Metacompact Space is Countably Metacompact

Theorem

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

Then $T$ is a countably metacompact space.


Proof

From the definition, $T$ is metacompact space iff every open cover of $S$ has an open refinement which is point finite.

This also applies to all countable open covers.

So every countable open cover of $S$ has an open refinement which is point finite.

This is precisely the definition for a countably metacompact space.

$\blacksquare$


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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