Neighborhood of bounded linear operator such that all operators are noninvertible

Question

Let $X$ be a Banach space, let $\mathscr{B}(X)$ be a set of bounded operators acting from $X$ to $X$, my question is: can an operator $T \in \mathscr{B}(X)$ have a neighborhood $U$ such that all operators in $U$ are noninvertible?And the same question for compact operator? I don't know how to come up to this question, what a good property is for $\mathscr{B}(X)$? I tried to construct an invertible operator for an arbitrary neighborhood, but it's also very unclear how to come up.

Answer 1

Let $T=0.$

Then $T$ is non- invertible. For $s \ne 0$ define $S=sI.$ Then $S$ is invertible. But

$$||T-S||= |s|.$$

Consequence: in each neigborhood of $T$, there is some invertible operator.

Answer 2

For the set $\mathscr{B}(X)$ @Fred construction works, and for $\mathscr{K}(X)$ it's also easy to proof, as i understood: If $X$ is infinite dimension, so then compact operator $T \in \mathscr{K}(X)$ can't be invertible. Suppose that it has an invertible, so $T^{-1}$ is also compact, then because $\mathscr{K}(X)$ is an two side ideal in algebra lie of $Hom(X,X)$, then $T^{-1}T = Id$ is a compact operator, but $Id(B_{1, 0})^{Cl}$ isn't compact, contradiction.