If $A:D(A)\subset H \rightarrow H$ is a selfadjoint Operator and possibly unbounded, $H$ a seperable Hilbert space. Does $A$ admit a non empty, atmost countable, pointspectrum?
I understand that, if there are finitely many, or no eigenvalues then there only countably many.
So lets just assume for now there are infinitely many of them.
Then there eigenspaces are orthogonal and we could decompose $H$ in a orthogonal sum of the nullspace and all the eigenspaces and some remainder.
i.e. $H=\overline{U}\oplus U^\perp$ where $U:=\bigoplus\limits_{\lambda \in \sigma_p(A)\cup\lbrace 0\rbrace}$ $\mathrm{ker}(\lambda- A)$
But why do we know that there are countably many eigenvalues and not overcountably many?
And how can we be sure that $\sigma_p(A)\neq \emptyset$.
To answer your updated question, we can certainly have $\sigma_p(A)=\emptyset$, even in the bounded case. A typical example is $T_x:L^2[0,1]\to L^2[0,1]$ given by $T_xf(x)=xf(x)$.
Having uncountably many eigenvalues would imply uncountably many orthogonal eigenvectors, violating the separability assumption.