Compactifications of limit ordinals

Question

I thought I knew that but it seems I don't.

Let $\alpha$ be a countable, limit ordinal $\alpha>\omega$. Give $\alpha$ its order topology. What is the Stone-Čech compactification of $\alpha$? Is there any reason why it should be $\beta \omega$?

Answer 1

If $\alpha>\omega$, then the subspace $\omega+1$ of $\alpha$ is compact. It is therefore a compact subset of $\beta\alpha$ and hence a closed subset. But every infinite closed subset of $\beta\omega$ contains a copy of $\beta\omega$, so $\beta\omega$ contains no set homeomorphic to $\omega+1$. Thus, $\beta\alpha$ cannot be homeomorphic to $\beta\omega$.

If $\alpha=\beta+\omega$ for some limit ordinal $\beta$, then $\alpha$ is homeomorphic to the disjoint union of $\beta+1$ and $\omega$, and since $\beta+1$ is compact, $\beta\alpha$ is homeomorphic to $(\beta+1)\sqcup\beta\omega$.

It’s not clear to me just what happens when $\alpha$ is more complicated, even for $\omega^2$.

Answer 2

If $\alpha > \omega$, then $\beta \alpha$ is not homeomorphic to $\beta \omega$. For $\omega + 1$ is a countably infinite subset of $\beta \alpha$ which is open and closed, while $\beta \omega$ contains no such subset.

Suppose $U \subset \beta \omega$ is infinite, open and closed. Set $V = U \cap \omega$. If $V$ is finite then it is closed, so $U \setminus V$ is a nonempty open set containing no point of the dense set $\omega$, which is absurd. Thus $V$ must be infinite. By the universal property of the Stone-Čech compactification, the closure $\bar{V}$ is homeomorphic to $\beta V$ which is uncountable. Hence as $\bar{V} \subset U$, $U$ is uncountable.