# Are most linear operators invertible?

Are most matrices invertible? discusses this question for matrices. All the answers implicitly used the (unique) vector topology on the space of $n \times n$ matrices. But my understanding (correct me if I'm wrong) is that infinite-dimensional linear operators can have multiple vector norms which induce inequivalent topologies, so the question becomes trickier. My (not entirely precise) question is, for an infinite-dimensional vector space, is the set of isomorphisms a dense open set under every "reasonable" operator topology? Under every operator norm topology induced by a "reasonable" norm on the vector space? My intuition says yes, but I'd be curious if anyone can make the words "reasonable" more precise.

(I assume that there's no natural way to generalize the Lebesgue-measure sense of "almost all matrices are invertible" to the infinite-dimensional case, due to the absence of an inifinite-dimensional Lebesgue measure, but correct me if I'm wrong.)

• Do you want the operators (and their inverses) to be continues? This is would seem more natural to me. – quid Jan 20 '18 at 20:57
• @quid Sure, either way. – tparker Jan 20 '18 at 20:58

Link to a Math.SE question and answer. Here it is shown that in the space $\mathcal B(E)$ of bounded operators on a Banach space $E$, equipped with the operator norm, invertible operators always form an open set that can fail to be dense.
The next interesting question is whether the same holds if $\mathcal B(E)$ is equipped with a weaker$^{[1]}$ topology, such as the strong operator topology. I have no idea whether this is true or not.
[1] I use the adjective "weaker" in the sense of analysis. That is, here a topology $\tau$ on $\mathcal B(E)$ is weaker than the operator norm topology $\tau_{\mathrm{op}}$ if and only if $\tau \subset \tau_{\mathrm{op}}$.