spectrum of an element of unital banach algebra is closed set

Question

Let $A$ be a unital Banach algebra and $a \in A$. Then the spectrum $\sigma(a)$ and set of invertible elements $\text{Inv}(A)$ are defined as: $$ \begin{align} &\text{Inv}(A):=\{a\in A~|~a \text{ is invertible}\}\\ &\sigma(a):=\{\lambda \in \mathbb C~|~\lambda 1-a \notin \text{Inv}(A)\}. \end{align} $$ I want to show that $\sigma(a)$ is closed subset of $\mathbb C$. But I already know that $\text{Inv}(A)$ is an open subset of $A$, since $A$ is a Banach algebra. I know there are many approaches to show that $\sigma(a)$ is closed, but I want to show by assuming $\text{Inv}(A)$ is an open subset of $A$.

Taking the set $\mathbb C-\sigma(a)$, we have to show that this set is open provided $\text{Inv}(A)$ is an open. Can you please give me some hint?

Answer 1

Hint. Consider the continuous(!) map $r\colon \mathbf C \to A$, $\lambda \mapsto \lambda 1 - a$. How can you express $\rho(a) := \mathbf C - \sigma(a)$ in terms of $r$ and $\mathrm{Inv}(A)$?

Answer 2

If $\lambda \in \rho(a)$, then $\lambda 1 - a \in \mathrm{Inv}(A)$. Because $\mathrm{Inv}(A)$ is open, an $\epsilon > 0$ exists such that $b \in \mathrm{Inv}(A)$ if $||b-\lambda 1 + a|| < \epsilon$. In particular, $b = \mu 1 - a \in \mathrm{Inv}(A)$ for $|\mu - \lambda| < \epsilon$ (assuming $||1|| = 1$). Thus, $\rho(a)$ is open.