Uncountable Open Ordinal Space is Sequentially 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 a sequentially compact space.

Proof

We have that:

Uncountable Open Ordinal Space is First-Countable

Uncountable Open Ordinal Space is Countably Compact

The result follows from First-Countable Space is Sequentially Compact iff Countably Compact.

$\blacksquare$

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 $[0, \Omega)$: $11$