I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа (p. 475 here) that 0 is an accumulation point for the spectrum of a completely continuous operator $A:H\to H$ where $A$ is a Hilbert space.
I know that the spectrum of $A$ is (at most) countable, as Kolmogorov-Fomin's shows, contains eigenvalues only and always contains 0. Moreover, I know that, for any $\rho>0$, $A$ has only a finite number of linearly independent eigenvectors corresponding to eigenvalues of absolute value greater than $\rho$, which implies that its eigenvalues can be ordered in a way such that $|\lambda_1|\ge|\lambda_2|\ge\ldots$
Please correct me if I am wrong, but, from these facts, I deduce that no eigenvalues $\lambda_k$ such that $|\lambda_k|>0$ can be an accumulation point of $\sigma(A)$. Nevertheless, I do not see why 0 necessarily is an accumulation point. In particular, Kolmogorov-Fomin's does not rule out the possibility that $\sigma(A)$ can be finite and I do not see how a finite set of complex numbers can have an accumulation point. Thank you for any answer!
on p.475 Kolmogorov-Fomin says: "$\dots0$ is the only possible accumulation point for the sequence $\{\mu_n\}\dots$"
Kolmogorov-Fomin does not exclude the possibility that $\sigma(A)$ is finite and there is no accumulation point for $\sigma(A).$