Spectrum of a self-adjoint compact operator

Question

Given a self-adjoint compact operator $A$ on a (separable) Hilbert space, is it true that the spectrum of $A$ is equal to the closure of the set of eigenvalues of $A$, or in symbols $$ \sigma(A) = \sigma_P(A)? $$

Answer 1

Yes. For a compact operator, any nonzero eigenvalue is an isolated point (because it has to have finite multiplicity). So the only point in the spectrum that may not be an eigenvalue is zero.

Now, a selfadjoint operator has no residual spectrum, so all points in the spectrum are either eigenvalues or approximate eigenvalues. So if $0$ is not an eigenvalue, it is an approximate eigenvalue; this means that $0=\lim\lambda_j$ for some $\lambda_j$, which being nonzero will necessarily be eigenvalues. So yes, $\sigma(A)=\overline{\sigma_P(A)}$; even more concretely, $$ \sigma(A)=\{0\}\cup\sigma_p(A). $$ No closure needed.