Matrices with n distinct eigenvalues form an open set

Idea. "Distinct eigenvalues" is a non-vanishing condition for a polynomial expression in the entries, hence open.

Step 1: the discriminant. For a monic polynomial $p(t)=t^n+c_{n-1}t^{n-1}+\dots+c_0$ the discriminant $\operatorname{disc}(p)=\pm\operatorname{Res}(p,p')$ (resultant, the determinant of the Sylvester matrix) is a polynomial in the coefficients $c_0,\dots,c_{n-1}$, and $\operatorname{disc}(p)\neq0$ iff $p$ has $n$ distinct roots in $\mathbb{C}$ (equivalently $\gcd(p,p')=1$).

Step 2: it is polynomial in the entries. The coefficients of $\chi_A(t)=\det(tI-A)$ are polynomials in the entries $a_{ij}$ (sums of principal minors, up to sign). Hence $\Delta(A)=\operatorname{disc}(\chi_A)$ is a polynomial in the entries, thus continuous.

Step 3: conclusion. The set in question is
$$\{A:\Delta(A)\neq0\}=\Delta^{-1}(\mathbb{R}\setminus\{0\}),$$
open as the preimage of an open set under a continuous map.

Example ($n=2$). $\Delta(A)=(\operatorname{tr}A)^2-4\det A$, and eigenvalues are distinct iff $\Delta\neq0$.

Remark.
- The complementary set $\{\Delta=0\}$ (a repeated eigenvalue) is closed with empty interior (see the problem on repeated eigenvalues below), so *generic matrices have distinct eigenvalues* and are therefore diagonalisable over $\mathbb{C}$.
- Equivalent proof: roots of a polynomial depend continuously on its coefficients (Part 3), so $n$ distinct roots stay distinct under small perturbations.