# Countable Closed Ordinal Space is Second-Countable

## Theorem

Let $\Omega$ denote the first uncountable ordinal.

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

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

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

## Proof

From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\closedint 0 \Gamma$ 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 $\closedint 0 \Gamma$.

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

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

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

$\blacksquare$