Theorem. the set of invertible linear operators [br98]

    Let $\Omega$ be the set of all invertible linear operators on $\mathbb{R}^{n}$.

    1. If $A \in \Omega, B \in L(\mathbb{R}^{n})$, and$$\lVert B - A \rVert \cdot \lVert A^{-1} \rVert < 1,$$then $B \in \Omega$.
    2. $\Omega$ is an open subset of $L(\mathbb{R}^{n})$, and the mapping $A \to A^{-1}$ is continuous on $\Omega$.

    Proof. sketch [br98A]

      1. Suppose $B$ is not invertible $\iff$ $B$ is not injective $\iff$ there exists $\mathbf{x} \in \mathbb{R}^{n}$ such that $\mathbf{x} \neq \mathbf{0}$ and $B\mathbf{x} = \mathbf{0}$. Since $A \in \Omega$, there exists $\mathbf{y} \in \mathbb{R}^{n}$ such that $\mathbf{y} \neq 0$ and $A^{-1}\mathbf{y} = \mathbf{x}$. Then,$$\begin{align*}\lvert (B - A)A^{-1} \mathbf{y} \rvert &= \lvert (B - A)\mathbf{x} \rvert = \lvert -A\mathbf{x} \rvert = \lvert \mathbf{y} \rvert \\ &\leq \lVert (B - A)A^{-1} \rVert \lvert \mathbf{y} \rvert < \lvert \mathbf{y} \rvert. \end{align*}$$This is impossible. So our assumption cannot be true. Therefore, $B \in \Omega$.
      2. Openness is trivial. Continuity can be immediately induced from the observation that$$B^{-1} - A^{-1} = B^{-1}(A - B)A^{-1}.$$