Fredholm Alternative and Compact operator I m working on the following problem: Let $K$ be a compact operator on a Hilbert space, $H$, and let $K^*$ be its adjoint. For each $\lambda \in \mathbb{C}$, define $$N_\lambda=N(\lambda I-K), N_{\lambda}^*=N(\lambda I-K^*)$$
I showed previously that $\dim N_\lambda =\dim N_\lambda^*$ (It's also part of Fredholm Alternative). So given $\epsilon>0$, show the set $S_\epsilon=\{|\lambda|>\epsilon: N_\lambda \neq \phi\}$ consists of finitely many points. 
So we want to show there exists only finitely many $\lambda$, eigenvalues, such that there exists $u$, $\lambda u=Ku$. How do you start this proof? I tried using the adjoint property, but did not get very far from there. Any hints on this problem?
 A: Suppose there are infinitely many distinct eigenvalues $\lambda_{1},\lambda_{2},\cdots \in S_{\epsilon}$. Let $S_{0}=\{ 0\}$ and let $S_{n}$ be the subspace generated by $\bigcup_{j=1}^{n}\mathcal{N}(A-\lambda_{j}I)$. Each $S_{n}$ is a finite-dimensional subspace. For each $j \ge 1$ choose a unit vector $e_{j}$ such that $e_{j} \in S_{j}\cap S_{j-1}^{\perp}$. Then $A : S_{j}\rightarrow S_{j}$ and
$$
                 \frac{1}{\lambda_{j}}(A-\lambda_{j}I)e_{j} \in S_{j-1},
$$
because $A-\lambda_{j}I$ annihilates any component in $S_{j}$. For $k > j$,
$$
\begin{align}
    \|A\frac{1}{\lambda_{k}}e_{k}-A\frac{1}{\lambda_{j}}e_{j}\|^{2} 
       & = \|e_{k}+(A-\lambda_{k}I)\frac{1}{\lambda_{k}}e_{k}-A\frac{1}{\lambda_{j}}e_{j}\|^{2} \\
      & = \|e_{k}\|^{2}+\|(A-\lambda_{k}I)\frac{1}{\lambda_{k}}e_{k}-A\frac{1}{\lambda_{j}}e_{j}\|^{2} \\
      & \ge 1.
\end{align}
$$
Because $\{\frac{1}{\lambda_{j}}e_{j} \}_{j=1}^{\infty}$ is a bounded sequence, the above contradicts the compactness of $A$. So the set $S_{\epsilon}$ is either finite or empty for all $\epsilon > 0$.
