Let $X$ be a $\mathbb C$-Banach space and $T\in\mathfrak L(X)$ with $\sigma(T)=\overline{\sigma_p(T)}$, where $\sigma_p(T)$ denotes the point spectrum of $T$. Does it necessarily follow that the residual spectrum $\sigma_r(T)$ is empty?
The situation I've got in mind is $X=\ell^2$ and $(Tx)_n:=\lambda_n x_n$ for some bounded $(\lambda_n)\subseteq\mathbb C$. In that situation, $\sigma_p(T)=\{\lambda_n:n\in\mathbb N\}$ and it is easy to show that $\mathbb C\setminus\overline{\sigma_p(T)}\subseteq\rho(T)$. On the other hand, $\overline{\sigma_p(T)}\subseteq\overline{\sigma(T)}=\sigma(T)$ so that we can conclude $\sigma(T)=\overline{\sigma_p(T)}$. Now I'd like to conclude the continuous spectrum $\sigma_c(T)$ is equal to $=\overline{\sigma_p(T)}\setminus\sigma_p(A)$.
EDIT: According to the answer below, the claim is wrong in general, but how do we see it in the particular situation described above?
No, take any operator of the form $T=T_1\oplus T_2$, on $X=X_1\oplus X_2$, with $\overline{\sigma_p(T_1)}=\sigma_r(T_2)$.