Proving that $ω_1$ is locally compact

Question

I'm trying to show that $ω_1$ is locally compact, but when doing so, I need to show something else, which got me a bit stuck on.

I'm taking a $\alpha\in ω_1$, so $\{\alpha\}$ is an open set. Since $ω_1$ is Hausdorff, the closure of $\{ \alpha \}$ is $\{ \alpha \}$ itself. Now I want to show that $\{ \alpha \}$ is compact, and this is where I got stuck - I want to show that for every cover of $\{ \alpha \}$ I can find a finite sub-cover, but I don't see how to prove this, what made me wonder if I'm not trying to prove something which is false?

Answer 1

In order to prove that the space $\omega_1$ is locally compact, you have to prove that every point in $\omega_1$ has a compact neighborhood. If $\alpha\in\omega_1$ then the one-point set $\{\alpha\}$ is obviously compact, but it is not necessarily a neighborhood of $\alpha$; namely, it's not a neighborhood if $\alpha$ is a limit ordinal, such as $\omega$.

Hint: Prove that for each ordinal $\alpha$ the closed interval $[0,\alpha]=\{\xi:\xi\le\alpha\}$ is a compact neighborhood of $\alpha$.

More hints: Either $[0,\alpha]$ is always compact, or else there is a least counterexample. Suppose $\alpha$ is the least ordinal for which $[0,\alpha]$ is not compact, and let $\mathcal U$ be an open cover of $[0,\alpha]$ with no finite subcover. Choose $U\in\mathcal U$ so that $\alpha\in U$. Since $U$ is open, there is an ordinal $\beta\lt\alpha$ such that $(\beta,\alpha]\subseteq U$. Now $[0,\beta]$ is compact and

Answer 2

I think you are on a right way, and in your proof you can use the fact that there is no infinite decreasing sequence of ordinals.

Answer 3

Every cover of $\{ a \}$ is actually $\{ a \}$ itself, so you can obviously take a finite sub-cover, which is simply $\{ a \}$! Thus, making $\omega_1$ locally compact.