# How to show this space $X$ is countably compact, first countable?

Consider the subspace $X$ of $(2^\omega)^+$, i.e., the smallest cardinal greater then $2^\omega$, equipped with the ordered topology consisting of all ordinals of countable cofinality.

How to show that $X$ is countably compact, first countable, and $|X| > 2^\omega$?

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$X$ is countably compact because it does not have an infinite closed, discrete subset. If $A\subseteq X$ is countably infinite, then $A$ is not cofinal in $X$, so let $\alpha=\min\{\xi\in(2^\omega)^+:|\xi\cap A|=\omega\}$; then $\xi\in X$, and $\xi$ is a limit point of $A$. (Alternatively, let $A=\{\alpha_\xi:\xi<\eta\}$ be a strictly increasing enumeration of $A$; clearly $\omega\le\eta<\omega_1$. Then $\sup\{\alpha_n:n\in\omega\}$ is in $X$ and is a limit point of $A$.)
If $\alpha\in X$, then there is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $(2^\omega)^+$ in $(2^\omega)^+$ converging to $\alpha$, and the sets $X\cap(\alpha_n,\alpha]$ are a countable local base at $\alpha$.
Finally, $X$ is cofinal in $(2^\omega)^+$, which is a regular cardinal, so $|X|=(2^\omega)^+$.
Why is $X$ here a regular cardinal? – Paul Apr 6 '13 at 7:02
@Paul, in $\sf ZFC$ successor cardinals are regular. – Asaf Karagila Apr 6 '13 at 7:30
@AsafKaragila: How could we see that $X$ here is a successor cardinal? – Paul Apr 6 '13 at 7:48
@Paul: $X$ is not a successor cardinal. $|X|$ is. Because $(2^\omega)^+$ is a successor cardinal. – Asaf Karagila Apr 6 '13 at 7:50