Countable ordinals property

Question

I was wondering how to prove that an ordinal $\alpha$ is countable if and only if the next holds:

there is an order-preserving injection from $(\alpha,\leq)$ into $(\mathbb Q,\leq)$.

Answer 1

=> Any countable linear ordering can be embedded into rational numbers -- enumerate $\alpha$ by natural numbers and construct images gradually in the order of the enumeration. Any embedding of a finite part can be extended by one element.

<= obvious since $\mathbb Q$ is countable. But even order-preserving embedding into reals guarantees countability of $\alpha$: there are disjoint intervals $(f(\lambda), f(\lambda+1))$ for all $\lambda\in\alpha$ with a possible exception of the last element.

Detailed proof of =>

Let $\alpha = \{a_n : n\in\omega\}$ be an infinite countable linearly ordered set. We will construct another sequence $f_n\in\mathbb Q$ such that $f_i<f_j$ iff $a_i<a_j$. The construction proceeds recursively. $f_0=0$.

In order of construction of $f_n$ where $n>0$ we find $$a_{k_1} = \max\{a_k: k<n, a_k<a_n\},$$ $$a_{k_2} = \min\{a_k: k<n, a_k>a_n\}.$$ Note that these sets are finite. So the only case when maximum or minimum does not exist occurs when one of these sets is empty. So we analyse these cases:

1. If $k_1$ does not exist, set $f_n=f_{k_2}-1$.
2. If $k_2$ does not exist, set $f_n=f_{k_1}+1$.
3. If both $k_1, k_2$ exist, set $f_n=(f_{k_1}+f_{k_2})/2$

By such construction of step $f_n$ we ensured that $f_n$ differs from all previous values $f_k$ and for all $k<n$ it holds $f_k < f_n$ iff $a_k < a_n$.

In the end, we define the desired function $f\colon\alpha\to\mathbb Q$ by $f(a_n)=f_n$.

Answer 2

Given $\alpha<\omega_1$ we construct an embedding into ${\bf Q}$ by transfinite recursion, from $\alpha$ into $\omega_1$. Actually we construct a bit more, we construct a sequence of embeddings $\langle f_\beta\mid\beta\leq\alpha\rangle$, such that $f_\beta\colon\beta\to\Bbb Q$ is an order embedding, and its range is bounded.

• $f_0=\varnothing$.
• Suppose that $f_\beta$ were defined (we don't care about the entire sequence at this point), and $q\in\Bbb Q$ is some upper bound to $\operatorname{rng}f_\beta$. Then $f_{\beta+1}=f_\beta\cup\{\langle\beta,q\rangle\}$ is an order embedding, as wanted.

• Suppose that $\beta$ is limit, and $\langle f_\gamma\mid\gamma<\beta\rangle$ were defined. We write $\beta$ as the union of intervals $[\gamma_n,\gamma_{n+1})$, each of order type $\beta_n$. Trivially, $\beta_n<\beta$, so we can embed it into $\Bbb Q$ using $f_{\beta_n}$.

  Let $g_n$ be the composition of $f_{\beta_n}$ with an order isomorphism of $\Bbb Q$ with $(0,1)\cap\Bbb Q$. We define the follow $g\colon\beta\to\Bbb Q$: $$g(\delta)=n+g_n(\varepsilon),\qquad\delta\in[\gamma_n,\gamma_{n+1}),\ \varepsilon<\beta_n,\ \delta=\gamma_n+\varepsilon.$$

  This is well-defined because each $\delta$ appears in a unique interval, and can be written uniquely as such sum of two ordinals. It is not hard to see that $g_n$ is indeed injective, and order preserving, but it is unbounded. Let $f_\beta$ be the composition of $g$ with the order-isomorphism of $\Bbb Q$ with $(0,1)\cap\Bbb Q$, then $f_\beta$ is an embedding whose range is bounded.

On the other hand, if $\alpha$ is uncountable, there cannot be an injection from $\alpha$ into the countable set ${\bf Q}$.