How to show that R is a closed ideal in B?

Question

Let $B$ be a commutative Banach algebra and $R=\left\{ f\in B:1+\lambda f\text{ is invertible for any }\lambda \in C\right\}$.
Show that R is a closed ideal in B
I used Gelfand representation to get $R$ is a ideal in $B$, but how to prove that it is "closed" ?

Answer 1

Note that $1+\lambda f$ is invertible if and only if $-\frac{1}{\lambda}\not\in\sigma(f)$ so $R=\{f\in B: \sigma(f)=\{0\}\}$.

Let $\Omega$ be the spectrum of $B$. Then $\sigma(f)=\{\omega(f):\omega\in\Omega\}$, so $\sigma(f)=\{0\}$ if and only if $\omega(f)=0$ for all $\omega\in\Omega$. In other words $R=\bigcap_{\omega\in\Omega}\ker(\omega)$, an intersection of closed sets.

Note that if $B$ is a $C^*$-algebra, then $\sigma(f)=\{0\}$ iff $f=0$ so $R=0$.