Countably Compact Space is Countably Paracompact

Theorem

Let $T = \struct {S, \tau}$ be a countably compact space.

Then $T$ is countably paracompact.


Proof

From the definition, $T$ is countably compact if and only if every countable open cover of $S$ has a finite subcover.

From Subcover is Refinement of Cover, it follows that every countable open cover of $S$ has an open refinement which is locally finite.


This is precisely the definition of countably paracompact.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
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