Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.