Compactness of the unit ball of the dual of $C(\omega_{1}+1)$

Question

In this article the authors give the following example:

Let $K=\omega_1 +1$ with the order topology and $E=C(K)^{*}$ the dual of $C(K)$ equipped with the weak$^{*}$ topology.

The space $E$ can be identified with the space of all regular Borel measures on $K$. And let $C$ be the subspace of $E$ consisting of probability measures.

The space $C$ is compact.

I know that $ K $ is compact, but not metrizable, so how can I show that $ C $ is compact?

Answer 1

$C(K)$ is a Banach space (in the sup norm) and $E$, its dual, is the space of regular Borel measures on $K$ (what I know as the Riesz representation theorem, in one of its forms). It can be given the weak$^\ast$ topology, as all duals spaces, and a standard result, the Banach-Alaoglu theorem tells us that the unit ball $B$ in $E$ (in this case all measures with norm $\le 1$) is compact in that topology.

And the (regular Borel) probability measures $C$ (having norm $1$) are a closed (and hence compact) subspace of that compact ball $B$. So it's all based on standard results (provided you did a reasonably advanced course in (standard) Banach spaces).