Uncountable Open Ordinal Space is not Separable

Theorem

Let $\Omega$ denote the first uncountable ordinal.

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

Then $\hointr 0 \Omega$ is not a separable space.

Proof

Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable.

Let $H \subseteq \hointr 0 \Omega$ be a countable subset of $\hointr 0 \Omega$.

Let $\sigma$ be the supremum of $H$.

As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable.

Thus $\sigma$ strictly precedes $\Omega$.

The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$.

By definition of the closure $H^-$ of $H$ as the smallest closed set of $\hointr 0 \Omega$ containing $H$, it follows that $H^- \subseteq \closedint 0 \sigma$.

Therefore, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$.

Thus the closure of $H$ does not equal $\hointr 0 \Omega$.

Thus $H$ is not everywhere dense in $\hointr 0 \Omega$.

Hence, by definition, $\hointr 0 \Omega$ is not separable.

$\blacksquare$

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 $[0, \Omega)$: $3$