If $T$ is bounded and $F$ has finite rank, what is the spectrum of $T+F$? Suppose that $T$ is a bounded operator with finite spectrum. What happens with the spectrum of $T+F$, where $F$ has finite rank? Is it possible that $\sigma(T+F)$ has non-empty interior? Is it always at most countable? 
Update: If $\sigma(T)=\{0\}$ then $0$ is in the essential spectrum of $T$ ($T$ is not invertible in the Calkin algebra),  hence for any compact $K$, $\sigma_{ess}(T+K)=\sigma_{ess}(T)=\{0\}$. For operators such that the essential spectrum is $\{0\}$, it is known that their spectrum is either finite or consists of a sequence converging to $0$. I think it should be the same for operators with finite spectrum, but I cannot find a proof or reference. 
 A: Short answer: $\sigma(T+F)$ is at most countable and so has empty interior.
Theorem (Spectral Theory of Linear Operators by V. Muller III.19.4):
Let $T\in\mathcal{B}(X)$, let $G$ be a component of $\mathbb{C}\setminus\sigma_e(T)$. Then either $G\subset\sigma(T)$ or $G\cap\sigma(T)$ consists of at most countably many isolated points.
Combining the above with the essential spectrum, as you note, we get:
$\sigma_e(T+F) = \sigma_e(T) \subset \sigma(T)$, which is finite.
Then $G = \mathbb{C}\setminus\sigma_e(T+F)$ is a connected component and $\sigma(T+F)\setminus\sigma_e(T+F)$ is at most countable and all isolated.
You might also find that the spectrum must be finite but I can't think how to prove it.
A: It is always true if $T$ is self-adjoint. Here is a theorem that you might be interested:
If $T$ is self-adjoint, a complex number is in the spectrum of $T$ but not in its essential spectrum iff it is an isolated eigenvalue of $T$ of finite multipliticity. 
The result can be found on page 32 of Analytic K-homology by Nigel Higson and John Roe.
