Let $A$ be a Banach algebra. Is there a Banach algebra $B$ which contains $A$ but the spectrum of each elements of $B$ has empty interior(as a subset of $\mathbb{C}$)?


The motivation comes from the fact that the spectrum of elements in a smaller algebra possibly loses its interior when we compute its spectrum in a larger algebra.(Rudin, Functional analysis)



Removing the interior of spectrums
The answer is *no*, in general. Here is a counterexample:


Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then the spectrum of $a$ is the closed unit disk $\overline{D}$, and the point spectrum of the operator $a$ is the open unit disk $D$.


Now we note that the notion "point spectrum" - which is defined for operators - can be translated into a notion that makes sense in Banach algebras: For each $\lambda$ in the point spectrum $D$ of $a$ there exists an element $a\_\lambda \in A \setminus \{0\}$ such that $(\lambda - a) a\_\lambda = 0$. Indeed, let $x\_\lambda \in \ell^2(\mathbb{N})$ be an eigenvector of the operator $a$ and choose an arbitrary non-zero functional $\varphi\_\lambda$ on $\ell^2(\mathbb{N})$. Then the operator $a\_\lambda := \varphi\_\lambda \otimes x\_\lambda \in A$, given by
$$
 a\_\lambda x = \langle \varphi\_\lambda, x \rangle x\_\lambda \qquad \text{for } x \in \ell^2(\mathbb{N}),
$$
satisfies $(\lambda - a)a\_\lambda = 0$.


Hence, if $B$ is a unital Banach algebra that contains $A$ as a subalgebra, we still have $(\lambda - a)a\_\lambda = 0$ in $B$, so $\lambda - a$ cannot be invertible in $B$ for any $\lambda \in D$.


