I am looking for a counter example to the claim
Let $H$ be a Hilbert space and $\mathcal{D} \subseteq H$. Then the spectrum of an unbounded linear operator $A: \mathcal{D} \rightarrow H$ decomposes into point-, continuous-, and residue-spectrum.
This is true when the operator $A$ is closed, since then $\forall z \in \mathbb{C}: A - z$ is also closed and thus is boundedly invertible if and only if it is bijective. So in particular I am looking for a non-closed linear operator $A: \mathcal{D} \rightarrow H$ s.t. $A - z$ is bijective for some $z \in \mathbb{C}$.
There may very well be an immediate example that I am missing, since I cannot come up with many non-closed operators to begin with.
Thank you in advance.