invertible elements in a Banach algebra form an open set

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

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

$$B_{{\parallel x^{-1}\parallel }^{-1}}(x)\subseteq G(𝒜)$$

and therefore $G(𝒜)$ is open in $𝒜$.

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

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

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

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