A simple proof of $\{D \in M_n(\mathbb C),\ D $ diagonalizable with distinct eigenvalues$\}$ $\subset$ Int$\{ D $ diagonalizable$\}$ What is the interior of $\{D \in M_n(\mathbb C),\ D $  diagonalizable$\}$ ?
Actually, it is $I = \{D \in M_n(\mathbb C),\ D $  diagonalizable with distinct eigenvalues$\}$. 
Int$(D) \subset I $ is quite natural but I know a proof of the reciprocal which is astute : 
$M \in I \iff \gcd (P_M, P'_M) = 1 \iff \exists \ A,B \neq 0$ s.t. $A P_M = BP_M'\ $ with conditions on degrees and then create a continuous function using $\det$ characterizing this. 
Would you have a more natural way to proof $I \subset$ Int$(D)$ ?  
 A: Equivalently you want to show that the closure of the space $X$ of matrices which are not diagonalizable is the space $Y$ of matrices which have a repeated eigenvalue. It is at most as large as $Y$ because the discriminant of the characteristic polynomial is continuous, vanishes on $X$, and has zero locus exactly $Y$. (This is equivalent to the direction you were asking about: a matrix has distinct eigenvalues iff the discriminant of the characteristic polynomial doesn't vanish.) 
Conversely, given a matrix $M \in Y$ with repeated eigenvalues, either it is already not diagonalizable or it can be diagonalized, and then it is easy to write down a sequence of nondiagonalizable matrices converging to $M$ with the same eigenvalues using "Jordan blocks" of the form $\left[ \begin{array}{cc} \lambda & \frac{1}{n} \\ 0 & \lambda \end{array} \right]$. 
A: One direction (which appears to be the one the question is about) amounts to showing that the set $I$ is open. Since it is the inverse image under the "characteristic polynomial" map $\chi:M_n(\Bbb C)\to V$, where $V$ is the subset of $\Bbb C[X]$ of monic polynomials of degree$~n$, of the subset $S\subset V$ of polynomials without multiple roots, this reduces (by continuity of$~\chi$) to showing that $S$ is open in$~V$. While I think is would be possible to show that $S$ is open by analytic means (showing that for any given polynomial $P$ in $S$ there is a ball around it where the polynomials have roots sufficiently close to each root of $P$ to ensure they remain separated), the algebraic approach seems easier here: the locus of polynomials with multiple roots is the zero locus of the discriminant, and therefore closed (as the discriminant of a polynomial$~P\in V$ is a polynomial expression in the coefficients of$~P$).
The other direction is also straightforward. To show that any diagonalisable matrix$~M$ whose characteristic polynomial has at least one multiple root is in the closure of the set of non-diagonalisable matrices, we can show the same property within a chosen subspace of $M_n(\Bbb C)$ that contains $M$. To this end take the subspace $X\subseteq M_n(\Bbb C)$ of matrices with the same characteristic polynomial as $M$ and the same generalised eigenspace for each eigenvalue. Then $M$ is the unique diagonalisable matrix in$~X$ (since in order for $N\in X$ to be diagonalisable, its restriction to the generalised eigenspace for$~\lambda$ must be multiplication by$~\lambda$, whence $N=M$), and since at least one generalised eigenspace has dimension${}>1$, one has $T\neq\{M\}$, and so $M\in\overline{X\setminus\{M\}}$ as desired.
