# Why is Isom(E,F) open in the set of bounded linear operators between E and F?

Let $E$ and $F$ be Banach spaces. According to the lecture notes I'm reading $Isom(E,F)$ (the set of continuous isomorphisms between $E$ and $F$ with continuous inverse) is open in the set of bounded linear operators between $E$ and $F$. Supposedly, it follows from the fact that the set of invertible elements of a Banach algebra is open. But I fail to see in what Banach algebra one could embed $Isom(E,F)$.

Any hints?

## 1 Answer

What you are asking essentially is that, given an invertible operator $T:E\to F$, then its neighboring operators $T+H$, for $H$ small enough, are also continuously invertible.

Consider $(T+H)T^{-1}=I+HT^{-1}$, which is an operator in $B(F)$. It is invertible when $\|HT^{-1}\|<1$; to ensure this, let $\|H\|<1/\|T^{-1}\|$, so that $\|HT^{-1}\|\le\|H\|\|T^{-1}\|<1$. Hence $T+H$ also has a continuous inverse.

Speaking more intuitively, note that the fact that $T$ has a continuous inverse means that $E$ is isomorphic to $F$, so $Isom(E,F)$ is essentially $Isom(E)$ which is embedded in $B(E)$.