For a compact Hausdorff space $K$, the topology of $K$ is metrisable whenever $C(K)$ is separable.

Question

The following result is Exercise 2.50 in Megginson's An Introduction to Banach Space Theory:

Let $K$ be a compact Hausdorff space. Suppose the space of continuous functions $C(K)$ is separable. Then the topology of $K$ is metrisable.

I am aware of a proof using Urysohn's metrisation theorem, as is given here: Show that if $C(K)$ is separable, then $K$ is metrisable, for $K$ compact and Hausdorff.

However, as the book does not contain Urysohn's metrisation theorem and assumes only a basic knowledge of general topology, I am curious if an alternative proof can be given. The exercise is part of a section on the initial topology, so a proof using such results is desirable.

Any comments are highly appreciated.

Answer 1

(In case I'm not the only one who didn't know the term: The initial topology on $K$ determined by a family of functions $F$ with domain $K$ is the weakest topology such that every $f\in F$ is continuous.)

It's hard to know what's going to count as a suitably "alternative" proof; seems to me more or less all the proofs of this are really the same at bottom. It's a fact that the initial topology determined by a countable family of real-valued functions is metrizable; if you know that you're more or less done.

Had nothing to say about this until I noticed the context was a book on Banach spaces. Here's a Banach-space version:

If $X$ is a Banach space the weak* topology on $X^*$ is the initial topology determined by the functions consisting of evaluation at points of $X$. The Banach-Alaoglu theorem says that the closed unit ball of $X^*$ is compact in the weak* topology. Also

If $X$ is a separable Banach space then the closed unit ball of $X^*$ is metrizable in the weak* topology.

Now if $K$ is a compact Hausdorff space define $$i:K\to C(K)^*$$by $$i(x)(f)=f(x).$$It's easy to see that $i$ is continuous, if we give $C(K)$ the weak* topology, and the fact that $C(K)$ separates points of $K$ shows that $i$ is injective. So $i$ is a homeomorphism onto a subset of the unit ball of $C(K)^*$; now if $C(K)$ is separable the result above shows that $K$ is metrizable.

That's Banach-spacy, but again it's really the same proof in disguise.