Let $X,Y$ be two Banach spaces.
A bounded operator $A$ is Fredholm if $\ker A$ and $\mathrm{coker} A$ are finite dimensional. Denote by $Fred(X,Y) \subset \mathcal{L}(X,Y)$ the space of Fredholm operators endowed with the subspace topology (hence the operator norm).


It is well known that $Fred(X,Y)$ is an open subset of $\mathcal{L}(X,Y)$.



> 
> Q. What is the closure of $Fred(X,Y)$ in $\mathcal{L}(X,Y)$? ($X,Y$ infinite dimensional)
> 
> 
> 


I would be happy to have an answer even for the case $X=Y$ separable Hilbert space.



Closure of the space of Fredholm operators
$\DeclareMathOperator\Ker{Ker}$



> 
> Let $\mathcal{H}$ be a separable Hilbert space. Then a bounded operator $T$ is **not** in the norm closure of Fredholm operators iff either $T$ or $T^\*$ has a finite-dimensional kernel and has a closed image of infinite codimension.
> 
> 
> 


(Note that in particular, the only operators that are in the closure of Fredholm operators but not in the closure of invertible operators, are Fredholm operators of nonzero index.)


*Proof:* Let $T$ not be in the norm closure of Fredholm operators. So it is not in the norm closure of invertible operators. By the [Bouldin result](https://www.ams.org/journals/proc/1990-108-03/S0002-9939-1990-1000147-9/S0002-9939-1990-1000147-9.pdf) mentioned previously, $T$ has a closed image and $\dim(\Ker(T))$ and $\dim(\Ker(T^\*))$ differ. If both are finite, $T$ is Fredhlom. So one is finite and the other is infinite.


Conversely, suppose that $T$ satisfies the condition. Up to adjoint, we can suppose that $T$ has closed range, has finite-dimensional kernel and infinite-dimensional cokernel. So there exists a block-decomposition under which we can write $T$ as
$$T=\begin{pmatrix} A\_0 & 0 \\ 0 & 0\end{pmatrix},$$
where the right-hand column is finite-dimensional, $A\_0$ is invertible, and both rows are infinite-dimensional. Let $T'$ be norm-close to $T$: write it as
$$T'=\begin{pmatrix} A & b \\ C & d\end{pmatrix},$$
so $A$ being close to $A\_0$, it is invertible. Suppose by contradiction that $T'$ is Fredholm. So there exists a bounded operator $S$ such that $TS-I$ has finite rank. Write $S=\begin{pmatrix} E & F \\ g & h\end{pmatrix}$ in the transpose block decomposition. (Small letters emphasize the fact that they have either finitely many rows or columns.) Then
$$T'S=\begin{pmatrix} AE+bg & AF+bh \\ CE+dg & CF+dh\end{pmatrix},$$
this time in the symmetric block decomposition corresponding to the row decomposition of $T$ (hence with both blocks infinite-dimensional. Hence $AF+bh$ has finite rank. Since $h$ has finite rank, we deduce that $AF$ has finite rank. Since $A$ is invertible, we deduce that $F$ has finite rank. But then $CF+dh$ has finite rank, which contradicts that $CF+dh-I$ has finite rank.


