Let $H$ be a complex Hilbertspace, let $ T \in B(H)$ and suppose that $T$ is not self-adjoint.
Show that there exists $\epsilon > 0$ such that $\{S \in B(H): ||T-S|| < \epsilon\}$ does not contain any self-adjoint bounded operator on H.
I know that an operator is self-adjoint if $S = S^*$. So I need to show that for all $S$ in this set this equality is not true.
Can anyone help how to do so?
Hint. The map $S \mapsto S - S^*$ is continuous on $\bigl(B(H), \|\cdot\|\bigr)$ and the selfadjoint operators are the kernel of this map.