Subparacompact Spaces 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?
 A: 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.
