Metacompact Countably Compact Space is Compact

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Theorem

Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.

Then $T$ is compact.


Proof

From the definition of countably compact space, every countable open cover of $S$ has a finite subcover.

Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.

Let $\UU_\alpha$ be an open cover of $S$.

Then let $\VV_\beta$ be the open refinement which is point finite, guaranteed by its metacompactness.


Let $x \in S$.

We have that $x$ is an element of only a finite number of the elements of $\VV_\beta$, as $\VV_\beta$ is point finite.

Consider all the subcovers of $\VV_\beta$, and order them by subset.

Consider any chain $\CC$ of such subcovers.

Aiming for a contradiction, suppose $x$ is not in the intersection of $\CC$.

Then it would fail to be covered by one of the elements of $\CC$.

This would be a contradiction of the fact that every element of $\CC$ is a cover of $\VV_\beta$.

So the intersection of a chain of such subcovers is itself a subcover.

Hence it must be that $\VV$ has a minimal subcover, $\VV_\gamma$, say.


Now each element $V_\gamma$ of $\VV_\gamma$ must itself contain a unique element $x_\gamma \in V_\gamma$ which belongs to no element of $\VV_\gamma$.

This is because $\VV_\gamma$ is minimal.

If $\VV_\gamma$ were infinite, the set of all $\set {x_\gamma}$ would be an infinite set without an $\omega$-accumulation point.

This can not be the case.


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Thus $\VV_\gamma$ is a finite cover.

So we have constructed a finite subcover for $\UU_\alpha$, demonstrating that $T$ is compact.

$\blacksquare$


Sources

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