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

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 $K$.

• Kevin, what is the connection between your two paragraphs? "Having a ball with countably many extreme points" is not a hereditary property. – Bill Johnson Sep 29 '16 at 14:42
• Guess I should have checked that. It seemed like it would be but if C(K) always has uncountably many extreme points it is clearly not. – Kevin Beanland Sep 29 '16 at 15:57
• Are there any conditions on the space (polyhedral?) that make the property hereditary? – Kevin Beanland Sep 29 '16 at 16:08
• Reflexivity. :) – Bill Johnson Sep 30 '16 at 13:55

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.
• Thanks for the quick answer. Counting clopen partitions of $K$ was not something I thought about. This is MO at its best. – Kevin Beanland Sep 29 '16 at 14:08
• The same argument shows that the unit ball of $C(K)$ has countably many extreme points when $K$ is any compact metric space (every clopen set is a finite union of sets from any given base for the topology). – Bill Johnson Sep 29 '16 at 14:51
• @BillJohnson alternatively we may use that $C(K)$ is separable, while the set of extreme points is 2-distant. – Fedor Petrov Sep 29 '16 at 16:05