Show that for each \(x \in S_\Omega\), the section of \(S_\Omega\) by \(x\) has a countable basis and hence is metrizable.
Conclude that \(S_\Omega\) is not paracompact.
Solution
- Fix \(x\in S_\Omega=[0,\Omega)\). Its section \[ S_x=\{y:y<x\} \] is countable, since every element of \(S_\Omega\) is a countable ordinal. The order topology on \(S_x\) has a basis consisting of open intervals and end intervals with endpoints in \(S_x\). Since \(S_x\) is countable, this basis is countable. An ordered space is regular, so the Urysohn metrization theorem implies that \(S_x\) is metrizable.
In particular, \(S_\Omega\) is locally metrizable: for each \(x<\Omega\), the open initial segment \[ [0,x+1)=S_{x+1} \] is a metrizable neighborhood of \(x\).
- Suppose \(S_\Omega\) were paracompact. Since it is locally metrizable, the Smirnov metrization theorem would imply that \(S_\Omega\) is metrizable. But \(S_\Omega\) is countably compact and not compact, whereas every countably compact metrizable space is compact. This contradiction shows that \(S_\Omega\) is not paracompact.