TITLE: Finding a well-ordering of the natural numbers of a given order type
QUESTION [5 upvotes]: Let $X$ be the set of all well-orderings of the set of natural numbers, and let $O$ be the set of countable ordinals, i.e. the set of ordinals that are order types of the well-orderings in $X$.  Then my question is, is it possible to use transfinite recursion to define a function $f:O\rightarrow X$ where for each ordinal $\alpha\in O$, $f(\alpha)$ is a well-ordering of the set of natural numbers with order type $\alpha$?
The base case is easy enough; we can let $f(\omega)$ be the standard order on $\mathbb{N}$.  And the successor case is easy; if we know what $f(\alpha)$ is, then to define $f(\alpha+1)$ we can apply (an analogue of) $f(\alpha)$ to $\mathbb{N}-0$ and then append 0 to the end of it. 
But how do we do the limit case, i.e. if $\lambda$ is a limit ordinal in $O$ and we know the values of $f(\alpha)$ for all $\alpha<\lambda$, how can we define $f(\lambda)$?

REPLY [4 votes]: I think you're having some confusion with Asaf's answer, so let me rephrase it a bit. The goal is to break our $\lambda$ into a bunch of blocks, and "paste together" fixed copies of each block to build a copy of $\lambda$ itself.
Walking up an ordinal
Suppose I have a countable limit ordinal $\lambda$. There is (exercise) a sequence $(\alpha_n)_{n\in\mathbb{N}}$ such that:

$\alpha_1<\alpha_2<\alpha_3<...$, and
$\lambda=\sup\{\alpha_n: n\in\mathbb{N}\}$. 

We call such a sequence a fundamental sequence of $\lambda$. Note that every element of a fundamental sequence of $\lambda$ is $<\lambda$.
Importantly, fundamental sequences are not unique, and indeed the fact that we have lots of choices is going to be where nonconstructivity/the axiom of choice creep in.
Clarification: choice is not needed to show that every countable ordinal has a fundamental sequence. However, it is required to get a map assigning each countable ordinal a fundamental sequence. Similarly, we need choice to prove that the $f$ you want exists, even though ZF proves that for each countable $\alpha$ there is a map assigning each ordinal $<\alpha$ to a relation on the naturals of that ordertype. As an extreme example of a way the $f$ you want can fail to exist in ZF, note that there are models of ZF in which $\omega_1$ is a countable union of countable sets; in such a model, the existence of an $f$ as you describe would imply the countability of $\omega_1$.
Examples

$\lambda=\omega+\omega$: the sequence $$\alpha_1=\omega, \alpha_2=\omega+1,\alpha_3=\omega+2,...,\alpha_{n+1}=\omega+n,...$$ is a fundamental sequence for $\lambda$. But so is the sequence $$\beta_1=5,\beta_2=13,\beta_3=\omega+4,\beta_4=\omega+6,\beta_5=\omega+8,...,\beta_{n+2}=\omega+2n,...$$
$\lambda=\omega^2$: the sequence $$\alpha_1=\omega,\alpha_2=\omega\cdot 2,\alpha_3=\omega\cdot 3,...,\alpha_n=\omega\cdot n,...$$ is the "most obvious" fundamental sequence for $\lambda$. However, there are lots of other possibilities, like $$\beta_1=\omega\cdot 2,\beta_2=\omega\cdot 4,\beta_3=\omega\cdot6,...,\beta_n=\omega\cdot 2n, ...$$

Breaking an ordinal into blocks
Now we come to the key step. Suppose $\lambda$ is a limit ordinal and $(\alpha_i)_{i\in\mathbb{N}}$ is a fundamental sequence for $\lambda$. Then we can break $\lambda$ into "blocks" given by the fundamental sequence: 

Given a countable limit ordinal $\lambda$ and a fundamental sequence $(\alpha_i)_{i\in\mathbb{N}}$ for $\lambda$, there is a unique sequence $(A_i)_{i\in\mathbb{N}}$ of ordinals such that $$\sum_{1\le i\le n}A_i=\alpha_n.$$ More importantly, we have $$\sum_{i\in\mathbb{N}}A_i=A_1+A_2+A_3+...=\sup\{\alpha_i: i\in\mathbb{N}\}=\lambda.$$

The proof is a good exercise. Intuitively, for the existence part (which is the only part we actually need here) we want "$A_{i+1}=\alpha_{i+1}-\alpha_i$," although obviously we have to be careful what we mean by that ...
For example, if we let $\lambda=\omega$ and $\alpha_i=2i$, then we have $A_i=2$ for each $i$: counting up to $\omega$ by even numbers cuts $\omega$ into a sequence of blocks each of length $2$.
Making the induction work
We can now prove the result you want:
Suppose I have a countable limit ordinal $\lambda$, and for each $\eta<\lambda$ I have a well-ordering $W_\lambda$ of (a subset of) $\mathbb{N}$ of ordertype $\eta$. (This is, after all, the only step of the induction you're missing.)
It's slightly easier to work with well-orderings of sets of natural numbers, as opposed to well-orderings of all of $\mathbb{N}$ - e.g. it means that finite blocks aren't a problem - but it's a purely cosmetic choice.
Fix a fundamental sequence $(\alpha_i)_{i\in\mathbb{N}}$ of $\lambda$, and let $(A_i)_{i\in\mathbb{N}}$ be the corresponding "sequence of blocks" as per the fact above. By the induction hypothesis - note that $A_i<\lambda$ for all $i$, since $A_i\le\alpha_i<\lambda$ - I get a sequence of well-orderings of (subsets of) $\mathbb{N}$ corresponding to the $A_i$s, namely $(W_i)_{i\in\mathbb{N}}$. Intuitively, I now want to add the $W_i$s together. I can do that as follows:

Let $W$ be the following well-ordering:

The underlying set $dom(W)$ of $W$ is $$\{\langle i, x\rangle: x\in dom(W_i)\}.$$ (Here "$\langle\cdot\cdot\rangle$" is your favorite pairing function $\mathbb{N}^2\cong\mathbb{N}$.)
The ordering relation on $W$ is given by $$\langle i,x\rangle\le \langle j,y\rangle\iff (i<j)\vee (i=j\wedge x\le_{W_i}y).$$


It's now easy to show that the ordertype of $W$ is $$A_1+A_2+A_3+...=\sup\{\alpha_n:n\in\mathbb{N}\}=\lambda.$$