Separability of an algebra is equivalent to separability of its spectrum

Question

Let $A$ be a commutative C*-algebra.

I would like to show that $A$ is separable (i.e. has a countable dense subset) if and only if the spectrum of $A$ (denoted by $\Omega(A)$) is separable.

---

Notes and reminders:

A C*-algebra is a Banach algebra over $\mathbb{C}$ with involution $*$ s.t. $\|x^*x\| = \|x\|^2$ for every $x \in A$.

Let me also remind that the spectrum $\Omega(A)$ is the set of the characters of $A$, that is $\Omega(A)\mathrel{:=} \{\chi:A\to \mathbb{C} : \text{$\chi$ is algebra homomorphism $\neq 0$}\}$. Also note that the topology on $\Omega(A)$ is the weak* (or w$^*$) topology. So the closure of any set which indicates separability is also taken under weak* topology.

Answer 1

It's not true.

For simplicity, suppose $A$ is unital, so that its spectrum is compact Hausdorff.

If $X$ is a compact Hausdorff space and $C(X)$ is separable, then you can show that $X$ is second countable. (Let $\{f_k\}$ be a countable dense subset of $C(X)$, and $\{U_n\}$ a countable basis of open sets in $\mathbb{C}$; then the countable collection of sets $\{f_k^{-1}(U_n)\}$ is a basis for $X$, using Urysohn's lemma.) In particular, such an $X$ would be metrizable (by the Urysohn metrization theorem).

So the forward direction of your claim is true, since a second countable space is always separable, but this also suggests how to find a counterexample to the reverse direction: take a compact Hausdorff $X$ which is separable but not metrizable. One standard example is $X = [0,1]^{[0,1]}$ with its product topology; considering it as the space of all functions from $[0,1]$ to itself, the set of polynomials with rational coefficients is dense. Another, as YCor and Jochen Wengenroth discussed, is $X = \beta \mathbb{N}$, the Stone-Cech compactification of $\mathbb{N}$.

The converse of the above statement is also true, as user131781 points out: if $X$ is a compact metric space, then $C(X)$ is separable.

So the correct version of your claim (including the non-unital case) would be that $A$ is separable iff $\Omega(A)$ is second countable iff $\Omega(A)$ is a separable metrizable space.