Embedding large countable ordinals into the complex plane

Question

Consider large countable ordinals (e.g. $\epsilon_0$ which is not "large", but still interesting).

• These are countable sets, so they inject into the complex plane ( or even the real line).
• They contain limit ordinals.

Can we construct such an embedding in such a way that the only limit points in the image are images of limit ordinals?

For ordinals like $\omega^n$ and even $\omega^\omega$ the answer is yes.

I guess the answer should be yes, because these ordinals are none else but well orderings of integers.

Answer 1

Yes, in fact every countable ordinal embeds into the rational numbers in this way, an order-preserving map as a closed set of rational numbers.

Let me give three proofs.

First, one can easily prove this by induction on ordinals. Having a closed copy of $\alpha$ in $\mathbb{Q}$, we can scale it down to a bounded interval and then add another point on top to get a copy of $\alpha+1$. And if we have copies of every $\alpha<\lambda$ for a countable limit ordinal, we can similarly scale to successive intervals in such a way that the combined embedding contains its limit points.

Second, alternatively, one can also prove it using the universal property of the rational order. For any countable ordinal $\alpha$, extend the order of $\alpha$ to a dense linear order by adding a copy of $\mathbb{Q}$ between every ordinal and its successor. This gives a countable dense linear order, which must be order isomorphic to an interval of $\mathbb{Q}$. The original order is closed in the larger order, so this gives the ordinal as a closed subset of $\mathbb{Q}$.

Third, here is another soft proof. Cantor proved that the rational order $\mathbb{Q}$ is universal for all countable linear orders. So every countable ordinal $\alpha$ admits an order-embedding into $\mathbb{Q}$. The image of this embedding might not be closed in the way you have asked. But let us simply add the limit points. The new set (the old copy of $\alpha$ plus the limits) will still be well ordered and thus be a copy of some ordinal $\beta\geq \alpha$. So taking the first $\alpha$ many elements of this set gives a closed copy of $\alpha$ in $\mathbb{Q}$.

Answer 2

Here is another construction derived from a solution a student submitted to a final exam for my undergraduate set theory course some years ago.

Let $f:\alpha \hookrightarrow \mathbb{N}$ be an injective function from $\alpha$ to the natural numbers. (This exists as $\alpha$ is countable.) Define a measure $\mu$ on $\alpha$ by assigning the point masses $\mu(\{ \beta \}) = 2^{-f(\beta)}$ for $\beta \in \alpha$. Then $\mu:\alpha \to \mathbb{R}$ is an order preserving embedding (if $\beta < \gamma$ with $\gamma \in \alpha$, then $\beta \subsetneq \gamma$ and because every point has positive measure, $\mu(\beta) < \mu(\gamma)$). Moreover, with the exception of $\mu(\alpha)$ itself when $\alpha$ is a limit ordinal, the only limits in of the image of $\mu$ are the images of the limit ordinals in $\alpha$. In the case that $\alpha$ is a limit, replace $\mu$ with $h \circ \mu$ where $h:[0,\mu(\alpha)) \to [0,\infty)$ is an order preserving homeomorphism.