Theorem - Let $\mathcal{A}$ be a Banach algebra with identity element $e$ and $G(\mathcal{A})$ be the set of invertible elements in $\mathcal{A}$ . Let $B_r(x)$ denote the open ball of radius $r$ centered in $x$ .

Then, for all $x \in G(\mathcal{A})$ we have that

and therefore $G(\mathcal{A})$ is open in $\mathcal{A}$ .

Proof : Let $x \in G(\mathcal{A})$ and $y \in B_{\|x^{-1}\|^{-1}}(x)$ . We have that

So, by the Neumann series we conclude that $e-(e-x^{-1}y)$ is invertible, i.e. $x^{-1}y \in G(\mathcal{A})$ .

As $G(\mathcal{A})$ is a group we must have $y \in G(\mathcal{A})$ .

So $B_{\|x^{-1}\|^{-1}}(x) \subseteq G(\mathcal{A})$ and the theorem follows. $\square$