The set of all invertible operators need not be dense

Question

I want to show that the set of all invertible operators $\mathcal{G}(\ell^2)$ is not dense in $\mathcal{B}(\ell^2)$.

Consider the right shift operator $T\in \mathcal{B}(\ell^2)$. We also know that $T\notin \mathcal{G}(\ell^2)$. I want to show now that $$B=\{R\in \mathcal{B}(\ell^2):\|R-T\|<1\}$$ is disjoint with $\mathcal{G}(\ell^2)$.

Suppose not. Then there exists $R\in \mathcal{G}(\ell^2)$ such that $\|R-T\|<1$. How to arrive at a contradiction from here? Any help is appreciated.

Answer 1

Let $S$ be the left shift operator, so that $ST = I$. Then $$\|I - SR\| = \|S(T - R)\| \le \|S\| \cdot \|T - R\| < \|S\| = 1.$$ Hence, $SR$ is invertible. If $R$ were invertible, then so too would be $S$, which it is not.