Metacompact Space is Countably Metacompact
Jump to navigation
Jump to search
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