Resolvent set of a bounded operator is open using OMT or Closed Graph Theorem I have seen the standard proof for the resolvent set of a bounded, linear operator being open. What I am wondering is if there is a way to show the same using some combination of Open Mapping or Closed Graph Theorems, especially if I move from the more general Banach to Hilbert spaces. 
Any insights would be highly appreciated.  
 A: Proving that a non-constant polynomial over $\mathbb{C}$ has a root is a non-trivial thing to prove. Even though this called the "Fundamental Theorem of Algebra," there is no purely algebraic proof! Analysis is required, and it turns out that the most powerful method of Analysis for proving this is Complex Analysis, where it is shown that a bounded entire function is a constant. Using this, a proof by contradiction is constructed by assuming that a non-constant polynomial $p(z)$ has no root in the finite plane. Then $1/p(z)$ is holomorphic and vanishes at $\infty$, which forces it to be a bounded entire function and, thus, equal to a constant $C$, which is a contradiction.
The most elegant proof that a bounded operator on a Banach space has non-empty spectrum follows the same pattern, which is no accident. In this case, one assumes that the resolvent $R(z)=(A-zI)^{-1}=1/(A-zI)$ is defined on all of $\mathbb{C}$, and one reaches the same contradiction--namely, $R(z)$ vanishes at $\infty$ and is holomorphic in $z$, which leads to a contradiction. So spectrum is non-empty for a bounded operator on a Banach space, or, more generally, for an element of a Banach algebra. But this requires the Complex field, just as it does for polynomials.
Open mapping and closed graph theorems do not require a Complex field. Somewhere Complex Analysis must enter into the picture. Even if you could restrict to special cases, analogous to restricting to real odd polynomials over $\mathbb{R}$, it seems very unlikely that a reasonable proof using only real techniques is going to be easy, if one exists at all.
