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Let $X$ be the subspace of $2^{2^\omega}$, consisting of all ordinals of countable cofinality, equipped with the ordered topology. Is $X$ countably compact? Is $X$ first countable?

Thanks for your help.

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Let $A\subseteq X$ be countably infinite. Let $\alpha_0=\min A$. Given $\alpha_n\in A$, let $$\alpha_{n+1}=\min\{\xi\in A:\xi>\alpha_n\}\;.$$ Then $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $A$. Let $\beta=\sup_n\alpha_n$; clearly $\operatorname{cf}\beta=\omega$, so $\beta\in X$, and $\beta$ is a limit point of $A$. Thus, $X$ has no infinite closed, discrete subset and is therefore compact.

$X$ is also first countable. Let $\alpha\in X$. If there is a sequence $\alpha_n:n\in\omega\rangle$ in $X$ such that $\alpha=\sup_n\alpha_n$, then the sets $X\cap(\alpha_n,\alpha]$ are a countable base at $\alpha$. If not, $\alpha$ is an isolated point of $X$, and $\{\alpha\}$ is a countable base at $\alpha$.

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