdot\eta + \omega^*$ (i.e., an $\omega$-sequence, followed by a bunch of copies of $\mathbb{Z}$ ordered like $\mathbb{Q}$, followed by a reverse $\omega$-sequence)?

Answer 1

It is true that any countable total order is isomorphic to a subset of $\mathbb{Q}$ (which implies that what you ask is true).

A proof, for instance, can be found in 5.6.2

Answer 2

Yes.

Let $C$ be the middle-thirds Cantor set, and let $\mathscr{I}$ be the set of deleted open intervals. The endpoints of an interval $I\in\mathscr{I}$ have the form

$$\sum_{k=1}^n\frac{a_k}{3^k}\qquad\text{and}\qquad\sum_{k=1}^n\frac{b_k}{3^k}\;,$$

where $a_k=b_k\in\{0,2\}$ for $k=1,\ldots,n-1$, $a_n=1$, and $b_n=2$. The map from $\mathscr{I}$ to $(0,1)$ that sends $I$ to

$$\sum_{k=1}^n\frac{b_k/2}{2^k}$$

is a an order-preserving bijection from $\mathscr{I}$ to the set of dyadic rationals in $(0,1)$, so $\mathscr{I}$ is ordered in type $\eta$. It’s easy to embed a set of type $\omega^*+\omega$ in each member of $\mathscr{I}$, and we can then add a set of type $\omega$ in $(\leftarrow,0)$ and a set of type $\omega^*$ in $(1,\to)$ to get a subset of $\Bbb R$ of the desired type.

Added: I should probably mention that this is a special case of a very general fact: $\eta$ is universal for countable order types. In this case, though, it happens to be quite easy to give a fairly explicit description of an embedding into $\Bbb R$.