Countably Paracompact Space is Countably Metacompact

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