Why is there an element with infinite spectrum in a commutative Banach algebra with infinitely many characters?

Question

Let $A$ be a commutative Banach algebra such that set of all characters is infinite. I want to prove that there exists an element in $A$ such that its spectrum is infinite.

Answer 1

Let $(\chi_{n})_{n \in \mathbb{N}}$ be a sequence of pairwise distinct characters.

For $m \neq n$ let $U_{m,n} = \{a \in A \mid \chi_{n}(a) \neq \chi_{m}(a)\}$. Since $U_{m,n}$ is the complement of the closed hyperplane $\ker{(\chi_{m} - \chi_{n})}$, the set $U_{m,n}$ is open and dense in $A$ whenever $m \neq n$. By the Baire category theorem, the countable intersection $$D = \bigcap_{m \neq n} U_{m,n}$$ is dense and of second category in $A$, so it is non-empty. Every $a \in D$ has infinite spectrum because $$\sigma(a) \supseteq \{\chi_{n}(a) \mid n \in \mathbb{N}\}$$ and because the set $\{\chi_{n}(a) \mid n \in \mathbb{N}\}$ is infinite by definition of $D$.