Uncountable Closed Ordinal Space is Sigma-Compact

Theorem

Let $\Omega$ denote the first uncountable ordinal.

Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.

Then $\closedint 0 \Omega$ is a $\sigma$-compact space.

Proof

We have:

Closed Ordinal Space is Compact

Compact Space is $\sigma$-Compact

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $10$