Uncountable Open Ordinal Space is not Sigma-Compact

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 $\sigma$-compact space.

Proof

Aiming for a contradiction, suppose $\hointr 0 \Omega$ is a $\sigma$-compact space.

From Sigma-Compact Space is Lindelöf, $\hointr 0 \Omega$ is a Lindelöf space.

But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf 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 $[0, \Omega)$: $10$