Let $V$ be a finite-dimensional Banach space, and $T \in \text{Hom}(V)$ be a bounded linear transformation. In Loomis & Sternberg's Advanced Calculus, it is mentioned that the following two facts may be deduced from the continuity of the determinant function $\Delta: \text{Hom}(V) \to \mathbb{R}:$ (but it is not actually shown how; they take a different approach)
- If $T$ has an inverse, then so does $S$ when $\lVert T-S\rVert$ is small enough;
- The mapping $T \to T^{-1}$ is continuous.
The first fact follows easily, since we may take the interval $(\det T-\epsilon,\det T+\epsilon),$ with $\epsilon$ sufficiently small so that the interval doesn't contain $0$. But how does the second fact follow? Thanks in advance.
Use the formula $$ T^{-1} =\det(T)^{-1} adj(T). $$ Here, $T\mapsto \det(T)^{-1}$ is continuous, $T\mapsto adj(T)$ is continuous, as the entries of $adj(T)$ are polynomial in the entries of $T$.