Uncountable Open Ordinal Space is not Metacompact

Theorem

Let $\Omega$ denote the first uncountable ordinal.

Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.

Then $\hointr 0 \Omega$ is not a metacompact space.

Proof

Aiming for a contradiction, suppose $\hointr 0 \Omega$ is a metacompact space.

From Open Ordinal Space is not Compact in Closed Ordinal Space we have that $\hointr 0 \Omega$ is a countably compact space.

From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space.

But from Open Ordinal Space is not Compact in Closed Ordinal Space this contradicts the fact that $\hointr 0 \Omega$ is not a compact space.

Hence the result by Proof by Contradiction.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $9$