Elementary Set Theory: Ordinals, Well Orderings and Isomorphisms I need to show that for any countable ordinal $\alpha$ there is a set A $\subseteq \mathbb{Q}$ such that (A, <) is isomorphic to ($\alpha, \in$).  To do it I am supposed to show the following stronger statement by induction on ordinals $\alpha$ < $\omega_{1}$:  Let P($\alpha$) be the statement "For every interval (a, b) with rational endpoints, there is an A $\subseteq$ (a,b) such that (A, <) is isomorphic to ($\alpha, \in$)."
I think that the purpose of breaking $\mathbb{Q}$ into intervals (a,b) is to make $\mathbb{Q}$ a countable set, but I don't see the significance of doing this or how it would make the proof easier.  I feel like I need to define a well-ordered set in (a,b) but I don't know how to do this since if a and b are both negative then there is no least element, so no well-ordering.
I believe that I am supposed to be using transfinite induction? But I don't know how P($\alpha$) implies P(S($\alpha$)) for S($\alpha$) successor of $\alpha$.
Any help would be greatly appreciated... thanks in advance.
 A: As with any transfinite induction over ordinals, we need to do this in three cases:

*

*$P(0)$. This is obvious.


*$P(\alpha) \implies P(S(\alpha))$. This might seem difficult, but it actually isn't. Take the interval $(a, b)$. $P(\alpha)$ implies that there is a subset $A_\alpha \subseteq (a, \frac{a+b}2)$ that is order isomorphic to $\alpha$. Now set $A_{\alpha + 1} = A_\alpha \cup \left\{\frac{a + 2b}{3}\right\}$, and you're done.


*$(\forall \alpha < \gamma(P(\alpha))\implies P(\gamma)$ for $\gamma$ a (countable) limit ordinal. This is the tricky part. The reason that it's tricky is that the different $A_\alpha$ might not converge in any way to a suitable candidate for $A_\gamma$. So we need some way to stabilize this. First, fix a infinite sequence $\{a_i\}_{i \in \omega_0}$ of rational numbers such that $a < a_1 < a_2 < a_3 <\cdots < b$. Also, fix some strictly increasing sequence $\{\alpha_i\}_{i \in \omega_0}$ of ordinals that converges to $\gamma$.
The idea is to make sure that both $A_{\alpha_i} \subseteq A_{\alpha_{i + 1}}$, and $A_{\alpha_i}\subseteq (a, a_i)$, so that one is able to take the "limit" (i.e. the union) and have it behave nicely so that we may use that as $A_\gamma$. We begin by using $P(\alpha_0)$ to establish an $A_{\alpha_0}\subseteq (a,a_0)$. Then, for any $i$, there is an ordinal $\beta_i$ so that $\alpha_i + \beta_i = \alpha_{i + 1}$ (which means $\beta_i \leq \alpha_{i+1} < \gamma$). Therefore we may use $P(\beta_i)$ to find an $A_{\beta_i}\subseteq (a_i, a_{i+1})$ which is order isomorphic to $\beta_i$, and thus $A_{\alpha_{i+1}} = A_{\alpha_i} \cup A_{\beta_i}$ is order isomorphic to $\alpha_{i+1}$. This finishes the proof.
It is worth noting where this fails for uncountable ordinals. The first uncountable ordinal, $\omega_1$ is a so-called regular cardinal, which means that if you try to apply step $3$ to it (it is, of course, a limit ordinal), then you cannot find an increasing sequence $\{\alpha_i\}_{i \in \omega_0}$ that converges to $\omega_1$. You need an uncountable sequence to reach it. And you cannot find an uncountable sequence $\{a_i\}_{i \in \omega_1}$ to match it, since there aren't enough rational numbers.
