Uncountable Closed Ordinal Space is not Separable

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Let $\Omega$ denote the first uncountable ordinal.

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

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


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

From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$.

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

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

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