Continuous spectrum can shrink to an isolated point

Let $A$ be a bounded linear operator in a Hilbert space $H$.

I had the misconception that the continuous spectrum of $A$ would necessarily have some "continuous" appearance: an interval, a union of intervals, or something like that. This is false as the following operator shows:

$$Vf(t)=\int_0^t f(s)\, ds,\qquad f \in L^2([0, 1])$$

(see Wikipedia) the spectrum of this operator is purely continuous and is reduced to $\{0\}$.

Question Is it true that, if $A$ is self-adjoint and $0$ is an isolated spectral value, then $0$ is not in the continuous spectrum (and so it is an eigenvalue)?
This is true if $A$ is compact, I believe: in that case, if $0$ is an isolated spectral value then $A$ only has a finite number of eigenvalues and so, by Hilbert-Schmidt theorem, it is a finite rank operator. Then any spectral value is an eigenvalue. But what about the general case? I suspect that in the general setting things are not so easy.
@Theo: Indeed the operator $V$ above is not normal (I just checked :-) ). So, at least for this class of operators the continuous spectrum is really something "continuous", having no isolated points. I noticed that the converse is not true, that is, the set of eigenvalues needs not be "granular" in nature. For example, if I'm not mistaken, the operator $$Uf(t)=f(t-1), \qquad f \in L^2(\mathbb{R})$$ is unitary, hence normal, its spectrum consist of the whole unit circle and every spectral value is an eigenvalue. Strange! – Giuseppe Negro Sep 10 '11 at 18:34
@Theo: Sorry to disturb you again, it is just to say that unfortunately the previous example I made is wrong! I erroneously assumed that $e^{i \theta t} \in L^2(\mathbb{R})$. And indeed, as I see in Rudin's Functional analysis, the point spectrum of a normal bounded operator in a separable Hilbert space is at most countable (Exercise 12-18.b pag.343). Anyway, your point in the last comment is very clear and it is exactly what I was looking for. It is important to know where terminology we use comes from. Thank you! – Giuseppe Negro Sep 10 '11 at 19:00