Show that every countable well-ordered set is order isomorphic to a subset of $\mathbb{Q}$

Question

I'm currently going through Nik Weaver's "Forcing for Mathematicians" and doing the exercises. I'm looking for a hint on this one:

Show that every countable well-ordred set is order isomorphic to a subset of $\mathbb{Q}$

I'm assuming that this means using the normal less than relation on $\mathbb{Q}$.

Here is what I've tried so far.

Attempt 1

Let $X$ be a countable well-ordred set. I tried to take $x_1 = min(X)$, $x_2=min(X-\{x_1\})$, etc.... but this process may not exhaust all of $X$ (e.g. if $X=\mathbb{N}\times\mathbb{N}$ ordered by $(a_1, b_1)<(a_2,b_2)$ if either (a) $a_1<a_2$; or (b) $a_1=a_2$ and $b_1<b_2$.

Attempt 2

I also tried to generalize this taking $x_1,x_2,...,x_\omega,x_{\omega+1}, ...$, etc... and mapping this to elements of $\mathbb{Q}$ but even this process may not exhaust all of $X$. E.g. if $X$ is set of all integer polynomials of the ordinals $\omega_1$ and $\omega_2$ this process will never get to polynomials with $\omega_2$.

Attempt 3

So I wasn't able to construct a subset of $\mathbb{Q}$ like this. I am trying to take a different approach. I was thinking I can use Theorem 3.6 in Weaver's book which states:

Let $V$ and $W$ be well-ordered sets. Then exactly one of the following is true: V is order isomorphic to an initial segment of $W$, $W$ is order isomorphic to an initial segment of $V$, or $V$ and $W$ are order isomorphic to each other.

But I'm not sure if this is a good approach and, if so, how to use it.

Any hints/pointers in the right direction/etc... are appreciated.

Answer 1

List the elements of a countable set $K = \{x_1, x_2, x_3, ...\}$ where $K$ has a linear order $<<=$.

Map $x_1$ to $0$.

Assume $K_n = \{x_1, x_2,.. x_n\}$ has been order embedded into $\mathbb Q$.
Let $f$ be the order embedding of $K_n$.

Extend $f$ to include $x = x_{n+1}$:

• If $x$ is $<<$ all the elements of $K_n$, then map $x$ to $\min f(K_n) - 1/n$.

• If $x$ is $>>$ all the elements of $K_n$, then map $x$ to $\max f(K_n) + 1/n$.

• Otherwise there is some $a,b$ in $K_n$ with $a << x << b$ with nothing in $K_n$ between $a$ and $x$ or between $x$ and $b$. Map $x$ to $(f(a) + f(b))/2$. Thus $K_{n+1}$ order embeds into $\mathbb Q$.

By induction, $K$ order embeds into $\mathbb Q$.

This establishes the theorem that every countable linear order order embeds into $\mathbb Q$. The proof can be refined to show every countable linear order order embeds into the diatic rationale.

Since countable well ordered sets are countable linear orders, they order embed into the rationals, even the positive rationals.

Answer 2

Here is an outline of how you could make a construction by transfinite induction: we prove that for every countable ordinal $\alpha$ there exists a subset of $(0, 1) \cap \mathbb{Q}$ which is order isomorphic to $\alpha$. For the base case $\alpha = 0$, use $\emptyset$ as the subset. For the case of a successor ordinal $\alpha = \beta + 1$, suppose we have a subset $S$ of $(0, 1)$ order isomorphic to $\beta$; then $\frac{1}{2} S \cup \{ \frac{1}{2} \}$ will be order isomorphic to $\alpha$.

For the case of a limit ordinal $\alpha$, use the fact that $\alpha$ is countable to express $\alpha$ as the supremum of an increasing sequence of ordinals $\beta_1 < \beta_2 < \cdots < \alpha$. Now, use the inductive hypothesis to find sets $S_1$ order isomorphic to $\beta_1$, $S_2$ order isomorphic to $\beta_2 - \beta_1$, $S_3$ order isomorphic to $\beta_3 - \beta_2$, etc. Then $\frac{1}{2} S_1 \cup (\frac{1}{4} S_2 + \frac{1}{2}) \cup (\frac{1}{8} S_3 + \frac{3}{4}) \cup \cdots$ will be order isomorphic to $\alpha$. (So a bit less formally, squash $S_1$ into $(0, 1/2)$, $S_2$ into $(1/2, 3/4)$, $S_3$ into $(3/4, 7/8)$, etc. and then stack them on top of each other.)