Let $\Lambda : X \to X$ be a bounded linear operator on a Banach space $X$. My question is whether the set $ \{\lambda \in \mathbb C: \lambda I - \Lambda \quad\text{is surjective} \} $ is necessarily open. The above set is similar to the resolvent set of $\Lambda$, which is defined to be the set of all $\lambda \in \mathbb C$ such that $\lambda I - \Lambda$ is invertible; I know that the resolvent set is always open. However, what about the set above?
For reference, it was a problem on a past qualifying exam (see problem 6) to prove that the set is in fact open. I'm not sure if they meant to indicate the resolvent set, or if the problem is correct as stated.