Spectrum can be an arbitrary subset. Given any subset $E$ of field $\mathbb{F}$ (real or complex), does there exist a normed linear space $X$ over $\mathbb{F}$ and a bounded linear operator $$A:X\rightarrow X$$ such that spectrum of $A$ is precisely the set $E$.
NOTE : It is known that this is true for compact sets as we can use their separability to construct such an operator.
 A: Yes.  If $E$ is empty, let $X=\{0\}$.  (By the Gelfand-Mazur theorem, this is the only possibility if $\mathbb{F}=\mathbb{C}$.)
If $E$ is not empty, we may assume without loss of generality that $0\in E$.  Now take a Banach space $Y$ with a bounded operator $T:Y\to Y$ that is quasinilpotent (i.e., its spectrum is $\{0\}$) but not nilpotent.  Let $B(Y)$ be the algebra of bounded operators on $Y$ and let $X\subseteq B(Y)$ be the subalgebra generated by $T$ and the elements $(T-\lambda I)^{-1}$ for all $\lambda\in \mathbb{F}\setminus E$ (here we use the assumption that $0\in E$, so that all these inverses must exist).  Note that the assumption that $T$ is not nilpotent means that that no nontrivial polynomial in $T$ is zero, so $X$ is isomorphic as an $\mathbb{F}$-algebra to the subalgebra of the field of rational functions $\mathbb{F}(x)$ generated by $x$ and $(x-\lambda)^{-1}$ for $\lambda\in\mathbb{F}\setminus E$, by mapping $T$ to $x$.  In particular, $T-\lambda I$ is not invertible in $X$ for any $\lambda\in E$.
Now let $A:X\to X$ be the operator given by multiplication by $T$.  Then $A$ is bounded since $X$ is a normed algebra.  Also, $A-\lambda I$ is invertible for any $\lambda\in\mathbb{F}\setminus E$ (the inverse is just multiplication by $(T-\lambda I)^{-1}$) but not for any $\lambda\in E$.  Thus, the spectrum of $A$ is $E$.
