# Is the continous spectrum of a self-adjoint operator always uncountable?

This is the natural continuation to this Is the point spectrum always countable?

Premise: Let $$T$$ be a linear densely defined symmetric/self-adjoint operator in a complex separable Hilbert space $$\frak{H}$$ (typical example in Quantum Mechanics).

Definition: Let $$T_l :=T-lI$$. The set of points $$B(T)$$ such that $$T_{l}^{-1}$$ exists and is an unbounded linear transformation with domain $$\frak{R}_l$$ everywhere dense in $$\frak{H}$$ is called the continous spectrum of $$T$$.

Is the following true?

Statement: The continous spectrum of $$T$$ is either uncountable of the empty set.

P.S. The definition I took is from M.H. Stone ""Linear Transformations in Hilbert Space and their Applications to Analysis", AMS, 1932, page 129.

The continuous spectrum can be finite, even when $$T$$ is bounded.
Let $$\{e_n\}$$ be an orthonormal basis of a separable Hilbert space $$H$$. Define a bounded operator $$T:H\to H$$ by $$Te_n=\tfrac1n\,e_n.$$ So $$T$$ is the selfadjoint operator that with respect to the aforementioned basis is diagonal with diagonal $$\{1,\tfrac12,\tfrac13,\ldots\}$$. The spectrum of $$T$$ is $$\sigma(T)=\{0\}\cup\{\tfrac1n:\ n\in\mathbb N\}.$$ Each $$\tfrac1n$$ is an eigenvalue. And the continuous spectrum of $$T$$ is $$\{0\}$$, since $$T$$ is injective with dense range.
Tweaking of the above example allows you to produce examples where the continuous spectrum is any finite (or countable, bounded, and discrete) subset of $$\mathbb C$$. Concretely, write $$T_0$$ for the operator above. Given $$\{q_n\}\subset\mathbb C$$ bounded, let $$T=\bigoplus_n (q_n I+ T_0).$$ Then $$\sigma(T)=\overline{\bigcup_n\sigma(q_n I+T_0)}=\overline{\bigcup_n\{q_n\}}$$.