Uncountable Open Ordinal Space is Countably 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 countably compact space.

Proof

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

From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.

So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.

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But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$.

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So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$.

This means that $\hointr 0 \Omega$ is 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 $[0, \Omega)$: $8$