Given an ordinal $(\alpha,\in)$, how do I find a subset of rationals which is isomorphic to it?

Question

How do I find a subset $E$ of the rationals such that $(E,<) \cong (\alpha,\in)$ where $<$ is the usual ordering of the rationals and $\alpha$ is an ordinal?

$(E,<) \cong (\alpha,\in)$ means that there is a bijective map $f: E \rightarrow \alpha$ such that $a<b \Leftrightarrow f(a) \in f(b)$.

If this question is too general, consider the following special cases:

(a) $\alpha=\omega + 2$

(b) $\alpha=\omega \cdot 3$

(c) $\alpha=\omega \cdot \omega$

(d) $\alpha=\omega^\omega$

$\omega$ is the set of all natural numbers which is an ordinal.

The first few are easy: ${-\frac1n\mid n\in\Bbb N}\cup{0,1}$, ${k-\frac1n\mid n\in\Bbb N, 1\le k\le 3}$, ${-\frac1{kn}-\frac1k\mid n,k\in\Bbb N}$ — Hagen von Eitzen, Oct 03 '18 at 22:31
@HagenvonEitzen Could you give some details of how you do it? I don't know how to construct the bijective map $f$, how did you come up with the answers? — bbw, Oct 03 '18 at 22:36

score 1 · Answer 1 · answered Oct 03 '18 at 22:34

A nice way to embed $\omega$ is as $1-\frac 1n$. That puts it into $(0,1)$, which is handy. Then to get $\omega+3$ just add $2,3,4$ to the list. To get $\omega \cdot 3$ just repeat three times. To get $\omega \cdot \omega$ there are a couple approaches. You can just put $\omega$ into $(n,n+1)$ for each $n$ or you can put $\omega$ into $(0,1), (1,\frac 32), (\frac 32, \frac 74)$ and so on. That might give you an idea for $\omega^\omega$

Given an ordinal $(\alpha,\in)$, how do I find a subset of rationals which is isomorphic to it?

1 Answers1