There is a well-known result in functional analysis saying that if $S$ is an invertible bounded operator on a Banach space $V$ and $T\in \mathcal L(V)$ verifies $\left\|S-T\right\|<\dfrac{1}{\left\|S^{-1}\right\|}$, then $T$ is also invertible. This result can be generalized to any associative Banach algebra with unity and is used to prove that the set of invertible elements is open.
I was wondering if one could find some sort of “converse” for it, that is if $\left(A,\left\|\cdot \right\|\right)$ is a Banach algebra and $y\in A$ is invertible, does there exist another $x\in A$ that is invertible and verifies $\left\|x-y\right\|<\dfrac{1}{\left\|x^{-1}\right\|}$ (with $x \neq y$)?
Yes. If $\|x - y\| < 1/\|y^{-1}\|$, $x$ is invertible. Moreover, the inversion function $x \to x^{-1}$ is continuous on the open set of invertible elements. Since $\lim_{x \to y} x^{-1} = y^{-1}$, while $\lim_{x \to y} (x-y) = 0$, if $x$ is close enough to $y$ we will have $\|x - y\| < 1/\|x^{-1}\|$.