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I've been trying to learn more about subparacompact spaces. While reading an article, I noticed a theorem that was stated without proof. It said: "A countably compact, subparacompact space $X$ is compact."

I am not seeing why this is true. Can anyone offer any suggestions?

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After giving it more thought I was able to come up with a solution.

Suppose $X$ is countably compact and subparacompact, and let $\alpha$ be an open cover of $X$. Since $X$ is subparacompact, every open cover of $X$ has a $\sigma$-discrete closed refinement.

I claim that in any countably compact space, any locally finite discrete collection of subsets must be finite. Otherwise, suppose not. Then, there is a locally finite discrete collection $\{D_1, D_2, \ldots , D_n , \dots \}$ that is infinite. Now, pick a point $x_i$ in each $D_i$. Then, this sequence will have no cluster point, which is a contradiction since every sequence in a countably compact space has a cluster point.

Therefore, if we have a discrete refinement of $\alpha$, it will be a countable collection of sets. Since we have a $\sigma$-discrete refinement, it will be a countable union of countable sets, which is countable. Thus, every open cover of $X$ has a countable refinement, which implies that $X$ is Lindelof. Thus, $X$ is compact since it is countably compact and Lindelof.

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You could be clearer at the end. You show first that if $\mathscr{U}$ is an open cover of $X$, it has a countable closed refinement $\{F_n:n\in\Bbb N\}$, which is fine. For clarity you should then point out that for each $n\in\Bbb N$ there is $U_n\in\mathscr{U}$ such that $F_n\subseteq U_n$, so that $\{U_n:n\in\Bbb N\}$ is a countable open subcover of $\mathscr{U}$. Now you can conclude that $X$ is Lindelöf and hence compact. –  Brian M. Scott Apr 25 '12 at 19:49

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