Closure of the invertible operators on a Banach space

Let $E$ be a Banach space, $\mathcal B(E)$ the Banach space of linear bounded operators and $\mathcal I$ the set of all invertible linear bounded operators from $E$ to $E$. We know that $\mathcal I$ is an open set, and if $E$ is finite dimensional then $\mathcal I$ is dense in $\mathcal B(E)$. It's not true that $\mathcal I$ is dense if we can find $T\in\mathcal B(E)$ injective, non surjective with $T(E)$ closed in $E$, since such an operator cannot be approximated in the norm on $\mathcal B(E)$ by elements of $\mathcal I$ (in particular $E$ has to be infinite dimensional).

So the question is (maybe a little vague): is there a nice characterization of $\overline{\mathcal I}^{\mathcal B(E)}$ when $E$ is infinite dimensional? Is the case of Hilbert space simpler?

In 1, we can find a characterization in the case of Hilbert spaces. For $T$ a bounded operator, let $T=U|T|$ be the polar decomposition of $T$, and $E(\cdot)$ be the spectral measure of $|T|$. Define $$\operatorname{ess\, nul}(T):=\inf\{\dim E[0,\varepsilon]H,\varepsilon>0\}.$$ Then $T$ is in the closure of invertible operators for the norm if and only if $\operatorname{ess\, nul}(T)=\operatorname{ess\, nul}(T^*)$.

1 Bouldin, Richard Closure of invertible operators on a Hilbert space. Proc. Amer. Math. Soc. 108 (1990), no. 3, 721–726.

There cannot be a nice characterisation, I believe, as for some Banach spaces the closure is everything. See my answer here.

A partial answer to your question would be that the set of Fredholm operators with index 0 is always in the closure of invertible operators.