Countable Open Ordinal Space is Second-Countable

Theorem

Let $\Omega$ denote the first uncountable ordinal.

Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.

Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.

Then $\hointr 0 \Gamma$ is a second-countable space.

Proof

From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form:

$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$

for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\hointr 0 \Gamma$.

As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\hointr 0 \Gamma$.

Each point in $\hointr 0 \Gamma$ has a countable basis.

From Countable Union of Countable Sets is Countable, it follows that $\hointr 0 \Gamma$ has a countable basis.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $40$. Open Ordinal Space $[0, \Gamma) \ (\Gamma < \Omega)$: $5$