# Density of Fredholm operators

Let $X$ be a Banach space. It is well known that an operator $T\in B(X)$ is Fredholm if and only if $\pi(T)$ is invertible in the Calkin algebra $B(X)/K(X)$.

Now suppose that the invertible elements in $B(X)/K(X)$ are dense. Can we conclude that Fredholm operators are dense in $B(X)$?

Note that there exist spaces for which every operator $T\in B(X)$ is Fredholm with index 0.

Yes, if the invertible elements in $B(X)/K(X)$ are dense then the Fredholm operators are dense.
Let $x \in B(X)$ and consider $\pi(x) \in B(X)/K(X)$. By density of the invertible elements $I$ in $B(X)/K(X)$, there exists a (Cauchy) sequence $y_n \in I$ such that $y_n \rightarrow \pi(x)$ as $n \rightarrow \infty$. Using an argument similar to the one that shows that the quotient of a Banach space by a closed subspace is a Banach, we can obtain a Cauchy sequence $x_n \in B(X)$ such that $\pi(x_n) = y_n$. Note that the $x_n \in B(X)$ are Fredholm since $\pi(x_n) \in I$. Completeness of $B(X)$ then implies that $x_n$ converges to some $x_\infty \in B(X)$ which must satisfy $\pi(x_\infty) = \pi(x)$. Thus, there exists some Fredholm operator $k \in K(X)$ such that $x_\infty + k =x$. It follows that the sequence $x_n + k$ is a sequence of Fredholm operators that converges to $x$.
• There are spaces $X$ for which the Calkin algebra is isomorphic to the algebra of continuous functions on the Cantor set. Take some non-trivial projection $p$ in that algebra and lift it to $B(X)$. How can we approximate it by Fredholm operators? – user512365 Apr 18 '17 at 22:56
• @RobertBarg: in response to why we can take the $x_n$ to be Cauchy, see for example the argument in the proof of theorem 3.1 on p. 7 of people.math.gatech.edu/~heil/6338/summer08/section6a.pdf Regarding your possible counter-example, are the invertibles dense in that Calkin algebra? – Maxwell Stolarski Apr 18 '17 at 23:24
• Yes, if a space is zero dimensional, then invertibles are dense in $C(X)$. – user512365 Apr 19 '17 at 8:47
• I think it is okay --- every projection $p$ can be approximated by invertible hence by image of a Fradholm operator. – user512365 Apr 19 '17 at 9:15