# Subsets of $\mathbb Q$ of order type $\omega^{\alpha}$ for each countable ordinal $\alpha$.

My introductory text in Set Theory (Stillwell) includes an exercise (6.3.1) asking for an explicit example of a subset of $\mathbb Q$ or order type $\omega^2$. This seems straight forward enough. I wish to generalize this, first by explicitly defining a sequence of sets $A_n \subset \mathbb Q$ such that $A_n$ has order type $\omega^{n}$ and then seeing where we can go from there. (See Details below.)

Taking the $limsup \{A_n\}$, I believe we get $A_{\omega} = \cup A_n$. While my intuition carries me through $A_n$ for finite $n$, it is not entirely clear to me what $A_{\omega}$ looks like, or indeed if $A_{\omega}$ has order-type $\omega^{\omega}$. I think it does. If so, we should be able to carry on to generate $A_{\alpha}$ for each countable ordinal $\alpha$.

Is this correct? This is self-learning, so I don’t have anyone to refer to. What is a better way of doing this so that intuition holds up past $\omega^{\omega}$?

I am reading an introductory text but my question is more intermediate than introductory. This is why I am unsure and seeking help. For example, the text does not cover set-theoretic limits.

DETAILS

This is what I have done so far :

First, let $A_1 = \mathbb N$. Clearly $A_1$ has order-type $\omega^1$.

To make clear my approach I shall write $A_2$ in long hand.$$A_2 = \{ 1 - \frac12, 1 - \frac13, 1 - \frac14, \dots, 2 - \frac12, 2 - \frac13, \dots \} .$$

To generalise, given $A_n$, write $A_n = \{ a_1, a_2, \dots \}$, where $a_n \lt a_{n+1}$.

Define $a_0 = 0$ and $d_i = a_i - a_{i-1}$ for $i \gt 1$.

Next, for each $i > 0$, let $$B_i = \{ a_i - (d_i \cdot \frac{1}{k}) : k \in \mathbb N \}.$$ Finally, we define $$A_{n+1} = \cup_i B_i$$

Defining our $A_n$ in this way should mean $A_n$ has order-type $\omega^n$. Does this extend to infinite ordinals?

• Every countable ordinal is isomorphic to a subset of $\mathbb Q$. But there is no explicit example for most of them. – azarel Oct 16 '14 at 17:44
• @azarel Yes, a later exercise asks us to prove this "indirectly". I was trying to understand how far we can push an explicit isomorphism. This seems correct for finite $n$. Is it not correct for $n = \omega$ using the limsup? – Epsilon Oct 16 '14 at 17:48
• $\bigcup_nA_n$ isn’t well-ordered. However, you can build a set isomorphic to $\omega^\omega$ by building a copy of $\omega$ in $[0,1)$ converging to $1$, a copy of $\omega^2$ in $[1,2)$ converging to $2$, etc. – Brian M. Scott Oct 16 '14 at 17:48
• @BrianM.Scott Excellent. That seems a much more clever approach than mine, and it also gives a reasonably clear picture of the resulting $\omega^{\omega}$ ordering. Thanks! – Epsilon Oct 16 '14 at 17:51
• You’re welcome! – Brian M. Scott Oct 16 '14 at 17:53