Is every countable ordinal completely metrizable?

Question

I'm trying to prove that every countable ordinal is completely metrizable (I do not know if it's true!). I've already proved that every countable successor is completely metrizable, because it is homeomorphic to a closed set of $\mathbb R$.

How can I solve this problem for limit ordinals? I will be grateful for all your hints.

Thank you!

Answer 1

Here's an approach: Every countable limit ordinal has cofinality $\omega$; that means that for any countable limit ordinal $\alpha$, there is a sequence $\beta_0 < \beta_1 < \beta_2 < \ldots$ so that $\sup\beta_n = \alpha$. Without loss of generality, these $\beta_n$ are successor ordinals (just add one to each one).

So here's what I'm thinking: Map $\beta_0$ homeomorphically into $[0,1]$. Map $\beta_1 \setminus \beta_0$ homeomorphically into $[1, 2]$, $\beta_2 \setminus \beta_1$ into $[2,3]$, and so on. The result will be a homeomorphic mapping of $\alpha$ into $\mathbb{R}$, with the additional property that $\alpha$ is "sent" to $\infty$ (I say "sent" in quotes because $\alpha$ isn't actually in the domain of this map, but you know what I mean). The induced metric should do what you want it to.

Now, there's a bunch of details to work out, like exactly how these homeomorphisms align with each other, but I think this is a good way to start.

Answer 2

Suppose $(\alpha_n)$ is an increasing sequence of ordinals of which you know, are homeomorphic to a closed subset of $\mathbb{R}$.

Now recursively map $\alpha'_{n+1} = \alpha_{n+1} \setminus \alpha_{n}$ onto a closed subset of $(n,n+1)$. The result is a map to a closed subset of $\mathbb{R}$.

Another proof if you know a bit of model theory:

Take the theory $T$ of dense linear orders without endpoints, add for every $\beta < \alpha$ a constant symbol $c_\beta $ to your language and add the axioms $\{ c_\beta < c_\gamma : \beta < \gamma < \alpha \}$. Now notice that for each limit $\gamma < \alpha$, $\{ c_\beta < x < c_\gamma : \beta < \gamma \}$ and also $\{c_\beta < x : \beta < \alpha \}$ is a non isolated type and there are countably many such types. Now by the omitting types theorem there is countable a model that omits all of those and by a typical back and forth argument you can identify it with $\mathbb{Q}$. The interpretations of the $c_\beta$'s give you your closed subset of $\mathbb{R}$.

Answer 3

This follows from general principles:

Every countable ordinal has a countable base in the order topology, and every order topology is normal. So Urysohn implies that it is metrisable. Successor ordinals are compact so automatically complete. The others are a $G_\delta$ in a compact one, so completely metrisable too.