I have this very simple question.
Premise: Let $A$ be a linear densely defined symmetric/self-adjoint operator in a complex separable Hilbert space $\mathcal H$ (typical example in Quantum Mechanics).
Definition: The set $\sigma_{\text{pp}}:=\{z| z\in\mathbb C,~ \nexists \left(A-z I\right)^{-1}:\mathcal H\to\mathcal H\} $ is called the pure point spectrum of A. (Definition from Stone, M.H. "Linear Transformations in Hilbert Space and their Applications to Analysis", AMS, 1932, page 129.)
Then is the following result true?
Statement: $\sigma_{\text{pp}}$ is countable (or an empty set).
Kindly provide me with a counterexample if not true, or with a rigorous proof if true.
Thank you,
If $Av=\lambda v$ and $Aw=\mu w$ with $v,w$ unit vectors and $\lambda\ne\mu$, then (using that $A$ is selfadjoint) $$ \lambda\langle v,w\rangle=\langle Av,w\rangle=\langle v,Aw\rangle=\mu\langle v,w\rangle $$ (note that $\lambda,\mu\in\mathbb R$). So $\langle v,w\rangle=0$. Thus the eigenspaces corresponding to distinct eigenvalues are orthogonal. If $H$ is separable, it can only have countably many pairwise orthogonal subspaces, so the set of eigenvalues of $A$ is at most countable.