Question regarding countable ordinals

Question

Feel free to suggest a better title.

We're going to regard $0$ as a limit ordinal, as people sometimes do. Let $L$ be the set of countable limit ordinals.

If $\alpha\in\omega_1$ there exist a unique $\beta_0\in L$ and $n_0\in\omega$ such that $$\alpha=\beta_0+n_0.$$

Now $L$ is evidently an uncountable well-ordered set such that every element has at most countably many predecessors, so there exists an order isomorphism $$\phi:\omega_1\to L.$$

So $\beta_0$ is $\phi$ of something, hence there exist a unique $\beta_1\in L$ and $n_1\in\omega$ with $$\beta_0=\phi(\beta_1+n_1).$$And so on. We've defined a map from $n:\omega_1\to\omega^\omega$ (where by $\omega^\omega$ I mean the set of all maps from $\omega$ to itself): $$n(\alpha)=(n_0,n_1,\dots).$$

Question: Is $n:\omega_1\to\omega^\omega$ injective?

(If yes then this gives a more elegant construction of sets serving the purpose of the $I_{n,j}\subset\omega_1$ here.)

Answer 1

It seems to be clear that $\phi$ is continuous, so we can construct a nonzero fixpoint for it in the usual way:

$$ \alpha = \bigcup_{k\in\omega} \phi^k(1) $$ Clearly $\alpha\le\omega_1$, but since it has cofinality $\omega$ it can't be $\omega_1$ itself. Thus $\alpha < \omega_1$ and $\phi(\alpha)=\alpha$.

Now $n(\alpha)=(0,0,0,\ldots) = n(0)$, so $n$ is not injective.

Answer 2

Every ordinal $\alpha$ has a unique expression in the form $$\alpha = \omega^{\gamma_{k-1}}c_{k-1} + \cdots + \omega^{\gamma_1}c_1 + \omega^{\gamma_0}c_0$$ where $k < \omega$, $0<c_i<\omega$ for each $i<k$ and $\gamma_{k-1} > \cdots > \gamma_1 > \gamma_0 \ge 0$ are ordinals. This is called Cantor normal form and is essentially a 'base-$\omega$' expression of $\alpha$. For $\alpha < \omega_1$, the ordinals $\gamma_i$ must also be countable.

Your construction acts on Cantor normal forms as follows: if $\alpha$ has CNF as above, then

• $n_0=c_0$ if $\gamma_0=0$, otherwise $n_0=0$;
• $\beta_0 = \omega^{\delta_{k-1}}c_{k-1} + \cdots + \omega^{\delta_1}c_1\ ({} + \omega^{\delta_0}c_0)$, where:
  • $\delta_i=\gamma_i$ if $\gamma_i$ is a limit ordinal;
  • $\delta_i=\gamma_i-1$ if $\gamma_i$ is a successor ordinal;
  • The last term is omitted if $\gamma_0=0$.

This process eventually stabilises, since if the CNF of an ordinal $\alpha$ has each $\gamma_i$ a positive limit ordinal then $\delta_i=\gamma_i$ for all $i$.

So in particular, we have $n(\omega^{\omega})=(0,0,\dots)$ and $n(\omega^{\omega^{\omega}}) = (0,0,\dots)$. So $n$ is not injective.