If $K$ is a countable compact metric space is the set of extreme point of $Ba(C(K))$ countable?

Question

The question is the title. The set $Ba(C(K))$ is the unit ball of $C(K)$. This has to be known, but I can't find the answer explicitly in the literature. There is some literature about polyhedral Banach spaces however I'm not sure if being polyhedral is sufficient to show the set of extreme points of the ball is countable. Note that these $C(K)$ spaces are isomorphic to $C(\omega^{\omega^\alpha}+1)$ for some $\alpha$.

The motivation for the question is that I spent some time proving each Banach space in certain class has countably many extreme points. I then discovered that each of these space isometrically embed into $C(K)$ for some countable $K$.

Answer 1

It is true (if you mean real spaces, for complex spaces it is obviously false). Note that if a function $f$ is extremal, it takes only values $\pm 1$. Indeed, if $|f(a)|<1$, then choose a small ball $B(a,r)$ on which $|f|<1-r$ and consider the functions $f(x)\pm \max(r-d(a,x),0)$. They belong to a unit ball and are different from $f$. So, our functions are in one-to-one correspondence with partitions of $K$ onto two disjoint compact subsets (preimages of $\pm 1$). There are countably many such clopen partitions. Indeed, any open set is a union of open balls with rational radii, if it is also closed, it is a finite union. There are countably many balls with rational radii and countably many finite subsets of them.