$X = [ 0,\omega_1 )$ is maximal countably compact

Question

(a) : why $X = [ 0,\omega_1 )$ is a first countable space? ($\omega_1$ is the first uncountable ordinal number.)

(b) : is $X = [ 0,\omega_1 )$ maximal countably compact?

Answer 1

HINTS:

(a) For $\alpha\in[0,\omega_1)$ let $\mathscr{B}(\alpha)=\{(\xi,\alpha]:\xi<\alpha\}$; show that $\mathscr{B}(\alpha)$ is a countable local base at $\alpha$.

(b) It’s well-known that $X$ is countably compact. Since $X$ is $T_1$, it suffices to show that every infinite subset of $X$ has a limit point, which is not hard to do: if $A\subseteq X$ is infinite, recursively construct a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $A$, and show that $\sup_{n\in\omega}\alpha_n$ is a limit point of $A$. Now let $\tau$ be the order topology on $X$, and let $\tau'$ be a strictly larger topology on $X$: $\tau\subsetneqq\tau'$. Let $U\in\tau'\setminus\tau$. Since $U\notin\tau$, there is some $\alpha\in U\setminus\operatorname{int}_\tau U$. This means that for every $\xi<\alpha$, $(\xi,\alpha]\nsubseteq U$.

• Show that $\alpha$ must be a limit ordinal.
• Show that there is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ such that $\sup_{n\in\omega}\alpha_n=\alpha$ and $\alpha_n\notin U$ for each $n\in\omega$.
• Conclude that $\{\alpha_n:n\in\omega\}$ is a closed, discrete set in $\langle X,\tau'\rangle$ and hence that $\langle X,\tau'\rangle$ is not countably compact.