Proposition 6.1.17 (General Linear Group is Open).label The group $G(A)$ is an open subset of $A$. More precisely for $x_{0}\in G(A)$ then $B\parens{x_0,\frac{1}{\norm{x_0\inv}{}}}\suf G(A)$ and for each $x\in B\parens{x_0,\frac{1}{\norm{x_0\inv}{}}}$
\begin{align*}x\inv=\parens{\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^n}x_{0}\inv\end{align*}
Proof. Let $x_{0}\in G(A)$ and $\norm{x-x_0}{}<\frac{1}{\norm{x_0\inv}{}}$ then
\begin{align*}\norm{1-x_0\inv x}{}=\norm{x_0\inv(x_0-x)}{}\leq \norm{x_0\inv}{}\norm{x_0-x}{}<1\end{align*}
so that $x_{0}\inv x$ has the inverse written by
\begin{align*}(x_{0}\inv x)\inv=\sums{\infty}{n=0}(1-x_{0}\inv x)^{n}=\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^{n}\end{align*}
Lastly $x=x_{0}(x_{0}\inv x)$ thus
\begin{align*}x\inv=\sums{\infty}{n=0}\braks{x_0\inv(x_0-x)}^{n}x_{0}\inv=x_{0}\inv+\sums{\infty}{n=1}\braks{x_0\inv(x_0-x)}^{n}x_{0}\inv\end{align*}
$\square$