Show that the minimal uncountable order with the order topology is not separable.
Solution
1. Let \(\omega_1\) denote the set of all countable ordinals with the order topology. Suppose \(D\subseteq\omega_1\) is countable.
This is the minimal uncountable ordinal/order.
2. The supremum \(\alpha=\sup D\) is still a countable ordinal, hence \(\alpha<\omega_1\).
A countable union of countable ordinals is countable, so the supremum of countably many countable ordinals is countable.
3. The open ray\[(\alpha,\omega_1)=\{\beta<\omega_1:\beta>\alpha\}\]is nonempty and disjoint from \(D\).
It is a basic open final interval in the order topology, and every element of \(D\) is at most \(\alpha\).
4. Therefore no countable subset of \(\omega_1\) is dense, so \(\omega_1\) is not separable.
Every countable subset misses the nonempty open set from 3.