# Countably Compact Space is Countably Paracompact

Jump to navigation
Jump to search

## 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