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Yesterday, Asaf Karagila's answer to my question sparked an extensive discussion on ways of proving that all countable ordinals are embeddable in $\Bbb Q$, and whether particular solutions to this use the Axiom of Choice.

Since the arguments used to prove this can be shady on their use of Choice when formulated tersely, fully fleshed out answers are appreciated.

To be concrete, this question asks for proofs of the following fact:

All countable ordinals $\alpha < \omega_1$ are order-embeddable in the rationals $\Bbb Q$.

Especially answers avoiding the Axiom of Choice are appreciated.

Of relevance and related interest is #123969 (but it's not a duplicate, since that question asks for explicit embeddings).

Martin Sleziak
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Lord_Farin
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    Is there a deeper reason why you restrict attention to ordinals and not just any countable linear ordering? – Martin Jun 01 '13 at 08:45
  • Ordinals are the application of immediate interest; I can imagine there are proofs relying on their structure that wouldn't follow through in the general case. Proofs that work for all countable linear orderings are welcome, of course. But there's nothing really deep to it, I guess. – Lord_Farin Jun 01 '13 at 08:51
  • Several links to proofs of this were given in this comment. – Martin Sleziak Jun 01 '13 at 18:42
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    @MartinSleziak Thank you; most of the proofs linked there seem to prove that it can't be done for uncountable sets (and in $\Bbb R$). I found two proofs of the desired statement, both along the lines of Brian's answer. – Lord_Farin Jun 01 '13 at 19:51
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    You can prove something stronger, without appealing to $\mathsf{AC}$: Let $(A,<)$ be a countable partial order. Then there is a linear order $\prec$ extending $<$, and an order embedding $f:(A,\prec)\to\mathbb Q$. – Andrés E. Caicedo Aug 20 '13 at 21:47

2 Answers2

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Finite linear orders are no problem, so let $\langle X,\preceq\rangle$ be any countably infinite linear order, and fix an enumeration $X=\{x_k:k\in\omega\}$. Also fix an enumeration of the rationals, $\Bbb Q=\{p_k:k\in\omega\}$. Define $f:X\to\Bbb Q$ as follows.

Let $f(x_0)=p_0$. Suppose that $0<n\in\omega$, and $f(x_k)$ has been defined for each $k<n$. Let $L_n=\{k<n:x_k\prec x_n\}$ and $R_n=\{k<n:x_n\prec x_k\}$.

  • If $L_n\ne\varnothing\ne R_n$, let $\ell(n)=\max\{x_k:k\in L_n\}$ and $r(n)=\min\{x_k:k\in R_n\}$. Then $x_{\ell(n)}\prec x_{r(n)}$, so $f(x_{\ell(n)})<f(x_{r(n)})$, and we can set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:f(x_{\ell(n)})<p_k<f(x_{r(n)})\}\;.$$

  • If $L_n=\varnothing\ne R_n$, let $r(n)=\min\{x_k:k\in R_n\}$, and set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:p_k<f(x_{r(n)})\}\;.$$

  • And if $L_n\ne\varnothing=R_n$, let $\ell(n)=\max\{x_k:k\in L_n\}$, and set $f(x_n)=p_m$, where $$m=\min\{k\in\omega:f(x_{\ell(n)})<p_k\}\;.$$

Given the enumerations, everything is explicitly defined.

Brian M. Scott
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  • The situation can arise where one of $L_n$ or $R_n$ is empty. Then one has to define $m$ slightly differently (by ignoring one of the inequalities). There is a small typo in the definition of $m$: It should be $p_k$ instead of $p_m$. – Martin Jun 01 '13 at 16:57
  • @Martin: Thanks for catching that. – Brian M. Scott Jun 01 '13 at 17:08
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The trick, as the discussion in the comments to my answer, is to notice the order of the quantifiers.

Given $\alpha<\omega_1$ we construct an embedding 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.

Asaf Karagila
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    At the limit stages of your induction, you need to have, for each relevant $\beta$, an increasing sequence $(\gamma_n)_{n\in\omega}$ with supremum $\beta$. To assign such a sequence to each limit $\beta$ would seem to use choice, but it can in fact be done explicitly, for all limits $\beta<\alpha$, if you begin by choosing a counting of $\alpha$. (This is where it's important that you have a fixed, countable $\alpha$ to start with.) – Andreas Blass Aug 20 '13 at 22:25
  • @AndreasBlass Because if we neglect what you say then Asaf's proof could be construed to "show" that $\omega_1$ embeds into $\mathbb{Q}$ (assuming $\omega_1$ is singular)? The reason then that this doesn't work is that we need a fixed, countable $\alpha$, or choice (but then $\omega_1$ isn't singular)? – Jori Jan 05 '20 at 01:24