I am writing a proof, trying to show that matrix inversion map is continuous w.r.t. $\sup$-norm. There is one step in my proof that might not be clear. I would appreciate any suggestion/advice.
Proof: Let $A$ be an invertible matrix. For any $\epsilon > 0$, we want to find $\delta>0$ such that $\left\| A^{-1} - B^{-1} \right\|_\infty<\epsilon$ whenever $\left\|A - B\right\| < \delta$ for invertible $B$.
Denote entry of $A^{-1}$ as $c_{ij}$, and entry of $B^{-1}$ as $d_{ij}$. First note that
\begin{align*} \left\|A^{-1} - B^{-1}\right\|_\infty<\epsilon \iff \left|c_{ij} - d_{ij}\right| < \epsilon \text{ for any }i,j\\ \left\|A - B\right\|_\infty < \delta \iff \left| a_{ij} - b_{ij}\right| < \delta \text{ for any }i,j \end{align*}
Consider the transform $g_{ij}$ from $\mathbb{R}^{d\times d}\to \mathbb{R}$ which gives us the $ij$-th entry of the inversed matrix. It is continuous.
(I thought it's obvious because we have determinant on denominator, and something like $a_{23} a_{33} - a_{23} a_{32}$ on numerator (in $3\times 3$ case). But I'm not sure if this part can be improved/made more clear?)
As a result, we can find $\delta > 0$ such that
$$\left\|\left(a_{11},a_{12},\cdots,a_{dd}\right)-\left(b_{11},b_{12},\cdots,b_{dd}\right) \right\|_2 < \delta \implies \left| c_{ij} - d_{ij}\right| < \epsilon$$
So it suffices to require that
$$\left| a_{ij} - b_{ij}\right| < \frac{\delta^2}{d^2} \text{ for all } ij$$