Set of bijective operators open in $L(X,X)$?

Question

Let $U=\{T \in L(X,X)$ s.t. $T$ bijective $\}$. $(X,||.||)$ be a normed $\mathbb R$ vector space. Is it true that $U$ is open in $L(X,X)$? Could someone explain me why it is true/not true?

Answer 1

Yes, $U$ is open.

We denote operator-norm on $L(X,X)$ also by $||.||$.

Let $T_0 \in U$ and $T \in L(X,X)$ and suppose that

(1) $||T-T_0||< \frac{1}{||T_0^{-1}||}$.

We want to show that $T \in U$. To this end verify that

(2) $T=T_0(I-T_0^{-1}(T_0-T))$.

The operator $I-T_0^{-1}(T_0-T)$ is invertible, since (1) holds (Neumann- Series !). Hence, by (2), $T$ is invertible.

Answer 2

Hint: prove first that $B(I,1)\subset U$. Given $I+H\in B(I,1)$, write $(I+H)^{-1}$ as a power series remembering the sum of a geometric series.