Isolated Eigenvalue What does it mean that an eigenvalue is "isolated"? My intuitive understanding says it is when one can find an open ball around it such that there is no other eigenvalue in that open ball.
However, I am reading a book ("Perturbation of Spectra in Hilbert Space" by Friedrichs) that says: 

...we assume that the undisturbed operator $H_0$ is Hermitian, possesses a spectral resolution, and has a single eigenvalue $\omega_0$ with an eigenvector $X_0$ [such that $X_0\neq 0$]. We assume this eigenvalue to be isolated, so that the equation $$ (H_0-\omega_0)X=\Psi $$ has a solution $X$ whenever the given right member $\Psi$ is orthogonal to $X_0$.

My question is: how does it follow from $\omega_0$ being isolated that the equation has a solution only if $<X_0,\Psi>=0$?
No matter if $\omega_0$ is isolated or not, we have that $$ <X_0, (H_0-\omega_0)X> = <(H_0-\omega_0)X_0, X> = <0,X> = 0$$so that if the equation $$ (H_0-\omega_0)X=\Psi $$ is satisfied by $X$ then $<X_0, \Psi>=0$.
What does the isolatedness of $\omega_0$ have to do with it?
 A: 
What does it mean that an eigenvalue is "isolated"?

An eigenvalue is isolated if it is an isolated point of the spectrum, i.e., it has a neighborhood in which there are no other points of the spectrum. This is a stronger property than not having other eigenvalues around. 
The claim is that the equation $ (H_0-\omega_0)X=\Psi $ has a solution for all $\Psi$ such that $\Psi\perp X_0$. Simply put, the range of $H_0-\omega_0I$ is $X_0^\perp$. 
You may recall that the closure of the range of a Hermitian operator is the orthogonal complement of the kernel. Since the kernel is the span of $X_0$, it follows that the range of $H_0-\omega_0I$ is dense in $X_0^\perp$. 
It remains to show that the range of $H_0-\omega_0I$ is closed. This is a consequence of the spectral theorem for normal operators: e.g., see Proposition IX.4.5 in Conway's Functional Analysis which says that for a normal operator $T$ the range of $T-\lambda I$ is closed if and only if $\lambda$ is not a limit point of $\sigma(T)$. 
A: I have good reasons to believe that the author of the book is treating a non-degenerate case but forgot to mention it. That is why he has treated the degenerate case separately in appendix I. It is evident because if there were another linearly independent eigenvector $X_1$ to the same eigenvalue, the statement must be modified to "the equation $(H_0−ω_0)X=Ψ$ has a solution $X$ whenever the given right member $\Psi$ is orthogonal to $X_0$ and $X_1$" which automatically satisfies "$\Psi$ is orthogonal to $X_0$", no problem there. But then you can apply Gram-Schmidt process and obtain a vector $X_1'$ that is orthogonal to $X_0$ and does not belong to Ran$(H_0 -\omega_0)$ which contradicts the original statement.
