A simple way to know whether a well-ordered set has a subset of a certain type

Question

Following my last question, Does $\Bbb R-\Bbb Q$ have a well ordered subset of type $\omega\cdot\omega$, I would like to have better tools to look at a set and know what order types can it have.

I already understood that for a set to have a subset of type $\omega+n$, it should have a bounded and infinite subset. One can prove $\Bbb Z$ doesn't have one, so $\Bbb Z$ doesn't have a subset of type $\omega+n$.

What about other types? For example, how can I know whether the negative rational numbers have a subset of type $\omega\cdot\omega$, or whether $\Bbb R$ has a subset of type $\omega^\omega$, without having to construct it?

Answer 1

From this question (ultimately, you're using transfinite recursion for countable ordinals; Asaf's answer is quite clear) we know that every countable ordinal order-embeds into $(\mathbb{Q},\leq)$, and so because $\mathbb{R}\setminus \mathbb{Q}$ contains a subset order-isomorphic to $\mathbb{Q}$ (say, $q\mapsto \pi+q$ is the embedding), we see that $\mathbb{R}\setminus \mathbb{Q}$ contains a copy of every countable ordinal.

Since $\omega^2$ and $\omega^\omega$ are countable ordinals, we know that $\mathbb{R}$ contains copies of each, along with $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$.

Answer 2

As I pointed in the previous question, every dense set embeds every countable linear ordering, in particular every countable ordinal.

You don't even have to use transfinite recursion for this.

Suppose that $(A,<)$ is a dense linear order without endpoints, and $(D,\prec)$ is a countable linear ordering, enumerate $D$ as $\{d_n\mid n<\omega\}$, by induction define an embedding $f(d_n)=x$ such that $\{f(d_k)\mid d < n\}\cup\{x\}$ is isomorphic (with $f$, extended using $d_n\mapsto x$) to $\{d_k\mid k\leq n\}$.

For example, you pick some $a_0$ so that $f(d_0)=a_0$, suppose that you mapped $d_0,\ldots,d_5$ already, look where $d_6$ lies in the order of $\prec$ with relation to $d_0,\ldots,d_5$. Because $A$ is dense, there is some $x$ which satisfies the same relation with $a_0,\ldots,a_5$. Choose one and continue.

If $A$ happened to be countable (or otherwise well-orderable), we can enumerate it, and choose the least possible index for such $x$, otherwise the axiom of choice is implicitly assumed.