# Countably Paracompact Space is Countably Metacompact

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## Theorem

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

Then $T$ is countably metacompact.

## Proof

From the definition, $T$ is countably paracompact if and only if every countable open cover of $X$ has an open refinement which is locally finite.

Consider some countable open cover $\UU$ of $X$.

Let $x \in X$.

Then there exists some neighborhood $\N_x$ of $x$ which intersects only finitely many elements of $\UU$.

Thus $x$ itself can be in only finitely many elements of $\UU$.

Hence $T$ must be, by definition, countably metacompact.

$\blacksquare$

## 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