During the study on invariant subspaces, the following question occurred to me.
Let $X$ be a complex Banach space, $B$ a closed subset of $\mathcal B(X)$ and $U$ an open subset of $B$. Can we say that $U+\mathbb CI$ is an open subset of $B+\mathbb CI$.
I think the only thing relevant is that you are dealing with a topological vector space $Z$. In that case, the map $x\mapsto y_0 +x$ is a homeomorphism for all $y_0\in Z$. Then for any open $U'\subseteq Z$ and $y_0\in Z$, the set $U'+y_0$ is open. Recall that $U$ is open in $B$, if there exists $U'\subseteq Z$ open such that $U=U'\cap B$. Then for any nonempty subset $Y\subseteq Z$, you have that
$$ U+Y= \cup_{y\in Y} \big( U+y \big). $$
Any element in $U+Y$ is of the form $u+y$ for some $u\in U$ and some $y\in Y$. There exists an open set $N_u$ satisfying $u\in N_u\subseteq U'$. Then $u+y\subseteq (N_u+y)\cap (B+Y)$. This shows that $u+y$ is also an interior point of $U+Y$ in the subspace topology on $B+Y$. Hence, $U+Y$ is open in $B+Y$.
In your case, $Z=\mathcal{B}(X)$ and $Y=\mathbb{C}I$.