Visually intuitive proof that set of invertible operators is open Let $H$ be a Hilbert space, and let $B(H)$ be the set of all bounded operators of $H$, equipped with the operator norm. It is well-known that the set of all invertible bounded operators forms an open subset of $B(H)$. The proof (which also applies to Banach algebras in general) seems to rely on mostly symbolic manipulation, in which given an operator $S + T$, with $S$ invertible and $T$ small, we consider the infinite sum $\sum_{n=0}^\infty ( S^{-1}T)^n$.
I would like to know if there is a intuitive (perhaps geometric) proof of why bounded operators which are close to an invertible bounded operator must also be invertible. My line of thought is something of the following: If $S$ is invertible and $T$ is close to $S$, then the behaviour of $T$ "can't differ too much" from that of $S$, and therefore must be invertible.

The picture I see is the following: Given two points $x, y \in H$, there exist balls of the same radius centered at $x$ and $y$ whose images under $S$ are disjoint and therefore a positive distance apart. Therefore, if you perturb $S$ by a map $T$ with sufficiently small operator norm, then $(S+T)x$ and $(S+T)y$ remain in small disjoint balls centered at $Sx$ and $Sy$ respectively. In particular, $(S+T)x \ne (S+T)y$.
If $S^{-1}$ is bounded, then the sufficiently small bound on $\|T\|$ is independent of $x$ and $y$. In other words, if $\|T\|$ is sufficiently small, then $S+T$ is injective.
