Given a (separable) Hilbert space **H** and an unbounded densely defined linear operator $T:{\cal D}(T) \to $**H** such that ${\cal D}$ is **diagonalizable** (it means $\exists$ an O.N.B. of **H** such that all basis elements are eigenvectors of $T$). As normal, take the point spectrum of $T$ to mean those $\lambda \in \sigma(T)$ such that $\lambda$ is an eigenvalue of $T$. Will the cloure of the point spectrum of $D$ be equal to the spectrum of $D$?



Closure of the point spectrum of an unbounded diagonalizable operator
From your assumption you can easily see that $T$ is unitarily similar to a multiplication operator on $\ell^2$ (and thus, $T$ is normal, by the way). This shows that the answer is "yes" (as it is easy to analyse the spectrum of multiplication operators on $\ell^2$).


