Why is the character space of a Banach algebra weak $\ast$ closed?

Question

Let $A$ denote a unital commutative Banach algebra and let $\Sigma(A)$ denote its character space.

Why is $\Sigma(A) \cup \{0\}$ weak $\ast$ closed in the closed unit ball of $A^\ast$? Is $\Sigma(A) \cup \{0\}$ weak $\ast$ closed in $A^\ast$?

Answer 1

$\Sigma(A) \cup \{0\}$ is precisely the set of multiplicative elements of $A^\ast$. Therefore

$$\Sigma(A)\cup \{0\} = \bigcap_{x,y\in A} \underbrace{\left\lbrace \varphi\in A^\ast : \varphi(xy) - \varphi(x)\varphi(y) = 0 \right\rbrace}_{M_{x,y}}$$

is weak* closed as the intersection of the weak* closed sets $M_{x,y}$.

To see that $M_{x,y}$ is weak* closed, for $x,y$ arbitrary but fixed, consider $F \colon A^\ast \to \mathbb{C}^3$, given by

$$F(\varphi) = \bigl(\varphi(xy),\varphi(x),\varphi(y) \bigr).$$

$F$ is weak* continuous by the definition of the weak* and product topologies (on $A^\ast$ resp. $\mathbb{C}^3$).

Let $P = \{ (u,v,w) \in \mathbb{C}^3 : u - v\cdot w = 0\}$. By the continuity of multiplication and subtraction, $P$ is closed.

Hence $M_{x,y} = F^{-1}(P)$ is weak* closed as the preimage of a closed set under a continuous map.