spectrum of an element of unital banach algebra is closed set 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?

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 $\mathrm{Inv}(A)$?
