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Why is the subspace of the ordinal $2^\omega+1$ consisting of all ordinal of countable confinality countably compact, first countable, and has cardinality $2^\omega+1$ ?

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Does $2^\omega$ represent ordinal exponentiation? –  azarel Mar 20 '12 at 5:29
Yes, and $2^\omega+1$ is the successor of $2^\omega$:P –  Paul Mar 20 '12 at 5:53
In this case $2^\omega=\sup_n 2^n=\omega$ so $2^\omega+1=\omega+1$ which is a convergent sequence. –  azarel Mar 20 '12 at 5:58

2 Answers 2

up vote 3 down vote accepted

Let $X=\{\alpha\le 2^\omega+1:\operatorname{cf}\alpha\le\omega\}$. Each $\alpha\in X$ is either a successor ordinal, in which case it is an isolated point of $X$, or the limit of a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$, in which case the sets $X\cap(\alpha_n,\alpha]$ for $n\in\omega$ are a countable base at $\alpha$. Thus, $X$ is first countable. It should also be clear that $|X|=2^\omega$, since the map $2^\omega\to X:\alpha\mapsto\alpha+1$ is an injection. The only slightly tricky bit is showing that $X$ is countable compact.

It suffices to show that $X$ does not contain an infinite closed discrete set. Let $S$ be an infinite subset of $X$. Clearly we can index $S=\{\sigma_\xi:\xi<\alpha\}$ for some ordinal $\alpha\ge\omega$ in such a way that $\sigma_\xi<\sigma_\zeta$ whenever $\xi<\zeta$. Now consider the sequence $\langle\alpha_n:n\in\omega\rangle$ in $S$: it’s strictly increasing, so it has a supremum $\alpha$ in $2^\omega$. (Remember, $\operatorname{cf}2^\omega>\omega$, so the supremum really is less than $2^\omega$.) But clearly $\operatorname{cf}\alpha=\omega$, so $\alpha\in X$, and it’s also clear that $\alpha$ is a limit point of $S$. Thus, $X$ has no infinite closed discrete subset and must therefore be countably compact.

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Let $X=\{\xi<2^{\aleph_0}+1: cof(\xi)=\omega\}$ with the order topology. In order to show that $X$ is countably compact, it suffices to prove that every infinite set has an accumulation point. Let $A$ be an infinite subset of $X$, by shrinking $A$ we may assume that $A$ has order-type $\omega$. It follows that $\sup(A)\in X$ and is an accumulation point of $A$, hence $X$ is countably compact.

For each $\xi\in X$, pick a strictly increasing sequence $\xi_n$ so that $\sup_n \xi_n=\xi$ then it is easy to see that $(\xi_n,\xi]$ is a countable basis at $\xi$, thus $X$ is first countable.

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