Give an example that the spectrum of a bounded self-adjoint operator is not closed.

Question

Let $T: V\to V$ be a bounded self-adjoint operator on a complex Hilbert space. Could anyone give an example that the spectrum of $T$ is not closed? Thanks.

Answer 1

If $B$ is a complex Banach space and $T:B \to B$ is a bounded linear operator, then the spectrum $\sigma(T)$ has the following properties:

$\sigma(T) \ne \emptyset$ and $\sigma(T)$ is compact (hence closed).

Answer 2

If $T-r_1$ has bounded inverse $R$ then $T-r_2=T-r_1+(r_1-r_2)=(T-r_1)(I+(r_1-r_2)R)$. If $|r_1-r_2|<\left\|R\right\|$, then $(I+(r_1-r_2)R)$ has inverse $\sum_{n}(r_1-r_2)^nR^n$. Therefore $T-r_2$ also has a bounded inverse $(I+(r_1-r_2)R)^{-1}R$.

This proves that the complement of the spectrum is open. Therefore the spectrum is always closed.