Let $X$ be a Banach space and let $A \in \mathcal{L}(X)$. Prove that the set of $\lambda \in \sigma(A)$ such that $\lambda I - A$ is injective and $\lambda I - A$ has closed range is an open subset of $\mathbb{C}$.
My (potentially wrong) attempt:
I guess what I need to show is there is a neighborhood $N = \{\lambda + r\ |\ r < \epsilon\}$ of $\lambda$ that satisfies:
$(\lambda + r)I - A$ is injective, closed, and not surjective.
By applying the theorem that bounded operators $T$ has closed range if and only if it is bounded below ($\|Tx\| \geq c\|x\|$). I can easily prove there is a $\epsilon$ keeps the range closed. But I don't know how to deal with injection and surjection.
Another (potentially right) strategy:
I received some advice to utilize the bounded below theorem, inverse mapping theorem, and open mapping theorem to first prove $\lambda I - A$ is open, (closed => bounded below => operator inverse bounded => operator bounded). But I don't know how to proceed with the remains. How to prove $\lambda$ is in an open set from here?
Any help is appreciated.