Uncountable Open Ordinal Space is not Metacompact
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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 $[0, \Omega)$: $9$