Paracompact Space is Metacompact

Theorem

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

Then $T$ is also metacompact.


Proof

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

Consider some open cover $\UU$ of $T$.

Let $x \in S$.

Then there exists some neighborhood $\exists N_x$ 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, 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
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.