Which of our four countability axioms does \(S_\Omega\) satisfy? What about \(\overline{S}_\Omega\)?
Solution
The four countability properties under discussion are first countability, second countability, Lindelöfness, and existence of a countable dense subset.
For \(S_\Omega=[0,\Omega)\):
- It is first countable. Every \(\alpha<\Omega=\omega_1\) is a countable ordinal. If \(\alpha\) is a successor, it has an obvious countable local basis. If \(\alpha\) is a limit ordinal, choose an increasing sequence \(\alpha_n\uparrow\alpha\); then intervals \((\alpha_n,\alpha]\) form a countable local basis at \(\alpha\).
- It is not separable. The closure of any countable subset \(D\subset S_\Omega\) is bounded below some countable ordinal \(\beta<\Omega\), since \(\sup D<\Omega\). Thus \(D\) cannot be dense.
- Therefore it is not second countable, since second countable spaces are separable.
- It is not Lindelöf. The open cover \(\{[0,\alpha):\alpha<\Omega\}\) has no countable subcover, since the supremum of countably many countable ordinals is still \(<\Omega\).
For \(\overline S_\Omega=[0,\Omega]\):
- It is compact, hence Lindelöf.
- It is not first countable at \(\Omega\). A countable local basis would give countably many left endpoints whose supremum is still \(<\Omega\), leaving a smaller neighborhood of \(\Omega\) not refined by the basis.
- It is not separable for the same boundedness reason as above: every countable subset has supremum \(<\Omega\), so it misses a final interval.
- Hence it is not second countable.
Thus \(S_\Omega\) satisfies only first countability, while \(\overline S_\Omega\) satisfies only Lindelöfness among the four axioms.