Closed Ordinal Space is Compact

Theorem

Let $\Gamma$ be a limit ordinal.

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

Then $\closedint 0 \Gamma$ is a compact space.

Proof

By definition, $\closedint 0 \Gamma$ is a linearly ordered space.

The result follows from Linearly Ordered Space is Compact iff Complete.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $6$