The spectrum of a commutative unital Banach algbra having linearly dense idempotents is totally disconnected I don't know how should I start to show:
If $A$ is an arbitrary abelian Banach algebra in which the idempotents have dense linear span, its specrum (the space of characters on $A$) is totally disconnected.
 A: The spectrum of a commutative Banach algebra is (equivalent to) the set of nonzero algebra homomorphisms (to $\mathbb{K}$, where $\mathbb{K} \in \{\mathbb{R},\,\mathbb{C}\}$) with the weak-$\ast$ topology.
Note that


*

*for any idempotent $p$ and any algebra homomorphism $\lambda$, we have $\lambda(p) = \lambda(p^2) = \lambda(p)^²$, hence $\lambda(p) \in \{0,\,1\}$, and

*any continuous linear map $A \to E$ is uniquely determined by its values on the idempotents, since those have dense span.


So let $\mathfrak{I}$ be the set of idempotents in $A$, and consider the mapping
$$\chi \colon \sigma(A) \to \{0,\,1\}^\mathfrak{I};\quad \chi(\lambda)(p) = \lambda(p).$$
By the denseness of the span of $\mathfrak{I}$, $\chi$ is injective, and since both $\sigma(A)$ and $\{0,\,1\}^\mathfrak{I}$ carry the topology of pointwise convergence, it is continuous (an embedding even, by Ascoli-Bourbaki, but we don't need that).
$\{0,\,1\}^\mathfrak{I}$ is totally disconnected.
If $f \colon X \to Y$ is a continuous injective map from a topological space $X$ into a totally disconnected space $Y$, then $X$ is totally disconnected.
