Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 4 / b / Solution

Because $A$ is a closed subalgebra of a Banach algebra, it is itself complete. If $\Vert a-1\Vert <1$, the Neumann series converges in $A$:Multiplying the partial sums by $a$ on either side gives $1-(1-a{)}^{N+1}$, which tends to $1$. Thus every such $a$ is invertible in $A$, and $1$ is an interior point of $G(A)$.